Appendix ASolutions to the Exercises

Solution

In each case, estimate the size $L_{\text{obj}}$ of the object itself and the characteristic length $L_{\text{motion}}$ of the motion, and decide by the ratio $L_{\text{obj}}/L_{\text{motion}}$ (the criterion of Section 1.1).

(a) The trajectory of a baseball thrown by a pitcher. The diameter of the ball is about $7\ \mathrm{cm}=0.07\ \mathrm m$, and the pitching distance (from the pitcher to the catcher) is about $18\ \mathrm m$. The ratio is $0.07/18\approx4\times10^{-3}$, which is far smaller than 1. Therefore, if we are interested only in the flight path of the ball (the shape of its trajectory), it may be treated as a particle (point mass). (However, in problems that focus on the rotation of the ball itself, such as the curving of a pitched ball caused by its spin, the particle approximation cannot be used.)

(b) A figure skater spinning in place on the ice. As we saw in Example 1.1(b), the radius of rotation (the length of an outstretched arm, $\sim0.6\ \mathrm m$) and the representative size of the body ($\sim1\ \mathrm m$) are of the same order of magnitude, so the ratio is of order 1. Moreover, the phenomenon itself, that the skater spins faster when the arms are pulled in, depends on the positional relationship among the parts of the body, so the skater cannot be treated as a particle. We must treat the parts of the body as a collection of many particles (a system of particles, Chapter 10). (Since the shape changes when the arms are pulled in, the body is not a rigid body either.)

(c) The orbital motion of the Moon as seen from the Earth. The radius of the Moon is about $1.7\times10^3\ \mathrm{km}$, and the mean distance from the Earth to the Moon (the orbital radius) is about $3.8\times10^5\ \mathrm{km}$. The ratio is $1.7\times10^3/3.8\times10^5\approx4.5\times10^{-3}$, which is sufficiently small compared with 1. Therefore, in considering the orbit of the Moon, the Moon may be treated as a particle (however, in problems concerned with the rotation of the Moon or its deformation by tidal forces, the particle approximation cannot be used).

To summarize: in (a) and (c), “the size of the object itself is negligible compared with the scale of the motion,” so the particle approximation is a good approximation; in (b), however, the ratio is of order 1 and, in addition, the internal structure of the object itself (the positions of the arms) bears on the essence of the problem, so the particle approximation cannot be used.

Solution

In components, the acceleration vector is $\bm a=[a_x,a_y]=[0,-g]$. By the definition (acceleration is the time derivative of velocity), $\dd v_x/\dd t=0$ and $\dd v_y/\dd t=-g$, so we integrate both sides with respect to $t$.

$x$ component: integrating $\dd v_x/\dd t=0$ gives $v_x(t)=C_1$ (a constant). From the initial condition $v_x(0)=v_0\cos\theta$, we get $C_1=v_0\cos\theta$. Hence $v_x(t)=v_0\cos\theta$ (the velocity in the $x$ direction does not change).

$y$ component: integrating $\dd v_y/\dd t=-g$ gives $v_y(t)=-gt+C_2$. From the initial condition $v_y(0)=v_0\sin\theta$, we get $C_2=v_0\sin\theta$. Hence $v_y(t)=v_0\sin\theta-gt$.

Therefore the velocity vector is

$$ \bm v(t)=[\,v_0\cos\theta,\ \ v_0\sin\theta-gt\,] $$

Furthermore, integrating $v_x=\dd x/\dd t,\ v_y=\dd y/\dd t$ once more under the initial conditions $x(0)=0,\ y(0)=0$, we obtain

$$ x(t)=\int_0^t v_0\cos\theta\,\dd t'=v_0\cos\theta\cdot t $$ $$ y(t)=\int_0^t(v_0\sin\theta-gt')\,\dd t'=v_0\sin\theta\cdot t-\frac12gt^2 $$

That is, the position vector is

$$ \bm r(t)=\left[\,v_0\cos\theta\cdot t,\ \ v_0\sin\theta\cdot t-\frac12gt^2\,\right] $$

This is nothing more than carrying out the “generalization to vectors” of Section 1.4 (applying Theorem 1.1 independently to each component), with the $x$ component as motion with constant acceleration $a_x=0$ (in fact, uniform motion) and the $y$ component as motion with constant acceleration $a_y=-g$; the result is exactly the standard formulas of projectile motion without air resistance (treated in earnest in Chapter 5).

Solution

Taking vertically upward as positive, with $v_0=0,\ a=-g$ ($g=9.8\ \mathrm{m/s^2}$), the velocity is $v(t)=-gt$.

The $v$-$t$ graph. Since $v(t)=-gt$ is a linear function of $t$, the $v$-$t$ graph is a straight line sloping downward to the right through the origin $(0,0)$. Its slope is $-g=-9.8\ \mathrm{m/s^2}$ (constant), which expresses that the (downward) speed keeps increasing with time. The intercept (the value at $t=0$) is 0.

The $a$-$t$ graph. Since the acceleration is $a=-g$ (a constant), the $a$-$t$ graph is a horizontal straight line on which the value on the vertical axis is constant at $-9.8\ \mathrm{m/s^2}$ (parallel to the $t$ axis).

Applying the relation “the slope is the derivative” of Section 1.6, the slope of the $v$-$t$ graph ($-g$, constant) is exactly the acceleration $a=\dd v/\dd t$, which is consistent with the $a$-$t$ graph being a horizontal line at the constant value $-g$. Also, from the relation “the area is the integral,” we can confirm that the area under the $v$-$t$ graph from $0$ to $t$ (carrying a negative sign because it lies below the $t$ axis) agrees with the displacement over that interval, $y(t)-y(0)=-\frac12gt^2$ (Eq. (1.6) of Chapter 1).

Solution

(a) Differentiate $x(t)=R\cos\omega t,\ y(t)=R\sin\omega t$ with respect to $t$, each according to the differentiation formula for composite functions. Since $\dd(\cos\omega t)/\dd t=-\omega\sin\omega t$ and $\dd(\sin\omega t)/\dd t=\omega\cos\omega t$,

$$ \bm v(t)=[\,\dot x,\dot y\,]=[\,-R\omega\sin\omega t,\ \ R\omega\cos\omega t\,] $$

Differentiating once more,

$$ \bm a(t)=[\,\ddot x,\ddot y\,]=[\,-R\omega^2\cos\omega t,\ \ -R\omega^2\sin\omega t\,]=-\omega^2\,[\,R\cos\omega t,\ R\sin\omega t\,]=-\omega^2\bm r(t) $$

That is, the acceleration is always directed opposite to the position vector $\bm r(t)$ (toward the origin), and its magnitude is $\omega^2$ times that of the position vector.

(b) The square of the speed is

$$ \abs{\bm v(t)}^2=(-R\omega\sin\omega t)^2+(R\omega\cos\omega t)^2=R^2\omega^2\sin^2\omega t+R^2\omega^2\cos^2\omega t=R^2\omega^2(\sin^2\omega t+\cos^2\omega t) $$

By the fundamental trigonometric identity $\sin^2\theta+\cos^2\theta=1$ (used with $\theta=\omega t$), the quantity in parentheses equals 1, so

$$ \abs{\bm v(t)}^2=R^2\omega^2\qquad\Longrightarrow\qquad \abs{\bm v(t)}=R\omega\ (\text{constant, }\ t\text{-independent}) $$

Thus the speed $\abs{\bm v}=R\omega$ is constant, but as we saw in (a), the acceleration $\bm a=-\omega^2\bm r$ is not $\bm 0$, because the direction of the velocity vector changes continuously; this is a concrete example of the statement in Section 1.2 that “even if the speed is constant, the acceleration is not necessarily 0.”

(c) The inner product (dot product) is the sum of the products of the components. From the result of (a),

$$ \bm v\cdot\bm a=(-R\omega\sin\omega t)(-R\omega^2\cos\omega t)+(R\omega\cos\omega t)(-R\omega^2\sin\omega t)=R^2\omega^3\sin\omega t\cos\omega t-R^2\omega^3\sin\omega t\cos\omega t=0 $$ $$ \bm r\cdot\bm v=(R\cos\omega t)(-R\omega\sin\omega t)+(R\sin\omega t)(R\omega\cos\omega t)=-R^2\omega\sin\omega t\cos\omega t+R^2\omega\sin\omega t\cos\omega t=0 $$

Here $\bm v\cdot\bm a=0$ expresses that the acceleration is perpendicular to the velocity (a characteristic of a motion in which only the direction changes while the speed does not), and $\bm r\cdot\bm v=0$ expresses that the velocity is perpendicular to the radial direction, that is, along the tangent to the circle. (Another way to see this: $\dd\abs{\bm v}^2/\dd t=2\bm v\cdot\bm a$, and if $\abs{\bm v}$ is constant the left side is 0, so $\bm v\cdot\bm a=0$.)

Solution

That the inner product is 0 is equivalent to the two vectors being orthogonal (Section 1.7). We compute the sum of the products of the components.

(a) $[1,2,3]\cdot[2,-1,0]=1\times2+2\times(-1)+3\times0=2-2+0=0$. Since the inner product is 0, the vectors are orthogonal.

(b) $[1,0,0]\cdot[0,1,1]=1\times0+0\times1+0\times1=0$. Since the inner product is 0, the vectors are orthogonal.

Since the inner product is 0 for both pairs, we have confirmed that both pairs are orthogonal.

Solution

In the component formula for the cross product, $\bm a\times\bm b=[a_yb_z-a_zb_y,\ a_zb_x-a_xb_z,\ a_xb_y-a_yb_x]$, substitute $\bm a=[1,0,0]$ and $\bm b=[0,1,0]$ to obtain

$$ \bm a\times\bm b=[\,0\times0-0\times1,\ \ 0\times0-1\times0,\ \ 1\times1-0\times0\,]=[\,0,\ 0,\ 1\,] $$

The result $[0,0,1]$ is none other than the unit vector $\bm e_z$ in the direction of the $z$ axis (Section 1.2). That is, $\bm e_x\times\bm e_y=\bm e_z$. This agrees with the direction in which the thumb points when the fingers of the right hand are curled from the positive direction of the $x$ axis ($\bm a$) toward the positive direction of the $y$ axis ($\bm b$), that is, the positive direction of the $z$ axis, so the result is exactly what the right-hand rule (Figure 1.5) predicts.

Solution

From the equation of motion $F=ma$, the dimension of force is the product of the dimension of mass and the dimension of acceleration. The dimension of mass is $[\mathrm M]$ and that of acceleration is $[\mathrm{L\,T^{-2}}]$ (Sections 1.2 and 1.3), so

$$ [F]=[\mathrm M][\mathrm{L\,T^{-2}}]=[\mathrm{M\,L\,T^{-2}}] $$

Hooke's law $F=kx$ must have equal dimensions on both sides by Law 1.1 (dimensional homogeneity), so $[F]=[k][x]$. Since $[x]=[\mathrm L]$,

$$ [k]=\frac{[F]}{[x]}=\frac{[\mathrm{M\,L\,T^{-2}}]}{[\mathrm L]}=[\mathrm{M\,T^{-2}}] $$

Written in SI base units, the dimension $[\mathrm{M\,T^{-2}}]$ becomes $\mathrm{kg/s^2}$, which is the unit of force, $\mathrm N=\mathrm{kg\,m/s^2}$, divided by the unit of length, $\mathrm m$; that is, it coincides with $\mathrm{N/m}$ ($\mathrm{kg/s^2}=(\mathrm{kg\,m/s^2})/\mathrm m=\mathrm{N/m}$). The reason the unit of the spring constant is usually written as $\mathrm{N/m}$ is precisely this result of dimensional analysis.

Solution

We set up the impulse–momentum theorem of Chapter 2 ($\bm p(t_1)+\int_{t_1}^{t_2}\bm F\,\dd t=\bm p(t_2)$) as a one-dimensional problem along the $x$ axis, with $t_1=0,\ t_2=t$. Since the force $F$ is constant in time, the integral for the impulse reduces to a simple multiplication:

$$ p(t)=p(0)+\int_0^t F\,\dd\tau=p(0)+F\big[\tau\big]_0^t=p(0)+Ft $$

Since the particle was at rest at time $t=0$ ($p(0)=0$), we obtain

$$ p(t)=Ft $$

For the velocity, the definition of momentum $p=m\dot x$ gives $\dot x=p/m$, so the velocity is

$$ \dot x(t)=\frac{p(t)}{m}=\frac{Ft}{m} $$

(This agrees exactly with what is obtained by substituting $a=F/m$ (the equation of motion) into the relation $v=at$ that we learned in high school physics as “an object under a constant force undergoes motion with constant acceleration.” We see that the impulse–momentum theorem gives the same result directly from the relation impulse = change in momentum, without going through the quantity called acceleration.)

Solution

The velocity formulas for an elastic collision ($e=1$) in Chapter 2 were

$$ v_1'=\frac{(m_1-m_2)v_1+2m_2v_2}{m_1+m_2},\qquad v_2'=\frac{2m_1v_1+(m_2-m_1)v_2}{m_1+m_2} $$

Substitute $m_1=m_2=m$ into these. First, for $v_1'$,

$$ v_1'=\frac{(m-m)v_1+2mv_2}{m+m}=\frac{0\times v_1+2mv_2}{2m}=\frac{2mv_2}{2m}=v_2 $$

(Note that the first term $(m_1-m_2)v_1$ of the numerator vanishes, becoming exactly 0 because $m_1=m_2$.) Similarly, for $v_2'$,

$$ v_2'=\frac{2mv_1+(m-m)v_2}{m+m}=\frac{2mv_1+0\times v_2}{2m}=\frac{2mv_1}{2m}=v_1 $$

(This time the second term $(m_2-m_1)v_2$ of the numerator vanishes.) Therefore we have shown that

$$ v_1'=v_2,\qquad v_2'=v_1 $$

In other words, when two bodies of equal mass collide elastically, their velocities are exchanged completely. The well-known phenomenon with billiard balls (whose masses are nearly equal and whose collisions are nearly elastic) is a concrete example of exactly this result: when one ball is sent straight into a ball at rest, the incoming ball nearly stops, and the ball that had been at rest sets off rolling with the original velocity of the incoming ball.

Solution

The magnitude of the cross product can be written, with $\theta$ ($0\leqq\theta\leqq\pi$) the angle between the two vectors $\bm r,\bm F$, as

$$ \abs{\bm N}=\abs{\bm r\times\bm F}=\abs{\bm r}\,\abs{\bm F}\sin\theta $$

Under the conditions $\bm r\neq\bm 0,\ \bm F\neq\bm 0$ (hence $\abs{\bm r}\neq0,\ \abs{\bm F}\neq0$), $\bm N=\bm 0$ ($\abs{\bm N}=0$) holds only when $\sin\theta=0$, that is, when $\theta=0$ ($\bm r$ and $\bm F$ point in the same direction) or $\theta=\pi$ ($\bm r$ and $\bm F$ point in exactly opposite directions). In other words, the torque is zero only when the direction of the force $\bm F$ is parallel to the position vector $\bm r$ (either the same direction or the opposite direction is allowed).

Put geometrically, the torque is zero when the “line of action” of the force $\bm F$ (the straight line drawn along the direction of the force vector) passes through the origin $O$. The more the line of action of the force is offset from the origin (that is, the more it fails to pass through the origin), the larger the torque (the effect of causing rotation) becomes, even for a force of the same magnitude—this agrees exactly with our experience, expressed by the principle of the lever, that “the farther from the fulcrum you push, the easier it is to turn.”

Solution

That the force $\bm F$ is a central force, that is, that $\bm F$ is always parallel to the position vector $\bm r$, means that it can be written, using some real-valued function (scalar function) $f(t)$, as

$$ \bm F(t)=f(t)\,\bm r(t) $$

(When $f\gt0$, $\bm F$ points in the same direction as $\bm r$, that is, it is repulsive; when $f\lt0$, $\bm F$ points opposite to $\bm r$, that is, it is attractive. Universal gravitation and the electrostatic force have this form.) Then the torque is

$$ \bm N=\bm r\times\bm F=\bm r\times\big(f\bm r\big)=f\,(\bm r\times\bm r) $$

(the scalar $f$ can be taken out of the cross product). Now, the cross product of a vector $\bm r$ with itself is always the zero vector, by the derivation in Section 2.5 of this chapter (for exactly the same reason as when we showed that $\bm v\times\bm v=\bm 0$): $\bm r\times\bm r=\bm 0$. Therefore,

$$ \bm N=f\cdot\bm 0=\bm 0 $$

holds at every time $t$. On the other hand, by the rotational equation of motion (Law 2.5) $\bm N=\dd\bm L/\dd t$,

$$ \frac{\dd}{\dd t}\bm L=\bm N=\bm 0 $$

That is, since the time derivative of the angular momentum $\bm L$ is identically zero, $\bm L$ is constant in time (it is conserved). It is precisely for this reason that angular momentum is conserved for a body moving under a central force (as in the motion of the planets).

Solution

The following two forces act on the book.

• Gravity $m\bm g$: the force with which the Earth pulls the book vertically downward ($m$ is the mass of the book, and $\bm g$ is the gravitational acceleration).
• Normal force $\bm R$: the force with which the desk pushes the book vertically upward.

The book stays at rest because these two forces have equal magnitudes and opposite directions, so that the resultant force is zero (equilibrium of forces): $m\bm g+\bm R=\bm 0$. However, this is not a pair under the third law (the law of action and reaction)—the law of action and reaction is a statement about “forces between two different bodies,” whereas $m\bm g$ and $\bm R$ are two different kinds of forces that both act on the same body (the book).

To find the action–reaction pairs correctly, for each force acting on the book, we ask: “What reaction does the book exert back on the body on the other side?”

• The partner of gravity $m\bm g$ is the force with which the book pulls the Earth vertically upward (by universal gravitation). This is a force acting on the Earth, not on the book.
• The partner of the normal force $\bm R$ is the force with which the book pushes the desk vertically downward. This is a force acting on the desk, and again not on the book.

To summarize: the reason the book is at rest (equilibrium of forces) is that two different forces acting on the single body, the book (gravity and the normal force), cancel each other, and this is a different matter from the law of action and reaction (two pairs, one between the book and the Earth and one between the book and the desk, each holding separately). Not confusing the two is an important point for understanding Newtonian mechanics correctly.

Solution

(1) Take the direction of motion before the collision (toward the wall) as the positive direction. The mass of the ball is $m=0.5\ \mathrm{kg}$. The momentum before the collision is $p_1=0.5\times8=4\ \mathrm{kg\,m/s}$, and after the rebound the direction is reversed, so $p_2=0.5\times(-6)=-3\ \mathrm{kg\,m/s}$. By the impulse–momentum theorem (Theorem 2.1), the impulse the ball received from the wall equals the change in momentum,

$$ I=p_2-p_1=-3-4=-7\ \mathrm{N\,s} $$

The magnitude of the impulse is $7\ \mathrm{N\,s}$, and its direction is the negative direction, that is, away from the wall (opposite to the direction of motion before the collision).

(2) Since the impulse equals “average force × contact time,” the magnitude of the average force is

$$ \bar F=\frac{\abs{I}}{\Delta t}=\frac{7\ \mathrm{N\,s}}{2.0\times10^{-3}\ \mathrm s}=3.5\times10^{3}\ \mathrm N $$

We see that a force as large as about $7\times10^{2}$ times the gravitational force $0.5\times9.8=4.9\ \mathrm N$ on the ball acts for a very short time.

(3) By Definition 2.6, since the velocity of the wall is $0$ both before and after the collision, the coefficient of restitution is

$$ e=-\frac{v_1'-0}{v_1-0}=-\frac{v_1'}{v_1}=-\frac{-6}{8}=0.75 $$

In general, if the velocity of the ball before the collision is $v_1=v\gt0$, the equation above gives the velocity after the collision as $v_1'=-ev$, and the change in the ball's momentum, that is, the impulse, is

$$ I=mv_1'-mv_1=-emv-mv=-(1+e)\,mv $$

whose magnitude is $(1+e)mv$. Substituting the values of this problem gives $(1+0.75)\times0.5\times8=7\ \mathrm{N\,s}$, in agreement with the result of (1). For $e=0$ (no rebound) the impulse is $mv$, and for $e=1$ (perfect rebound) it is $2mv$; the impulse is larger when the ball rebounds.

Solution

(1) Substituting $h=1.00\ \mathrm m,\ h'=0.64\ \mathrm m$ into $e=\sqrt{h'/h}$, which was derived in the Application in Section 2.6, we obtain

$$ e=\sqrt{\frac{0.64}{1.00}}=0.80 $$

This is the coefficient of restitution of the ball.

(2) Let $v$ be the speed just before the bounce. The speed just after the bounce is $ev$, and the height reached is proportional to the square of the speed, from $v'^2=2gh'$. Therefore the height is multiplied by $e^2$ at each bounce. The speed just before the first bounce is determined by free fall from the height $h$, so $h_1=e^2h$; then, falling from $h_1$ and bouncing a second time gives $h_2=e^2h_1$, and so on, so that

$$ h_n=e^{2n}h $$

holds. We have $h_1=0.80^2\times1.00=0.64\ \mathrm m$ (consistent with the statement of the problem) and $h_2=0.80^4\times1.00=0.4096\ \mathrm m\approx0.41\ \mathrm m$.

Solution

(1) The momentum is $\bm p=m\bm v=1.5\times(0,2,0)=(0,3,0)\ \mathrm{kg\,m/s}$. Applying the component expression of the cross product $\bm a\times\bm b=(a_yb_z-a_zb_y,\ a_zb_x-a_xb_z,\ a_xb_y-a_yb_x)$ to $\bm r=(2,0,0),\ \bm p=(0,3,0)$, we obtain

$$ \bm L=\bm r\times\bm p=\big(0\times0-0\times3,\ 0\times0-2\times0,\ 2\times3-0\times0\big)=(0,\,0,\,6)\ \mathrm{kg\,m^2/s} $$

(If we turn the fingers of the right hand from the $x$ axis toward the $y$ axis, the thumb points in the positive direction of the $z$ axis, so the $z$ component is positive.)

(2) Since $\bm r=(2,0,0)$ and $\bm F=(-4,0,0)$ are parallel, the torque is

$$ \bm N=\bm r\times\bm F=\big(0\times0-0\times0,\ 0\times(-4)-2\times0,\ 2\times0-0\times(-4)\big)=(0,0,0)=\bm 0 $$

By Law 2.5, $\dd\bm L/\dd t=\bm N=\bm 0$. If the force $\bm F$ is parallel to the position vector $\bm r$ (a central force directed toward the origin), $\bm N=\bm r\times\bm F=\bm 0$ always holds, so the angular momentum $\bm L$ is kept constant in time (Exercise 2.4). The value $\bm L=(0,0,6)\ \mathrm{kg\,m^2/s}$ at this instant is conserved and stays the same from then on.

Solution

Differentiating each component, we find that the velocity and acceleration of the curve $\bm r(t)=[t,\ t^2,\ 0]$ are

$$ \bm v(t)=[1,\ 2t,\ 0],\qquad \bm a(t)=[0,\ 2,\ 0] $$

(the acceleration is constant, independent of the time $t$). The speed is

$$ v(t)=\abs{\bm v(t)}=\sqrt{1^2+(2t)^2}=\sqrt{1+4t^2} $$

To use the formula for the radius of curvature (Formula 3.3 of Chapter 3), $\rho=v^3/\abs{\bm v\times\bm a}$, we first compute $\bm v\times\bm a$. Since this is planar motion (the $z$ component is always 0), the cross product has only a $z$ component, whose value is $v_xa_y-v_ya_x$:

$$ \bm v\times\bm a=[\,0,\ 0,\ (1)(2)-(2t)(0)\,]=[0,\ 0,\ 2] \qquad\therefore\quad \abs{\bm v\times\bm a}=2 $$

(That $\abs{\bm v\times\bm a}$ turned out to be constant, independent of $t$, is due to a circumstance peculiar to this curve: the acceleration $\bm a$ is constant, and the $x$ component of the velocity is also constant.) Substituting into Formula 3.3 gives

$$ \rho(t)=\frac{v(t)^3}{\abs{\bm v\times\bm a}}=\frac{(1+4t^2)^{3/2}}{2} $$

which is the required general formula.

Check at $t=1$: $\rho(1)=\dfrac{(1+4)^{3/2}}{2}=\dfrac{5^{3/2}}{2}=\dfrac{5\sqrt5}{2}$, which agrees exactly with $\rho=5\sqrt5/2\ \mathrm m$ found in Example 3.4 of Chapter 3 (by the method of decomposing the acceleration into tangential and normal components). Thus two independent methods (the method of decomposition into components and the method of the cross-product formula) have reached the same answer for the same curve and the same point.

Solution

First we convert the units to $\mathrm{m/s}$:

$$ 72\ \mathrm{km/h}=\frac{72\times1000\ \mathrm m}{3600\ \mathrm s}=20\ \mathrm{m/s} $$

Since the radius of curvature of a circular-arc curve equals the radius of the circle (Section 3.4), $\rho=50\ \mathrm m$.

(a) By Theorem 3.2 of Chapter 3, the magnitude of the normal acceleration is

$$ \frac{v^2}{\rho}=\frac{(20\ \mathrm{m/s})^2}{50\ \mathrm m}=\frac{400}{50}\ \mathrm{m/s^2}=8.0\ \mathrm{m/s^2} $$

(the unit is $\mathrm{(m/s)^2/m=m/s^2}$, which agrees with the unit of acceleration). Its direction is toward the center of the curve.

(b) Let the mass of the car be $m$. The centripetal force needed to go around the curve is $m v^2/\rho$ (Newton's second law), and this is supplied by the static friction from the horizontal road surface. The maximum static friction is $\mu mg$, so the condition for not skidding sideways is

$$ \frac{mv^2}{\rho}\le\mu mg \qquad\therefore\quad \mu\ge\frac{v^2}{\rho g}=\frac{8.0}{9.8}\approx0.82 $$

(the mass $m$ cancels from both sides and disappears; that is, the condition is the same for a light car and for a heavy car). Therefore, unless the coefficient of static friction is at least about $0.82$, the car skids sideways. If the speed doubles, $v^2$ becomes four times as large, so the required $\mu$ also becomes four times as large.

(c) Since the car is decelerating, the tangential acceleration is the rate of change of the speed, $\dd v/\dd t=-2.0\ \mathrm{m/s^2}$ (opposite to the direction of travel). The normal acceleration is the $8.0\ \mathrm{m/s^2}$ of (a), and since the tangential and normal components are perpendicular to each other, their magnitudes can be combined by the Pythagorean theorem:

$$ \abs{\bm a}=\sqrt{\left(\frac{\dd v}{\dd t}\right)^2+\left(\frac{v^2}{\rho}\right)^2}=\sqrt{(-2.0)^2+8.0^2}\ \mathrm{m/s^2}=\sqrt{68}\ \mathrm{m/s^2}\approx8.2\ \mathrm{m/s^2} $$

Even with the deceleration, the normal component accounts for most of the magnitude of the acceleration.

Solution

Since $\bm e_x,\bm e_y$ are constant vectors independent of time, when we differentiate $\bm e_r=\cos\theta\,\bm e_x+\sin\theta\,\bm e_y$ with respect to time, only the coefficients $\cos\theta,\sin\theta$ are differentiated. By the rule for differentiating a composite function, $\dfrac{\dd}{\dd t}\cos\theta=-\dot\theta\sin\theta$ and $\dfrac{\dd}{\dd t}\sin\theta=\dot\theta\cos\theta$, so

$$ \frac{\dd\bm e_r}{\dd t}=-\dot\theta\sin\theta\,\bm e_x+\dot\theta\cos\theta\,\bm e_y=\dot\theta\,(-\sin\theta\,\bm e_x+\cos\theta\,\bm e_y)=\dot\theta\,\bm e_\theta $$

(the last equality is the definition of $\bm e_\theta$, Formula 3.2 of Chapter 3). Similarly, differentiating $\bm e_\theta=-\sin\theta\,\bm e_x+\cos\theta\,\bm e_y$ gives

$$ \frac{\dd\bm e_\theta}{\dd t}=-\dot\theta\cos\theta\,\bm e_x-\dot\theta\sin\theta\,\bm e_y=-\dot\theta\,(\cos\theta\,\bm e_x+\sin\theta\,\bm e_y)=-\dot\theta\,\bm e_r $$

as required.

Next, we differentiate $\bm r=R\,\bm e_r$ ($R$ is constant) with respect to time. Since $R$ is a constant, it can be taken outside, and using $\dot\theta=\omega$ we obtain

$$ \bm v=\frac{\dd\bm r}{\dd t}=R\,\frac{\dd\bm e_r}{\dd t}=R\dot\theta\,\bm e_\theta=R\omega\,\bm e_\theta $$

The velocity points in the direction of $\bm e_\theta$ (the tangential direction of the circle), and the speed is $R\omega$. This also agrees with the Outlook in Section 3.4 (for uniform circular motion, $\bm e_t=\bm e_\theta$ and $\dd\bm e_t/\dd t=-\omega\bm e_r$, so that $\bm e_n=-\bm e_r$).

Solution

By Formula 3.2 of Chapter 3, the components of the polar-coordinate basis vectors are

$$ \bm e_r=[\cos\theta,\ \sin\theta],\qquad \bm e_\theta=[-\sin\theta,\ \cos\theta] $$

The inner product (dot product) is computed as the sum of the products of corresponding components (the definition of the inner product in University Mathematics, Section 15.2):

$$ \bm e_r\cdot\bm e_\theta=(\cos\theta)\times(-\sin\theta)+(\sin\theta)\times(\cos\theta) $$

Expanding the right-hand side gives

$$ =-\sin\theta\cos\theta+\sin\theta\cos\theta $$

and since the same term $\sin\theta\cos\theta$ appears exactly twice with opposite signs, these cancel each other, and

$$ \bm e_r\cdot\bm e_\theta=0 $$

This calculation uses no specific value of the angle $\theta$ at all (all the terms containing $\theta$ cancelled), so it always holds for any value of $\theta$—that is, we have shown that the polar-coordinate basis vectors $\bm e_r,\bm e_\theta$ are always orthogonal to each other, regardless of the position of the point (the angle $\theta$).

Solution

Let $f(x)=\sin x$. Then

$$ f'(x)=\cos x,\qquad f''(x)=-\sin x $$

Substituting into Formula 3.4 of Chapter 3, $\rho=(1+f'(x)^2)^{3/2}/\abs{f''(x)}$, we obtain

$$ \rho(x)=\frac{(1+\cos^2x)^{3/2}}{\abs{-\sin x}}=\frac{(1+\cos^2x)^{3/2}}{\abs{\sin x}} $$

which is the required general formula (we used $\abs{-\sin x}=\abs{\sin x}$).

Value at $x=\pi/2$ (the top of a peak): Substituting $\cos(\pi/2)=0,\ \sin(\pi/2)=1$, we get

$$ \rho\left(\frac\pi2\right)=\frac{(1+0^2)^{3/2}}{\abs{1}}=\frac{1^{3/2}}{1}=1 $$

That is, at the top of a peak (or the bottom of a trough) of the graph of $y=\sin x$, the radius of curvature is exactly 1. This means that near the top of a peak, the graph of $y=\sin x$ overlaps best with a circle of radius 1 (this is also convincing if we use the Taylor expansion $y=1-\dfrac{(x-\pi/2)^2}{2}+\cdots$, which shows that the behavior of $y=\sin x$ near $x=\pi/2$ has almost the same shape as the neighborhood of the top of a circle of radius 1).

Solution

The velocity and acceleration of the helical motion $\bm r(t)=[\cos t,\ \sin t,\ t]$ of Example 3.6 of Chapter 3 were

$$ \bm v(t)=[-\sin t,\ \cos t,\ 1],\qquad \bm a(t)=[-\cos t,\ -\sin t,\ 0] $$

Substituting $t=0$ ($\sin0=0,\ \cos0=1$), we obtain

$$ \bm v(0)=[-\sin0,\ \cos0,\ 1]=[0,\ 1,\ 1],\qquad \bm a(0)=[-\cos0,\ -\sin0,\ 0]=[-1,\ 0,\ 0] $$

The speed is $v(0)=\abs{\bm v(0)}=\sqrt{0^2+1^2+1^2}=\sqrt2$. The components of the cross product are

$$ \bm v(0)\times\bm a(0)=[\,(1)(0)-(1)(0),\ \ (1)(-1)-(0)(0),\ \ (0)(0)-(1)(-1)\,]=[0,\ -1,\ 1] $$

and $\abs{\bm v(0)\times\bm a(0)}=\sqrt{0^2+(-1)^2+1^2}=\sqrt2$. By Formula 3.3,

$$ \rho(0)=\frac{v(0)^3}{\abs{\bm v(0)\times\bm a(0)}}=\frac{(\sqrt2)^3}{\sqrt2}=\frac{2\sqrt2}{\sqrt2}=2 $$

and we have confirmed that this indeed agrees, without contradiction even at the specific time $t=0$, with the result “$\rho=2$ (constant, independent of the time $t$)” found in Example 3.6.

Solution

Recall the definition of the inner product (dot product) $\bm a\cdot\bm b=\abs{\bm a}\abs{\bm b}\cos\theta$ ($\theta$ is the angle between the two vectors $\bm a,\bm b$). Since we assume $\bm a=\bm F\neq\bm0$ and $\bm b=\dd\bm r\neq\bm0$, we have $\abs{\bm a}\neq0$ and $\abs{\bm b}\neq0$. Therefore, the product $\abs{\bm a}\abs{\bm b}\cos\theta$ is 0 only when the remaining factor $\cos\theta$ is 0, that is, only when $\theta=90^\circ$ ($\bm F$ and $\dd\bm r$ are perpendicular to each other).

In other words, a force does no work when its direction is exactly perpendicular to the direction in which the particle moves (the direction of the displacement). This is because, no matter how large the force is, its component perpendicular to the displacement has no influence whatsoever on the “direction” of the motion (the intuition of Section 4.1 of the main text: the inner product measures the “magnitude of the influence” of the force in the direction of the displacement, and the influence of the perpendicular component is exactly 0).

We apply this result to the centripetal force in uniform circular motion. The centripetal force is a force that always points toward the center of the circle. On the other hand, the velocity of a particle in uniform circular motion (that is, the direction of the displacement $\dd\bm r/\dd t$) points along the tangent to the circle at the particle's position at that instant. The radial direction of the circle (the direction of the centripetal force, which points toward the center) and the tangential direction are always orthogonal geometrically (that the radius and the tangent meet at right angles is a basic property of the circle). Therefore, since the direction of the centripetal force and the direction of the displacement always make an angle of $90^\circ$, the centripetal force never does work during the motion—this is also the reason why the kinetic energy (and hence the speed) remains constant in uniform circular motion (the work–energy theorem, Theorem 4.1: if the work is 0, the change in kinetic energy is also 0).

Solution

(1) For $\bm F=[F_x,F_y,F_z]=[2xy,\ x^2,\ 0]$, we compute each component of the defining formula of the curl (Section 4.4.2). Since $F_z=0$ does not depend on $x,y$, the first and second components are immediately seen to be 0:

$$ \pdiff{F_z}{y}-\pdiff{F_y}{z}=0-0=0, \qquad \pdiff{F_x}{z}-\pdiff{F_z}{x}=0-0=0 $$

The third component is

$$ \pdiff{F_y}{x}-\pdiff{F_x}{y}=\pdiff{}{x}(x^2)-\pdiff{}{y}(2xy)=2x-2x=0 $$

Therefore $\rot\bm F=[0,0,0]=\bm0$, and by Theorem 4.2, $\bm F$ is a conservative force.

(2) From $\bm F=-\nabla U$ we have $F_x=-\pdiff Ux=2xy$, so $\pdiff Ux=-2xy$. Integrating this with respect to $x$ (we integrate regarding $y$ as a constant, so the constant of integration is in general a function $g(y)$ of $y$), we obtain

$$ U=-x^2y+g(y) $$

Next, we determine $g(y)$ by using the fact that this $U$ must also satisfy $F_y=-\pdiff Uy=x^2$. Differentiating $U=-x^2y+g(y)$ partially with respect to $y$ gives $\pdiff Uy=-x^2+g'(y)$. On the other hand, from $F_y=-\pdiff Uy=x^2$ we must have $\pdiff Uy=-x^2$. Comparing the two,

$$ -x^2+g'(y)=-x^2 \quad\Longrightarrow\quad g'(y)=0 \quad\Longrightarrow\quad g(y)=(\text{constant}) $$

Taking the constant to be 0, we get

$$ U(x,y)=-x^2y $$

As a check, $-\pdiff{}{x}(-x^2y)=2xy=F_x$ and $-\pdiff{}{y}(-x^2y)=x^2=F_y$, so indeed $\bm F=-\nabla U$ is satisfied.

Solution

(1)(i) The general formula shown in the “Discussion” of Section 4.3 says that when a particle is pulled slowly by a tangential force along a curve $y=f(x)$ ($0\le x\le X$) (a path along which $x$ only increases), the work done by the frictional force is

$$ W_{\text{friction}}=-\mu mgX $$

($X$ is the horizontal distance traveled; it does not depend on the shape of the curve in between). For $y=x^2\ (0\le x\le1)$ the horizontal distance is $X=1$, so simply substituting this gives

$$ W_{\text{friction}}=-\mu mg\times1=-\mu mg $$

as the work done by the frictional force.

(ii) We compute directly. The unit tangent vector is $\bm e_t=[1,f'(x)]/\sqrt{1+f'(x)^2}$, and since $f(x)=x^2$ we have $f'(x)=2x$, so

$$ \bm e_t=\frac{[1,2x]}{\sqrt{1+4x^2}} $$

From the equilibrium in the normal direction (the particle is pulled by a tangential force, so the normal component of gravity and the normal force balance each other), by the same derivation as in Illustration 4.1 of the main text, the magnitude of the normal force is $N=mg/\sqrt{1+f'(x)^2}=mg/\sqrt{1+4x^2}$, the kinetic friction is $\bm F_2=-\mu N\bm e_t$, and the arc-length element is $\dd s=\sqrt{1+4x^2}\,\dd x$. Since $\dd\bm r=\bm e_t\,\dd s$, the work done by the frictional force is

$$ W_{\text{friction}}=\int_0^1\bm F_2\cdot\bm e_t\,\dd s=\int_0^1(-\mu N)\,\dd s=-\int_0^1\mu\cdot\frac{mg}{\sqrt{1+4x^2}}\cdot\sqrt{1+4x^2}\,\dd x=-\int_0^1\mu mg\,\dd x=-\mu mg $$

(that the square-root factor cancels between the numerator and the denominator is exactly the same mechanism as in the generalization of Illustration 4.1). In both (i) and (ii) we obtain $W_{\text{friction}}=-\mu mg$, and the two results agree.

(2) The work done by friction while the particle is pulled from $x=0$ to $x=1$ is $-\mu mg\cdot1$ by (1). While it is pulled back ($x$ decreasing from $1$ to $0.5$), the kinetic friction is again directed opposite to the motion, and its magnitude is $\mu N$. Writing $\dd\bm r=\bm e_t'\,\dd s$ in this case ($\bm e_t'$ is the unit vector in the direction of travel, which is now the direction in which $x$ decreases), we have $\bm F_2=-\mu N\bm e_t'$, and $\bm F_2\cdot\dd\bm r=-\mu N\,\dd s$ is unchanged. Since the increment of the arc length $s$ is the positive quantity $\dd s=\sqrt{1+4x^2}\,\abs{\dd x}$,

$$ W_{\text{pull-back}}=-\int\mu N\,\dd s=-\int_{0.5}^{1}\mu\frac{mg}{\sqrt{1+4x^2}}\sqrt{1+4x^2}\,\dd x=-\mu mg(1-0.5)=-0.5\,\mu mg $$

(when integrating over $x$ from $1$ to $0.5$, we use $\abs{\dd x}$ so that the length $0.5$ of the interval of integration is taken as positive). Therefore the work done by friction over the whole trip is

$$ W_{\text{total}}=-\mu mg-0.5\,\mu mg=-1.5\,\mu mg $$

On the other hand, if the particle is pulled directly from $x=0$ to $x=0.5$, the general formula of (1) with $X=0.5$ gives $W=-0.5\,\mu mg$. Although the starting point $x=0$ and the endpoint $x=0.5$ are the same for both paths, the work done by friction differs: $-1.5\,\mu mg$ and $-0.5\,\mu mg$. Since the work done by a conservative force must be determined by the starting point and the endpoint alone, independent of the path ($W_c=U(P_1)-U(P_2)$ in Section 4.4.1), the frictional force is not a conservative force. (For a path along which $x$ only increases, the work is determined by the horizontal distance $X$ alone, but if $x$ goes back and forth, the horizontal distance covered in the round trip is added.)

Solution

(1) Take the direction in which the object slides down along the incline as positive. The component of gravity along the incline is $mg\sin\alpha$ (in the sliding direction, positive), and the object moves a distance $L$, so by $W=\bm F\cdot\Delta\bm r$ of Section 4.1 (for straight-line motion),

$$ W_{\text{gravity}}=mg\sin\alpha\times L=mgL\sin\alpha $$

(positive work: gravity helps the motion along). The magnitude of the normal force is $N=mg\cos\alpha$ from the equilibrium of forces perpendicular to the incline, and the magnitude of the kinetic friction is $\mu N=\mu mg\cos\alpha$, directed so as to oppose the motion (opposite to the direction of sliding down the incline). Therefore,

$$ W_{\text{friction}}=-\mu mg\cos\alpha\times L=-\mu mgL\cos\alpha $$

(negative work).

(2) The object started to slide gently (initial speed 0), so the initial kinetic energy is $K_1=0$. The work done by the net force is the sum of the work done by gravity, the frictional force, and the normal force (the normal force does no work because it is perpendicular to the direction of motion), so $W=W_{\text{gravity}}+W_{\text{friction}}$. By the work–energy theorem (Theorem 4.1),

$$ W_{\text{gravity}}+W_{\text{friction}}=K_2-K_1=\frac12mv^2-0 $$

Solving this for $v$, we obtain

$$ v=\sqrt{\frac{2(W_{\text{gravity}}+W_{\text{friction}})}{m}}=\sqrt{\frac{2\left(mgL\sin\alpha-\mu mgL\cos\alpha\right)}{m}}=\sqrt{2gL\left(\sin\alpha-\mu\cos\alpha\right)} $$

(Note that the mass $m$ cancels out. This result is meaningful only when $\sin\alpha>\mu\cos\alpha$, that is, when $\tan\alpha>\mu$; in other words, only when the incline overcomes friction so that the object starts to slide at all.)

Solution

From the defining formula of the curl (Section 4.4.2), writing $\rot\bm F=\bm0$ in components gives three simultaneous conditions:

$$ \pdiff{F_z}{y}-\pdiff{F_y}{z}=0,\qquad \pdiff{F_x}{z}-\pdiff{F_z}{x}=0,\qquad \pdiff{F_y}{x}-\pdiff{F_x}{y}=0 $$

This is the general form in three dimensions. Now consider a two-dimensional force for motion only in the $xy$ plane, that is, $F_z=0$ and $F_x,F_y$ independent of $z$ ($\pdiff{F_x}{z}=\pdiff{F_y}{z}=0$), and substitute these conditions into the three equations.

First equation: $\pdiff{F_z}{y}-\pdiff{F_y}{z}=\pdiff{}{y}(0)-0=0-0=0$—this is satisfied automatically because $F_z=0$ (a constant).

Second equation: $\pdiff{F_x}{z}-\pdiff{F_z}{x}=0-\pdiff{}{x}(0)=0-0=0$—this too is satisfied automatically.

Third equation: $\pdiff{F_y}{x}-\pdiff{F_x}{y}=0$—since $F_x,F_y$ remain as functions of $x,y$, this is not satisfied automatically, and it remains as a genuinely meaningful condition.

Therefore, for a two-dimensional force, the condition for being conservative simplifies to effectively just one of the three equations,

$$ \pdiff{F_y}{x}-\pdiff{F_x}{y}=0 \qquad\Bigl(\text{that is}\ \pdiff{F_y}{x}=\pdiff{F_x}{y}\Bigr) $$

This expresses the fact that, of the three-dimensional $\rot\bm F$, only the component perpendicular to the $xy$ plane (the $z$ component) is meaningful, and the other two components vanish identically in two-dimensional motion. (The condition used in Exercise 4.2 is this two-dimensional version (the third equation); in Example 4.4 we computed all three components.)

(3) For $\bm F=[y^2,\ 2xy,\ z]$, we compute the three components of $\rot\bm F$:

$$ \pdiff{F_z}{y}-\pdiff{F_y}{z}=0-0=0,\qquad \pdiff{F_x}{z}-\pdiff{F_z}{x}=0-0=0,\qquad \pdiff{F_y}{x}-\pdiff{F_x}{y}=2y-2y=0 $$

Hence $\rot\bm F=\bm0$, and by Theorem 4.2, $\bm F$ is a conservative force. We now find the potential. From $F_x=-\pdiff Ux=y^2$ we get $\pdiff Ux=-y^2$, and integrating with respect to $x$ gives $U=-xy^2+g(y,z)$ (the constant of integration is a function of $y,z$). Differentiating this partially with respect to $y$ gives $\pdiff Uy=-2xy+\pdiff gy$, and comparing with $F_y=-\pdiff Uy=2xy$, that is, $\pdiff Uy=-2xy$, we find $\pdiff gy=0$; that is, $g$ is a function $g(z)$ of $z$ only. Furthermore, from $F_z=-\pdiff Uz=z$ we have $\pdiff Uz=-z$, so $g'(z)=-z$ and $g(z)=-\dfrac{z^2}2$ (taking the constant to be 0). Therefore

$$ U(x,y,z)=-xy^2-\frac{z^2}2 $$

As a check, $-\pdiff Ux=y^2$, $-\pdiff Uy=2xy$, and $-\pdiff Uz=z$, which agree with $\bm F$.

Solution

(1) The equilibrium points are the points that satisfy $\dd U/\dd x=0$. Differentiating $U(x)=\dfrac{x^4}{4}-\dfrac{x^2}{2}$, we obtain

$$ \diff Ux=x^3-x=x(x^2-1)=x(x-1)(x+1) $$

This vanishes at the three points $x=0,\ 1,\ -1$.

(2) To judge stability, we examine the sign of the second derivative, as in Theorem 4.4. We have $\dd^2U/\dd x^2=3x^2-1$.

• $x=0$: $3(0)^2-1=-1\lt0$ ⟹ unstable equilibrium.
• $x=1$: $3(1)^2-1=2\gt0$ ⟹ stable equilibrium.
• $x=-1$: $3(-1)^2-1=2\gt0$ ⟹ stable equilibrium.

In other words, $U(x)$ has two left–right symmetric valleys (stable equilibria) at $x=\pm1$, and a hill (unstable equilibrium) at $x=0$ between them; it is a so-called “double-well” potential.

(3) By the idea of Section 4.6.2 ($U(x)\le E$), the range in which motion with mechanical energy $E=0$ is possible is the range of $x$ satisfying $U(x)\le0$. Since $U(x)=\dfrac{x^4}{4}-\dfrac{x^2}{2}=\dfrac{x^2}{4}\left(x^2-2\right)$ can be factored, $U(x)\le0$ holds when $x^2\le2$ (because $x^2/4\ge0$ always holds, the sign is determined by the factor $(x^2-2)$), that is,

$$ -\sqrt2\ \le\ x\ \le\ \sqrt2 $$

This is the energetically allowed region (the range satisfying $U(x)\le E$). The turning points for $E=0$ are $x=\pm\sqrt2$ ($U=E$ and $U'=x^3-x\neq0$; Definition 4.5).

However, a particle does not necessarily reach every part of the allowed region. At $x=0$ we have $U(0)=0=E$, and moreover $U'(0)=0$ and $U''(0)=-1\lt0$, so this is an unstable equilibrium point (the top of the hill) and not a turning point. Near $x=0$, the speed of a particle with $E=0$ approaches $0$, and the particle merely approaches $x=0$ over an infinitely long time and cannot pass through that point in a finite time (asymptotic approach to an unstable equilibrium point). Therefore, the range in which a particle starting from the valley at $x\gt0$ can actually move is $0\lt x\le\sqrt2$, and for a particle starting from the valley at $x\lt0$ it is limited to $-\sqrt2\le x\lt0$. If $E$ exceeds $0$ even slightly, the particle can cross $x=0$ and go back and forth between the left and right valleys. $E=0$ is at the same height as the top of the central hill, and it marks the boundary between motion that “crosses between the wells” and motion that is “confined to one of the wells.”

Solution

For $\bm F=[-y,\ x,\ 0]$ we have $\bm F\cdot\dd\bm r=-y\,\dd x+x\,\dd y$. We compute the four sides of the square in order.

• Side 1: $(1,-1)\to(1,1)$. Here $x=1$ is constant, $\dd x=0$, and $y$ goes from $-1$ to $1$. Since $\bm F\cdot\dd\bm r=x\,\dd y=\dd y$, we have $\displaystyle\int_{-1}^{1}\dd y=2$.
• Side 2: $(1,1)\to(-1,1)$. Here $y=1$ is constant, $\dd y=0$, and $x$ goes from $1$ to $-1$. Since $\bm F\cdot\dd\bm r=-y\,\dd x=-\dd x$, we have $\displaystyle\int_{1}^{-1}(-\dd x)=-(-1-1)=2$.
• Side 3: $(-1,1)\to(-1,-1)$. Here $x=-1$ is constant, and $y$ goes from $1$ to $-1$. Since $\bm F\cdot\dd\bm r=x\,\dd y=-\dd y$, we have $\displaystyle\int_{1}^{-1}(-\dd y)=2$.
• Side 4: $(-1,-1)\to(1,-1)$. Here $y=-1$ is constant, and $x$ goes from $-1$ to $1$. Since $\bm F\cdot\dd\bm r=-y\,\dd x=\dd x$, we have $\displaystyle\int_{-1}^{1}\dd x=2$.

Therefore, the work around one loop is

$$ \oint\bm F\cdot\dd\bm r=2+2+2+2=8\neq0 $$

On the other hand, by the result of Example 4.7, the third component of $\rot\bm F$ is $\pdiff{F_y}{x}-\pdiff{F_x}{y}=2$, and the area of the square is $2\times2=4$, so their product is $2\times4=8$, which exactly coincides with the work around one loop (this is a concrete example of Stokes's theorem (University Mathematics, Section 20.5), “the line integral around a closed path equals the surface integral of the curl”). Since the work around one loop is not 0, and for a conservative force the work around one loop would have to be 0, this force is not conservative.

Solution

(1) Symmetry of the range By the result of Section 5.1, the horizontal range of projectile motion without air resistance is, for initial speed $v_0$ and launch angle $\theta$,

$$ R(\theta)=\frac{v_0^2\sin2\theta}{g} $$

If the launch angle is changed to $90^\circ-\theta$, then $2(90^\circ-\theta)=180^\circ-2\theta$, so using $\sin(180^\circ-x)=\sin x$ (on the unit circle, the points at angles $x$ and $180^\circ-x$ have equal $y$ coordinates), we obtain

$$ R(90^\circ-\theta)=\frac{v_0^2\sin(180^\circ-2\theta)}{g}=\frac{v_0^2\sin2\theta}{g}=R(\theta) $$

Hence the horizontal range is the same for $\theta$ and $90^\circ-\theta$ ($\theta=45^\circ$ is its own partner, since $90^\circ-45^\circ=45^\circ$, so it has no separate pair; it is the launch angle at which $\sin2\theta$ takes its maximum value 1).

(2) Time and height of the highest point The vertical velocity in projectile motion without air resistance is given by the general expression obtained by integrating the equation of motion (Section 5.1):

$$ v_y(t)=v_0\sin\theta-gt $$

The highest point is the instant at which the ball turns from rising to falling, that is, the time at which the vertical velocity is exactly 0. So we set $v_y(t)=0$ and solve for $t$:

$$ v_0\sin\theta-gt_{\text{top}}=0 \qquad\therefore\ t_{\text{top}}=\frac{v_0\sin\theta}{g} $$

Substituting $v_0=15\ \mathrm{m/s}$, $\theta=45^\circ$ ($\sin45^\circ=\dfrac{\sqrt2}{2}\approx0.7071$), and $g=9.8\ \mathrm{m/s^2}$ gives

$$ t_{\text{top}}=\frac{15\times0.7071}{9.8}=\frac{10.607}{9.8}\approx1.082\ \mathrm{s} $$

Next, the height $H$ of the highest point is found by substituting $t=t_{\text{top}}$ into the expression for the vertical position, $y(t)=v_0t\sin\theta-\frac12gt^2$ (the same procedure as in the derivation in Section 5.1):

$$ H=v_0t_{\text{top}}\sin\theta-\frac12g\,t_{\text{top}}^2 =\frac{v_0^2\sin^2\theta}{g}-\frac12\cdot\frac{v_0^2\sin^2\theta}{g} =\frac{v_0^2\sin^2\theta}{2g} $$

Substituting the numerical values gives

$$ H=\frac{15^2\times0.7071^2}{2\times9.8}=\frac{225\times0.5}{19.6}=\frac{112.5}{19.6}\approx5.740\ \mathrm{m} $$

Therefore, the time at which the highest point is reached is about $1.082\ \mathrm s$, and the height at that moment is about $5.740\ \mathrm m$.

Solution

Under air resistance proportional to the velocity, the vertical velocity is given by (Section 5.2)

$$ v_y(t)=\Big(v_{0y}+\frac{g}{b}\Big)e^{-bt}-\frac{g}{b} $$

We consider the limit $t\to\infty$. Since $b\gt0$, the exponential function $e^{-bt}$ approaches 0 as $t\to\infty$ (for any initial velocity $v_{0y}$, the coefficient $\big(v_{0y}+\dfrac gb\big)$ is a finite constant, so the whole term obtained by multiplying it by $e^{-bt}$, which approaches 0, also approaches 0). Therefore,

$$ \lim_{t\to\infty}v_y(t)=\underbrace{\Big(v_{0y}+\frac gb\Big)\times0}_{\to\,0}-\frac gb=-\frac gb $$

Thus the limiting value $-g/b$ is not affected at all by the initial velocity $v_{0y}$ with which the object was thrown (whether large or small, upward or downward)—the information about the initial velocity is “forgotten” through the decay $e^{-bt}\to0$.

The physical meaning of this can be understood from the balance of forces. The object is acted on by gravity $mg$ downward and by air resistance $B|v_y|$ in the direction opposite to the velocity. When the object has accelerated enough and its falling speed has become large, the air resistance also grows, and eventually it balances gravity exactly:

$$ mg=B|v_y| \qquad\therefore\ |v_y|=\frac{mg}{B}=\frac gb $$

This speed, at which the net force (and hence the acceleration) becomes 0, is the vertically downward terminal velocity (Definition 5.1). However the initial velocity is chosen, once enough time has passed the object always settles into this state of force balance (the terminal velocity), so it is only natural that the limit as $t\to\infty$ does not depend on the initial velocity. (If the initial velocity is already faster than the terminal velocity, the resistance is larger than gravity, the object decelerates, and again approaches the same terminal velocity.)

Solution

(1) From the second derivative of the position vector (the method of Section 5.4 of this chapter)

The position vector of uniform circular motion with radius $r$ and (constant) angular velocity $\omega$ is $\bm r(t)=[r\cos\omega t,\,r\sin\omega t]$, and differentiating it twice (the derivation in Section 5.4) gives the acceleration

$$ \bm a(t)=-\omega^2\bm r(t) $$

Its magnitude is $|\bm a|=\omega^2|\bm r(t)|=r\omega^2$ ($|\bm r(t)|=r$ is constant). On the other hand, the magnitude of the velocity (the speed) is $v=|\bm v|=r\omega$, since $\bm v(t)=\dot{\bm r}(t)=r\omega[-\sin\omega t,\cos\omega t]$, so substituting $\omega=v/r$ into $|\bm a|=r\omega^2$ gives

$$ |\bm a|=r\left(\frac vr\right)^2=\frac{v^2}{r} $$

Thus we obtain $|\bm a|=v^2/r=r\omega^2$.

(2) From the decomposition into tangential and normal components (the method of Chapter 3)

In Chapter 3 we learned that the acceleration in general curvilinear motion can be decomposed into a component along the direction of motion (the tangential direction, unit vector $\bm e_t$) and a component in the direction perpendicular to it (the principal normal direction, unit vector $\bm e_n$, pointing to the side toward which the curve bends):

$$ \bm a=\dot v\,\bm e_t+\frac{v^2}{\rho}\,\bm e_n $$

($v$ is the speed and $\rho$ is the radius of curvature). In uniform circular motion, the speed $v=r\omega$ is constant in time, so the tangential component is $\dot v=0$. Also, for motion on a circle, the bending of the curve (the circle) is characterized by the radius of the circle itself, so the radius of curvature is $\rho=r$. Substituting these, we obtain

$$ \bm a=0\cdot\bm e_t+\frac{v^2}{r}\,\bm e_n=\frac{v^2}{r}\,\bm e_n $$

Thus the magnitude of the acceleration is $|\bm a|=v^2/r$, and its direction is the principal normal direction $\bm e_n$ (the direction toward the center of the circle). This is exactly the same result, in both magnitude and direction, as the centripetal acceleration found in (1), and substituting $v=r\omega$ again gives $v^2/r=r\omega^2$.

The two approaches—(1) differentiating the position vector twice as a vector in its rectangular (Cartesian) components, and (2) decomposing the velocity and acceleration into the tangential and normal directions along the “direction” of the motion—differ in their starting points and in their computational tools, but they reach the same conclusion about the same physical phenomenon (the centripetal acceleration of uniform circular motion). This is one concrete example of the fact that physical laws do not depend on the choice of coordinate system.

Solution

(1) Orbital speed in symbols When an artificial satellite (mass $m$) moves in uniform circular motion in a circular orbit of radius $r$ around the Earth (mass $M$), the only force acting on the satellite is the gravitational attraction from the Earth. By Newton's law of universal gravitation, its magnitude is

$$ F_{\text{grav}}=\frac{GMm}{r^2} $$

and its direction always points toward the center of the Earth (the center of the circular orbit). On the other hand, the magnitude of the centripetal force required for this satellite to move in uniform circular motion of radius $r$ and speed $v$ is obtained by rewriting the centripetal-force expression of Chapter 5 (Definition 5.2), $F=m\omega^2r$ (the form with angular velocity $\omega$), using the speed $v=r\omega$ (that is, $\omega=v/r$): $F=m(v/r)^2r=mv^2/r$. Hence

$$ F_{\text{centripetal}}=\frac{mv^2}{r} $$

The force that keeps the satellite in its circular orbit is precisely this gravitational attraction itself, so the two forces are equal:

$$ \frac{mv^2}{r}=\frac{GMm}{r^2} $$

Dividing both sides by the mass $m$ (the satellite's own mass; since $m$ appears in the same form in both the gravitational-force expression and the centripetal-force expression, it cancels), we obtain

$$ \frac{v^2}{r}=\frac{GM}{r^2} \qquad\therefore\ v^2=\frac{GM}{r} $$

Choosing $v\gt0$ and taking the square root, the orbital speed is

$$ v=\sqrt{\frac{GM}{r}} $$

The fact that the satellite's own mass $m$ does not appear in the final answer is an important result: it means that once the orbital radius $r$ is fixed, the satellite circles at the same speed however heavy it is (regardless of its mass). (This is for the same reason that the acceleration of free fall does not depend on mass—both the gravitational attraction and the two sides of the equation of motion are proportional to the mass $m$.)

(2) Numerical values Substitute $G=6.674\,30\times10^{-11}\ \mathrm{N\,m^2/kg^2}$, $M=5.972\times10^{24}\ \mathrm{kg}$, and $r=6.371\times10^{6}\ \mathrm m$ into $v=\sqrt{GM/r}$. First, $GM=6.674\,30\times10^{-11}\times5.972\times10^{24}\approx3.986\times10^{14}\ \mathrm{N\,m^2/kg}=\mathrm{m^3/s^2}$, and

$$ v=\sqrt{\frac{3.986\times10^{14}}{6.371\times10^{6}}}=\sqrt{6.256\times10^{7}}\approx7.91\times10^{3}\ \mathrm{m/s}=7.91\ \mathrm{km/s} $$

(The units, $\mathrm{\sqrt{(m^3/s^2)/m}}=\mathrm{m/s}$, are correct.) The period is found by dividing the length of one revolution, $2\pi r$, by the speed $v$:

$$ T=\frac{2\pi r}{v}=\frac{2\pi\times6.371\times10^{6}}{7.91\times10^{3}}\approx5.06\times10^{3}\ \mathrm s\approx84\ \text{min} $$

Thus the speed of a circular orbit just skimming the Earth's surface is about $7.9\ \mathrm{km/s}$ (the first cosmic velocity), and the time for one revolution is about 84 minutes.

Solution

Under air resistance proportional to the velocity, the horizontal position is given by (Section 5.2)

$$ x(t)=\frac{v_{0x}}{b}\big(1-e^{-bt}\big) $$

Taking the limit $t\to\infty$, since $e^{-bt}\to0$ ($b\gt0$), we have

$$ \lim_{t\to\infty}x(t)=\frac{v_{0x}}{b}\big(1-0\big)=\frac{v_{0x}}{b} $$

That is, it converges to a finite value. This is in contrast to the vertical position $y(t)$, which (because of the term $-\dfrac gb\,t$ in its expression) keeps decreasing without bound as $t\to\infty$.

The physical reason for this difference is that the horizontal and vertical directions differ in whether there is a “balancing force.” In the horizontal direction gravity does not act, and only air resistance acts (equation of motion $\dot v_x=-bv_x$). Therefore, the horizontal velocity approaches 0 exponentially with time, the object keeps decelerating indefinitely, and finally comes to a stop (its speed becomes small enough to be regarded as 0 for practical purposes). When the velocity approaches 0, the object scarcely advances any further horizontally, so the horizontal position approaches a particular value $v_{0x}/b$ and levels off. In the vertical direction, on the other hand, gravity keeps acting all the time, so the velocity (instead of approaching 0 as in the horizontal direction) approaches the terminal velocity $-g/b$, which is a finite constant value (it does not become 0). As long as the object keeps falling at the terminal velocity, the vertical position keeps decreasing forever (at a constant rate), so $y(t)$ diverges. To summarize: “because of air resistance, the horizontal motion eventually stops, but in the vertical direction the object is pulled by gravity and keeps falling (at the terminal velocity)”—this is the physical meaning of $x(t)$ converging while $y(t)$ diverges, and it is also consistent with the obvious fact that, even with air resistance, the object cannot avoid reaching the ground (or the water surface) itself.

Solution

Under resistance proportional to the square of the velocity, the terminal velocity is given by (Formula 5.3)

$$ v_{\text t}=\sqrt{\frac{mg}{c}},\qquad c=\frac12C_{\mathrm d}\rho_{\text{air}}A $$

Using the same $m=2.0\ \mathrm{kg}$, $C_{\mathrm d}=1.05$, and $\rho_{\text{air}}=1.2\ \mathrm{kg/m^3}$ as in Example 5.5, and changing only the cross-sectional area to $A=0.50\ \mathrm{m^2}$ (twice the $A=0.25\ \mathrm{m^2}$ of Example 5.5), we recompute the drag coefficient:

$$ c=\frac12\times1.05\times1.2\times0.50=0.315\ \mathrm{kg/m} $$

This is exactly twice the $c\approx0.1575\ \mathrm{kg/m}$ of Example 5.5 (from the form $c=\frac12C_{\mathrm d}\rho_{\text{air}}A$ one can read off directly that doubling $A$ doubles $c$). Substituting this into the expression for the terminal velocity gives

$$ v_{\text t}=\sqrt{\frac{2.0\times9.8}{0.315}}=\sqrt{62.22}\approx7.89\ \mathrm{m/s} $$

How many times its former value the terminal velocity becomes when the cross-sectional area is doubled can be found without redoing the calculation, from the relations $c\propto A$ and $v_{\text t}\propto1/\sqrt c$ (proportionalities that can be seen from the form of the formulas): if $A$ doubles, $c$ also doubles, and $v_{\text t}$ is inversely proportional to the square root of the drag coefficient, so

$$ \frac{v_{\text{t,new}}}{v_{\text{t,old}}}=\sqrt{\frac{c_{\text{old}}}{c_{\text{new}}}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt2}\approx0.707 $$

That is, the terminal velocity becomes about 0.707 times its former value. Indeed, multiplying $v_{\text t}\approx11.16\ \mathrm{m/s}$ of Example 5.5 by $1/\sqrt2\approx0.707$ gives $11.16\times0.707\approx7.89\ \mathrm{m/s}$, which agrees with the value computed directly above. Physically, this agrees with the intuition that when a parachute is opened and the cross-sectional area is increased, the object experiences stronger air resistance even at the same speed, so the speed at which the resistance balances gravity (the terminal velocity) becomes smaller.

Solution

By Formula 5.2 of Section 5.2, the velocity and position in projectile motion under resistance proportional to the velocity are

$$ v_x=v_{0x}e^{-bt},\qquad v_y=\Big(v_{0y}+\frac gb\Big)e^{-bt}-\frac gb, $$ $$ x=\frac{v_{0x}}{b}\big(1-e^{-bt}\big),\qquad y=\frac1b\Big(v_{0y}+\frac gb\Big)\big(1-e^{-bt}\big)-\frac gb\,t $$

With $v_{0x}=12\ \mathrm{m/s}$, $v_{0y}=16\ \mathrm{m/s}$, $b=0.50\ \mathrm{s^{-1}}$, and $g=9.8\ \mathrm{m/s^2}$, we have $g/b=19.6\ \mathrm{m/s}$ and $v_{0y}+g/b=35.6\ \mathrm{m/s}$. At $t=1.0\ \mathrm s$, $bt=0.50$, $e^{-0.50}\approx0.6065$, and $1-e^{-0.50}\approx0.3935$, so

$$ \begin{aligned} x&=\frac{12}{0.50}\times0.3935=24\times0.3935\approx9.44\ \mathrm m,\\ y&=\frac{35.6}{0.50}\times0.3935-19.6\times1.0=71.2\times0.3935-19.6\approx28.02-19.6\approx8.42\ \mathrm m \end{aligned} $$

Also, the velocities are

$$ v_x=12\times0.6065\approx7.28\ \mathrm{m/s},\qquad v_y=35.6\times0.6065-19.6\approx21.59-19.6\approx1.99\ \mathrm{m/s} $$

(Since $v_y\gt0$, the object is still rising at $t=1.0\ \mathrm s$.) Without air resistance, the position at the same time is $x=v_{0x}t=12\ \mathrm m$, $y=v_{0y}t-\frac12gt^2=16-4.9=11.1\ \mathrm m$. Therefore, with resistance, the horizontal position decreases from $12\ \mathrm m$ to about $9.4\ \mathrm m$, and the vertical position from $11.1\ \mathrm m$ to about $8.4\ \mathrm m$; both become smaller.

Solution

In Chapter 6 we saw that the relation between the basis vectors can be written in matrix form as $[\bm e_r\ \bm e_\theta]=[\bm e_x\ \bm e_y]R(\theta)$ (where $R(\theta)=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}$). To solve this for $[\bm e_x\ \bm e_y]$, it suffices to multiply both sides by $R(\theta)^{-1}$ from the right:

$$ [\bm e_x\ \bm e_y]=[\bm e_r\ \bm e_\theta]\,R(\theta)^{-1} $$

As we confirmed in Section 6.1 (ii), the inverse matrix of a rotation matrix equals its transpose ($R(\theta)^{-1}=R(\theta)^{\mathrm T}=\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}$). Substituting this gives

$$ [\bm e_x\ \bm e_y]=[\bm e_r\ \bm e_\theta]\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix} $$

Writing out this matrix product column by column (the first column gives $\bm e_x$, and the second column gives $\bm e_y$), we obtain

$$ \bm e_x=\cos\theta\,\bm e_r-\sin\theta\,\bm e_\theta,\qquad \bm e_y=\sin\theta\,\bm e_r+\cos\theta\,\bm e_\theta $$

This is the inverse relation we were looking for. As a check, if we substitute $\bm e_r=\cos\theta\bm e_x+\sin\theta\bm e_y,\ \bm e_\theta=-\sin\theta\bm e_x+\cos\theta\bm e_y$ into the right-hand side, we get $\bm e_x=\cos\theta(\cos\theta\bm e_x+\sin\theta\bm e_y)-\sin\theta(-\sin\theta\bm e_x+\cos\theta\bm e_y)=(\cos^2\theta+\sin^2\theta)\bm e_x+(\cos\theta\sin\theta-\sin\theta\cos\theta)\bm e_y=\bm e_x$, so it is indeed an identity (the same holds for $\bm e_y$).

Solution

All we need to do is substitute $\theta=0$ ($\cos0=1$) and $\theta=\pi$ ($\cos\pi=-1$) into Formula 6.3 of Chapter 6, $r(\theta)=\dfrac{k}{1+\varepsilon\cos\theta}$:

$$ r(0)=\frac{k}{1+\varepsilon\cos0}=\frac{k}{1+\varepsilon},\qquad r(\pi)=\frac{k}{1+\varepsilon\cos\pi}=\frac{k}{1-\varepsilon}\quad(0\le\varepsilon\lt1) $$

$r(0)$ is the distance to the point at $\theta=0$ (the point closest to the focal point, which is the perihelion in the planetary motion treated in Chapter 8), and $r(\pi)$ is the distance to the point farthest from the focal point (the aphelion). The larger $\varepsilon$ is, the smaller $r(0)$ and the larger $r(\pi)$ become—we see that the larger the eccentricity of an orbit, the more closely it approaches the focal point at perihelion and the farther away it goes at aphelion (as $\varepsilon\to1$, $r(\pi)\to\infty$, and the ellipse approaches an open parabola).

These two points are the two ends of the major axis, so the length of the major axis is $2a=r(0)+r(\pi)$:

$$ 2a=\frac{k}{1+\varepsilon}+\frac{k}{1-\varepsilon}=\frac{k(1-\varepsilon)+k(1+\varepsilon)}{(1+\varepsilon)(1-\varepsilon)}=\frac{2k}{1-\varepsilon^2} \qquad\Longrightarrow\qquad a=\frac{k}{1-\varepsilon^2} $$

This agrees with $a=k/(1-\varepsilon^2)$ in Theorem 6.1 of Chapter 6. We can also write $r(0)=a(1-\varepsilon)$ and $r(\pi)=a(1+\varepsilon)$ ($\dfrac{k}{1+\varepsilon}=\dfrac{k(1-\varepsilon)}{1-\varepsilon^2}$).

Solution

Substituting $\varepsilon=0.8=\dfrac45,\ k=2$ into the general form $(1-\varepsilon^2)x^2+2k\varepsilon x+y^2-k^2=0$ in rectangular (Cartesian) coordinates derived in Chapter 6, we obtain

$$ \left(1-\frac{16}{25}\right)x^2+2\times2\times\frac45\,x+y^2-4=0 \qquad\Longrightarrow\qquad \frac{9}{25}x^2+\frac{16}{5}x+y^2-4=0 $$

We complete the square in $x$. Since $\dfrac{9}{25}x^2+\dfrac{16}{5}x=\dfrac{9}{25}\left(x^2+\dfrac{16}{5}\times\dfrac{25}{9}x\right)=\dfrac{9}{25}\left(x^2+\dfrac{80}{9}x\right)=\dfrac{9}{25}\left[\left(x+\dfrac{40}{9}\right)^2-\left(\dfrac{40}{9}\right)^2\right]$,

$$ \frac{9}{25}\left(x+\frac{40}{9}\right)^2-\frac{9}{25}\times\frac{1600}{81}+y^2-4=0 \qquad\Longrightarrow\qquad \frac{9}{25}\left(x+\frac{40}{9}\right)^2+y^2=4+\frac{1600}{225}=\frac{900+1600}{225}=\frac{100}{9} $$

Dividing both sides by $\dfrac{100}{9}$, we get

$$ \frac{\left(x+\dfrac{40}{9}\right)^2}{\dfrac{2500}{81}}+\frac{y^2}{\dfrac{100}{9}}=1 $$

This is the standard form of an ellipse with center $\left(-\dfrac{40}{9},0\right)$, semi-major axis $a=\dfrac{50}{9}\approx5.56$, and semi-minor axis $b=\dfrac{10}{3}\approx3.33$. As a check, the distance from the center to the focal point is $c=\sqrt{a^2-b^2}=\sqrt{\dfrac{2500}{81}-\dfrac{900}{81}}=\sqrt{\dfrac{1600}{81}}=\dfrac{40}{9}$, which coincides exactly with the distance from the center to the origin, $\left|-\dfrac{40}{9}\right|=\dfrac{40}{9}$—this is consistent with the fact that Formula 6.3 is constructed with the origin as the focal point.

Let us also compare with the values given by Theorem 6.1. We have $a=\dfrac{k}{1-\varepsilon^2}=\dfrac{2}{1-16/25}=\dfrac{2}{9/25}=\dfrac{50}{9}$, $b=\dfrac{k}{\sqrt{1-\varepsilon^2}}=\dfrac{2}{3/5}=\dfrac{10}{3}$, and the center is at $-a\varepsilon=-\dfrac{50}{9}\times\dfrac45=-\dfrac{40}{9}$; all of them agree.

Solution

Preparation: verifying $\dot{\bm e}_r=\dot\theta\bm e_\theta,\ \dot{\bm e}_\theta=-\dot\theta\bm e_r$. We differentiate the expression for the basis vector $\bm e_r=\cos\theta\bm e_x+\sin\theta\bm e_y$ ($\bm e_x,\bm e_y$ are constant vectors independent of time, and only $\theta=\theta(t)$ is a function of time) with respect to $t$. By the chain rule, $\dfrac{\dd}{\dd t}\cos\theta=-\dot\theta\sin\theta,\ \dfrac{\dd}{\dd t}\sin\theta=\dot\theta\cos\theta$, so

$$ \dot{\bm e}_r=-\dot\theta\sin\theta\,\bm e_x+\dot\theta\cos\theta\,\bm e_y=\dot\theta\left(-\sin\theta\,\bm e_x+\cos\theta\,\bm e_y\right)=\dot\theta\,\bm e_\theta $$

(the quantity in the last parentheses is exactly $\bm e_\theta$, by the equations of Chapter 6). Similarly, differentiating $\bm e_\theta=-\sin\theta\bm e_x+\cos\theta\bm e_y$ gives

$$ \dot{\bm e}_\theta=-\dot\theta\cos\theta\,\bm e_x-\dot\theta\sin\theta\,\bm e_y=-\dot\theta\left(\cos\theta\,\bm e_x+\sin\theta\,\bm e_y\right)=-\dot\theta\,\bm e_r $$

so both relations are verified.

Derivation of the velocity. The position vector can be written as $\bm r=r\bm e_r$ ($r,\bm e_r$ are both functions of time). Using the product rule for differentiation, we obtain

$$ \bm v=\dot{\bm r}=\frac{\dd}{\dd t}(r\bm e_r)=\dot r\,\bm e_r+r\,\dot{\bm e}_r=\dot r\,\bm e_r+r\left(\dot\theta\,\bm e_\theta\right)=\dot r\,\bm e_r+r\dot\theta\,\bm e_\theta $$

(we substituted $\dot{\bm e}_r=\dot\theta\bm e_\theta$). This is exactly the same result as Formula 6.1 of Chapter 6, $\bm v=\dot r\bm e_r+r\dot\theta\bm e_\theta$, so we have derived it without using the rotation matrix.

Derivation of the acceleration. We differentiate the velocity once more. This time we apply the product rule to each of the two terms of $\bm v=\dot r\bm e_r+r\dot\theta\bm e_\theta$. The first term is the product of $\dot r$ and $\bm e_r$, so

$$ \frac{\dd}{\dd t}\left(\dot r\,\bm e_r\right)=\ddot r\,\bm e_r+\dot r\,\dot{\bm e}_r=\ddot r\,\bm e_r+\dot r\dot\theta\,\bm e_\theta $$

The second term is a product of the three factors $r,\dot\theta,\bm e_\theta$, so we apply the product rule twice (the derivative of $fgh$ is $f'gh+fg'h+fgh'$). Letting $f=r,\ g=\dot\theta,\ h=\bm e_\theta$, we have $f'=\dot r,\ g'=\ddot\theta,\ h'=\dot{\bm e}_\theta=-\dot\theta\bm e_r$, so

$$ \frac{\dd}{\dd t}\left(r\dot\theta\,\bm e_\theta\right)=\dot r\dot\theta\,\bm e_\theta+r\ddot\theta\,\bm e_\theta+r\dot\theta\left(-\dot\theta\,\bm e_r\right)=-r\dot\theta^2\,\bm e_r+\left(\dot r\dot\theta+r\ddot\theta\right)\bm e_\theta $$

Adding the two results, we get

$$ \bm a=\dot{\bm v}=\left[\ddot r\,\bm e_r+\dot r\dot\theta\,\bm e_\theta\right]+\left[-r\dot\theta^2\,\bm e_r+\left(\dot r\dot\theta+r\ddot\theta\right)\bm e_\theta\right] $$

Collecting the coefficients of $\bm e_r$ and the coefficients of $\bm e_\theta$ separately, we have

$$ \bm a=\left(\ddot r-r\dot\theta^2\right)\bm e_r+\left(\dot r\dot\theta+\dot r\dot\theta+r\ddot\theta\right)\bm e_\theta=\left(\ddot r-r\dot\theta^2\right)\bm e_r+\left(2\dot r\dot\theta+r\ddot\theta\right)\bm e_\theta $$

which is exactly the same result as Formula 6.2 of Chapter 6. We have thus confirmed that two paths that look entirely different—the comparison of rotation matrices (the method of Chapter 6) and the time derivatives of the basis vectors (the method of this exercise)—lead to the same conclusion. This derivation also shows that the breakdown of $2\dot r\dot\theta$ is the sum of the $\dot r\dot\theta$ coming from the derivative of $\dot r\bm e_r$ and the $\dot r\dot\theta$ coming from the derivative of $r\dot\theta\bm e_\theta$.

Solution

Substituting $\varepsilon=1$ into the general form $(1-\varepsilon^2)x^2+2k\varepsilon x+y^2-k^2=0$ in rectangular coordinates derived in Chapter 6, we have $(1-1^2)=0$, so the $x^2$ term vanishes exactly, leaving

$$ 2kx+y^2-k^2=0 $$

in which only a linear expression in $x$ remains. Arranging this into the form $y^2=\cdots$, we get

$$ y^2=k^2-2kx=-2k\left(x-\frac{k}{2}\right) $$

This is none other than the standard form $y^2=-2k\left(x-\dfrac{k}{2}\right)$ of a parabola with vertex $\left(\dfrac{k}{2},0\right)$ that opens in the negative $x$ direction (toward the origin). (It corresponds to taking the standard form $y^2=4px$ of High School Mathematics, Chapter 8, translating the vertex to $x=k/2$, and setting $4p=-2k$, that is, $p=-k/2$.) So we have indeed confirmed that the curve is a parabola. That the vertex lies at a distance $k/2$ from the focal point (the origin) is also consistent with the fact that, for $\varepsilon=1$, the perihelion distance is $r(0)=k/(1+1)=k/2$ by the result of Exercise 6.2.

Solution

Differentiating $r(t)=1+t^2$ gives $\dot r=2t\ [\mathrm{m/s}]$ and $\ddot r=2\ [\mathrm{m/s^2}]$. From $\theta(t)=t$ we have $\dot\theta=1\ \mathrm{rad/s}$ and $\ddot\theta=0$. Substituting $t=1\ \mathrm s$ gives

$$ r=2\ \mathrm m,\quad \dot r=2\ \mathrm{m/s},\quad \ddot r=2\ \mathrm{m/s^2},\quad \dot\theta=1\ \mathrm{rad/s},\quad \ddot\theta=0 $$

By Formula 6.2,

$$ a_r=\ddot r-r\dot\theta^2=2-2\times1^2=0\ \mathrm{m/s^2} $$ $$ a_\theta=2\dot r\dot\theta+r\ddot\theta=2\times2\times1+2\times0=4\ \mathrm{m/s^2} $$

Therefore the acceleration is only $4\ \mathrm{m/s^2}$ in the angular direction. Here $\ddot r=2\ \mathrm{m/s^2}\neq0=a_r$, so $\ddot r$ and $a_r$ are not equal—$\ddot r=2$ exactly cancels $-r\dot\theta^2=-2$, giving $a_r=0$. As a check, if we differentiate $x=(1+t^2)\cos t,\ y=(1+t^2)\sin t$ twice in rectangular coordinates, substitute $t=1$, and take the inner product (dot product) with $\bm e_r=(\cos1,\sin1),\ \bm e_\theta=(-\sin1,\cos1)$, we again obtain $a_r=0,\ a_\theta=4\ \mathrm{m/s^2}$.

Solution

(1) From $r=\sqrt{b^2+u^2t^2}$ we have $r^2=b^2+u^2t^2$. Differentiating both sides with respect to $t$ gives $2r\dot r=2u^2t$, that is, $r\dot r=u^2t$, so $\dot r=\dfrac{u^2t}{r}=\dfrac{u^2t}{\sqrt{b^2+u^2t^2}}$. The angle is $\theta=\tan^{-1}\dfrac{ut}{b}$, and from $\dfrac{\dd}{\dd t}\tan^{-1}\dfrac{ut}{b}=\dfrac{1}{1+(ut/b)^2}\cdot\dfrac ub=\dfrac{bu}{b^2+u^2t^2}$ we obtain $\dot\theta=\dfrac{bu}{r^2}$.

(2) Differentiating $r\dot r=u^2t$ once more with respect to $t$ (product rule), we get $\dot r^2+r\ddot r=u^2$. Therefore

$$ \ddot r=\frac{u^2-\dot r^2}{r}=\frac{u^2}{r}\left(1-\frac{u^2t^2}{r^2}\right)=\frac{u^2}{r}\cdot\frac{r^2-u^2t^2}{r^2}=\frac{u^2b^2}{r^3} $$

(we used $r^2-u^2t^2=b^2$). On the other hand, $r\dot\theta^2=r\cdot\dfrac{b^2u^2}{r^4}=\dfrac{b^2u^2}{r^3}$, so $\ddot r=r\dot\theta^2$ holds. Hence $a_r=\ddot r-r\dot\theta^2=0$.

Next, differentiating $\dot\theta=bu\,r^{-2}$ gives $\ddot\theta=bu\cdot(-2)r^{-3}\dot r=-\dfrac{2bu\,\dot r}{r^3}$, so

$$ r\ddot\theta=-\frac{2bu\,\dot r}{r^2}=-2\dot r\cdot\frac{bu}{r^2}=-2\dot r\dot\theta $$

and thus $r\ddot\theta=-2\dot r\dot\theta$ holds. Hence $a_\theta=2\dot r\dot\theta+r\ddot\theta=0$.

This confirms, from the polar-coordinate formulas as well, that $a_r=a_\theta=0$ (that is, the acceleration of uniform rectilinear motion is $\bm 0$). However, since $\dot r\neq0,\ \dot\theta\neq0$ ($t\neq0$), the four terms $\ddot r$, $r\dot\theta^2$, $2\dot r\dot\theta$, and $r\ddot\theta$ are not individually zero; only their sums are zero. For example, for $b=u=1$ (in units of m and m/s) and $t=1\ \mathrm s$, we have $r=\sqrt2$, $\dot r=\dfrac{1}{\sqrt2}$, $\dot\theta=\dfrac12$, and $\ddot r=\dfrac{\sqrt2}{4}=r\dot\theta^2$, $2\dot r\dot\theta=\dfrac{\sqrt2}{2}=-r\ddot\theta$ (all in units of $\mathrm{m/s^2}$).

Solution

In Formula 6.3, $r\to\infty$ occurs when the denominator $1+\varepsilon\cos\theta$ approaches $0$. The angle at which $\cos\theta=-\dfrac1\varepsilon$ (since $\varepsilon\gt1$, we have $-1\lt-\dfrac1\varepsilon\lt0$, so a solution exists) is $\theta_\infty$, and in the range $0\lt\theta_\infty\lt\pi$ we have

$$ \theta_\infty=\cos^{-1}\left(-\frac1\varepsilon\right)\qquad(\pi/2\lt\theta_\infty\lt\pi) $$

(the curve has $r\gt0$ in the range $-\theta_\infty\lt\theta\lt\theta_\infty$). Then $\sin\theta_\infty=\sqrt{1-\dfrac1{\varepsilon^2}}=\dfrac{\sqrt{\varepsilon^2-1}}{\varepsilon}$ (because $\sin\gt0$ for $0\lt\theta_\infty\lt\pi$), so

$$ \tan\theta_\infty=\frac{\sin\theta_\infty}{\cos\theta_\infty}=\frac{\sqrt{\varepsilon^2-1}/\varepsilon}{-1/\varepsilon}=-\sqrt{\varepsilon^2-1} $$

For $\varepsilon=\dfrac53$, we have $\cos\theta_\infty=-\dfrac35$, $\sin\theta_\infty=\dfrac45$, and $\tan\theta_\infty=-\dfrac43$ ($\theta_\infty\approx126.9^\circ$).

By Theorem 6.1, the magnitude of the slope of the asymptotes of the hyperbola is $\dfrac ba=\dfrac{k/\sqrt{\varepsilon^2-1}}{k/(\varepsilon^2-1)}=\sqrt{\varepsilon^2-1}$, which for $\varepsilon=\dfrac53$ is $\sqrt{\dfrac{25}{9}-1}=\dfrac43$. Since $\left|\tan\theta_\infty\right|=\sqrt{\varepsilon^2-1}=\dfrac ba$, the direction $\theta_\infty$ in which the curve goes off to infinity coincides with the direction of the asymptotes of the hyperbola (whose slope with respect to the $x$-axis has magnitude $b/a$).

Solution

Solving the characteristic equation $m\lambda^2+n\lambda+k=0$ of the equation of motion $m\ddot x+n\dot x+kx=0$ ($m,n,k\gt 0$) with the quadratic formula, we obtain

$$ \lambda=\frac{-n\pm\sqrt{n^2-4mk}}{2m}=\frac{-n\pm\sqrt{D}}{2m},\qquad D=n^2-4mk $$

The sign of the radicand $D$ determines whether the roots are real or complex—this is the reason the sign of $D$ corresponds to the type of damping.

• When $D\lt 0$: Since the radicand is negative, $\sqrt{D}=i\sqrt{-D}$ ($i$ is the imaginary unit), and the two roots become the mutually conjugate complex numbers $\lambda=\alpha\pm i\omega_d$ ($\alpha=-n/(2m)$, $\omega_d=\sqrt{-D}/(2m)$). Having an imaginary part $\omega_d\neq0$ means that oscillating terms of the form $\cos\omega_d t,\sin\omega_d t$ appear in the general solution, and since the real part satisfies $\alpha\lt 0$, the amplitude decays exponentially. Decaying while oscillating—this is damped oscillation.
• When $D\gt 0$: Since the radicand is positive, $\sqrt{D}$ is real, and the two roots $\lambda_1,\lambda_2$ are both real. Moreover, since $m,n,k\gt 0$ gives $\sqrt{D}=\sqrt{n^2-4mk}\lt n$, both $\lambda_{1,2}=(-n\pm\sqrt{D})/(2m)$ are negative. The general solution $x=C_1e^{\lambda_1t}+C_2e^{\lambda_2t}$ is a sum of two real exponential functions and contains no trigonometric functions, so it does not oscillate. Since an exponential function with a negative real exponent approaches 0 monotonically in both cases, $x(t)$ also approaches 0 monotonically, without oscillating—this is overdamping.
• When $D=0$: Since the radicand is 0, the two roots coincide in the single value $\lambda_0=-n/(2m)$ (a double root). In this case the form of the general solution for two real roots cannot be used as it stands (because only one linearly independent fundamental solution can be built), and we must use $t\,e^{\lambda_0t}$ as the other fundamental solution. The general solution $x=(C_1+C_2t)e^{\lambda_0t}$ also does not oscillate, but because it lies exactly on the boundary between the case $D\gt 0$ and the case $D\lt 0$ (the boundary where the discriminant is 0), it is called critical damping.

To summarize, the sign of $D=n^2-4mk$ determines whether the characteristic roots are real or complex: if they are complex, the motion decays while oscillating ($D\lt 0$, damped oscillation); if they are real, it decays without oscillating ($D\gt 0$, overdamping); and the boundary where the roots just become a double root is critical damping ($D=0$).

Solution

The energy conservation law for simple harmonic motion is $\frac12m\dot x^2+\frac12kx^2=C$ (constant), and substituting $x=A,\dot x=0$ (the instant of momentary rest at the position of the amplitude) gives $C=\frac12kA^2$ (Law 7.1 in the main text). Therefore, at any time,

$$ \frac12m\dot x^2+\frac12kx^2=\frac12kA^2 $$

holds. Since the right-hand side is constant, the magnitude of the velocity $|\dot x|$ is largest when the second term $\frac12kx^2$ on the left-hand side is smallest ($x=0$). Substituting $x=0$, we obtain

$$ \frac12m\dot x_{\max}^2=\frac12kA^2 \qquad\therefore\ \dot x_{\max}=\sqrt{\frac{k}{m}}\,A=\omega A $$

($\omega=\sqrt{k/m}$ is the angular frequency). This agrees exactly with the maximum value $|\dot x|_{\max}=A\omega$ (attained when $|\sin\omega t|=1$) of $\dot x(t)=-A\omega\sin\omega t$, which is obtained by differentiating $x(t)=A\cos\omega t$.

For the maximum acceleration, we use the equation of motion $m\ddot x=-kx$, which gives the relation $\ddot x=-(k/m)x=-\omega^2x$ (proportional to the position and opposite in direction). Since $|x|$ is largest at $x=\pm A$ (the positions of the amplitude),

$$ |\ddot x|_{\max}=\omega^2A $$

holds. To summarize, the maximum velocity is $\omega A$ (at the instant of passing through $x=0$), and the maximum acceleration is $\omega^2A$ (at the instant of reaching $x=\pm A$).

Solution

Critical damping is the case in which the discriminant of the characteristic equation is 0 (Chapter 7, Theorem 7.2 (iii)):

$$ D=n^2-4mk=0 \qquad\therefore\ n^2=4mk $$

Since $n$ is the drag coefficient and must physically satisfy $n\gt 0$, we choose the positive square root:

$$ n=\sqrt{4mk}=2\sqrt{mk} $$

This is the drag coefficient that makes a system with mass $m$ and spring constant $k$ critically damped. For example, for the same system as in Example 7.5 of Chapter 7, with $m=1\ \mathrm{kg},k=5\ \mathrm{N/m}$, we have $n=2\sqrt{1\times5}=2\sqrt5\approx4.47\ \mathrm{N\cdot s/m}$; if $n$ is smaller than this, the system is overdamped ($D\gt 0$), and if it is larger, the system undergoes damped oscillation ($D\lt 0$).

Solution

The steady-state response of the forced oscillation found in Chapter 7 has the form $x_0(t)=X_0\cos(\omega t-\delta)$, and its amplitude is

$$ X_0=\frac{f_0}{\sqrt{(k-m\omega^2)^2+(n\omega)^2}} $$

Consider the limit $\omega\to0$. In the denominator, $(k-m\omega^2)^2$ becomes $(k-0)^2=k^2$ at $\omega=0$, and $(n\omega)^2$ becomes $0$ at $\omega=0$. Therefore,

$$ \lim_{\omega\to0}X_0=\frac{f_0}{\sqrt{k^2+0}}=\frac{f_0}{\sqrt{k^2}}=\frac{f_0}{k} $$

(since $k\gt 0$, $\sqrt{k^2}=k$). This agrees with the intuition that a driving force of very small angular frequency is close to “a constant force $f_0$ that hardly varies with time,” and the mass stays (almost statically) near the position $x_{\text{st}}=f_0/k$ at which the spring force balances the constant force, as given by Hooke's law of the spring, $f_0=kx_{\text{st}}$.

Solution

The correspondence between the operator method of University Mathematics, Chapter 38, and the operator method used in the “Math Tool” of Chapter 7 is organized in the table below.

Item	AD 02, Chapter 38	Chapter 7 (this chapter)
Independent variable	$x$ (a mathematical variable)	$t$ (time)
Definition of the differential operator	$D\equiv\dfrac{\dd}{\dd x}$ (Section 38.1)	$D\equiv\dfrac{\dd}{\dd t}$ (the same symbol carried over unchanged)
Notation for operators up to second order	$Dy=y',\ D^2y=y''$	$Dx=\dot x,\ D^2x=\ddot x$
Polynomial operator	$\Phi(D)=D^2+aD+b$ (leading coefficient 1, from Section 38.3 onward)	$\Phi(D)=mD^2+nD+k$ (leading coefficient $m$; the form used as it stands, without dividing both sides of the equation by $m$)
Substitution formula for exponential functions	$\Phi(D)y=e^{\alpha x}$ has the particular solution $y=\dfrac{e^{\alpha x}}{\Phi(\alpha)}$ ($\Phi(\alpha)\neq0$, Section 38.4)	$\Phi(D)x=f_0e^{i\omega t}$ has the particular solution $x=\dfrac{f_0e^{i\omega t}}{\Phi(i\omega)}$ ($\Phi(i\omega)\neq0$)
Value substituted in the exponent	An arbitrary real (or complex) number $\alpha$	A purely imaginary number $\alpha=i\omega$ (to treat a trigonometric external force as a complex exponential function)

There is only one essential difference: in AD 02, Chapter 38, the leading coefficient is normalized to 1 and the operator is handled in the form $\Phi(D)=D^2+aD+b$, whereas in Chapter 7 both sides of the equation of motion are not divided by the mass $m$, and $\Phi(D)=mD^2+nD+k$ (leading coefficient $m$) is used as it stands. The formula $x=e^{\alpha t}/\Phi(\alpha)$ itself holds regardless of the leading coefficient of $\Phi(D)$ (as we checked in Example 7.6, as long as $\Phi(\alpha)\ne0$, substituting $x=e^{\alpha t}/\Phi(\alpha)$ directly confirms that it satisfies $\Phi(D)x=e^{\alpha t}$), so it can be used in the same way with either normalization.

Solution

Maximizing the amplitude $X_0(\omega)=f_0/\sqrt{(k-m\omega^2)^2+(n\omega)^2}$ is the same as minimizing the quantity under the square root in the denominator,

$$ D(\omega)\equiv(k-m\omega^2)^2+(n\omega)^2 $$

(since the numerator $f_0$ is a positive constant independent of $\omega$, and the square root and its radicand increase and decrease in the same direction). We differentiate $D(\omega)$ with respect to $\omega$. First, differentiating the first term by the chain rule for composite functions, we obtain

$$ \frac{\dd}{\dd\omega}(k-m\omega^2)^2=2(k-m\omega^2)\cdot\frac{\dd}{\dd\omega}(k-m\omega^2)=2(k-m\omega^2)\cdot(-2m\omega)=-4m\omega(k-m\omega^2) $$

The second term gives

$$ \frac{\dd}{\dd\omega}(n\omega)^2=\frac{\dd}{\dd\omega}(n^2\omega^2)=2n^2\omega $$

and hence

$$ D'(\omega)=-4m\omega(k-m\omega^2)+2n^2\omega=2\omega\left[-2m(k-m\omega^2)+n^2\right]=2\omega\left(2m^2\omega^2-2mk+n^2\right) $$

$D'(\omega)=0$ holds either at $\omega=0$ (this is a local maximum point, and since $D(0)=k^2\gt 0$ it is not the minimum—indeed, as we saw in Section 7.4, $X_0\to f_0/k$, a finite value, as $\omega\to0$, and as $\omega$ is increased the amplitude becomes still larger near resonance, so $\omega=0$ is not the minimum point), or when

$$ 2m^2\omega^2-2mk+n^2=0 \qquad\therefore\ \omega^2=\frac{2mk-n^2}{2m^2}=\frac{k}{m}-\frac{n^2}{2m^2}=\omega_0^2-\frac{n^2}{2m^2} $$

(where $\omega_0=\sqrt{k/m}$). Only when this right-hand side is positive ($\omega_0^2\gt n^2/(2m^2)$, that is, when the resistance is weak) does a real solution with $\omega\neq0$,

$$ \omega_{\text{res}}=\sqrt{\omega_0^2-\frac{n^2}{2m^2}} $$

exist. This is the resonance angular frequency given by the equation in the main text (Eq. (7.20) of Chapter 7). (That this $\omega$ is actually the minimum point of $D(\omega)$, that is, the maximum point of $X_0(\omega)$, can be confirmed from the fact that $D(\omega)$ takes large values both as $\omega\to0$ and as $\omega\to\infty$, and that the only stationary point in between is this $\omega_{\text{res}}$.)

Solution

Equation ② that appeared in the derivation in Section 8.2 was $\dfrac1r\dfrac{\dd}{\dd t}(r^2\dot\theta)=0$. Multiplying both sides by $r$ ($\ne0$) gives $\dfrac{\dd}{\dd t}(r^2\dot\theta)=0$, which means that $r^2\dot\theta$ takes a constant value independent of the time $t$. We therefore let this constant value be $C_1$:

$$ r^2\dot\theta=C_1=\text{constant} $$

On the other hand, when the position vector of the particle (point mass) rotates through an angle $\dd\theta$ between time $t$ and $t+\dd t$, the infinitesimal area it sweeps out can be written, as the area of a circular sector of radius $r$ and central angle $\dd\theta$, as $\dd A=\dfrac12r^2\,\dd\theta$. Dividing both sides by $\dd t$, the areal velocity is

$$ \frac{\dd A}{\dd t}=\frac12r^2\frac{\dd\theta}{\dd t}=\frac12r^2\dot\theta=\frac12(r^2\dot\theta)=\frac{C_1}{2} $$

Since $C_1$ is a constant independent of time, the areal velocity $\dd A/\dd t=C_1/2$ is also constant in time. In other words, the content of Eq. ②, “$r^2\dot\theta$ is constant,” states exactly the same thing as Kepler's second law (the law of constant areal velocity), “the line segment joining the particle and the center sweeps out equal areas in equal times”—for the areal velocity itself is nothing but the quantity $r^2\dot\theta$ multiplied by $\dfrac12$. This result also agrees with $L=mr^2\dot\theta=$ constant, which we derived from conservation of angular momentum in Section 8.1, and the two constants are related by $C_1=L/m$.

Solution

Solving Kepler's third law (Chapter 8, Law 8.4) $T^2=\dfrac{4\pi^2}{GM}a^3$ for $GM$ gives

$$ GM=\frac{4\pi^2a^3}{T^2} $$

Substituting Jupiter's data $a=5.203\ \mathrm{AU}=5.203\times1.496\times10^{11}\ \mathrm{m}=7.784\times10^{11}\ \mathrm{m}$ and $T=11.86\ \mathrm{yr}=11.86\times3.156\times10^{7}\ \mathrm{s}=3.743\times10^{8}\ \mathrm{s}$, we obtain

$$ GM_\odot=\frac{4\pi^2\times(7.784\times10^{11}\ \mathrm{m})^3}{(3.743\times10^{8}\ \mathrm{s})^2} =\frac{4\pi^2\times4.716\times10^{35}\ \mathrm{m^3}}{1.401\times10^{17}\ \mathrm{s^2}} \approx1.329\times10^{20}\ \mathrm{m^3/s^2} $$

The value obtained from the Earth's data in Example 8.3(2) of Chapter 8 was $GM_\odot\approx1.3270\times10^{20}\ \mathrm{m^3/s^2}$ (which agreed well with the literature value), and its relative difference from the value obtained from Jupiter's data, $1.329\times10^{20}\ \mathrm{m^3/s^2}$ (more precisely, $1.3288\times10^{20}$), is about $0.14\ \%$. This difference is due not only to rounding error but also to a deviation from the approximation that “the Sun is at rest” (strictly, $M$ in Law 8.4 must be replaced by $M+m$), since Jupiter's mass is about $1/1000$ that of the Sun. From the data of the Earth, an inner planet, and Jupiter, an outer planet—data with completely different semi-major axes and periods—the same solar mass (multiplied by $G$) is obtained with good agreement. This is a confirmation, by observational data, that the proportionality constant $4\pi^2/(GM)$ of Kepler's third law does not change from planet to planet (it is determined only by the central body).

Solution

We express the position vector of the particle, using the unit vector $\bm e$ along the tunnel and the unit vector $\bm e_\perp$ perpendicular to it, as $\bm r=p\,\bm e_\perp+s\,\bm e$ ($p$ is the length of the perpendicular dropped from the center $\mathrm O$ to the tunnel, which is constant, and $s$ is the coordinate along the tunnel). Extending Law 8.6 of Chapter 8 to the form of a three-dimensional vector, the gravitational force inside a uniform sphere ($|\bm r|\lt R$) on a particle of mass $m$ is

$$ \bm F(\bm r)=-\frac43\pi G\rho\,m\,\bm r $$

(its magnitude is proportional to the distance $|\bm r|$ from the center, and its direction always points toward the center $\mathrm O$). The component of this force in the tunnel direction (the $\bm e$ direction), obtained by substituting $\bm r=p\,\bm e_\perp+s\,\bm e$, is

$$ \bm F\cdot\bm e=-\frac43\pi G\rho\,m\,\bm r\cdot\bm e =-\frac43\pi G\rho\,m\,\big(p\,\bm e_\perp\cdot\bm e+s\,\bm e\cdot\bm e\big) $$

Since $\bm e_\perp$ and $\bm e$ are, by definition, the direction perpendicular to the tunnel and the direction along the tunnel, they are orthogonal to each other, and $\bm e_\perp\cdot\bm e=0$ (by the definition of the inner product, or dot product, the inner product of two orthogonal vectors is 0). Also, since $\bm e$ is a unit vector, $\bm e\cdot\bm e=|\bm e|^2=1$. Therefore the term containing $p$ vanishes, and

$$ \bm F\cdot\bm e=-\frac43\pi G\rho\,m\,s $$

This shows that the component of the force in the tunnel direction does not depend at all on the perpendicular length $p$ (how far the tunnel is from the center); it is determined only by the coordinate $s$ along the tunnel. The tunnel wall is smooth and frictionless, and the normal force that the wall exerts on the particle cancels only the component in the $\bm e_\perp$ direction (it exerts no force along the tunnel), so for the equation of motion in the $\bm e$ direction we need only consider the $\bm e$ component of the force:

$$ m\ddot s=\bm F\cdot\bm e=-\frac43\pi G\rho\,m\,s \qquad\therefore\ \ddot s+\omega^2s=0,\qquad \omega=\sqrt{\frac43\pi G\rho} $$

This is an equation of simple harmonic motion with exactly the same angular frequency $\omega$ as the equation of motion (Section 8.5.3) of the tunnel through the center in Figure 8.8 (the special case $p=0$). Therefore the period of oscillation in the chord-shaped tunnel, $T=2\pi/\omega$, does not depend at all on how far the tunnel is from the center (the value of $p$, that is, the length of the tunnel), and is the same as for the tunnel through the center, about 84 minutes.

Solution

From the force $F(r)=-\dfrac43\pi G\rho\,mr$ of Law 8.6 of Chapter 8, we find the potential energy $U(r)$ (with $U(0)=0$ as the reference) by the inverse operation (integration) of $F=-\dd U/\dd r$:

$$ U(r)=-\int_0^rF(r')\,\dd r'=-\int_0^r\left(-\frac43\pi G\rho\,mr'\right)\dd r' =\frac43\pi G\rho\,m\int_0^rr'\,\dd r' =\frac43\pi G\rho\,m\left[\frac{r'^2}{2}\right]_0^r =\frac23\pi G\rho\,m\,r^2 $$

The mechanical energy is $E=\frac12m\dot r^2+U(r)=\frac12m\dot r^2+\frac23\pi G\rho\,mr^2$. Substituting the initial conditions $r(0)=R,\ \dot r(0)=0$, we obtain

$$ E=\frac12m\times0^2+\frac23\pi G\rho\,mR^2=\frac23\pi G\rho\,mR^2 $$

Next, differentiating the solution $r(t)=R\cos\omega t$ ($\omega=\sqrt{\frac43\pi G\rho}$) of the equation of motion gives $\dot r(t)=-R\omega\sin\omega t$, so

$$ \dot r^2=R^2\omega^2\sin^2\omega t=R^2\omega^2\big(1-\cos^2\omega t\big)=\omega^2\big(R^2-R^2\cos^2\omega t\big)=\omega^2(R^2-r^2) $$

(we used $\sin^2\omega t=1-\cos^2\omega t$ and $R\cos\omega t=r(t)$). Substituting this into the expression for the energy gives

$$ \frac12m\dot r^2+U(r)=\frac12m\,\omega^2(R^2-r^2)+\frac23\pi G\rho\,mr^2 $$

Substituting $\omega^2=\frac43\pi G\rho$ and simplifying, we get

$$ =\frac12m\cdot\frac43\pi G\rho\,(R^2-r^2)+\frac23\pi G\rho\,mr^2 =\frac23\pi G\rho\,m\,R^2-\frac23\pi G\rho\,mr^2+\frac23\pi G\rho\,mr^2 =\frac23\pi G\rho\,mR^2 $$

The terms $-\frac23\pi G\rho\,mr^2$ and $+\frac23\pi G\rho\,mr^2$ cancel each other, leaving the constant value $\frac23\pi G\rho\,mR^2$, which does not depend on $r$. This agrees exactly with $E=\frac23\pi G\rho\,mR^2$, which we found earlier from the initial conditions. Therefore the solution $r(t)=R\cos\omega t$ satisfies the law of conservation of energy $\frac12m\dot r^2+U(r)=E$ (constant) at every instant, and we have confirmed that it is consistent as a solution of the equation of motion.

Solution

Substituting the data for Mars, $GM_{\text{Mars}}=4.283\times10^{13}\ \mathrm{m^3/s^2}$ and $R_{\text{Mars}}=3.390\times10^6\ \mathrm{m}$, into the escape velocity formula $v_{\rm esc}=\sqrt{2GM/R}$ of Chapter 8, we obtain

$$ v_{\rm esc,\text{Mars}}=\sqrt{\frac{2\times4.283\times10^{13}\ \mathrm{m^3/s^2}}{3.390\times10^6\ \mathrm{m}}} =\sqrt{2.527\times10^{7}\ \mathrm{m^2/s^2}} \approx5.03\times10^3\ \mathrm{m/s}\approx5.03\ \mathrm{km/s} $$

Comparing this with the escape velocity of the Earth, $\approx11.2\ \mathrm{km/s}$, and that of the Moon, $\approx2.38\ \mathrm{km/s}$, found in Example 8.6 of Chapter 8, we see that the escape velocity of Mars lies between them (about 45% of the Earth's and about 2.1 times the Moon's). This reflects the fact that both the mass and the radius of Mars are smaller than the Earth's but larger than the Moon's ($GM_{\text{Mars}}/GM_\oplus\approx0.107$, $R_{\text{Mars}}/R_\oplus\approx0.532$). It is because of this difference in escape velocity that an actual Mars probe (for example, in a Mars sample-return mission) needs a smaller velocity to take off from Mars than is needed for a launch from the Earth.

Solution

A geostationary satellite revolves around the Earth in a circular orbit (eccentricity $\varepsilon=0$, semi-major axis $a=r_0$) with the same period as the Earth's rotation period, $T_{\rm sid}=86164\ \mathrm{s}$. Solving Kepler's third law (Chapter 8, Law 8.4) $T^2=\dfrac{4\pi^2}{GM}a^3$ for $a^3$ gives

$$ a^3=\frac{GM\,T^2}{4\pi^2} $$

Substituting the Earth's $GM_\oplus=3.986\times10^{14}\ \mathrm{m^3/s^2}$ and $T_{\rm sid}=86164\ \mathrm{s}$, we obtain

$$ r_0^3=a^3=\frac{3.986\times10^{14}\ \mathrm{m^3/s^2}\times(86164\ \mathrm{s})^2}{4\pi^2} \approx\frac{3.986\times10^{14}\times7.424\times10^9}{39.48}\ \mathrm{m^3} \approx7.497\times10^{22}\ \mathrm{m^3} $$

Taking the cube root,

$$ r_0\approx4.216\times10^7\ \mathrm{m}=42164\ \mathrm{km} $$

(Converting to the altitude above the Earth's surface gives $42164-6371\approx35793\ \mathrm{km}$, which agrees with the well-known altitude of a geostationary satellite, “about 36,000 km.”) Next, using the circular-orbit speed $v_{\rm circ}=\sqrt{GM/r_0}$ (Section 8.4; from the balance of the centripetal force, $GMm/r^2=mv^2/r$), the angular momentum per unit mass can be written as

$$ \frac{L}{m}=r_0v_{\rm circ}=r_0\sqrt{\frac{GM}{r_0}}=\sqrt{GM\,r_0} $$

Substituting numerical values,

$$ \frac{L}{m}=\sqrt{3.986\times10^{14}\ \mathrm{m^3/s^2}\times4.216\times10^{7}\ \mathrm{m}}\approx1.296\times10^{11}\ \mathrm{m^2/s} $$

That this satisfies $r_0=L^2/(GMm^2)=(L/m)^2/(GM)$ of Chapter 8 can be confirmed from $(1.296\times10^{11})^2/(3.986\times10^{14})\approx4.216\times10^{7}\ \mathrm{m}$ (it agrees with the value of $r_0$).

Solution

Since $r$ is a function $r(\theta)$ of $\theta$, and $\theta$ is a function $\theta(t)$ of $t$, the chain rule gives $\dfrac{\dd}{\dd t}=\dot\theta\,\dfrac{\dd}{\dd\theta}$. Using $\dot\theta=C_1/r^2$ (from $r^2\dot\theta=C_1$), we obtain

$$ \dot r=\frac{\dd r}{\dd\theta}\,\dot\theta=\frac{C_1}{r^2}\frac{\dd r}{\dd\theta} $$

Letting $u=1/r$, we have $\dfrac{\dd u}{\dd\theta}=\dfrac{\dd u}{\dd r}\dfrac{\dd r}{\dd\theta}=-\dfrac1{r^2}\dfrac{\dd r}{\dd\theta}$, so $\dfrac1{r^2}\dfrac{\dd r}{\dd\theta}=-\dfrac{\dd u}{\dd\theta}$, and therefore

$$ \dot r=-C_1\frac{\dd u}{\dd\theta} $$

Next, we substitute the above $\dot r$ into $\ddot r=\dfrac{\dd}{\dd t}\dot r=\dot\theta\dfrac{\dd}{\dd\theta}\dot r$ ($C_1$ is a constant):

$$ \ddot r=\frac{C_1}{r^2}\frac{\dd}{\dd\theta}\left(-C_1\frac{\dd u}{\dd\theta}\right)=-\frac{C_1^2}{r^2}\frac{\dd^2u}{\dd\theta^2} $$

We substitute these into $\ddot r\,r^2-C_1^2/r=-GM$. Noting that $C_1^2/r=C_1^2u$, we have

$$ r^2\left(-\frac{C_1^2}{r^2}\frac{\dd^2u}{\dd\theta^2}\right)-C_1^2u=-GM \quad\Longrightarrow\quad -C_1^2\frac{\dd^2u}{\dd\theta^2}-C_1^2u=-GM $$

Dividing both sides by $-C_1^2\ (\ne0)$, we obtain $\dfrac{\dd^2u}{\dd\theta^2}+u=\dfrac{GM}{C_1^2}$.

Solution

The Earth tunnel obeys $m\ddot r=-\dfrac43\pi G\rho\,m\,r$, and the spring–mass oscillator obeys $m\ddot x=-k_{\rm s}x$. Comparing the coefficients, under the correspondence $x\leftrightarrow r$ we have $k_{\rm s}\leftrightarrow\dfrac43\pi G\rho\,m$. Accordingly, the angular frequency and the period correspond as follows.

Quantity	Spring–mass oscillator	Earth tunnel
Position	$x$	$r$ (signed distance from the center)
Restoring force	$-k_{\rm s}x$	$-\frac43\pi G\rho\,m\,r$
Spring constant	$k_{\rm s}$	$\frac43\pi G\rho\,m$
Angular frequency	$\omega=\sqrt{k_{\rm s}/m}$	$\omega=\sqrt{\frac43\pi G\rho}$
Period	$T=2\pi\sqrt{m/k_{\rm s}}$	$T=2\pi\sqrt{3/(4\pi G\rho)}$

The numerical values, for $m=1\ \mathrm{kg}$ and $\rho=5500\ \mathrm{kg/m^3}$, are

$$ k_{\rm s}=\frac43\pi G\rho\,m=\frac43\pi\times6.674\times10^{-11}\ \mathrm{N\,m^2/kg^2}\times5500\ \mathrm{kg/m^3}\times1\ \mathrm{kg}\approx1.537\times10^{-6}\ \mathrm{N/m} $$

(The unit is $\mathrm{N\,m^2/kg^2\cdot kg/m^3\cdot kg}=\mathrm{N/m}$, which agrees with the unit of the spring constant.) This is an extremely soft spring (stretching it by $1\ \mathrm m$ gives a force of only $1.5\ \mathrm{\mu N}$), but because the distance the particle moves is very large, the Earth's radius $6.4\times10^6\ \mathrm m$, the maximum force becomes $k_{\rm s}R\approx9.8\ \mathrm N$ (which is just the gravitational force acting on a particle of $1\ \mathrm{kg}$). Note also that here $\omega=\sqrt{k_{\rm s}/m}=\sqrt{1.537\times10^{-6}}\ \mathrm{s^{-1}}\approx1.24\times10^{-3}\ \mathrm{s^{-1}}$, which agrees with the value in the main text.

Solution

(1) Let $s_{\max}$ be the distance from the midpoint of the tunnel (the foot of the perpendicular) to an endpoint. Since the endpoint lies on the surface of the sphere, the Pythagorean theorem for the right triangle formed by the foot of the perpendicular, the center, and the endpoint gives $s_{\max}^2+p^2=R^2$.

$$ s_{\max}=\sqrt{R^2-p^2}=\sqrt{(6371\ \mathrm{km})^2-(3000\ \mathrm{km})^2}=\sqrt{3.1590\times10^{7}}\ \mathrm{km}\approx5620\ \mathrm{km} $$

The length of the tunnel is $2s_{\max}\approx1.124\times10^4\ \mathrm{km}$.

(2) By the result of Exercise 8.3, the motion along the tunnel obeys $\ddot s=-\omega^2s$ ($\omega=\sqrt{\frac43\pi G\rho}=1.240\times10^{-3}\ \mathrm{s^{-1}}$), and if the particle is released from rest at one end $s=-s_{\max}$, then $s(t)=-s_{\max}\cos\omega t$. It reaches the opposite end $s=+s_{\max}$ when $\omega t=\pi$, which is half a period:

$$ t=\frac T2=\frac{\pi}{\omega}=\frac{\pi}{1.240\times10^{-3}\ \mathrm{s^{-1}}}\approx2.53\times10^{3}\ \mathrm s\approx42.2\ \mathrm{min} $$

This is exactly the same as the one-way time ($T/2$) of the tunnel through the center. The tunnel is much shorter, but the force (the right-hand side of the equation of motion) is correspondingly smaller, so the time does not change.

(3) The speed is greatest at the midpoint $s=0$, where $v_{\max}=\omega s_{\max}$:

$$ v_{\max}=1.240\times10^{-3}\ \mathrm{s^{-1}}\times5.620\times10^{6}\ \mathrm m\approx6.97\times10^{3}\ \mathrm{m/s}\approx7.0\ \mathrm{km/s} $$

(This is smaller than the $7.9\ \mathrm{km/s}$ for the tunnel through the center, because the amplitude is only $s_{\max}$.)

Solution

Take the direction of travel of the train as forward. The train moves forward while decelerating with magnitude $a$, so the acceleration vector of the train (the reference frame) points opposite to the direction of travel (backward) and has magnitude $a$. Working in the non-inertial frame fixed to the train, three forces act on the strap (mass $m$): gravity $mg$ (downward), the tension $T$ of the cord (along the cord), and the fictitious force (inertial force), which points opposite to the acceleration of the frame, that is, forward, with magnitude $ma$. The strap is at rest, in equilibrium.

Suppose the cord is tilted forward from the vertical by an angle $\alpha$. From the balance in the vertical direction, $T\cos\alpha=mg$, and from the balance in the horizontal direction, $T\sin\alpha=ma$ (since the fictitious force pushes the strap forward, the cord tilts forward from the vertical). Taking the ratio of the two equations,

$$ \tan\alpha=\frac{T\sin\alpha}{T\cos\alpha}=\frac{ma}{mg}=\frac{a}{g} $$

Substituting the numerical values,

$$ \tan\alpha=\frac{3.0}{9.8}\approx0.3061 \qquad\therefore\ \alpha=\arctan(0.3061)\approx17.0^\circ $$

The strap comes to equilibrium tilted from the vertical by about $17^\circ$ in the direction of travel of the train (this is the same reasoning as why passengers' bodies lurch forward under sudden braking). $\tan\alpha$ does not depend on $m$.

Solution

(1) In the frame that rotates with the bucket with angular velocity $\omega$, the water is at rest. A small element of water at the surface (mass $\Delta m$) is acted on by gravity $\Delta m\,g$ (downward), the centrifugal force $\Delta m\,\omega^2\rho$ (outward from the axis of rotation, horizontal), and a force due to the pressure, directed perpendicular to the water surface. Since the pressure force is perpendicular to the surface, the force balance along the surface means that the component along the surface of the resultant of gravity and the centrifugal force is zero, that is, that the resultant is perpendicular to the water surface.

Consider the cross section described by the distance $\rho$ from the axis of rotation and the vertically upward coordinate $z$. The resultant force (per unit mass) is $(\omega^2\rho,\ -g)$. The tangent direction of the water surface $z(\rho)$ is $(1,\ \dd z/\dd\rho)$, so the condition that this be perpendicular to the resultant force (inner product, or dot product, equal to zero) is

$$ \omega^2\rho\times1+(-g)\times\diff{z}{\rho}=0 \qquad\therefore\quad \diff{z}{\rho}=\frac{\omega^2\rho}{g} $$

Integrating this with $z=z_0$ at $\rho=0$ gives

$$ z(\rho)=z_0+\frac{\omega^2}{2g}\rho^2 $$

so the water surface is a paraboloid of revolution.

(2) $60\ \mathrm{rpm}$ is one revolution per second, so $\omega=2\pi\ \mathrm{rad/s}\approx6.283\ \mathrm{rad/s}$. The difference in height between the rim $\rho=a=0.15\ \mathrm m$ and the center is

$$ z(a)-z_0=\frac{\omega^2a^2}{2g}=\frac{(6.283)^2\times(0.15)^2}{2\times9.8}=\frac{39.48\times0.0225}{19.6}\approx0.0453\ \mathrm m $$

that is, about $4.5\ \mathrm{cm}$ (the units are $\mathrm{s^{-2}\cdot m^2/(m/s^2)}=\mathrm m$). The faster the rotation ($\propto\omega^2$), the higher the water is at the rim and the lower it is at the center.

Solution

(1) Take $\bm e_r'$ outward along the rod and $\bm e_\theta'$ in the counterclockwise direction perpendicular to it in the horizontal plane. Here $\bm\omega=\omega\bm e_z$ ($\omega=2.0\ \mathrm{rad/s}$, vertically upward). Since the bead moves only along the rod, $\bm v_{\text{rel}}=\dot r\,\bm e_r'$. The forces acting on the bead (in the rotating frame) are as follows.

• Centrifugal force: $m\omega^2r\,\bm e_r'$ (outward along the rod).
• Coriolis force: $-2m\bm\omega\times\bm v_{\text{rel}}=-2m\omega\dot r\,(\bm e_z\times\bm e_r')=-2m\omega\dot r\,\bm e_\theta'$ (perpendicular to the rod, to the right of the direction of motion).
• Gravity, and the vertical normal force with which the rod supports the bead (these balance). Since the rod is smooth, the horizontal force exerted by the rod is perpendicular to the rod and can be written $N\bm e_\theta'$.

The equation of motion along the rod is

$$ m\ddot r=m\omega^2r\qquad\therefore\quad\ddot r=\omega^2r $$

The general solution is $r=Ce^{\omega t}+De^{-\omega t}$. From the initial conditions $r(0)=r_0$ and $\dot r(0)=0$ we get $C+D=r_0$ and $\omega(C-D)=0$, so $C=D=r_0/2$, and therefore

$$ r(t)=\frac{r_0}{2}\left(e^{\omega t}+e^{-\omega t}\right)=r_0\cosh\omega t $$

The bead is accelerated outward exponentially.

(2) Setting $r=1.0\ \mathrm m$, we have $\cosh\omega t=1.0/0.10=10$, so $\omega t=\cosh^{-1}10=\ln(10+\sqrt{99})\approx2.993$. Therefore

$$ t=\frac{2.993}{2.0}\approx1.5\ \mathrm s $$

(3) In the direction perpendicular to the rod (the $\bm e_\theta'$ direction) the bead does not move, so the forces in this direction balance: $N-2m\omega\dot r=0$. Therefore

$$ N=2m\omega\dot r $$

In other words, the normal force of the rod has exactly the magnitude needed to balance the Coriolis force. For example, at the instant when $r=1.0\ \mathrm m$ is reached, $\dot r=r_0\omega\sinh\omega t=0.10\times2.0\times\sqrt{99}\approx1.99\ \mathrm{m/s}$, so $N=2m\times2.0\times1.99\approx8.0\,m\ \mathrm{N}$ ($m$ is the numerical value of the mass measured in kg).

(4) In the inertial frame of reference (polar coordinates, origin O), the angle of the rod is $\theta=\omega t$ ($\dot\theta=\omega$, $\ddot\theta=0$). The only horizontal force on the bead is the normal force $N\bm e_\theta$ perpendicular to the rod, so by the formula for the acceleration in polar coordinates in Chapter 6, the radial component of the equation of motion is

$$ m(\ddot r-r\dot\theta^2)=0\qquad\therefore\quad \ddot r=\omega^2r $$

which agrees with (1). In the $\theta$ direction it is

$$ m(2\dot r\dot\theta+r\ddot\theta)=N\qquad\therefore\quad N=2m\omega\dot r $$

which agrees with (3). The centrifugal term $m\omega^2r$ of the rotating frame is the acceleration term $-r\dot\theta^2$ of the inertial frame moved to the right-hand side, and the Coriolis term corresponds to $2\dot r\dot\theta$.

Solution

(1) From the period $T=8.0\ \mathrm s$, the angular velocity is

$$ \omega=\frac{2\pi}{T}=\frac{2\pi}{8.0}\approx0.7854\ \mathrm{rad/s} $$

Substituting $m=25\ \mathrm{kg}$ and $\rho=3.0\ \mathrm m$ into the formula of Chapter 9 (the centrifugal force $F_{\text{centrifugal}}=m\omega^2\rho$) gives

$$ F_{\text{centrifugal}}=25\times(0.7854)^2\times3.0\approx25\times0.6169\times3.0\approx46.3\ \mathrm N $$

Its direction is from the axis of rotation toward the child (outward). The child feels an outward force whose magnitude is about 19% of the child's weight (about $25\times9.8\approx245\ \mathrm N$).

(2) When the child moves outward with speed $v$, the magnitude of the Coriolis force is $2m\omega v$. Setting it equal to the centrifugal force $m\omega^2\rho$,

$$ m\omega^2\rho=2m\omega v\qquad\therefore\quad v=\frac{\omega\rho}{2}=\frac{0.7854\times3.0}{2}\approx1.18\ \mathrm{m/s} $$

The mass $m$ cancels, so the result is the same no matter who moves. At an ordinary walking speed (about $1.2\ \mathrm{m/s}$) the Coriolis force is comparable to the centrifugal force, so when you walk on a merry-go-round you are pushed around strongly not only by the centrifugal force but also by the Coriolis force.

Solution

Take a local coordinate system with $\bm e_x$ pointing east, $\bm e_y$ north, and $\bm e_z$ vertically upward. By the approximation of Section 9.4.1 (for motion in the horizontal plane, only the vertical component $\Omega_z=\omega_{\text{Earth}}\sin\varphi\ (\gt0,\ \text{Northern Hemisphere})$ of the angular velocity vector of the Earth's rotation matters), $\bm\omega\approx\Omega_z\bm e_z$, and $\bm v_{\text{rel}}=v\bm e_y$ ($v\gt0$, due north). The Coriolis force is

$$ \bm F_{\text{Co}}=-2m\,\bm\omega\times\bm v_{\text{rel}}=-2m\Omega_z v\,(\bm e_z\times\bm e_y) $$

Here, since $\bm e_x,\bm e_y,\bm e_z$ form a right-handed orthonormal basis, $\bm e_z\times\bm e_y=-\bm e_x$ (from the cyclic relations starting from $\bm e_x\times\bm e_y=\bm e_z$). Therefore,

$$ \bm F_{\text{Co}}=-2m\Omega_z v\,(-\bm e_x)=2m\Omega_z v\,\bm e_x $$

so the Coriolis force points east (the $+\bm e_x$ direction). For an object moving due north, east lies on the right-hand side of the direction of motion (north), which agrees with the rule of Theorem 9.2 of Chapter 9 (the rule for a counterclockwise-rotating frame), “the Coriolis force acts to the right of the direction of motion.”

Solution

Using the formula of Chapter 9 ($g_{\text{eff}}(\varphi)\approx g_0-\omega_{\text{Earth}}^2R_{\text{Earth}}\cos^2\varphi$), we have

$$ g_{\text{eff}}(0^\circ)-g_{\text{eff}}(90^\circ) =\left[g_0-\omega_{\text{Earth}}^2R_{\text{Earth}}\cos^20^\circ\right]-\left[g_0-\omega_{\text{Earth}}^2R_{\text{Earth}}\cos^290^\circ\right] $$

Since $\cos0^\circ=1,\ \cos90^\circ=0$, the $g_0$ terms cancel in the difference, and

$$ g_{\text{eff}}(0^\circ)-g_{\text{eff}}(90^\circ)=-\omega_{\text{Earth}}^2R_{\text{Earth}}\times1+\omega_{\text{Earth}}^2R_{\text{Earth}}\times0=-\omega_{\text{Earth}}^2R_{\text{Earth}} $$

The sign indicates that $g_{\text{eff}}$ is smaller at the equator than at the poles (the minus sign means “the equator is smaller”). The magnitude is

$$ \omega_{\text{Earth}}^2R_{\text{Earth}}=(7.292\times10^{-5})^2\times6.371\times10^6\approx0.0339\ \mathrm{m/s^2} $$

That is, the centrifugal effect alone makes the apparent gravitational acceleration at the equator smaller than at the poles by about $0.034\ \mathrm{m/s^2}$. As stated in the Caution in Section 9.4.2, the actually measured difference in gravitational acceleration between the equator and the poles (about $0.052\ \mathrm{m/s^2}$) is larger than this, and the remaining difference of about $0.018\ \mathrm{m/s^2}$ is due to the fact that the Earth is not a perfect sphere but an ellipsoid of revolution bulging in the equatorial direction (the net effect combining the decrease in attraction because the equator is farther from the Earth's center and the increase in attraction due to the mass of the equatorial bulge).

Solution

The formula of Chapter 9 (the local components of the Coriolis acceleration) is, for $\bm v=(v_{\mathrm E},v_{\mathrm N},v_z)$, eastward $a_{\mathrm E}=2\omega_{\text{Earth}}(\sin\varphi\,v_{\mathrm N}-\cos\varphi\,v_z)$ and upward $a_z=2\omega_{\text{Earth}}\cos\varphi\,v_{\mathrm E}$.

(1) From $h=\tfrac12gt^2$, the fall time is $t=\sqrt{2h/g}=\sqrt{600/9.8}\approx7.82\ \mathrm s$. During the fall the velocity is $v_z\approx-gt$, $v_{\mathrm N}\approx0$, so $a_{\mathrm E}=2\omega_{\text{Earth}}g\cos\varphi\,t$. Integrating twice with respect to time (the initial eastward velocity and displacement are zero),

$$ x_{\mathrm E}=\frac13\,\omega_{\text{Earth}}g\cos\varphi\,t^3 =\frac13\times7.292\times10^{-5}\times9.8\times\cos45^\circ\times(7.82)^3 $$

Substituting $\cos45^\circ\approx0.7071$ and $(7.82)^3\approx478.2\ \mathrm{s^3}$,

$$ x_{\mathrm E}\approx\frac13\times7.146\times10^{-4}\times0.7071\times478.2\approx8.1\times10^{-2}\ \mathrm m $$

that is, the body is displaced eastward by about $8\ \mathrm{cm}$ (the units are $\mathrm{s^{-1}\cdot m/s^2\cdot s^3}=\mathrm m$).

(2) When moving horizontally due east with speed $v=100\ \mathrm{m/s}$ ($v_{\mathrm E}=100\ \mathrm{m/s}$), the upward Coriolis acceleration is

$$ a_z=2\omega_{\text{Earth}}\cos\varphi\,v_{\mathrm E}=2\times7.292\times10^{-5}\times0.7071\times100\approx1.03\times10^{-2}\ \mathrm{m/s^2} $$

Since it is directed upward, the apparent gravitational acceleration is decreased by $1.03\times10^{-2}\ \mathrm{m/s^2}$. Relative to the gravitational acceleration $9.8\ \mathrm{m/s^2}$ the fraction is $1.03\times10^{-2}/9.8\approx1.05\times10^{-3}$; that is, the body becomes about 0.1% lighter. When moving westward, it becomes heavier by the same amount.

Solution

In the Southern Hemisphere the latitude $\varphi$ is negative ($\varphi\lt0$), so the vertical component of the angular velocity vector of the Earth's rotation, $\Omega_z=\omega_{\text{Earth}}\sin\varphi$, is negative (its direction is opposite to that in the Northern Hemisphere; one may also say that in the Southern Hemisphere the vertical component of the rotation axis as seen from the ground points downward). As found in Section 9.4.5, the direction of the Coriolis force also reverses when the sign of $\Omega_z$ reverses—the Coriolis force that was “to the right of the direction of motion” in the Northern Hemisphere becomes “to the left of the direction of motion” in the Southern Hemisphere.

Therefore, a wind trying to blow in toward a low-pressure system (cyclone) in the Southern Hemisphere is bent to the left again and again. The winds are in equilibrium when the pressure-gradient force (directed toward the center) and the Coriolis force to the left of the direction of motion (directed outward) balance each other, that is, when the air circulates with the center on its right-hand side, and overall the air blows in toward the center while spiraling clockwise (as seen from above). This is exactly opposite to the typhoons and extratropical cyclones of the Northern Hemisphere (counterclockwise). Indeed, cyclones that form near Australia have been observed to spiral clockwise.

Solution

(1) We carry out the integral in the definition directly. Take a coordinate $r$ ($0\le r\le\ell$) with its origin at one end of the rod; the axis is then at the position $r=\ell/4$. The distance from the axis is $|r-\ell/4|$, and since it is squared, the absolute-value sign is unnecessary:

$$ I=\int_0^{\ell}\left(r-\frac{\ell}{4}\right)^2\frac{M}{\ell}\,\dd r $$

Expanding the integrand gives $\left(r-\dfrac{\ell}{4}\right)^2=r^2-\dfrac{\ell}{2}r+\dfrac{\ell^2}{16}$, so

$$ I=\frac{M}{\ell}\int_0^{\ell}\left(r^2-\frac{\ell}{2}r+\frac{\ell^2}{16}\right)\dd r =\frac{M}{\ell}\left[\frac{r^3}{3}-\frac{\ell}{4}r^2+\frac{\ell^2}{16}r\right]_0^{\ell} $$

Substituting $r=\ell$ gives

$$ I=\frac{M}{\ell}\left(\frac{\ell^3}{3}-\frac{\ell^3}{4}+\frac{\ell^3}{16}\right) =M\ell^2\left(\frac13-\frac14+\frac{1}{16}\right) $$

We bring the fractions to a common denominator (take the denominator to be 48: $\frac13=\frac{16}{48},\ \frac14=\frac{12}{48},\ \frac{1}{16}=\frac{3}{48}$):

$$ \frac13-\frac14+\frac1{16}=\frac{16-12+3}{48}=\frac{7}{48} $$

Therefore,

$$ I=\frac{7}{48}M\ell^2 $$

(2) A method using the parallel-axis theorem. The distance from the center of mass ($r=\ell/2$) to the axis ($r=\ell/4$) is $d=\ell/2-\ell/4=\ell/4$. Using Theorem 10.3 of Chapter 10 ($I=I_{\mathrm{cm}}+Md^2$) and the moment of inertia about the center of mass, $I_{\mathrm{cm}}=\dfrac{1}{12}M\ell^2$ (Section 10.4), we obtain

$$ I=\frac{1}{12}M\ell^2+M\left(\frac{\ell}{4}\right)^2=\frac{1}{12}M\ell^2+\frac{1}{16}M\ell^2 $$

Bringing the fractions to a common denominator (48): since $\dfrac{1}{12}=\dfrac{4}{48},\ \dfrac{1}{16}=\dfrac{3}{48}$,

$$ I=\frac{4+3}{48}M\ell^2=\frac{7}{48}M\ell^2 $$

which agrees exactly with the result of the direct integration in (1).

Solution

First, we find the total mass $M$ from the linear density $\rho(r)=\rho_0\left(1+\dfrac{r}{\ell}\right)$:

$$ M=\int_0^{\ell}\rho(r)\,\dd r=\rho_0\int_0^{\ell}\left(1+\frac{r}{\ell}\right)\dd r =\rho_0\left[r+\frac{r^2}{2\ell}\right]_0^{\ell}=\rho_0\left(\ell+\frac{\ell}{2}\right)=\frac{3}{2}\rho_0\ell $$

Solving this for $\rho_0$ gives

$$ \rho_0=\frac{2M}{3\ell} $$

Next, we compute the moment of inertia about one end ($r=0$):

$$ I=\int_0^{\ell}r^2\rho(r)\,\dd r=\rho_0\int_0^{\ell}r^2\left(1+\frac{r}{\ell}\right)\dd r =\rho_0\int_0^{\ell}\left(r^2+\frac{r^3}{\ell}\right)\dd r $$ $$ =\rho_0\left[\frac{r^3}{3}+\frac{r^4}{4\ell}\right]_0^{\ell}=\rho_0\left(\frac{\ell^3}{3}+\frac{\ell^3}{4}\right)=\rho_0\ell^3\left(\frac13+\frac14\right)=\rho_0\ell^3\cdot\frac{7}{12}=\frac{7}{12}\rho_0\ell^3 $$

Finally, we substitute $\rho_0=\dfrac{2M}{3\ell}$:

$$ I=\frac{7}{12}\cdot\frac{2M}{3\ell}\cdot\ell^3=\frac{14}{36}M\ell^2=\frac{7}{18}M\ell^2 $$

Compared with a rod of uniform density (the result of Chapter 10, $I_{\text{end}}=\dfrac13M\ell^2=\dfrac{6}{18}M\ell^2$), $\dfrac{7}{18}M\ell^2$ is slightly larger—this is because the density $\rho(r)=\rho_0(1+r/\ell)$ becomes larger the closer one gets to the side far from the end ($r=0$), namely $r=\ell$ (the linear density doubles), so that, even for the same total mass, more of the mass is distributed on the side far from the axis, which has the effect of increasing the moment of inertia.

Solution

(1) Ring about its central axis. The total circumference of the ring is $2\pi R$, so the linear density is $\lambda=\dfrac{M}{2\pi R}$. Dividing the ring into small pieces of length $\dd s$ (mass $\dd m=\lambda\,\dd s$), every piece lies at distance $R$ from the axis of rotation (the axis through the center of the circle perpendicular to its plane), so by Definition 10.3,

$$ I=\int R^2\,\dd m=R^2\int\dd m=R^2M=MR^2 $$

(We used the total mass $\int\dd m=M$. One does not even need to integrate: the result follows immediately from the fact that the distance from the axis is $R$ everywhere.)

(2) Cylinder about its central axis. Slice the cylinder (radius $R$, length $h$) into disks of thickness $\dd z$ perpendicular to the central axis. The mass of a single disk is $\dd m=\dfrac{M}{h}\dd z$, and by the derivation for the disk in Section 10.6, the moment of inertia of this disk about its central axis is $\tfrac12\,\dd m\,R^2$. Adding them all up,

$$ I=\int\frac12R^2\,\dd m=\frac12R^2\int\dd m=\frac12MR^2 $$

We obtain the same value as for a disk, regardless of the length $h$ of the cylinder.

(3) Cylinder about a diameter. Take the center of the cylinder as the origin, the central axis as the $z$ axis, and a diametral direction as the $x$ axis. The mass of the disk of thickness $\dd z$ at height $z$ ($-h/2\le z\le h/2$) from the center is $\dd m=\dfrac{M}{h}\dd z$, and by the perpendicular-axis theorem the moment of inertia of this disk about a diameter through its own center (center of mass) is $\tfrac14\,\dd m\,R^2$ (Section 10.7.2). The $x$ axis is a parallel axis at distance $z$ from the center of mass of this disk, so by the parallel-axis theorem (Theorem 10.3), the moment of inertia of this disk about the $x$ axis is

$$ \dd I=\frac14\,\dd m\,R^2+\dd m\,z^2=\left(\frac14R^2+z^2\right)\frac{M}{h}\,\dd z $$

We integrate this from $z=-h/2$ to $h/2$:

$$ I=\frac{M}{h}\int_{-h/2}^{h/2}\left(\frac14R^2+z^2\right)\dd z=\frac{M}{h}\left[\frac14R^2z+\frac{z^3}{3}\right]_{-h/2}^{h/2} =\frac{M}{h}\left(\frac14R^2h+\frac{h^3}{12}\right) $$

(The difference between the upper and lower limits of $z^3/3$ is $2\cdot\dfrac{(h/2)^3}{3}=\dfrac{h^3}{12}$, and the difference for the first term is $\dfrac14R^2\cdot h$.) Therefore,

$$ I=\frac14MR^2+\frac1{12}Mh^2 $$

This is the result. (Check: in the limit $h\to0$ of a thin disk it gives $\tfrac14MR^2$ (the disk about a diameter), and in the limit $R\to0$ of a thin rod it gives $\tfrac1{12}Mh^2$ (the rod about its center of mass, Section 10.4.2), so it agrees with both known results.)

Solution

(1) Substituting the law of action and reaction $\bm F_{21}=-\bm F_{12}$, we obtain

$$ \bm r_1\times\bm F_{12}+\bm r_2\times\bm F_{21}=\bm r_1\times\bm F_{12}-\bm r_2\times\bm F_{12}=(\bm r_1-\bm r_2)\times\bm F_{12} $$

(The cross product satisfies the distributive law with respect to its first factor.)

(2) When the internal forces lie along the line joining the two particles, $\bm F_{12}$ is parallel to $\bm r_1-\bm r_2$ ($\bm F_{12}=f\,(\bm r_1-\bm r_2)$, where $f$ is a scalar). Since the cross product of parallel vectors is $\bm 0$,

$$ (\bm r_1-\bm r_2)\times\bm F_{12}=f\,(\bm r_1-\bm r_2)\times(\bm r_1-\bm r_2)=\bm 0 $$

Universal gravitation, the Coulomb force, and the elastic force of a spring all act in a direction along the line joining the two particles (and their magnitudes are functions of the distance alone), so they satisfy this condition. Therefore the torques (moments of force) of the internal forces cancel pair by pair, and for the system of particles as a whole $\dd\bm L/\dd t=\bm N_{\mathrm{ext}}$ (the sum of the torques of the external forces) holds.

(3) For example, let $\bm r_1-\bm r_2=(\ell,0,0)$ and take the internal forces to be $\bm F_{12}=(0,f,0)$, $\bm F_{21}=(0,-f,0)$ (equal in magnitude and opposite in direction, so they satisfy the law of action and reaction; but the forces are perpendicular to the line joining the two particles (the $x$ direction)). Then, by the result of (1), the sum of the torques of the internal forces is

$$ (\bm r_1-\bm r_2)\times\bm F_{12}=(\ell,0,0)\times(0,f,0)=(0,0,\ell f)\neq\bm 0 $$

which is not zero (we used the components of the cross product, $(a_yb_z-a_zb_y,\ a_zb_x-a_xb_z,\ a_xb_y-a_yb_x)$, and the $z$ component is $\ell f-0=\ell f$). In other words, the forces that the two particles exert on each other alone make the angular momentum of the whole system change at the rate $\dd L_z/\dd t=\ell f$. Therefore, for the conservation of angular momentum (the vanishing of the torques of the internal forces) to hold, the condition that “the internal forces lie along the line joining the two particles” is needed in addition to the law of action and reaction (this is called the strong form of the law of action and reaction). Note that some forces, such as the electromagnetic force between moving charged particles, do not satisfy this condition (in that case, angular momentum is conserved once the angular momentum carried by the electromagnetic field is included; this is studied in Part V).

Solution

We use the general formula for the acceleration of rolling motion obtained in Chapter 10, $a=\dfrac{g\sin\alpha}{1+c}$ ($I=cMR^2$, where $c$ is a dimensionless coefficient determined by the shape). A thin-walled hollow cylinder has $I=MR^2$, so $c=1$, and

$$ a_{\text{hollow cyl.}}=\frac{g\sin\alpha}{1+1}=\frac12g\sin\alpha $$

On the other hand, as found in Chapter 10, a uniform solid sphere has $c=2/5$ and $a_{\text{sphere}}=\dfrac57g\sin\alpha\approx0.714\,g\sin\alpha$, and a uniform solid cylinder has $c=1/2$ and $a_{\text{solid cyl.}}=\dfrac23g\sin\alpha\approx0.667\,g\sin\alpha$. Comparing the magnitudes of the numbers,

$$ a_{\text{sphere}}\left(\frac57\approx0.714\right)\ \gt\ a_{\text{solid cyl.}}\left(\frac23\approx0.667\right)\ \gt\ a_{\text{hollow cyl.}}\left(\frac12=0.5\right) $$

and the larger the acceleration, the sooner the body finishes descending the incline. Therefore, if the bodies roll down the same incline at the same time, the solid sphere reaches the bottom of the incline first, the solid cylinder next, and the hollow cylinder last. This corresponds to the fact that for a shape with a smaller $c=I/(MR^2)$ (a dimensionless coefficient expressing the magnitude of the moment of inertia for the same mass and radius), the mass is concentrated nearer the axis of rotation and a smaller fraction of the energy is taken up by the rotation, so the translational speed increases faster ($c$ is smallest for the solid sphere and largest for the hollow cylinder, whose mass is all concentrated on the outer circumference, the place farthest from the axis).

Solution

We use the parallel-axis theorem (Theorem 10.3 of Chapter 10, $I=I_{\mathrm{cm}}+Md^2$). Describe the position of the axis by the distance $x_0$ measured from one end ($0\le x_0\le\ell$); the distance from the center of mass ($x=\ell/2$) to the axis ($x=x_0$) is then $|x_0-\ell/2|$, and since it is squared, the absolute value is unnecessary:

$$ I(x_0)=I_{\mathrm{cm}}+M\left(x_0-\frac{\ell}{2}\right)^2=\frac{1}{12}M\ell^2+M\left(x_0-\frac{\ell}{2}\right)^2 $$

Differentiating $I(x_0)$ with respect to $x_0$,

$$ \frac{\dd I}{\dd x_0}=2M\left(x_0-\frac{\ell}{2}\right) $$

Setting this equal to 0,

$$ 2M\left(x_0-\frac{\ell}{2}\right)=0\qquad\therefore\ x_0=\frac{\ell}{2} $$

(Since $M\neq0$, we may divide by $M$.) The second derivative is $\dfrac{\dd^2I}{\dd x_0^2}=2M\gt0$ (a constant, always positive), so $x_0=\ell/2$ gives a local minimum (in this case, since $I(x_0)$ is a quadratic function of $x_0$—a parabola opening upward—this local minimum is at the same time the minimum over the whole interval $[0,\ell]$). The minimum value itself is

$$ I\left(\frac{\ell}{2}\right)=\frac{1}{12}M\ell^2+M\cdot0^2=\frac{1}{12}M\ell^2=I_{\mathrm{cm}} $$

which is precisely the moment of inertia about the center of mass. This expresses a result that is natural physically—in the parallel-axis theorem $I=I_{\mathrm{cm}}+Md^2$, $d^2\ge0$ is always nonnegative, so $I(x_0)\ge I_{\mathrm{cm}}$ always holds, and equality (the minimum) is attained only when the axis passes through the center of mass ($x_0=\ell/2$, that is, the distance between the axis and the center of mass is zero). In other words, for any body, whichever axis it is rotated about, the moment of inertia about the axis through the center of mass is the smallest among all parallel axes—this is nothing other than a generalization of the concrete example seen in Section 10.4, that “it is harder to rotate about an end ($I=M\ell^2/3$) than about the center of mass ($I=M\ell^2/12$).”

Solution

Let $v$ be the speed of the center of mass and $\omega$ the angular velocity. The condition for rolling without slipping (the rolling-contact condition of Section 10.9.1) is $v=R\omega$. Writing the moment of inertia as $I=cMR^2$, the translational kinetic energy and the rotational kinetic energy are

$$ K_{\mathrm{trans}}=\frac12Mv^2,\qquad K_{\mathrm{rot}}=\frac12I\omega^2=\frac12\,cMR^2\left(\frac{v}{R}\right)^2=\frac12\,c\,Mv^2 $$

(substituting $\omega=v/R$ cancels $R^2$). Therefore the ratio, independent of the shape, is

$$ \frac{K_{\mathrm{rot}}}{K_{\mathrm{trans}}}=c $$

which is determined by $c$ alone.

(1) Disk. Since $c=1/2$, $K_{\mathrm{trans}}:K_{\mathrm{rot}}=1:\tfrac12=2:1$. Of the total kinetic energy $K=K_{\mathrm{trans}}+K_{\mathrm{rot}}=\tfrac32K_{\mathrm{trans}}$, the fraction taken up by rotation is $\dfrac{1/2}{1+1/2}=\dfrac13$.

(2) Solid sphere and hollow cylinder. The solid sphere has $c=2/5$, so $K_{\mathrm{trans}}:K_{\mathrm{rot}}=1:\tfrac25=5:2$, and the fraction taken up by rotation is $\dfrac{2/5}{1+2/5}=\dfrac27\approx0.286$. The hollow cylinder (ring) has $c=1$, so $K_{\mathrm{trans}}:K_{\mathrm{rot}}=1:1$, and the fraction taken up by rotation is $\dfrac{1}{1+1}=\dfrac12$. To summarize, the fraction taken up by rotation increases in the order solid sphere ($2/7\approx0.29$) $\lt$ disk ($1/3\approx0.33$) $\lt$ ring ($1/2$). The farther from the axis the mass is distributed (the larger $c$ is), the more energy is diverted to rotation for the same speed of advance, and the smaller the acceleration when descending an incline (Section 10.9).

Solution

(1) The pivot is a point at distance $d$ from the center of mass. From the parallel-axis theorem (Theorem 10.3) and the moment of inertia of the rod about its center of mass, $I_{\mathrm{cm}}=M\ell^2/12$, the moment of inertia about the pivot is $I=\dfrac{M\ell^2}{12}+Md^2$. By Theorem 10.7,

$$ T(d)=2\pi\sqrt{\frac{I}{Mgd}}=2\pi\sqrt{\frac{\ell^2/12+d^2}{g\,d}} $$

(2) Minimizing $T$ is equivalent to minimizing the quantity under the square root, $f(d)=\dfrac{\ell^2/12+d^2}{d}=\dfrac{\ell^2}{12d}+d$. Differentiating with respect to $d$,

$$ f'(d)=-\frac{\ell^2}{12d^2}+1 $$

From $f'(d)=0$ we get $d^2=\ell^2/12$, that is, $d=\dfrac{\ell}{\sqrt{12}}=\dfrac{\ell}{2\sqrt3}\approx0.289\,\ell$. Since the second derivative $f''(d)=\dfrac{\ell^2}{6d^3}\gt0$, this is a local minimum (and the minimum). At this point $f=\dfrac{\ell^2}{12}\cdot\dfrac{\sqrt{12}}{\ell}+\dfrac{\ell}{\sqrt{12}}=\dfrac{2\ell}{\sqrt{12}}=\dfrac{\ell}{\sqrt3}$, so the minimum period is $T_{\min}=2\pi\sqrt{\dfrac{\ell}{\sqrt3\,g}}$. (At this point the length of the equivalent simple pendulum is $\ell_{\mathrm{eq}}=d+\dfrac{\ell^2}{12d}=2d=\dfrac{\ell}{\sqrt3}$, exactly twice the distance $d$ from the pivot to the center of mass.)

(3) Substituting $\ell=1\ \mathrm m$ and $g=9.8\ \mathrm{m/s^2}$,

$$ T_{\min}=2\pi\sqrt{\frac{1}{\sqrt3\times9.8}}=2\pi\sqrt{\frac{1}{16.97}}=2\pi\times0.2428\approx1.53\ \mathrm s $$

For the case in which one end is used as the pivot ($d=\ell/2=0.5\ \mathrm m$),

$$ T=2\pi\sqrt{\frac{1/12+0.25}{9.8\times0.5}}=2\pi\sqrt{\frac{0.3333}{4.9}}=2\pi\times0.2608\approx1.64\ \mathrm s $$

Taking the pivot at the optimal position (about $0.29\ \mathrm m$ from the center of mass) makes the period about $7\%$ shorter than using an end as the pivot. If the pivot is brought too close to the center of mass ($d\to0$), the restoring torque becomes small and the period becomes arbitrarily long.

Solution

Since the disk is uniform, mass is proportional to area. Let $M_0$ be the mass of the disk (radius $R$) before the hole is cut out. From the area ratio $\dfrac{\pi(R/2)^2}{\pi R^2}=\dfrac14$, the mass of the disk-shaped piece corresponding to the hole (radius $R/2$) is $m_{\text{hole}}=\dfrac14M_0$. The mass of the remaining plate is $M=M_0-m_{\text{hole}}=\dfrac34M_0$.

Take $O$ as the origin and the $x$ axis in the direction of the center of the hole. The center of mass of the disk before cutting is at $x=0$, the center of mass of the disk-shaped piece corresponding to the hole is its center $x=R/2$, and the center of mass of the remaining plate is denoted by $x_{\mathrm{cm}}$. The disk before cutting is made up of two parts, the “remaining plate” and the “disk-shaped piece corresponding to the hole,” so applying Definition 10.1 (the $x$ coordinate) to the two parts separately,

$$ 0=\frac{M\,x_{\mathrm{cm}}+m_{\text{hole}}\cdot\dfrac R2}{M_0} $$

(the left-hand side is the center of mass $x=0$ of the disk before cutting). Solving this for $x_{\mathrm{cm}}$:

$$ M\,x_{\mathrm{cm}}=-m_{\text{hole}}\cdot\frac R2\qquad\therefore\ x_{\mathrm{cm}}=-\frac{\frac14M_0}{\frac34M_0}\cdot\frac R2=-\frac13\cdot\frac R2=-\frac R6 $$

The center of mass lies on the side opposite the hole (the negative side of $x$), at a distance $R/6$ from the center (since the mass of the hole is missing, the center of mass shifts away from the hole. The same result is obtained by the method of “regarding the hole as a negative mass and using the definition of the center of mass,” writing $x_{\mathrm{cm}}=\dfrac{M_0\cdot0-m_{\text{hole}}\cdot R/2}{M_0-m_{\text{hole}}}$).

Solution

Multiplying both sides of the equation of motion $m\ddot x=-\kappa(x-x_0)$ ($x_0$ is a constant) by the velocity $\dot x=v$, we obtain

$$ m\dot v\cdot v=-\kappa(x-x_0)\cdot v \qquad\text{that is}\qquad mv\frac{\dd v}{\dd t}=-\kappa(x-x_0)\frac{\dd x}{\dd t} $$

By the rule for differentiating a composite function, the left-hand side is equal to $\dfrac{\dd}{\dd t}\!\left(\dfrac12mv^2\right)=mv\dfrac{\dd v}{\dd t}$. For the right-hand side as well, noting that $x_0$ is a constant, the following equality holds:

$$ \frac{\dd}{\dd t}\left[\frac12\kappa(x-x_0)^2\right]=\kappa(x-x_0)\cdot\frac{\dd}{\dd t}(x-x_0)=\kappa(x-x_0)\cdot\left(\frac{\dd x}{\dd t}-0\right)=\kappa(x-x_0)\frac{\dd x}{\dd t} $$

(Differentiating $x_0$ as a constant gives $0$, so $\dd(x-x_0)/\dd t=\dd x/\dd t$.) Therefore,

$$ \frac{\dd}{\dd t}\left(\frac12mv^2\right)=-\frac{\dd}{\dd t}\left[\frac12\kappa(x-x_0)^2\right] \qquad\therefore\qquad \frac{\dd}{\dd t}\left[\frac12mv^2+\frac12\kappa(x-x_0)^2\right]=0 $$

holds. That the time derivative is identically $0$ means that the quantity inside the brackets is constant, independent of time, so, using a constant of integration $E$,

$$ \frac12mv^2+\frac12\kappa(x-x_0)^2=E $$

holds (this is exactly the same procedure as the derivation for the case $x_0=0$ in Section 11.1 of the main text, except that the equilibrium position is displaced by $x_0$).

Solution

Let $g$ be an arbitrary (twice differentiable) function of one variable, let $w\equiv x+vt$, and consider $y(x,t)=g(w)=g(x+vt)$. Here $w$ is a function of $x,t$, with $\pdiff{w}{x}=1,\ \pdiff{w}{t}=v$ (note that the sign is opposite to that of $\partial u/\partial t=-v$ for $u=x-vt$). By the formula for partial derivatives of a composite function,

$$ \pdiff{y}{t}=\frac{\dd g}{\dd w}\cdot\pdiff{w}{t}=g'(w)\cdot v=vg'(w) $$

Differentiating partially with respect to $t$ once more, since $g'(w)$ is also a composite function that depends on $t$ through $w$,

$$ \pdiff{^2y}{t^2}=\pdiff{}{t}\bigl[vg'(w)\bigr]=v\cdot g''(w)\cdot\pdiff{w}{t}=v\cdot g''(w)\cdot v=v^2g''(w) $$

we obtain (in the case of $u$ the factor $-v$ was multiplied twice, so the signs canceled and gave $+v^2$; here the factor $+v$ is multiplied twice, so again we get $+v^2$—we see that even though the signs of the first derivatives are opposite, the second derivatives have the same sign). On the other hand, differentiating partially with respect to $x$, since $\partial w/\partial x=1$, we have

$$ \pdiff{y}{x}=g'(w),\qquad \pdiff{^2y}{x^2}=g''(w)\cdot\pdiff{w}{x}=g''(w) $$

Therefore,

$$ \pdiff{^2y}{t^2}=v^2g''(w)=v^2\pdiff{^2y}{x^2} $$

and we have confirmed that $y=g(x+vt)$ also satisfies the one-dimensional wave equation (regardless of the specific form of $g$). Together with the case $y=f(x-vt)$ in the main text (Section 11.2), this verifies that both terms of the general solution $y=f(x-vt)+g(x+vt)$ satisfy the wave equation.

Solution

Comparing $y=0.03\sin(15t-3x)\ [\mathrm{m}]$ with the form $y=A\sin(\alpha t+\beta x+\phi)$ given in Theorem 11.2 of Chapter 11, we find

$$ A=0.03\ \mathrm{m},\qquad \alpha=15\ \mathrm{rad/s},\qquad \beta=-3\ \mathrm{rad/m},\qquad \phi=0 $$

(since $-3x=+(-3)x$, we have $\beta=-3$). The product of the coefficients is

$$ \alpha\beta=15\times(-3)=-45\lt 0 $$

By Theorem 11.2, when $\alpha\beta\lt 0$ the wave travels in the positive direction of the $x$ axis. From the relation $\dot x=-\alpha/\beta$ in the proof of Theorem 11.2, the phase velocity is

$$ \dot x=-\frac{\alpha}{\beta}=-\frac{15}{-3}=5\ \mathrm{m/s} $$

Since the sign is positive, this is, as expected, consistent with the wave traveling in the $+x$ direction. The speed is $5\ \mathrm{m/s}$ (this wave has exactly the form $y=A\sin(\omega t-kx)$ with $\omega=15\ \mathrm{rad/s}$ and $k=3\ \mathrm{rad/m}$, and computing the phase velocity directly as $v=\omega/k=15/3=5\ \mathrm{m/s}$ gives the same value).

Solution

The phase difference (the phase of $y_B$ $-$ the phase of $y_A$) between $y_A=a_A\sin\{\omega(t-\overline{\mathrm{AP}}/v)+\phi\}$ and $y_B=a_B\sin\{\omega(t-\overline{\mathrm{BP}}/v)+\phi+\pi/2\}$ is, as in the main text (Section 11.6.1),

$$ \Delta\theta=\frac{\omega}{v}\bigl(\overline{\mathrm{AP}}-\overline{\mathrm{BP}}\bigr)+\frac{\pi}{2}=\frac{2\pi}{\lambda}\bigl(\overline{\mathrm{AP}}-\overline{\mathrm{BP}}\bigr)+\frac{\pi}{2} $$

(the first term is the phase difference due to the difference in the distances traveled by the waves, and the second term is the phase difference of the sources themselves). The waves reinforce each other when the total phase difference $\Delta\theta$ is $2n\pi$, and cancel each other when it is $(2n+1)\pi$ (the “Key point” box in Section 11.6.1).

Condition for reinforcement: From $\Delta\theta=2n\pi$ ($n\in\mathbb{Z}$),

$$ \frac{2\pi}{\lambda}\bigl(\overline{\mathrm{AP}}-\overline{\mathrm{BP}}\bigr)=2n\pi-\frac{\pi}{2} \qquad\therefore\quad \overline{\mathrm{AP}}-\overline{\mathrm{BP}}=\left(n-\frac14\right)\lambda $$

Condition for cancellation: From $\Delta\theta=(2n+1)\pi$,

$$ \frac{2\pi}{\lambda}\bigl(\overline{\mathrm{AP}}-\overline{\mathrm{BP}}\bigr)=(2n+1)\pi-\frac{\pi}{2}=2n\pi+\frac{\pi}{2} \qquad\therefore\quad \overline{\mathrm{AP}}-\overline{\mathrm{BP}}=\left(n+\frac14\right)\lambda $$

These are intermediate conditions, neither those for sources in phase (reinforcement at $n\lambda$, cancellation at $(n+1/2)\lambda$) nor those for sources in antiphase (reinforcement at $(n-1/2)\lambda$, cancellation at $n\lambda$). For example, for $n=0$, the waves reinforce each other at points where $\overline{\mathrm{AP}}-\overline{\mathrm{BP}}=-\lambda/4$ (points where $\mathrm{B}$ is farther away than $\mathrm{A}$ by $\lambda/4$), and cancel each other at the points where the difference is $+\lambda/4$. We can understand this as follows: because the phase of source $\mathrm{B}$ is advanced, the wave from $\mathrm{B}$ comes to be in phase with the wave from $\mathrm{A}$ only after it has traveled an extra $\lambda/4$.

Solution

Let $y_1=A\sin(\omega t-kx)$ and $y_2=A\sin(\omega t+kx+\pi)$. First, we simplify $y_2$ using $\sin(\theta+\pi)=-\sin\theta$ (the periodicity of the sine; see the “Key point” box in Section 11.6.1), which gives

$$ y_2=A\sin(\omega t+kx+\pi)=-A\sin(\omega t+kx) $$

Therefore,

$$ y=y_1+y_2=A\sin(\omega t-kx)-A\sin(\omega t+kx) $$

holds. Now we expand each of $y_1,\ -y_2'$ ($y_2'\equiv A\sin(\omega t+kx)$) using the addition formulas (the same procedure as in Section 11.7):

$$ A\sin(\omega t-kx)=A(\sin\omega t\cos kx-\cos\omega t\sin kx) $$ $$ A\sin(\omega t+kx)=A(\sin\omega t\cos kx+\cos\omega t\sin kx) $$

Subtracting side by side ($y=$ the upper expression $-$ the lower expression), the terms in $\sin\omega t\cos kx$ cancel each other, and

$$ y=A(\sin\omega t\cos kx-\cos\omega t\sin kx)-A(\sin\omega t\cos kx+\cos\omega t\sin kx)=-2A\cos\omega t\sin kx $$

is obtained. Therefore,

$$ y=-2A\cos\omega t\sin kx $$

holds. This differs from the main text's $y=2A\sin\omega t\cos kx$ (a combination of $\sin\omega t$ and $\cos kx$): it is a standing wave in which $\cos\omega t$ and $\sin kx$ appear in place of those. Since it is again completely separated into a part depending on time and a part depending on space, this too is a standing wave. In the main text, the positions of the nodes (where the amplitude is always $0$) were given by $\cos kx=0$ ($x=(2m+1)\lambda/4$), whereas here they are the positions where $\sin kx=0$ ($kx=m\pi$, that is, $x=m\lambda/2$)—exactly interchanged with the positions of the antinodes in the main text's case. Likewise, the positions of the antinodes coincide with the positions of the nodes in the main text ($x=(2m+1)\lambda/4$). In other words, shifting the phase of the source by $\pi$ shifts the nodes and antinodes of the standing wave by exactly half of a half wavelength ($\lambda/4$).

Solution

By Definition 11.5 of Chapter 11 and the Caution in Section 11.9, diffraction occurs regardless of the width of the gap, but the spreading of the wave around becomes pronounced when the gap width is comparable to the wavelength or smaller. The width of the gap between the buildings is $20\ \mathrm{m}$.

FM radio waves ($\lambda\approx3\ \mathrm{m}$): The gap width ($20\ \mathrm{m}$) is much larger than the wavelength ($3\ \mathrm{m}$) (about $20/3\approx6.7$ times as large). Therefore, this falls outside the condition for diffraction to become “pronounced” in Definition 11.5; FM radio waves do not spread around much and tend to weaken in the shadow of buildings (indeed, it is known that FM broadcasts and television signals are received poorly in the shadow of buildings).

AM radio waves ($\lambda\approx300\ \mathrm{m}$): The gap width ($20\ \mathrm{m}$) is much smaller than the wavelength ($300\ \mathrm{m}$) (the ratio is about $20/300\approx0.067$, so the gap width is less than one tenth of the wavelength). Since the gap width is far smaller than the wavelength, the condition of Definition 11.5 is well satisfied, and AM radio waves spread around well and are comparatively easy to receive even in the shadow of buildings.

In summary, AM radio waves, which have a long wavelength, spread around better at the same gap between buildings than FM radio waves, which have a short wavelength. This is the same principle as the phenomenon seen in Example 11.11 of Chapter 11—sound (long wavelength) can be heard well even behind a wall, whereas light (extremely short wavelength) casts a shadow behind a wall—that is, the ratio of the wavelength to the width of the gap (or of the obstacle) determines how conspicuous diffraction is.

Solution

(1) By Formula 11.4, the amplitude of the resultant wave is $a=\sqrt{a_A^2+a_B^2+2a_Aa_B\cos\Delta\theta}$. Substituting $a_A=3.0\ \mathrm{mm},\ a_B=1.0\ \mathrm{mm},\ \Delta\theta=\pi/3$ ($\cos(\pi/3)=1/2$), we obtain

$$ a=\sqrt{3.0^2+1.0^2+2\times3.0\times1.0\times\dfrac12}\ \mathrm{mm}=\sqrt{9+1+3}\ \mathrm{mm}=\sqrt{13}\ \mathrm{mm}\approx3.6\ \mathrm{mm} $$

(2) For $\Delta\theta=0$ ($\cos0=1$), $a=\sqrt{9+1+6}\ \mathrm{mm}=\sqrt{16}\ \mathrm{mm}=4.0\ \mathrm{mm}$, which is equal to $a_A+a_B$ (the waves reinforce each other). For $\Delta\theta=\pi$ ($\cos\pi=-1$), $a=\sqrt{9+1-6}\ \mathrm{mm}=\sqrt{4}\ \mathrm{mm}=2.0\ \mathrm{mm}$, which is equal to $|a_A-a_B|$ (the waves cancel each other). Therefore the $3.6\ \mathrm{mm}$ in (1) lies between $2.0\ \mathrm{mm}$ and $4.0\ \mathrm{mm}$. As the phase difference increases from $0$ to $\pi$, the resultant amplitude decreases smoothly from $4.0\ \mathrm{mm}$ to $2.0\ \mathrm{mm}$ (because $\cos\Delta\theta$ decreases from $1$ to $-1$).

Solution

(1) Consider the moment when one end $\mathrm{O}$ of the wavefront of the incident wave (the segment $\mathrm{OQ}$, inclined at an angle $\theta_i$ to the boundary) reaches the boundary (a horizontal straight line). Let $\Delta t$ be the time it takes the other end $\mathrm{Q}$ to reach the point $\mathrm{R}$ on the boundary, and let $v$ be the speed of the wave; then $\overline{\mathrm{QR}}=v\Delta t$. The elementary wavelet emitted from $\mathrm{O}$ during this time spreads at speed $v$ in the same medium and becomes a semicircle of radius $v\Delta t$ centered at $\mathrm{O}$. If we draw the tangent from $\mathrm{R}$ to this semicircle and let $\mathrm{T}$ be the point of tangency, then $\mathrm{RT}$ is the wavefront of the reflected wave. The right triangles $\mathrm{OQR}$ (right angle at $\mathrm{Q}$) and $\mathrm{RTO}$ (right angle at $\mathrm{T}$) share the hypotenuse $\mathrm{OR}$ and have $\overline{\mathrm{QR}}=\overline{\mathrm{OT}}=v\Delta t$, so they are congruent. Hence $\angle\mathrm{QOR}=\angle\mathrm{TRO}$; that is, the angle between the incident wavefront and the boundary is equal to the angle between the reflected wavefront and the boundary. Since a wavefront is perpendicular to the direction of travel and the boundary is perpendicular to the normal, these angles are equal to the angle of incidence $\theta_i$ and the angle of reflection $\theta_r$, respectively, and therefore $\theta_i=\theta_r$.

(2) When a plane wave passes through a gap of width $a=\lambda/2$, each point lined up in the gap becomes a source of an elementary wavelet. However, since the gap width is smaller than the wavelength, the whole gap may be regarded as a single point source, and elementary wavelets (wavefronts) in the form of semicircles of radius $v\Delta t,\ 2v\Delta t,\dots$ centered at the center of the gap are formed. The wavefront tangent to all of these elementary wavelets is a semicircle centered on the gap (a cross section of a spherical wave). Therefore, the wave also goes around into the shadow of the plate and spreads almost uniformly over a wide region behind the gap (this is the same situation as in Figure 11.15 of the main text).

Solution

(1) The frequency does not change across the boundary, so it remains $f=25\ \mathrm{Hz}$. From $v=f\lambda$,

$$ \lambda_A=\frac{v_A}{f}=\frac{8.0\ \mathrm{m/s}}{25\ \mathrm{Hz}}=0.32\ \mathrm{m},\qquad \lambda_B=\frac{v_B}{f}=\frac{5.0\ \mathrm{m/s}}{25\ \mathrm{Hz}}=0.20\ \mathrm{m} $$

(the ratio $\lambda_A/\lambda_B=0.32/0.20=1.6$ agrees with $v_A/v_B=8.0/5.0=1.6$).

(2) By the law of refraction $\dfrac{\sin\theta_i}{\sin\theta_t}=\dfrac{v_A}{v_B}$,

$$ \sin\theta_t=\sin\theta_i\times\frac{v_B}{v_A}=\sin30^\circ\times\frac{5.0}{8.0}=0.5\times0.625=0.3125 $$ $$ \therefore\quad\theta_t=\arcsin(0.3125)\approx18.2^\circ $$

Since the wave enters the slower medium $B$, the angle of refraction $18.2^\circ$ is smaller than the angle of incidence $30^\circ$, and the wave bends toward the normal.

Solution

(1) The energy passing per second through a sphere of radius $r$ centered on the sound source (area $4\pi r^2$) is equal to the total power $P$ of the source (since no energy is lost). The intensity is the power per unit area, so $P=I\times4\pi r^2$.

$$ P=4.0\times10^{-6}\ \mathrm{W/m^2}\times4\pi\times(10\ \mathrm m)^2=4.0\times10^{-6}\times400\pi\ \mathrm W\approx5.0\times10^{-3}\ \mathrm W $$

(the units agree: $\mathrm{W/m^2}\times\mathrm{m^2}=\mathrm W$).

(2) The same total power $P$ spreads over a sphere of radius $r'=40\ \mathrm m$, so

$$ I'=\frac{P}{4\pi r'^2}=I\times\left(\frac{r}{r'}\right)^2=4.0\times10^{-6}\times\left(\frac{10}{40}\right)^2=2.5\times10^{-7}\ \mathrm{W/m^2} $$

(the intensity is inversely proportional to the square of the distance). By Formula 11.3, $I\propto A^2$ (the frequency $f$ being the same), so the ratio of the amplitudes is

$$ \frac{A'}{A}=\sqrt{\frac{I'}{I}}=\sqrt{\frac{1}{16}}=\frac14 $$

When the distance becomes four times as large, the intensity becomes $1/16$ and the amplitude becomes $1/4$ (the amplitude is inversely proportional to the distance).

Solution

From the formula for the speed of sound in Chapter 12, $V=331.5+0.6t$, the speed of sound at an air temperature of $10\,^\circ\mathrm{C}$ is

$$ V=331.5+0.6\times10=331.5+6.0=337.5\ \mathrm{m/s} $$

We use the Doppler effect formula (Formula 12.1 of Chapter 12) $f'=\dfrac{V-v_0}{V-v_s}f$. The observer is at rest, so $v_0=0$ (in both the approaching and the receding case).

(1) When the train approaches the observer, the direction of the source's velocity is the same as the direction in which the sound propagates (from the source toward the observer), so $v_s=+25\ \mathrm{m/s}$.

$$ f'=\frac{337.5-0}{337.5-25}\times600=\frac{337.5}{312.5}\times600\approx648.0\ \mathrm{Hz} $$

(2) When the train passes the observer and moves away, $v_s=-25\ \mathrm{m/s}$.

$$ f''=\frac{337.5-0}{337.5-(-25)}\times600=\frac{337.5}{362.5}\times600\approx558.6\ \mathrm{Hz} $$

Therefore, the observer hears about $648.0\ \mathrm{Hz}$ while the train approaches and about $558.6\ \mathrm{Hz}$ while it recedes.

Solution

The relation between the beat frequency $f_{\text{beat}}$ and the period $T$ (Chapter 12) is $f_{\text{beat}}=1/T=\abs{f_1-f_2}$. Since $T=0.25\ \mathrm s$,

$$ \abs{f_1-f_2}=\frac{1}{T}=\frac{1}{0.25}=4\ \mathrm{Hz} $$

Therefore, the difference between the frequencies of the two tuning forks is $4\ \mathrm{Hz}$. Since one of them is known to be $f_1=392\ \mathrm{Hz}$, the frequency $f_2$ of the other is

$$ f_2=f_1\pm4=392\pm4 $$

That is, either $f_2=396\ \mathrm{Hz}$ or $f_2=388\ \mathrm{Hz}$. The beat frequency (the magnitude of the frequency difference) alone does not determine whether $f_2$ is higher or lower than $f_1$—to decide this, as in Example 12.6, one must change the frequency slightly, for example by adding a small mass to one of the tuning forks, and observe how the beat frequency changes.

Solution

(1) The natural frequencies of an open pipe are $f_n=nV/(2L)$ and the wavelengths are $\lambda_n=2L/n$ (the open-pipe formulas of Chapter 12; both ends are antinodes, and $L=n\lambda_n/2$). Substituting $L=0.80\ \mathrm m$ and $V=343.5\ \mathrm{m/s}$, the fundamental frequency is

$$ f_1=\frac{V}{2L}=\frac{343.5}{2\times0.80}=\frac{343.5}{1.60}\approx214.7\ \mathrm{Hz} $$

The wavelength and the frequency of the third harmonic ($n=3$) are

$$ \lambda_3=\frac{2L}{3}=\frac{1.60}{3}\approx0.533\ \mathrm m,\qquad f_3=3f_1=\frac{3\times343.5}{1.60}\approx644.1\ \mathrm{Hz} $$

(This also agrees with $f_3=V/\lambda_3=343.5/0.5333\approx644.1\ \mathrm{Hz}$.)

(2) The natural frequencies of a closed pipe of the same length are $f^{(m)}=mV/(4L)$ ($m=1,3,5,\ldots$), so the fundamental frequency is

$$ f^{(1)}=\frac{V}{4L}=\frac{343.5}{4\times0.80}=\frac{343.5}{3.20}\approx107.3\ \mathrm{Hz} $$

(half that of the open pipe). The resonances at or below $20\ \mathrm{kHz}=20000\ \mathrm{Hz}$ are those satisfying $m\times107.34\le20000$, that is, $m\le20000/107.34\approx186.3$; these are the odd numbers $m=1,3,5,\ldots,185$. The number of odd numbers $1,3,\ldots,185$ is $(185+1)/2=93$.

For an open pipe of the same length, $n\times214.69\le20000$ gives $n\le93.2$, that is, $n=1,2,\ldots,93$, which is $93$ values. Therefore, the number of resonances at or below $20\ \mathrm{kHz}$ is the same, $93$, for the closed pipe and the open pipe (the closed pipe has half the fundamental frequency and so reaches lower notes, but only odd multiples appear, so the count comes out exactly the same).

Solution

(1) From the formula of Chapter 12, $\sin\theta=V/v_s=1/M$, the half-angle of the Mach cone is

$$ \sin\theta=\frac{1}{1.5}=\frac23\qquad\therefore\ \theta=\arcsin\frac23\approx41.8^\circ $$

(2) The speed of the airplane is $v_s=MV=1.5\times340=510\ \mathrm{m/s}$. When the surface of the Mach cone reaches the observer, the observer lies on a generatrix of the cone, and the airplane (the apex of the cone) has moved horizontally a distance $x$ beyond the point directly above the observer. Since the half-angle of the cone is $\theta$, the straight line joining the apex and the observer (a generatrix) makes the angle $\theta$ with the flight path extended backward, and in terms of the height $h$ we have $\tan\theta=h/x$. Therefore,

$$ x=\frac{h}{\tan\theta},\qquad \tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{2/3}{\sqrt5/3}=\frac{2}{\sqrt5}\quad\Bigl(\cos\theta=\sqrt{1-\tfrac49}=\tfrac{\sqrt5}{3}\Bigr) $$ $$ x=\frac{8000\sqrt5}{2}=4000\sqrt5\approx8944\ \mathrm m $$

The time the airplane takes to cover this distance is $t_1$, so

$$ t_1=\frac{x}{v_s}=\frac{8944}{510}\approx17.5\ \mathrm s $$

That is, about $17.5\ \mathrm{s}$ after the airplane passes directly overhead, the shock wave (the “boom” sound) arrives.

(3) The engine sound emitted when the airplane was directly overhead travels straight down at the speed of sound $V$ and arrives after $t_2=h/V=8000/340\approx23.5\ \mathrm s$. Therefore, the delay between the arrival of the shock wave and the arrival of the engine sound is

$$ t_2-t_1\approx23.53-17.54\approx6.0\ \mathrm s $$

In other words, the engine sound is heard this much later than the shock wave. In general, $t_2-t_1=\dfrac hV\left(1-\dfrac{\sqrt{M^2-1}}{M}\right)$ (we used $\cot\theta=\sqrt{M^2-1}$).

Solution

We use the angle form of Formula 12.2 (Chapter 12), $f'=f\,\dfrac{V-v_{\mathrm O}\cos\theta_{\mathrm O}}{V-v_{\mathrm S}\cos\theta_{\mathrm S}}$. Here $\theta_{\mathrm S},\theta_{\mathrm O}$ are the angles between the direction of the sound ray (from the source toward the observer) and the velocities of the source and of the observer, respectively.

(1) The observer is at rest ($v_{\mathrm O}=0$), and the source has $v_{\mathrm S}=20\ \mathrm{m/s}$ and $\theta_{\mathrm S}=60^\circ$, so $v_{\mathrm S}\cos\theta_{\mathrm S}=20\times\tfrac12=10\ \mathrm{m/s}$.

$$ f'=\frac{340}{340-10}\times1000=\frac{340}{330}\times1000\approx1030\ \mathrm{Hz} $$

(The component of the source's velocity along the line of sight is $+10\ \mathrm{m/s}$; since the source has a component in the direction of approach, the sound is heard higher than the frequency actually emitted.)

(2) The source is at rest ($v_{\mathrm S}=0$), and the observer has $v_{\mathrm O}=10\ \mathrm{m/s}$ and $\theta_{\mathrm O}=120^\circ$, so $v_{\mathrm O}\cos\theta_{\mathrm O}=10\times(-\tfrac12)=-5\ \mathrm{m/s}$ (a component toward the source).

$$ f'=\frac{340-(-5)}{340}\times1000=\frac{345}{340}\times1000\approx1015\ \mathrm{Hz} $$

(Since the observer has a component toward the source, the sound is heard higher than the frequency actually emitted.)

Solution

(1) In the resonance of a closed pipe, the effective length $l+\Delta\ell$ (from the water surface, which is the closed end and a node, to the antinode at the open end) is an odd multiple of $\lambda/4$. For the first resonance (fundamental mode) and the next resonance (third harmonic),

$$ l_1+\Delta\ell=\frac{\lambda}{4},\qquad l_2+\Delta\ell=\frac{3\lambda}{4} $$

Subtracting these two equations eliminates the unknown $\Delta\ell$, and we obtain

$$ l_2-l_1=\frac{\lambda}{2}\qquad\therefore\ \lambda=2(l_2-l_1)=2\times(0.510-0.160)=0.700\ \mathrm m $$

The speed of sound is $V=f\lambda=500\times0.700=350\ \mathrm{m/s}$.

(2) From $l_1+\Delta\ell=\lambda/4$,

$$ \Delta\ell=\frac{\lambda}{4}-l_1=0.175-0.160=0.015\ \mathrm m=1.5\ \mathrm{cm} $$

(This also agrees with $\Delta\ell=0.525-0.510=0.015\ \mathrm m$ obtained from the other equation $l_2+\Delta\ell=3\lambda/4$. Also, using $\Delta\ell\approx0.6r_0$, the radius of the tube is estimated to be $r_0\approx2.5\ \mathrm{cm}$.)

(3) From Definition 12.1 (the formula $V=331.5+0.6t$),

$$ t=\frac{V-331.5}{0.6}=\frac{350-331.5}{0.6}\approx30.8\,^\circ\mathrm{C} $$

That is, the air temperature is about $31\,^\circ\mathrm{C}$.

Solution

(1) Solving the formula for the speed of a wave on a string (Chapter 12), $v=\sqrt{S/\rho}$, for the tension $S$ gives $S=\rho v^2$.

$$ S=\rho v^2=1.0\times10^{-3}\times250^2=1.0\times10^{-3}\times6.25\times10^4=62.5\ \mathrm{N} $$

(2) Substituting $L=0.60\ \mathrm m$ and $v=250\ \mathrm{m/s}$ into the formula for the natural frequencies of a string fixed at both ends (Formula 12.3 of Chapter 12), $f_n=nv/(2L)$, we obtain

$$ f_1=\frac{v}{2L}=\frac{250}{1.20}\approx208.3\ \mathrm{Hz},\qquad f_2=2f_1\approx416.7\ \mathrm{Hz},\qquad f_3=3f_1=625\ \mathrm{Hz} $$

(3) The displacement of the third harmonic is $y\propto\sin(3\pi x/L)$ ($\lambda_3=2L/3$, $k_3=3\pi/L$). The antinodes are the positions where $\abs{\sin(3\pi x/L)}=1$, that is, $3\pi x/L=\pi/2,\ 3\pi/2,\ 5\pi/2$, so

$$ x=\frac{L}{6},\ \frac{L}{2},\ \frac{5L}{6}=0.10\ \mathrm m,\ 0.30\ \mathrm m,\ 0.50\ \mathrm m $$

(The nodes are at $x=0,\ L/3,\ 2L/3,\ L$, that is, $0,\ 0.20,\ 0.40,\ 0.60\ \mathrm m$, and the antinodes lie midway between them.)

Solution

If the tuning fork is mounted axially (the direction in which its arms open and close is parallel to the string), the arm merely moves one end of the string back and forth along the string and does not directly shake the string sideways. Instead, when the arm moves in the direction that pulls on the string (outward), the string is pulled taut and the tension increases, and when the arm moves inward, the string slackens and the tension decreases. During one back-and-forth motion of the tuning-fork arm (open → close → open), the tension reaches its maximum only once, so the tension varies periodically at the same frequency $f_{\text{fork}}$ as the tuning fork.

Next, consider the transverse vibration of the string. When the string undergoes a small transverse vibration $y(x,t)$, its total length increases from $L$ by

$$ \Delta\ell_{\text{elongation}}\approx\frac12\int_0^L\left(\pdiff{y}{x}\right)^2\dd x $$

This is proportional to the square of the slope $\partial y/\partial x$, that is, to the square of the magnitude of the displacement, and it takes the same value whether the displacement is positive (a crest) or negative (a trough). For example, for $y=a\sin(2\pi x/L)\cos\omega_{\text{string}}t$ (a standing wave of one wavelength), the elongation is $\dfrac{\pi^2a^2}{L}\cos^2\omega_{\text{string}}t$, which is largest when the displacement is largest (crest or trough) and zero when the string is straight ($y=0$). The more the string stretches, the larger the tension becomes, so the tension of the string varies as follows: large when the string is at a crest, small when it is straight, large when it is at a trough, small when it is straight, and so on. During one vibration of the string, the tension reaches its maximum twice (at the crest and at the trough).

It is only when the tuning fork vibrates exactly twice (so that the tension maximum occurs twice) during one vibration of the string (during which the tension maximum occurs twice) that the times at which the tuning fork changes the tension of the string match the times at which the motion of the string requires the tension to change. Then the tension reaches its maximum each time the string is at a crest or a trough, energy is continually supplied to the vibration of the string, and the string resonates and vibrates with a large amplitude. Therefore,

$$ f_{\text{fork}}=2f_{\text{string}} $$

holds. In the horizontal arrangement, on the other hand, the arm shakes the string directly sideways, so one vibration of the tuning fork corresponds to one transverse vibration of the string, and $f_{\text{fork}}=f_{\text{string}}$. (This is the mechanism of parametric excitation, the same as in pumping a swing: the swing grows larger if you stand up and crouch twice during one back-and-forth swing.)

Solution

When unpolarized (natural) light passes through the first polarizer, vibrations in all directions are averaged and the intensity is halved:

$$ I_1=\frac{I_0}{2} $$

The transmission axis of the second polarizer is tilted by $60^{\circ}$ from the direction of vibration just after passing through the first, so applying Malus's law $I=I_0\cos^2\theta$ gives

$$ I_2=I_1\cos^2 60^{\circ}=\frac{I_0}{2}\times\left(\frac12\right)^2=\frac{I_0}{2}\times\frac14=\frac{I_0}{8} $$

Therefore, the intensity of the light after passing through the second polarizer is $I_2=\dfrac{I_0}{8}$ ($12.5\ \%$ of the incident unpolarized light).

Solution

Let the transmission axis of the first polarizer be at angle $0^{\circ}$ and that of the second polarizer, which is in the crossed-Nicols arrangement, at angle $90^{\circ}$, and insert between them a third polarizer at angle $45^{\circ}$ (making $45^{\circ}$ with each of the two). Let the intensity of the unpolarized light be $I_0$. At each polarizer, we apply Malus's law, taking the direction of the transmission axis of the preceding polarizer as the new “direction of vibration.”

Immediately after the first polarizer (which transmits the unpolarized light):

$$ I_a=\frac{I_0}{2} $$

Immediately after the third polarizer (angle $45^{\circ}$, an angle difference of $45^{\circ}$ from the first):

$$ I_b=I_a\cos^2 45^{\circ}=\frac{I_0}{2}\times\frac12=\frac{I_0}{4} $$

Immediately after the second polarizer (angle $90^{\circ}$, an angle difference of $45^{\circ}$ from the third):

$$ I_c=I_b\cos^2 45^{\circ}=\frac{I_0}{4}\times\frac12=\frac{I_0}{8} $$

Therefore, the intensity of the light finally transmitted is $I_c=\dfrac{I_0}{8}$ ($12.5\ \%$).

Before the third polarizer is inserted, the first (angle $0^{\circ}$) and the second (angle $90^{\circ}$) are directly in a crossed-Nicols relationship, so $I=I_0\cos^2 90^{\circ}=0$ and no light passes through at all. Yet when the third polarizer at angle $45^{\circ}$ is inserted between them, light of intensity $I_0/8$ passes through after all—this is because a polarizer is not merely a device that “absorbs and weakens light” but a device that projects the vibration of the light onto the direction of the transmission axis and extracts only that component. After passing through the first polarizer, the light is linearly polarized in the $0^{\circ}$ direction. After passing through the third ($45^{\circ}$), its amplitude is multiplied by $\cos45^{\circ}$ and it is linearly polarized in the $45^{\circ}$ direction. This light again has a $45^{\circ}$ component with respect to the transmission axis of the second polarizer ($90^{\circ}$), so it can pass through. It is a seemingly counterintuitive result: the polarizers do not “rotate” the direction of vibration; rather, the extraction of components is split into two steps, $0^{\circ}\to45^{\circ}\to90^{\circ}$, and that is why the light gets through.

Solution

(1) Presence or absence of dispersion. The speed at which a wave propagates through a medium is determined by what kind of restoring force the medium exerts against displacement (its elasticity). For sound waves and light waves, the mechanism by which this restoring force arises is entirely different.

Sound waves (why there is almost no dispersion in the audible range). The restoring force for a sound wave in a gas originates from the macroscopic elasticity of pressure (the compressibility) exhibited by the collective of gas molecules. The speed of sound is essentially of the form $V=\sqrt{\gamma RT/M}$ ($\gamma$: heat capacity ratio, $M$: molar mass, $T$: absolute temperature), which is determined solely by the macroscopic properties of the gas, namely its temperature and composition, and the frequency of the sound appears nowhere in this expression. The periods in the audible range (roughly $20\ \mathrm{Hz}\sim20\ \mathrm{kHz}$) are far longer than the time scale over which collisions between gas molecules produce this macroscopic elasticity, so the gas responds to a sound wave as “a continuum that behaves elastically in the same way regardless of frequency.” This is why sound in the audible range shows almost no dispersion.

Light waves (why they show dispersion in transparent substances). When light (an electromagnetic wave) propagates through a transparent substance, the restoring force of the substance originates from the forced oscillation, driven by the electric field of the light, of the individual electrons bound to atoms and molecules (they are shaken like small spring pendulums with natural angular frequency $\omega_0$). Because the response of the forced oscillation of an electron depends on the relation between the driving angular frequency $\omega$ (= the frequency of the light) and the natural angular frequency $\omega_0$, specifically on the form $1/(\omega_0^2-\omega^2)$, the refractive index $n(\omega)$ takes a different value at each frequency (wavelength)—this is dispersion.

In short, the difference is this: the restoring force of a sound wave originates from a macroscopic elasticity that does not depend on frequency (a property of the medium as a whole), so almost no dispersion occurs, whereas the restoring force of a light wave originates from the resonant forced oscillation of individual electrons that respond sensitively to frequency, so dispersion inherently occurs.

(2) Polarization and a summary of the properties. Polarization means that the direction of vibration of a transverse wave is restricted to a single direction in the plane perpendicular to the direction of propagation. The direction of vibration of a transverse wave can take various directions in the plane perpendicular to the direction of propagation (there are two degrees of freedom), so a distinction arises between selecting and not selecting that direction. In contrast, for a sound wave in a gas, which is a longitudinal wave, the direction of vibration is always fixed on the single straight line along the direction of propagation, and since there is only one direction to choose from, the distinction of a “directional bias” does not exist. That light exhibits polarization is evidence that light is a transverse wave.

The differences in properties are summarized in the following table.

Item	Light	Sound (in gases and liquids)
Type of wave	Transverse wave (oscillation of the electric and magnetic fields) → polarization exists	Longitudinal wave (oscillation of density and pressure) → no polarization
What determines the speed	The medium and the wavelength (constant at $c_0$ in vacuum)	The medium and the temperature
Dispersion	Yes (colors are separated by a prism)	Almost none in the audible range
Order of magnitude of the wavelength	$10^{-7}\sim10^{-6}\ \mathrm{m}$ → diffraction is inconspicuous	$10^{-2}\sim10\ \mathrm{m}$ → diffraction is pronounced

Solution

Using the defined value of the speed of light in vacuum, $c_0=299\,792\,458\ \mathrm{m/s}$, and the defined value of the astronomical unit, $1\ \mathrm{au}=1.495\,978\,707\times10^{11}\ \mathrm{m}$, we have

$$ \Delta t=\frac{1\ \mathrm{au}}{c_0}=\frac{1.495\,978\,707\times10^{11}\ \mathrm{m}}{2.997\,924\,58\times10^8\ \mathrm{m/s}}\approx499.0\ \mathrm{s}\approx8.32\ \mathrm{min} $$

This is exactly half of “the time for light to cross the diameter of Earth's orbit ($2\ \mathrm{au}$)” (about $998.0\ \mathrm{s}\approx16.6$ minutes) found in Example 13.5 of Chapter 13. This is of course the expected result—the diameter of the orbit is exactly twice the distance of the radius ($1\ \mathrm{au}$), and since the speed of light is constant, the time required to travel it is exactly twice as well.

Solution

The light “first became completely invisible” when the gear wheel turned by exactly half a pitch during the round trip. Therefore, substituting $n=25.0\ \mathrm{/s}$, $N=600$, and $L=5.00\times10^3\ \mathrm{m}$ into Formula 13.1 of Chapter 13 (the formula for the speed of light from Fizeau's experiment), $c_0=4nNL$, we obtain

$$ c_0=4\times25.0\ \mathrm{/s}\times600\times5.00\times10^3\ \mathrm{m}=3.00\times10^8\ \mathrm{m/s} $$

This is a value very close to the defined value of the speed of light in vacuum, $c_0=2.997\,924\,58\times10^8\ \mathrm{m/s}$.

Check: the round-trip time is $2L/c_0=2\times5.00\times10^3/(3.00\times10^8)\approx3.33\times10^{-5}\ \mathrm{s}$, and the time for the half-pitch rotation is $1/(2nN)=1/(2\times25.0\times600)\approx3.33\times10^{-5}\ \mathrm{s}$; the two agree.

The light next appears brightest when the gear wheel turns exactly one pitch during the round trip; the condition is $2L/c_0=1/(n'N)$, that is, $c_0=2n'NL$. Therefore

$$ n'=\frac{c_0}{2NL}=\frac{3.00\times10^8}{2\times600\times5.00\times10^3}\ \mathrm{/s}=50.0\ \mathrm{/s} $$

which is exactly twice the rotation rate at which the light first disappeared ($25.0\ \mathrm{/s}$).

Solution

Solving the definition of the absolute refractive index, $n=c_0/v$, for $v$ gives $v=c_0/n$. Substituting $n\approx2.42$ gives

$$ v=\frac{c_0}{n}=\frac{2.997\,924\,58\times10^8\ \mathrm{m/s}}{2.42}\approx1.239\times10^8\ \mathrm{m/s} $$

The time to pass through diamond of thickness $d=1.0\times10^{-2}\ \mathrm{m}$ is, from $t=d/v$,

$$ t=\frac{1.0\times10^{-2}\ \mathrm{m}}{1.239\times10^8\ \mathrm{m/s}}\approx8.07\times10^{-11}\ \mathrm{s}=80.7\ \mathrm{ps} $$

Comparing this with the time it takes light to travel the same $1.0\ \mathrm{cm}$ in vacuum ($d/c_0\approx33.4\ \mathrm{ps}$), we see that in diamond, which has a large refractive index, light takes exactly $n\approx2.42$ times as long (this is natural since, from $v=c_0/n$, we have $t=d/v=nd/c_0$).

Solution

(1) The time for light to make a round trip between the rotating mirror and the concave mirror is $t=2R/c_0$, and the angle through which the mirror turns during this time is $\theta=\omega t=2\pi n\times2R/c_0=4\pi nR/c_0$. Using the approximate value $c_0\approx2.998\times10^8\ \mathrm{m/s}$ to estimate $\theta$, we get

$$ \theta=\frac{4\pi\times400.0\times10.0}{2.998\times10^8}\ \mathrm{rad}=\frac{5.027\times10^4}{2.998\times10^8}\ \mathrm{rad}\approx1.68\times10^{-4}\ \mathrm{rad} $$

and the deviation in the direction of the returning light is $2\theta\approx3.35\times10^{-4}\ \mathrm{rad}$. From $\tan x=x+x^3/3+\cdots$, the relative error of the approximation $\tan2\theta\approx2\theta$ is $(2\theta)^2/3\approx4\times10^{-8}$, so the approximation is quite good.

(2) Substitute $\ell=1.006\ \mathrm{cm}=1.006\times10^{-2}\ \mathrm{m}$ into Formula 13.2, $c_0=8\pi nRL/\ell$. The numerator is $8\pi\times400.0\times10.0\times30.0=3.016\times10^{6}\ \mathrm{m^2/s}$, so

$$ c_0=\frac{3.016\times10^{6}}{1.006\times10^{-2}}\ \mathrm{m/s}\approx3.00\times10^8\ \mathrm{m/s} $$

(Check: $\ell=L\times2\theta=30.0\times3.35\times10^{-4}\approx1.006\times10^{-2}\ \mathrm{m}$.)

(3) Since $\ell=8\pi nRL/c_0$ is proportional to $n$, doubling the rotation rate to $800.0\ \mathrm{/s}$ doubles the deviation to $\ell\approx2.01\ \mathrm{cm}$. The higher the rotation rate, the larger the deviation and the easier it is to measure (Foucault's experiment also used a rapidly rotating mirror).

Solution

(1) $n=\dfrac{c_0}{4NL}=\dfrac{2.998\times10^8}{4\times720\times100}\ \mathrm{/s}\approx1.04\times10^{3}\ \mathrm{/s}$; that is, about 1040 revolutions per second (about 60,000 revolutions per minute) are required. Spinning a gear wheel with 720 teeth at this speed is unrealistic with the technology of the nineteenth century (and, for a large gear wheel, even today). In addition, the round-trip time $2L/c_0=2\times100/(2.998\times10^8)\approx6.7\times10^{-7}\ \mathrm{s}$ is too short, so it is difficult to achieve a rotation so fast that the gear wheel turns half a pitch in that interval and to measure the rotation rate accurately.

(2) To make $n=c_0/(4NL)$ smaller, it suffices to make the product $NL$ larger. Either increase the number of teeth $N$ (make the teeth finer) or lengthen the distance $L$ to the mirror, or do both. For example, to make $n=12.6\ \mathrm{/s}$ or so requires $NL=c_0/(4n)\approx5.9\times10^6\ \mathrm{m}$, and with $N=720$ this gives $L\approx8.3\ \mathrm{km}$ (almost the same as in Fizeau's experiment). However, if $L$ is made too long, the light making the round trip spreads out and weakens, making it difficult to judge whether the light has disappeared.

(3) In $c_0=4nNL$, $N$ is an integer and has no error. If the errors in $n$ and $L$ are independent, the relative errors add, and

$$ \frac{\Delta c_0}{c_0}\approx\frac{\Delta n}{n}+\frac{\Delta L}{L} $$

(the relative error of a quantity in product form can be estimated as the sum of the relative errors of the factors). For example, if $\Delta n/n=1\ \%$ and $\Delta L/L=0.1\ \%$, then $\Delta c_0/c_0\approx1.1\ \%$, and most of the error comes from the measurement of the rotation rate and from how the rotation rate at which the light is judged to have “disappeared” is determined (since the light darkens continuously toward extinction, there is a spread in the judgment). Therefore, to measure the speed of light with good precision, it is important to improve the precision of the rotation-rate measurement and of the judgment of extinction (or of maximum brightness).

Solution

Let the incident wave be $y_i=A\sin(\omega t-kx+\delta)$ and the reflected wave be $y_r=A\sin(\omega t+kx+\delta+\phi)$ ($\phi$ is the phase shift relative to $y_i$). Substituting $x=0$ into the fixed-end boundary condition $(y_i+y_r)|_{x=0}=0$ gives

$$ A\sin(\omega t+\delta) + A\sin(\omega t+\delta+\phi) = 0 $$

Now let $\theta\equiv\omega t+\delta$ (since $\delta$ is a constant, $\theta$ ranges over all real numbers when $t$ does). Then this equation can be rewritten as

$$ A\sin\theta + A\sin(\theta+\phi) = 0 $$

This has exactly the same form as the equation treated in the main text of Chapter 14 with $\theta=\omega t$. Using the sum-to-product formula $\sin\alpha+\sin\beta=2\sin\frac{\alpha+\beta}2\cos\frac{\alpha-\beta}2$, we obtain

$$ A\sin\theta+A\sin(\theta+\phi)=2A\sin\left(\theta+\frac\phi2\right)\cos\frac\phi2=0 $$

Since $A\neq0$, we have $\sin(\theta+\phi/2)\cos(\phi/2)=0$. This equation must hold for every value of $\theta$ (and hence of $t$). If $\cos(\phi/2)\neq0$, it would require $\sin(\theta+\phi/2)=0$ for every $\theta$; but $\sin$ is a function whose value changes if $\theta$ is moved even slightly, so this cannot hold. Therefore

$$ \cos\frac\phi2=0 \qquad\therefore\ \frac\phi2=\left(n+\frac12\right)\pi\ (n\in\mathbb Z) \qquad\therefore\ \phi=(2n+1)\pi $$

In the range $0\leqq\phi\lt2\pi$, only $\phi=\pi$ (with $n=0$) satisfies the condition.

That is, no matter what initial phase $\delta$ is given to the incident wave, the relative phase shift at fixed-end reflection is exactly the same as in the main text, $\phi=\pi$. This is also a natural result—$\delta$ merely changes “where to place the origin of time,” and the physical phenomenon of reflection itself (how energy is thrown back at the boundary) should not depend on the choice of the time origin.

Solution

The speed of a wave on a string is $v=\sqrt{S/\rho}$, so

$$ v_1=\sqrt{\frac{10}{1.0}}=3.162\ \mathrm{m/s},\qquad v_2=\sqrt{\frac{10}{4.0}}=1.581\ \mathrm{m/s} $$

The angular frequency is common to both strings, $\omega=2\pi f=2\pi\ \mathrm{rad/s}$, and the wavenumber is $k=\omega/v$, so

$$ k_1=\frac{2\pi}{v_1}=1.987\ \mathrm{rad/m},\qquad k_2=\frac{2\pi}{v_2}=3.974\ \mathrm{rad/m} $$

Since $\rho_1=1.0\ \mathrm{kg/m}\lt\rho_2=4.0\ \mathrm{kg/m}$, Case [3] of Section 14.2 ($(\alpha,\beta)=(0,\pi)$, $\rho_1\lt\rho_2$) applies. The formulas for Case [3] are

$$ B=\frac{2k_1}{k_1+k_2}A,\qquad C=\frac{k_2-k_1}{k_1+k_2}A $$

Here $k_2/k_1=v_1/v_2=\sqrt{\rho_2/\rho_1}=\sqrt{4.0}=2.0$, so $k_2=2k_1$, a simple ratio. Substituting this gives

$$ B=\frac{2k_1}{k_1+2k_1}A=\frac23A=\frac23\times2.0=1.33\ \mathrm{cm} $$ $$ C=\frac{2k_1-k_1}{k_1+2k_1}A=\frac13A=\frac13\times2.0=0.67\ \mathrm{cm} $$

In Case [3] the phase shift of the reflected wave is $\beta=\pi$, which has the same form as the fixed-end reflection of Section 14.1 of the main text ($\phi=\pi$), so this reflection is fixed-end-like (this agrees with the intuition that a wave going from the light string $\rho_1$ to the heavy string $\rho_2$ behaves much like the fixed-end reflection of a wave “hitting a wall”).

Solution

Solving the law of refraction of Chapter 14, $\sin\theta_1/\sin\theta_2=n_2/n_1$, for $\sin\theta_2$ gives

$$ \sin\theta_2=\frac{n_1}{n_2}\sin\theta_1=\frac{1.0}{1.33}\times\sin40^\circ=\frac{1.0}{1.33}\times0.6428=0.4833 $$ $$ \theta_2=\arcsin(0.4833)\fallingdotseq28.9^\circ $$

Since $n_2=1.33\gt n_1=1.0$ (the light goes from the smaller absolute refractive index to the larger), we have $\theta_2\lt\theta_1$ ($28.9^\circ\lt40^\circ$), and the refracted ray bends toward the normal. This is the same as the rule “on entering a denser medium, the ray bends toward the normal” confirmed in Example 14.4.

Solution

The optical path length is $L(x)=n_1\sqrt{x^2+h_1^2}+n_2\sqrt{(d-x)^2+h_2^2}$ ($n_1=1.33,\ h_1=4.0,\ n_2=1.0,\ h_2=1.0,\ d=6.0$). Differentiating by exactly the same procedure as in the derivation in Chapter 14 gives

$$ L'(x)=\frac{n_1x}{\sqrt{x^2+h_1^2}}-\frac{n_2(d-x)}{\sqrt{(d-x)^2+h_2^2}}=n_1\sin\theta_1-n_2\sin\theta_2 $$

Set $L'(x)=0$ and solve the resulting equation

$$ \frac{1.33\,x}{\sqrt{x^2+16.0}}=\frac{1.0\,(6.0-x)}{\sqrt{(6.0-x)^2+1.0}} $$

Squaring both sides gives a quartic equation, which is tedious to solve by hand, so instead we examine the sign change of $f(x)\equiv L'(x)$ and bracket the root. For example, at $x=3.7$,

$$ f(3.7)=\frac{1.33\times3.7}{\sqrt{3.7^2+16.0}}-\frac{2.3}{\sqrt{2.3^2+1.0}}=\frac{4.921}{5.4489}-\frac{2.3}{2.5080}=0.9031-0.9171=-0.0139 $$

and at $x=3.8$,

$$ f(3.8)=\frac{1.33\times3.8}{\sqrt{3.8^2+16.0}}-\frac{2.2}{\sqrt{2.2^2+1.0}}=\frac{5.054}{5.5172}-\frac{2.2}{2.4166}=0.9160-0.9104=+0.0057 $$

$x\ /\ \mathrm{m}$	$0$	$3.0$	$3.7$	$3.8$	$6.0$
$f(x)$	$-0.986$	$-0.151$	$-0.014$	$+0.006$	$+1.107$

Since $f(3.7)\lt0$ and $f(3.8)\gt0$, the sign changes, so $3.7\lt x_0\lt3.8$. Treating $f$ as a straight line in this interval and interpolating linearly, we obtain

$$ x_0\fallingdotseq3.8-0.0057\times\frac{0.1}{0.0057+0.0139}=3.8-0.029\fallingdotseq3.77\ \mathrm{m} $$

At this point,

$$ \theta_1=\arctan\frac{x_0}{h_1}=\arctan\frac{3.77}{4.0}\fallingdotseq43.3^\circ,\qquad \theta_2=\arctan\frac{d-x_0}{h_2}=\arctan\frac{2.23}{1.0}\fallingdotseq65.8^\circ $$

and therefore

$$ n_1\sin\theta_1=1.33\times\sin43.3^\circ\fallingdotseq0.912,\qquad n_2\sin\theta_2=1.0\times\sin65.8^\circ\fallingdotseq0.912 $$

The two agree, so we confirm that Snell's law is satisfied. Moreover, $L(3.0)=9.812\ \mathrm{m}$, $L(3.77)=9.754\ \mathrm{m}$, and $L(4.0)=9.760\ \mathrm{m}$, so $L$ is indeed smallest at $x_0$. Since $n_1=1.33\gt n_2=1.0$ (the light goes from the denser to the less dense medium), we have $\theta_2\gt\theta_1$, which shows that the refracted ray bends away from the normal (this is exactly the configuration in which the total internal reflection studied right afterward in Section 14.7 can occur).

Solution

Substituting $n_1=2.42$ (diamond) and $n_2=1.0$ (air) into the definition of the critical angle, $\sin\theta_c=n_2/n_1$, gives

$$ \sin\theta_c=\frac{1.0}{2.42}=0.413 $$ $$ \theta_c=\arcsin(0.413)\fallingdotseq24.4^\circ $$

This is considerably smaller than the critical angle of glass (Example 14.9, $\theta_c\fallingdotseq41.8^\circ$). A small critical angle means that light traveling even slightly obliquely inside the diamond undergoes total internal reflection unless it strikes the boundary at an angle close to perpendicular, so inside a cut diamond much of the incident light is totally internally reflected many times before it emerges. These “many total internal reflections” are one reason a diamond appears to sparkle so strongly (the cutting techniques of gem polishing design the angles so that total internal reflection occurs effectively).

Solution

(1) The core of an optical fiber is made with a larger absolute refractive index than the surrounding cladding. When light that has entered the core strikes the core–cladding boundary at an angle equal to or greater than the critical angle, total internal reflection occurs, and the light is reflected and keeps traveling without leaving the core. Because this is repeated many times inside the fiber, the light can travel long distances with almost no loss.

(2) The critical angle follows from $\sin\theta_c=n_2/n_1$:

$$ \sin\theta_c=\frac{1.44}{1.48}=0.973 \qquad\therefore\ \theta_c\fallingdotseq76.6^\circ $$

The numerical aperture is

$$ \mathrm{NA}=\sqrt{n_1^2-n_2^2}=\sqrt{1.48^2-1.44^2}=\sqrt{2.1904-2.0736}=\sqrt{0.1168}\fallingdotseq0.342 $$

Compared with Example 14.10 (core $1.50$, cladding $1.46$, $\theta_c\fallingdotseq76.7^\circ$, $\mathrm{NA}\fallingdotseq0.344$), the difference between core and cladding is the same in both fibers, $0.04$. However, since the refractive index values themselves are slightly smaller for this fiber, the ratio $n_2/n_1$ is slightly smaller, $0.9730$ (versus $0.9733$ in Example 14.10), and the critical angle is also slightly smaller, $76.6^\circ$ (versus $76.7^\circ$ in Example 14.10). The numerical aperture is likewise slightly smaller, $0.342$ versus $0.344$, because $n_1+n_2$ in $\mathrm{NA}=\sqrt{(n_1-n_2)(n_1+n_2)}$ is smaller, and the range of angles over which light can be accepted is correspondingly very slightly narrower.

Solution

As confirmed in Figure 14.8, electromagnetic waves are called by different names depending on their wavelength, but they are essentially the same kind of wave. The order from longer to shorter wavelength is

radio waves > infrared > visible light > ultraviolet > X-rays > $\gamma$-rays

Since frequency is inversely proportional to wavelength ($f=c_0/\lambda$, where $c_0$ is the speed of light in vacuum, a constant), this order coincides exactly with the order from lower to higher frequency—that is, $\gamma$-rays have the highest frequency (the highest energy), and radio waves have the lowest frequency (the lowest energy).

Within visible light, red ($\lambda_1\fallingdotseq7.7\times10^{-7}\ \mathrm m$) has a longer wavelength than blue-violet ($\lambda_2\fallingdotseq3.8\times10^{-7}\ \mathrm m$). Therefore, even within visible light, red has the lower frequency and blue-violet the higher frequency (as calculated in Example 14.7, $f_1\fallingdotseq3.89\times10^{14}\ \mathrm{Hz}\lt f_2\fallingdotseq7.89\times10^{14}\ \mathrm{Hz}$).

Solution

Taking the partial derivative of $y_i+y_r$ with respect to $x$ gives

$$ \frac{\partial}{\partial x}(y_i+y_r)=-kA\cos(\omega t-kx+\delta)+kA\cos(\omega t+kx+\delta+\phi) $$

When we substitute $x=0$ and let $\theta\equiv\omega t+\delta$, the boundary condition becomes

$$ -kA\cos\theta+kA\cos(\theta+\phi)=0 $$

Using the sum-to-product formula $\cos\alpha-\cos\beta=-2\sin\dfrac{\alpha+\beta}{2}\sin\dfrac{\alpha-\beta}{2}$ ($\alpha=\theta+\phi,\ \beta=\theta$), we obtain

$$ kA\left[\cos(\theta+\phi)-\cos\theta\right]=-2kA\sin\left(\theta+\frac\phi2\right)\sin\frac\phi2=0 $$

Since $kA\neq0$, this equation must hold for every $\theta$ (and hence at all times $t$). If $\sin(\phi/2)\neq0$, then $\sin(\theta+\phi/2)=0$ would have to hold for every $\theta$, which is impossible. Therefore $\sin(\phi/2)=0$, that is, $\phi/2=n\pi$ ($n$ an integer), so $\phi=2n\pi$. In the range $0\leqq\phi\lt2\pi$, only $\phi=0$ (with $n=0$) satisfies the condition.

The initial phase $\delta$ does not appear in this result. Since the initial phase $\delta$ merely changes the choice of the time origin, the phase shift at free-end reflection is $\phi=0$, the same as in the main text, regardless of $\delta$.

Solution

(1) By the law of reflection (Law 14.1), the angle of reflection equals the angle of incidence, $50^\circ$. The angle of refraction follows from the law of refraction (Law 14.2), $\sin\theta_2=\dfrac{n_1}{n_2}\sin\theta_1$:

$$ \sin\theta_2=\frac{1.00}{1.33}\times\sin50^\circ=\frac{0.7660}{1.33}=0.5760\qquad\therefore\ \theta_2=\arcsin0.5760\fallingdotseq35.2^\circ $$

Since $n_2\gt n_1$, we have $\theta_2\lt\theta_1$ ($35.2^\circ\lt50^\circ$), and the refracted ray bends toward the normal. The construction is done as follows.

1. Draw a horizontal line as the boundary, with air above and water below.
2. Take a point $\mathrm{O}$ on the boundary and draw the vertical normal (dashed line) through $\mathrm{O}$, extending both above and below.
3. On the upper side, draw the incident ray, with an arrow, heading toward $\mathrm{O}$ from the direction $50^\circ$ to the left of the normal.
4. On the upper side, draw the reflected ray leaving $\mathrm{O}$ in the direction $50^\circ$ to the right of the normal (symmetric to the incident ray about the normal).
5. On the lower side, draw the refracted ray leaving $\mathrm{O}$ in the direction $35.2^\circ$ to the right of the normal (closer to the normal than the extension of the incident ray).

(2) For incidence from water into air, $n_1=1.33,\ n_2=1.00$ and $\theta_1=50^\circ$, so the law of refraction gives

$$ \sin\theta_2=\frac{n_1}{n_2}\sin\theta_1=\frac{1.33}{1.00}\times0.7660=1.019\gt1 $$

Since $\sin\theta_2$ exceeds $1$, no angle of refraction $\theta_2$ satisfies this equation. This corresponds to the fact that the angle of incidence $50^\circ$ is larger than the critical angle $\theta_c=\arcsin(1.00/1.33)\fallingdotseq48.8^\circ$. Therefore no refracted light is produced, and all of the light is reflected at the boundary (total internal reflection). The angle of reflection is, as before, equal to the angle of incidence, $50^\circ$.

Solution

(1) When light goes from air ($n_1\fallingdotseq1$) into glass ($n_2$), the law of refraction becomes $\sin\theta_2=\dfrac{1}{n_2}\sin\theta_1$. For the same angle of incidence $\theta_1$, the larger $n_2$ is, the smaller $\sin\theta_2$ is, and the smaller the angle of refraction $\theta_2$ is (the ray bends toward the normal to a greater degree, and is deflected more from its incident direction). In glass, “the shorter the wavelength, the larger the absolute refractive index,” so blue-violet light, with its shorter wavelength, has a larger $n_2$ than red light and a smaller $\theta_2$—that is, it is bent more. In a prism, the light bends in the same direction at both the entrance face and the exit face (toward the side opposite the apex angle, that is, the base side), so the color-dependent difference in bending accumulates over the two refractions and grows larger, and the colors separate clearly.

(2) Substitute $\sin60^\circ=0.8660$ into the law of refraction $\sin\theta_2=\sin60^\circ/n$.

$$ \sin\theta_{2,\text{red}}=\frac{0.8660}{1.514}=0.5720\ \Longrightarrow\ \theta_{2,\text{red}}\fallingdotseq34.89^\circ $$ $$ \sin\theta_{2,\text{blue}}=\frac{0.8660}{1.522}=0.5690\ \Longrightarrow\ \theta_{2,\text{blue}}\fallingdotseq34.68^\circ $$

The deviation from the incident direction is $60^\circ-34.89^\circ=25.11^\circ$ for red light and $60^\circ-34.68^\circ=25.32^\circ$ for blue light, so blue light approaches the normal more, by $0.21^\circ$ (its angle of refraction is smaller by $0.21^\circ$).

Solution

Place an object of height $h_A$ at a position $a\gt f$ in front of a convex lens of focal length $f$. Consider two light rays emerging from the tip of the object. The first is the ray that passes through the front focal point $F$ and enters the lens. For a thin lens, a ray passing through the front focal point $F$ travels parallel to the optical axis after refraction (this is the parallel ray that passes through the back focal point, with the path of the light traced in the reverse direction). The second is the ray that passes through the center of the lens and goes straight on (because the region near the center of a thin lens can be regarded as a thin plate whose front and back surfaces are nearly parallel, so the change of direction caused by refraction is negligible). The point where these two rays intersect is the tip of the image, and the foot of the image lies on the optical axis. The image is a real image (inverted, since $a\gt f$), and we let its height (its distance from the optical axis) be $h_B$.

First focus on the first ray. Since this ray becomes parallel to the optical axis after refraction, the height above the optical axis at which it leaves the lens equals the height $h_B$ of the tip of the image. Compare the following two right triangles.

• The right triangle formed by three points: the foot of the object, the tip of the object, and the front focal point $F$. The right angle is at the foot of the object, and the two legs are $a-f$ (the distance from the foot to $F$) and $h_A$.
• The right triangle formed by three points: the center of the lens, the point where this ray leaves the lens (at height $h_B$ above the optical axis), and $F$. The right angle is at the center of the lens, and the two legs are $f$ (the distance from the center of the lens to $F$) and $h_B$.

The ray forms a single straight line from the tip of the object through $F$ to the point where it leaves the lens, and $F$ lies on that line between the two; hence the angles of the two triangles at $F$ are equal as vertical angles. Since each triangle also has a right angle with respect to the optical axis, they are similar by the AA similarity criterion, and the ratios of corresponding sides are equal, so we have

$$ h_A:h_B=(a-f):f \qquad\cdots\text{①} $$

Next focus on the second ray (the ray that passes through the center and goes straight). The right triangle formed by the tip of the object, the foot of the object, and the center of the lens, and the right triangle formed by the tip of the image, the foot of the image, and the center of the lens, have equal angles at the center of the lens as vertical angles, and each has a right angle with the optical axis; hence they are similar by the AA similarity criterion, and

$$ a:b=h_A:h_B \qquad\cdots\text{②} $$

Since the ratios in ① and ② are both equal to $h_A:h_B$, we obtain

$$ a:b=(a-f):f $$

By “the product of the means equals the product of the extremes,” $a\,f=b\,(a-f)$. Dividing both sides by $abf$ (the quantities $a,b,f$ are all positive, none of them zero), we obtain

$$ \frac{af}{abf}=\frac{b(a-f)}{abf} \qquad\therefore\quad \frac{1}{b}=\frac{a-f}{af}=\frac{a}{af}-\frac{f}{af}=\frac{1}{f}-\frac{1}{a} $$

Transposing terms, we obtain

$$ \frac{1}{a}+\frac{1}{b}=\frac{1}{f} $$

This is the same lens equation as the one in the main text of Section 15.2 (the derivation using a parallel ray and a ray through the center). We have confirmed that whichever two reference rays we choose, we obtain the same image position.

Solution

Place an object of height $h_A$ at a distance $a$ in front of a concave lens whose focal length has magnitude $\abs{f}$. A ray from the tip of the object that is incident parallel to the optical axis diverges after refraction as if it had spread out from the front focal point $F$ (on the object side of the lens, at distance $\abs{f}$) (a basic property of a concave lens). Another ray, the one passing through the center of the lens, goes straight. These two rays do not intersect behind the lens, but if we extend the ray that was parallel to the axis backward, it passes through $F$ and meets the (actual path of the) ray through the center at a single point. This intersection is the tip of the virtual image, and for the time being we denote the distance from the lens to the virtual image by the positive quantity (length) $\abs{b}$.

Consider the similar triangles formed by the ray through the center (the triangle with vertices at the tip of the object, the foot of the object, and the center of the lens, and the triangle with vertices at the tip of the virtual image, the foot of the virtual image, and the center of the lens). The object and the virtual image are on the same side of the lens, so the angles at the center of the lens are equal as a common angle formed by the same straight line. Each also has a right angle with the optical axis, so they are similar by the AA similarity criterion, and

$$ a:\abs{b}=h_A:h_B $$

holds. Also, consider the right triangle formed by the point where the parallel ray passes through the lens (at height $h_A$), the center of the lens, and the front focal point $F$, and the right triangle formed by the tip of the virtual image, the foot of the virtual image, and $F$. Their angles at $F$ are equal as a common angle (formed by the same single dashed line), and each also has a right angle with the optical axis, so they satisfy the AA similarity criterion, and

$$ \abs{f}:(\abs{f}-\abs{b})=h_A:h_B $$

holds (the distance from $F$ to the virtual image is $\abs{f}-\abs{b}$ because $F$ is at distance $\abs{f}$ from the lens and the virtual image is at distance $\abs{b}$ from the lens; and since $\abs{b}\lt\abs{f}$ always holds for a concave lens, $\abs{f}-\abs{b}\gt0$. We will confirm this again at the end of the calculation). Since the right-hand sides of the two proportions are both $h_A:h_B$, we obtain

$$ a:\abs{b}=\abs{f}:(\abs{f}-\abs{b}) $$

and, by “the product of the means equals the product of the extremes,” we have

$$ \abs{b}\,\abs{f}=a\,(\abs{f}-\abs{b}) $$

Dividing both sides by $a\abs{b}\abs{f}$, we obtain

$$ \frac{1}{a}=\frac{\abs{f}-\abs{b}}{\abs{b}\abs{f}}=\frac{\abs{f}}{\abs{b}\abs{f}}-\frac{\abs{b}}{\abs{b}\abs{f}}=\frac{1}{\abs{b}}-\frac{1}{\abs{f}} $$

Subtracting $1/\abs{b}$ from both sides, we obtain

$$ \frac{1}{a}-\frac{1}{\abs{b}}=-\frac{1}{\abs{f}} $$

The right-hand side is negative, so $1/a\lt 1/\abs{b}$, that is, $a\gt\abs{b}$, and the magnitude $\abs{b}/a$ of the magnification is always less than 1.

Now, to match the sign convention of Table 15.1 of Chapter 15, we write the image distance of the virtual image as the signed quantity $b=-\abs{b}$ ($b\lt 0$) and the focal length of the concave lens as $f=-\abs{f}$ ($f\lt 0$). Since $-1/\abs{b}=1/b$ and $-1/\abs{f}=1/f$, we obtain

$$ \frac{1}{a}+\frac{1}{b}=\frac{1}{f}\qquad(b\lt 0,\ f\lt 0) $$

This has exactly the same form as Formula 15.2 of Chapter 15 (the lens equation for a concave lens, the unified formula).

Finally, solving the above equation for $\abs{b}$ gives $\abs{b}=a\abs{f}/(a+\abs{f})$, and comparing this with $\abs{f}$, we obtain

$$ \abs{f}-\abs{b}=\abs{f}-\frac{a\abs{f}}{a+\abs{f}}=\frac{\abs{f}(a+\abs{f})-a\abs{f}}{a+\abs{f}}=\frac{\abs{f}^2}{a+\abs{f}}\gt0 $$

so indeed $\abs{b}\lt\abs{f}$ ($\abs{f}-\abs{b}\gt0$) holds, which confirms what we assumed along the way. Furthermore,

$$ a-\abs{b}=a-\frac{a\abs{f}}{a+\abs{f}}=\frac{a^2}{a+\abs{f}}\gt0 $$

so $a\gt\abs{b}$; that is, we can confirm that the magnitude of the magnification satisfies $\abs{b}/a\lt1$ always, regardless of the values of the object distance $a$ and the magnitude $\abs{f}$ of the focal length.

Solution

Place a convex lens $L_1$ of focal length $f_1=10\ \mathrm{cm}$ and a convex lens $L_2$ of focal length $f_2=15\ \mathrm{cm}$ a distance $d=5\ \mathrm{cm}$ apart, and put an object at $a_1=30\ \mathrm{cm}$ in front of $L_1$. Below, all lengths are computed in $\mathrm{cm}$.

Step 1 Find the position $b_1$ of the image $P_1$ formed by $L_1$ alone. Since $a_1=30\gt f_1=10$, an inverted real image should form. By the unified formula $1/a+1/b=1/f$,

$$ \frac{1}{30}+\frac{1}{b_1}=\frac{1}{10} \qquad\therefore\quad \frac{1}{b_1}=\frac{1}{10}-\frac{1}{30}=\frac{3-1}{30}=\frac{2}{30}=\frac{1}{15} $$

we get $b_1=15\ \mathrm{cm}$. The inverted real image formed by $L_1$ lies $15\ \mathrm{cm}$ behind $L_1$.

Step 2 Find the object distance $a_2$ as seen from $L_2$. The distance between $L_1,L_2$ is $d=5\ \mathrm{cm}$, and the image formed by $L_1$ would have formed $15\ \mathrm{cm}$ behind $L_1$, that is, $15-5=10\ \mathrm{cm}$ behind $L_2$. In reality, however, the light is intercepted before that (at the position of $L_2$), so for $L_2$ the point $P_1$ is a virtual object, and

$$ a_2=-(b_1-d)=-(15-5)=-10\ \mathrm{cm} $$

This is the object distance for the second lens.

Step 3 Find the position $b_2$ of the image formed by $L_2$ (a convex lens, $f_2=15\ \mathrm{cm}$).

$$ \frac{1}{-10}+\frac{1}{b_2}=\frac{1}{15} \qquad\therefore\quad \frac{1}{b_2}=\frac{1}{15}+\frac{1}{10}=\frac{2+3}{30}=\frac{5}{30}=\frac{1}{6} $$

we get $b_2=6\ \mathrm{cm}$. Since $b_2\gt0$, the final image is a real image $6\ \mathrm{cm}$ behind $L_2$.

The overall magnification is the product of the (signed) magnifications of the individual lenses, $m_1=-b_1/a_1$ and $m_2=-b_2/a_2$:

$$ m=m_1m_2=\left(-\frac{15}{30}\right)\left(-\frac{6}{-10}\right)=\left(-\frac{1}{2}\right)\times\frac{3}{5}=-\frac{3}{10}=-0.3 $$

Since $m\lt0$, the image is inverted, and its size is $\abs{m}=0.3$ times that of the object. Looking at each lens separately, the image of $L_1$ is inverted relative to the object ($m_1=-1/2\lt 0$); for $L_2$, the object ($P_1$) is a virtual object ($a_2\lt0$) and the image is a real image ($b_2\gt0$)—the signs being opposite—so the image is upright relative to $P_1$ ($m_2=+3/5\gt0$), and the final image remains inverted relative to the original object.

Ans. An inverted real image, $0.3$ times the size of the object, forms $6\ \mathrm{cm}$ behind $L_2$.

Solution

(1) By the definition of power, $D=1/f$ ($f$ in meters), taking $f_1=25\ \mathrm{cm}=0.25\ \mathrm{m}$ for the convex lens and, noting the sign, $f_2=-50\ \mathrm{cm}=-0.50\ \mathrm{m}$ for the concave lens, we obtain

$$ D_1=\frac{1}{f_1}=\frac{1}{0.25}=4.0\ \mathrm{D}, \qquad D_2=\frac{1}{f_2}=\frac{1}{-0.50}=-2.0\ \mathrm{D} $$

These are the powers of the two lenses.

(2) With the method that uses Formula 15.3 ($1/f_{\mathrm{c}}=1/f_1+1/f_2$), keeping the lengths in $\mathrm{cm}$, we have

$$ \frac{1}{f_{\mathrm{c}}}=\frac{1}{25}+\frac{1}{-50}=\frac{2}{50}-\frac{1}{50}=\frac{1}{50} \qquad\therefore\quad f_{\mathrm{c}}=50\ \mathrm{cm}=0.50\ \mathrm{m} $$

On the other hand, with the method that finds the result from the sum of the powers, we have

$$ D=D_1+D_2=4.0+(-2.0)=2.0\ \mathrm{D} \qquad\therefore\quad f_{\mathrm{c}}=\frac{1}{D}=\frac{1}{2.0}=0.50\ \mathrm{m}=50\ \mathrm{cm} $$

Thus the two methods indeed agree: both give $f_{\mathrm{c}}=50\ \mathrm{cm}$ (reading both sides of Formula 15.3 in terms of power gives $D=D_1+D_2$ itself, so this agreement is only natural). Since $f_{\mathrm{c}}\gt0$, the combined lens acts as a convex lens.

Solution

As we confirmed in Definition 15.3 of Chapter 15, a real image is “an image formed when the light emerging from the object actually converges to a single point after passing through the lens.” For example, in Figure 15.3 (a convex lens with $a\gt f$), the rays emerging from the tip of the object (the green and blue solid lines) really do meet, after passing through the lens, at a single point behind the lens (the tip of the image). If a screen or a sheet of paper is placed at this point, light energy actually gathers there, so a sharp image appears. Countless rays pass through the lens, but all the light emerging from one point of the object (for a thin lens, under the paraxial approximation) converges to this same single point, so if the screen is placed at that position, every ray arrives at the same point and a bright image forms.

In contrast, as we confirmed in Figure 15.4 (a convex lens with $0\lt a\lt f$) and Figure 15.6 (a concave lens), in the case of a virtual image the rays after passing through the lens (solid lines) only spread farther apart from each other behind the lens, and do not actually converge to any single point. The position of the virtual image is found as the point where the extensions of these rays meet when they are extended backward with dashed lines, in a direction that the light does not actually travel (behind the lens, toward the object side). However, these extended parts are not paths along which light actually travels—they are merely hypothetical lines used in the construction, meaning “if the light extended this far, it would look as though it had started from this point.”

Therefore, even if a screen is placed at the position of the virtual image, no actual light converges to a single point there (the light merely passes by, diverging behind the lens), so nothing appears. Wherever the screen is moved behind the lens, no image appears, for the same reason. When the eye looks at this diverging light, the human visual system, through the brain's mechanism of interpreting light as having traveled in a straight line, perceives the object as if it were at the position of the virtual image (an apparent image seen only when looking into the lens), but this is a phenomenon that occurs only inside the eye, and it has no effect on a physical detector such as a screen. This is the reason why “a real image appears on a screen, but a virtual image does not.”

Solution

For the objective lens, with focal length $f_{\text{obj}}=1.0\ \mathrm{cm}$ and object distance $a_1=1.2\ \mathrm{cm}$ (since $a_1\gt f_{\text{obj}}$, an inverted real image forms), using Formula 15.1 of Chapter 15 (the unified formula) with lengths in $\mathrm{cm}$, we have

$$ \frac{1}{1.2}+\frac{1}{b_1}=\frac{1}{1.0} \qquad\therefore\quad \frac{1}{b_1}=1-\frac{1}{1.2}=1-\frac56=\frac16 $$

so $b_1=6.0\ \mathrm{cm}$. The magnitude of the magnification of the objective lens is

$$ \abs{m_{\text{obj}}}=\left|\frac{b_1}{a_1}\right|=\frac{6.0}{1.2}=5.0 $$

This is the magnitude of the magnification of the objective lens.

The eyepiece observes this intermediate image $P_1$ as a magnifying glass. When we observe so that the virtual image forms at the distance of distinct vision $D_0=25\ \mathrm{cm}$, the angular magnification is, by Formula 15.4 of Chapter 15,

$$ M_{\text{eye}}=1+\frac{D_0}{f_{\text{eye}}}=1+\frac{25}{5.0}=1+5.0=6.0 $$

This is the angular magnification of the eyepiece. The total magnification of the whole microscope is, by Eq. (15.5) of Chapter 15 ($M_{\text{total}}=\abs{m_{\text{obj}}}\times M_{\text{eye}}$),

$$ M_{\text{total}}=5.0\times6.0=30 $$

That is, this microscope magnifies the object under observation by a factor of 30.

Solution

(1) Let the object distance be $a_1$. For $L_1$ (focal length $f_1$), the unified formula gives

$$ \frac{1}{a_1}+\frac{1}{b_1}=\frac{1}{f_1}\qquad\cdots\text{①} $$

The gap between the lenses is $d=0$, so the object distance of the object $P_1$ for $L_2$ is $a_2=-(b_1-d)=-b_1$ ($P_1$ would have formed at a position $b_1$ behind $L_2$, so it is a virtual object). Writing the unified formula for $L_2$ (focal length $f_2$), we get

$$ \frac{1}{-b_1}+\frac{1}{b_2}=\frac{1}{f_2}\qquad\cdots\text{②} $$

Adding ① and ② side by side, $1/b_1$ and $-1/b_1$ cancel, and we obtain

$$ \frac{1}{a_1}+\frac{1}{b_2}=\frac{1}{f_1}+\frac{1}{f_2} $$

The left-hand side is exactly $1/(\text{object distance})+1/(\text{image distance})$ when the whole system is regarded as a single lens of combined focal length $f_{\mathrm{c}}$, so comparing with $1/a_1+1/b_2=1/f_{\mathrm{c}}$, we obtain

$$ \frac{1}{f_{\mathrm{c}}}=\frac{1}{f_1}+\frac{1}{f_2} $$

This is the desired result.

(2) As we saw in the box “Why do parallel rays converge to a single point?” in Section 15.1, a thin lens of focal length $f$ bends a ray passing at height $h$ above the optical axis toward the axis by the angle $h/f$ (paraxial approximation). Follow the ray that enters $L_1$ parallel to the optical axis at height $h$. We measure the slope in the direction toward the optical axis (if the slope is $s$, then after advancing a distance $x$ the height gets closer to the optical axis by $sx$).

• Slope just after leaving $L_1$: $s_1=0+\dfrac{h}{f_1}=\dfrac{h}{f_1}$.
• Height at which the ray enters $L_2$: while the ray travels the distance $d$ from $L_1$, its height gets closer to the optical axis by $s_1d$, so $h_2=h-s_1d=h\left(1-\dfrac{d}{f_1}\right)$.
• Slope just after leaving $L_2$: $s_2=s_1+\dfrac{h_2}{f_2}=\dfrac{h}{f_1}+\dfrac{h}{f_2}\left(1-\dfrac{d}{f_1}\right)$.

If the whole lens system is regarded as a single lens of focal length $f_{\mathrm{c}}$, a parallel ray at height $h$ should be bent with the slope $s_2=h/f_{\mathrm{c}}$, so

$$ \frac{1}{f_{\mathrm{c}}}=\frac{s_2}{h}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1f_2} $$

has been shown. For $d=0$ it agrees with the result of (1). When $f_1=15\ \mathrm{cm}$, $f_2=10\ \mathrm{cm}$, and $d=5\ \mathrm{cm}$,

$$ \frac{1}{f_{\mathrm{c}}}=\frac{1}{15}+\frac{1}{10}-\frac{5}{15\times10}=\frac{10+15-5}{150}=\frac{20}{150}=\frac{2}{15} \qquad\therefore\quad f_{\mathrm{c}}=\frac{15}{2}=7.5\ \mathrm{cm} $$

(Verification: by the sequential-imaging method, parallel rays coming from infinity try to converge at $b_1=15\ \mathrm{cm}$ behind $L_1$, and $a_2=d-b_1=-10\ \mathrm{cm}$ and $1/b_2=1/f_2-1/a_2=1/10-1/(-10)=1/5$ give $b_2=5\ \mathrm{cm}$. This $5\ \mathrm{cm}$ is the distance to the back focal point measured from $L_2$, and it agrees with $f_{\mathrm{c}}(1-d/f_1)=7.5\times(1-1/3)=5\ \mathrm{cm}$.)

Solution

(1) Draw the two reference rays emerging from the tip of the object. Let the height of the object be $h_A$ and let $a=f$.

• The ray parallel to the optical axis: it goes from the point at height $h_A$ on the lens surface toward the back focal point $F'$ (on the optical axis, at distance $f$ from the lens), so its slope after leaving the lens (in the direction approaching the optical axis) is $h_A/f$.
• The ray through the center: it goes straight from the tip of the object (at distance $a=f$ in front of the lens, at height $h_A$) to the center of the lens, so its slope (in the direction approaching the optical axis) is $h_A/a=h_A/f$.

Since the slopes of the two rays are equal ($h_A/f$), the two rays after leaving the lens are exactly parallel and never intersect, however far they go. That is, the light emerging from the tip of the object does not converge to a single point after passing through the lens but proceeds as parallel rays. For the image to “disappear” means that the position $b$ where the image would form jumps to infinity ($b=\infty$ from $1/b=1/f-1/a=0$), so that no image is formed wherever a screen is placed.

(2) We compute lengths in $\mathrm{cm}$. When $f=10$ and $a=10.5$,

$$ \frac{1}{b}=\frac{1}{10}-\frac{1}{10.5}=\frac{10.5-10}{105}=\frac{0.5}{105}=\frac{1}{210} \qquad\therefore\quad b=210\ \mathrm{cm},\quad m=-\frac{b}{a}=-\frac{210}{10.5}=-20 $$

and an inverted real image magnified 20 times forms $210\ \mathrm{cm}$ behind the lens. When $a=9.5$,

$$ \frac{1}{b}=\frac{1}{10}-\frac{1}{9.5}=\frac{9.5-10}{95}=-\frac{0.5}{95}=-\frac{1}{190} \qquad\therefore\quad b=-190\ \mathrm{cm},\quad m=-\frac{b}{a}=+\frac{190}{9.5}=+20 $$

and an upright virtual image magnified 20 times forms $190\ \mathrm{cm}$ in front of the lens. Therefore, when the object is brought closer from outside the focal point, the real image rapidly moves away to the positive side ($b\to+\infty$); at $a=f$ the light becomes parallel rays and the image vanishes; and once the object enters the inside of the focal point, the virtual image comes back from the side near negative infinity. That the curve for the convex lens in Figure 15.7 has an asymptote at $a=f$, and has $b\lt0$ for $a\lt f$ and $b\gt0$ for $a\gt f$, expresses this switch.

Solution

Let $\ell_1=\mathrm{S}_1\mathrm{P},\ \ell_2=\mathrm{S}_2\mathrm{P}$ be the distances from the slits $\mathrm{S}_1,\mathrm{S}_2$ (separation $d$) to a point $\mathrm{P}$ on the screen, respectively. Let $\ell$ be the distance from the slit plate to the screen, let $\mathrm{O}$ be the center of the screen (directly in front of the midpoint $\mathrm{S}$ of $\mathrm{S}_1,\mathrm{S}_2$), let $x$ be the distance from $\mathrm{O}$ to $\mathrm{P}$, and let $\theta$ be the angle between $\mathrm{SO}$ and $\mathrm{SP}$.

Under the condition $d\ll\ell$ (the slit separation is much smaller than the slit-to-screen distance), the three lines $\mathrm{S}_1\mathrm{P},\mathrm{S}_2\mathrm{P},\mathrm{SP}$ can be regarded as almost parallel. Drop a perpendicular from $\mathrm{S}_1$ to the line $\mathrm{S}_2\mathrm{P}$ and let $\mathrm{H}$ be its foot. Since $\mathrm{P}$ is far away, $\mathrm{H}$ and $\mathrm{S}_1$ are at almost the same distance from $\mathrm{P}$, so the extra distance traveled by the light from $\mathrm{S}_2$ is $\mathrm{S}_2\mathrm{H}$: $|\ell_2-\ell_1|\fallingdotseq\mathrm{S}_2\mathrm{H}$.

The right triangle $\mathrm{S}_1\mathrm{S}_2\mathrm{H}$ has hypotenuse $\mathrm{S}_1\mathrm{S}_2=d$, and its angle at $\mathrm{S}_1$ is $\theta$ (since $\mathrm{S}_1\mathrm{S}_2\perp\mathrm{SO}$ and $\mathrm{S}_1\mathrm{H}\perp\mathrm{SP}$, each pair of perpendicular lines is obtained from the other by rotating through the same angle, so the angle between $\mathrm{S}_1\mathrm{S}_2$ and $\mathrm{S}_1\mathrm{H}$ also equals the angle $\theta$ between $\mathrm{SO}$ and $\mathrm{SP}$). In a right triangle, “opposite side $=$ hypotenuse $\times\sin(\text{opposite angle})$,” so since the hypotenuse is $d$ and the opposite angle is $\theta$, we obtain

$$ |\ell_2-\ell_1|\fallingdotseq d\sin\theta $$

Next, from the condition $x\ll\ell$ (the range viewed on the screen is much smaller than the slit-to-screen distance), the angle $\theta$ is small. When angles are measured in radians, the three values $\sin\theta,\tan\theta,\theta$ approach one another as the angle approaches 0 (a basic approximation from radian measure in high school mathematics), so we may make the replacement

$$ \sin\theta\fallingdotseq\tan\theta $$

On the other hand, considering the right triangle $\mathrm{SOP}$ ($\mathrm{SO}=\ell$, $\mathrm{OP}=x$), we have $\tan\theta=(\text{opposite})/(\text{adjacent})=x/\ell$. Therefore,

$$ \sin\theta\fallingdotseq\tan\theta=\frac{x}{\ell} $$

Substituting this into the previous equation gives

$$ |\ell_2-\ell_1|\fallingdotseq d\sin\theta\fallingdotseq d\cdot\frac{x}{\ell}=\frac{dx}{\ell} $$

and Eq. (16.1) of Chapter 16 (Section 16.2) has been derived. We used two approximations: (1) “$\mathrm{S}_1\mathrm{P},\mathrm{S}_2\mathrm{P},\mathrm{SP}$ are almost parallel,” which follows from $d\ll\ell$, and (2) “$\sin\theta\fallingdotseq\tan\theta$,” which follows from $x\ll\ell$ (and hence $\theta$ is small).

Solution

Consider monochromatic light (wavelength $\lambda$ in vacuum) incident almost perpendicularly (angle of incidence $\theta_{\mathrm{i}}\fallingdotseq0$, and hence angle of refraction inside the film $\theta_{\mathrm{t}}\fallingdotseq0$) on a thin film of refractive index $n$ and thickness $d$ (surrounded by air). We compare two rays: the light reflected at the top surface of the film, and the light that enters the film, is reflected at the bottom surface, and then returns to the top surface. The latter travels an extra optical path length of $n\times2d$ inside the film (we used the fact that inside a medium of refractive index $n$ light travels an optical path length $n$ times that in vacuum, and that at normal incidence the round-trip distance is simply $2d$), so the optical path difference between the two rays is

$$ 2nd $$

(For a general angle of incidence this becomes $2nd\cos\theta_{\mathrm{t}}$, but here we assumed almost normal incidence and set $\cos\theta_{\mathrm{t}}\fallingdotseq1$.)

Next, we check the phase changes on reflection. The reflection at the top surface of the film occurs at the boundary going from air (small refractive index) to the film (large refractive index; assume $n\gt1$), so it is a reflection at a fixed end, and the phase changes by $\pi$ (half a wavelength). The reflection at the bottom surface of the film occurs at the boundary going from the film (large refractive index) to the air beneath it (small refractive index), so it is a reflection at a free end, and the phase does not change.

Therefore only the light reflected at the top surface carries an extra phase $\pi$, which corresponds to a phase shift equivalent to a path difference of half a wavelength, $\lambda/2$. For the two reflected rays actually to reinforce each other (crests overlap crests, troughs overlap troughs), the “effective path difference,” obtained by combining the optical path difference $2nd$ with this extra half wavelength $\lambda/2$, must be an integer multiple of the wavelength:

$$ 2nd+\frac{\lambda}{2}=m\lambda\qquad(m=1,2,3,\ldots) $$

(Taking $m=0$ would give $d\lt0$, which is not admissible, so $m$ starts from 1.) Subtracting $\lambda/2$ from both sides and rearranging, we get

$$ 2nd=m\lambda-\frac{\lambda}{2}=\left(m-\frac12\right)\lambda $$

Now let us redefine $m'=m-1$. Then $m-\tfrac12=m'+\tfrac12$, and $m'=0,1,2,\ldots$. Therefore the bright-fringe condition is

$$ 2nd=\left(m'+\frac12\right)\lambda\qquad(m'=0,1,2,\ldots) $$

(Below, we write $m'$ as $m$ again.)

Conversely, the condition for the rays to cancel each other (a crest overlaps a trough) is that the effective path difference $2nd+\lambda/2$ be a half-integer multiple of the wavelength:

$$ 2nd+\frac{\lambda}{2}=\left(m+\frac12\right)\lambda\ \Rightarrow\ 2nd=m\lambda\qquad(m=0,1,2,\ldots) $$

This agrees with Law 16.2 of the main text (for a general angle of incidence, the optical path difference is replaced by $2nd\cos\theta_{\mathrm{t}}$). In ordinary interference without phase changes, the conditions for bright and dark fringes are the reverse of these (a path difference of $m\lambda$ gives a bright fringe, and $(m+\frac12)\lambda$ gives a dark fringe). Comparing with this, we see that because “only one of the two reflections is at a fixed end and shifts the phase by $\pi$,” the conditions are exactly interchanged.

Solution

Let $d$ be the thickness of the air layer at horizontal distance $x$ from the apex of the wedge. The dark-fringe condition (Law 16.3) can then be written $2d=m\lambda$ ($m=0,1,2,\ldots$). The apex ($d=0$) is the $m=0$ dark fringe, which, as stated in the problem, is not counted, so the 32nd dark fringe counted from the apex corresponds to $m=32$. The end where the foil is inserted (where the air layer has thickness $d_{\mathrm{f}}$) is exactly this 32nd fringe, so

$$ 2d_{\mathrm{f}}=32\lambda $$

Substituting $\lambda=500\ \mathrm{nm}=500\times10^{-9}\ \mathrm{m}$, we obtain

$$ d_{\mathrm{f}}=\frac{32\lambda}{2}=\frac{32\times(500\times10^{-9}\ \mathrm{m})}{2}=8.00\times10^{-6}\ \mathrm{m}=8.00\ \mathrm{\mu m} $$

The thickness of the metal foil is thus found to be about 8 micrometers.

Since the wedge angle is $\theta\fallingdotseq d_{\mathrm{f}}/L=(8.00\times10^{-6}\ \mathrm{m})/(0.0800\ \mathrm{m})=1.00\times10^{-4}\ \mathrm{rad}$, Eq. (16.11) of the main text gives the spacing of the dark fringes as

$$ \Delta x=\frac{\lambda}{2\theta}=\frac{500\times10^{-9}\ \mathrm{m}}{2\times1.00\times10^{-4}}=2.5\times10^{-3}\ \mathrm{m}=2.5\ \mathrm{mm} $$

This agrees with $L/32=80.0\ \mathrm{mm}/32=2.5\ \mathrm{mm}$, and is consistent with the situation in which 32 dark fringes line up over the length $L$.

Solution

We regard the spherical surface of the plano-convex lens as part of a sphere with center $\mathrm{O}$ and radius $R$. Let $\mathrm{A}$ be the topmost point of the spherical surface and $\mathrm{B}$ the point of contact between the lens and the flat glass plate ($\mathrm{A},\mathrm{O},\mathrm{B}$ lie on the same vertical line, and $\mathrm{AB}=2R$ is a diameter of the sphere). On this axis $\mathrm{AB}$, let $\mathrm{D}$ be the point at height $d$ (the thickness of the air layer) above $\mathrm{B}$ ($\mathrm{DB}=d$, $\mathrm{AD}=2R-d$), and let $\mathrm{C}$ be the point where the horizontal line through $\mathrm{D}$ meets the spherical surface. Since the thickness of the air layer at horizontal distance $r$ from the axis is $d$, we have $\mathrm{DC}=r$.

(1) Since $\mathrm{AB}$ is a diameter of the circle, the inscribed angle subtending a diameter is a right angle (high school mathematics), so $\angle\mathrm{ACB}=90^{\circ}$.

(2) Since $\mathrm{CD}$ is perpendicular to the axis $\mathrm{AB}$, the angle at $\mathrm{D}$ is a right angle in both $\triangle\mathrm{ACD}$ and $\triangle\mathrm{CBD}$.

(3) Since the acute angles of the right triangle $\mathrm{ACB}$ sum to $90^{\circ}$, $\angle\mathrm{CAD}+\angle\mathrm{CBD}=90^{\circ}$. In the right triangle $\mathrm{CBD}$ as well, $\angle\mathrm{DCB}+\angle\mathrm{CBD}=90^{\circ}$. Therefore $\angle\mathrm{CAD}=\angle\mathrm{DCB}$.

(4) Hence the two right triangles have two equal angles, so $\triangle\mathrm{ACD}\sim\triangle\mathrm{CBD}$ ($\mathrm{A}\leftrightarrow\mathrm{C}$, $\mathrm{C}\leftrightarrow\mathrm{B}$, $\mathrm{D}\leftrightarrow\mathrm{D}$). Taking the ratio of corresponding sides gives

$$ \mathrm{AD}:\mathrm{DC}=\mathrm{CD}:\mathrm{DB}\ \Rightarrow\ (2R-d):r=r:d $$

Using the property of proportions (the product of the means $=$ the product of the extremes), we get

$$ r\times r=(2R-d)\times d $$ $$ \therefore\ r^2=2dR-d^2 $$

Since $d$ is far smaller than the radius of curvature $R$ of the lens, which is on the order of magnitude of a few meters ($d\sim10^{-6}\ \mathrm{m}$ against $R\sim1\ \mathrm{m}$ or so), the $d^2$ term can be neglected compared with the $2dR$ term:

$$ r^2\fallingdotseq2dR $$

This is the desired equation. Therefore the optical path difference (the round-trip distance in the air layer) is $2d\fallingdotseq r^2/R$.

(Alternative solution 1: the intersecting chords theorem. The chord $\mathrm{CC'}$ ($\mathrm{C'}$ is the point on the spherical surface on the opposite side of the axis from $\mathrm{C}$) and the diameter $\mathrm{AB}$ meet at $\mathrm{D}$, so $\mathrm{DC}\cdot\mathrm{DC'}=\mathrm{DA}\cdot\mathrm{DB}$. Since $\mathrm{DC}=\mathrm{DC'}=r$, we get $r^2=(2R-d)\,d$.)

(Alternative solution 2: a method using coordinates. Take the contact point $\mathrm{B}$ as the origin, the $y$ axis vertically upward, and the $x$ axis horizontal. The center of the sphere is $(0,R)$, so the spherical surface satisfies $x^2+(y-R)^2=R^2$. The position where the thickness of the air layer is $d$ is $y=d$, and the horizontal distance $x=r$ from the axis there satisfies $r^2+(d-R)^2=R^2$, so

$$ r^2=R^2-(d-R)^2=2Rd-d^2 $$

and we obtain the same result.)

Solution

The case of reflected Newton's rings: The two reflected rays that reach the observer are (a) the ray reflected at the lower spherical surface of the lens (the glass → air boundary) and (b) the ray reflected at the upper surface of the flat glass plate (the air → glass boundary). Since (a) is a reflection going from a medium of large refractive index to one of small refractive index, it is a reflection at a free end (no phase change), and since (b) is a reflection going from a medium of small refractive index to one of large refractive index, it is a reflection at a fixed end (the phase changes by $\pi$). Because only one of the two rays acquires the phase shift $\pi$, where the path difference is 0 (the center, $r=0$) the two rays meet in antiphase (crest and trough) and cancel each other—the center is a dark fringe.

The case of transmitted Newton's rings: The two transmitted rays that reach the observer are (c) the ray that passes straight through the lens, the air layer, and the glass plate, and (d) the ray that is first reflected at the upper surface of the flat glass plate (a fixed-end reflection at the air → glass boundary, with a phase change of $\pi$), is then reflected again at the lower spherical surface of the lens (this too is a reflection at the air → glass boundary, hence a fixed-end reflection, with a further phase change of $\pi$), and is then transmitted. The ray (d) undergoes two fixed-end reflections along its path, so the total phase change is $\pi+\pi=2\pi$. A phase change of $2\pi$ (one full period) is completely indistinguishable, as far as the shape of the wave is concerned, from no change at all ($\cos(\phi+2\pi)=\cos\phi$). Therefore the two transmitted rays (c) and (d) effectively meet in phase, and where the path difference is 0 (the center, $r=0$) they reinforce each other—the center is a bright fringe.

From the viewpoint of energy conservation: The energy of the light incident at the center should be transmitted if it is not reflected. Indeed, in reflection the center is a dark fringe (cancellation = little reflected light), whereas in transmission the center is a bright fringe (reinforcement = much transmitted light), and the relation that transmission is strengthened by exactly as much as reflection is weakened is consistent with the law of conservation of energy. Rather, the fact that the conditions for reflection and transmission are interchanged is a result that the law of conservation of energy naturally demands.

Solution

The reflection at the top surface of the film (air $n=1.0$ → film $n_{\mathrm{f}}=1.45$) is a reflection at a fixed end (phase change $\pi$), and the reflection at the bottom surface of the film (film $n_{\mathrm{f}}=1.45$ → glass $n_{\mathrm{glass}}=1.52$) is also a reflection at a fixed end (phase change $\pi$) because $n_{\mathrm{f}}\lt n_{\mathrm{glass}}$. Since both are fixed-end reflections and the phase shifts by the same amount, the phase difference between the two reflected rays is determined by the optical path difference ($2n_{\mathrm{f}}d$) alone, and there is no extra phase change. Therefore the condition under which the reflected light cancels most completely (favorable for antireflection) is

$$ 2n_{\mathrm{f}}d=\left(m+\frac12\right)\lambda\qquad(m=0,1,2,\ldots) $$

For the thinnest film ($m=0$), $2n_{\mathrm{f}}d=\lambda/2$, so

$$ d=\frac{\lambda}{4n_{\mathrm{f}}} $$

Substituting $\lambda=520\ \mathrm{nm}$ and $n_{\mathrm{f}}=1.45$, we obtain

$$ d=\frac{520\ \mathrm{nm}}{4\times1.45}=\frac{520}{5.80}\ \mathrm{nm}\approx89.7\ \mathrm{nm} $$

Thus a film thickness of about 90 nanometers, much thinner than the wavelength of visible light, is required.

Solution

The interference intensity for $N$ slits is, by Eq. (16.6) of Chapter 16,

$$ I(\delta)=I_0\left(\frac{\sin(N\delta/2)}{N\sin(\delta/2)}\right)^2 $$

The principal maximum is at $\delta=0$ (or at its integer multiples $\delta=2\pi m$), where $I(0)=I_0$. We seek the point $\delta_1$ (the smallest of those with $\delta_1>0$) at which the intensity first becomes 0 slightly away from the principal maximum. $I(\delta)=0$ occurs when the numerator satisfies $\sin(N\delta/2)=0$, and at the same time the denominator must satisfy $\sin(\delta/2)\neq0$ (otherwise we get the indeterminate form $0/0$, which in fact turns out to be the principal maximum itself).

Among the $\delta>0$ satisfying $\sin(N\delta/2)=0$, the smallest is obtained by solving $N\delta/2=\pi$ (the smallest positive case in which $N\delta/2$ is an integer multiple of $\pi$):

$$ \delta_1=\frac{2\pi}{N} $$

At this point $\delta_1/2=\pi/N$, and as long as $N\ge2$ we have $0\lt\pi/N\lt\pi$, so we can also confirm that $\sin(\delta_1/2)\neq0$ (the denominator does not vanish). Therefore the “one-sided width” from the principal maximum ($\delta=0$) to the first point where the intensity is 0 is $2\pi/N$, and the principal maximum lies within the range $-2\pi/N\lt\delta\lt2\pi/N$—that is, the full width of the principal maximum (the width between the first zeros on both sides) is

$$ \Delta\delta=\frac{4\pi}{N} $$

From this equation it follows directly that the width $\Delta\delta$ of the principal maximum is inversely proportional to the number of slits $N$ ($\Delta\delta\propto1/N$). The larger $N$ is (the more grooves there are, as in a diffraction grating, where the number of grooves is very large), the narrower and sharper the bright line of the principal maximum becomes—this is the mathematical confirmation of how the bright fringes grow sharper as the number of slits is increased through $N=2,5,20$ in Figure 16.7 of Chapter 16. (We also confirmed numerically, for $N=2,5,20,100$, that the intensity vanishes at $\delta_1=2\pi/N$.)

Solution

Solving the Bragg condition $2d\sin\theta=n\lambda$ for $d$, we get

$$ d=\frac{n\lambda}{2\sin\theta} $$

Substituting $n=1$ (first order), $\lambda=0.15406\ \mathrm{nm}$, and $\theta=14.22^{\circ}$, we obtain

$$ d=\frac{1\times0.15406\ \mathrm{nm}}{2\sin14.22^{\circ}}=\frac{0.15406\ \mathrm{nm}}{2\times0.24567}\approx0.3136\ \mathrm{nm} $$

This value agrees very well with the interplanar spacing of the (111) lattice planes of a real silicon crystal (from the crystal structure with lattice constant $a=0.5431\ \mathrm{nm}$, $d_{111}=a/\sqrt3\approx0.3136\ \mathrm{nm}$), and it is the Bragg angle of the (111) reflection, which appears most strongly in the powder X-ray diffraction pattern of silicon. In this way, from measurements of the X-ray wavelength and the Bragg angle alone, the spacing between atomic planes in a crystal can be determined with a precision finer than a nanometer.

Solution

The optical path length is the product $nt$ of the refractive index $n$ and the physical thickness $t$ (Definition 16.1). Since the optical path length of each layer is to be $\lambda_0/4$, we have $nt=\lambda_0/4$, that is, $t=\lambda_0/(4n)$. With $\lambda_0=550\ \mathrm{nm}$,

$$ t_{\mathrm{TiO_2}}=\frac{550\ \mathrm{nm}}{4\times2.4}=\frac{550}{9.6}\ \mathrm{nm}\approx57.3\ \mathrm{nm} $$ $$ t_{\mathrm{SiO_2}}=\frac{550\ \mathrm{nm}}{4\times1.46}=\frac{550}{5.84}\ \mathrm{nm}\approx94.2\ \mathrm{nm} $$

A ray making one round trip through each layer travels a physical distance of $2t$, so its optical path length is $n\times2t=2\times\lambda_0/4=\lambda_0/2$. In terms of phase, since the phase advances by $2\pi$ over one wavelength $\lambda_0$, an optical path length of $\lambda_0/2$ corresponds to a phase of $2\pi\times(\lambda_0/2)/\lambda_0=\pi$. To this is added the difference $\pi$ between the phase changes on reflection at adjacent boundaries (a fixed-end reflection from low to high refractive index gives $\pi$; a free-end reflection from high to low gives 0), giving a total of $2\pi$, so the rays reflected from all the boundaries reinforce each other in phase.

Solution

Since the initial displacement is $u_0\equiv0$, only the second term of d'Alembert's formula remains, giving

$$ y(x,t)=\frac{1}{2v}\int_{x-vt}^{x+vt}w_0(s)\,\dd s $$

Since $w_0(s)=w_0=2.0\ \mathrm{m/s}$ ($-a\le s\le a$, $a=0.020\ \mathrm{m}$) and it is $0$ elsewhere, the integral can be written as $y=\dfrac{w_0\,\ell}{2v}$ in terms of “the length $\ell$ of the part of the interval $[x-vt,\ x+vt]$ that overlaps $[-a,a]$.”

(1) At $t=1.0\times10^{-4}\ \mathrm{s}$, $vt=100\times1.0\times10^{-4}=0.010\ \mathrm{m}$. For $x=0$ the interval is $[-0.010,\,0.010]$, which lies entirely within $[-0.020,\,0.020]$, so the overlap length is $\ell=0.020\ \mathrm{m}$. Therefore

$$ y=\frac{2.0\times0.020}{2\times100}=2.0\times10^{-4}\ \mathrm{m}=0.20\ \mathrm{mm} $$

(2) At $t=5.0\times10^{-4}\ \mathrm{s}$, $vt=0.050\ \mathrm{m}$. For $x=0.050\ \mathrm{m}$ the interval is $[0,\ 0.100]$, and its overlap with $[-0.020,\,0.020]$ is $[0,\,0.020]$, of length $\ell=0.020\ \mathrm{m}$. Therefore $y=\dfrac{2.0\times0.020}{2\times100}=2.0\times10^{-4}\ \mathrm{m}=0.20\ \mathrm{mm}$.

(3) After a sufficiently long time, $vt\gg a$, and the interval $[-vt,\,vt]$ for $x=0$ completely contains $[-a,a]$. The overlap length is $\ell=2a=0.040\ \mathrm{m}$, and so

$$ y=\frac{2.0\times0.040}{2\times100}=4.0\times10^{-4}\ \mathrm{m}=0.40\ \mathrm{mm} $$

We see that, even though the initial displacement is zero, a flat plateau of height $w_0a/v=0.40\ \mathrm{mm}$ forms between the waves spreading to the left and right from the struck portion (this is for an infinitely long string; for a real, finite string it holds only until the waves reflected at the ends return).

Solution

Since $y(-t)=y(t)$ (an even function), $b_n=0$. The mean value $a_0/2$ is $0$, since the area over one period is $2\int_0^{T/2}A(1-4t/T)\,\dd t=2A\bigl(T/2-T/2\bigr)=0$. With $\omega_1=2\pi/T$, the cosine coefficients are

$$ a_n=\frac4T\int_0^{T/2}A\left(1-\frac{4t}{T}\right)\cos n\omega_1t\,\dd t =\frac{4A}{T}\left[\int_0^{T/2}\cos n\omega_1t\,\dd t-\frac4T\int_0^{T/2}t\cos n\omega_1t\,\dd t\right] $$

The first integral is $\dfrac{\sin(n\omega_1T/2)}{n\omega_1}=\dfrac{\sin n\pi}{n\omega_1}=0$. For the second integral, integration by parts gives

$$ \int_0^{T/2}t\cos n\omega_1t\,\dd t=\left[\frac{t\sin n\omega_1t}{n\omega_1}\right]_0^{T/2}-\int_0^{T/2}\frac{\sin n\omega_1t}{n\omega_1}\,\dd t=0+\left[\frac{\cos n\omega_1t}{(n\omega_1)^2}\right]_0^{T/2}=\frac{(-1)^n-1}{(n\omega_1)^2} $$

Since $\omega_1=2\pi/T$, we have $(n\omega_1)^2=4\pi^2n^2/T^2$, and therefore

$$ a_n=\frac{4A}{T}\cdot\left(-\frac4T\right)\cdot\frac{(-1)^n-1}{(n\omega_1)^2}=\frac{16A}{T^2}\cdot\frac{1-(-1)^n}{4\pi^2n^2/T^2}=\frac{4A\,[1-(-1)^n]}{\pi^2n^2} $$

That is, $a_n=\dfrac{8A}{\pi^2n^2}$ ($n$ odd) and $a_n=0$ ($n$ even), and $y(t)=\dfrac{8A}{\pi^2}\displaystyle\sum_{n\,\text{odd}}\dfrac{\cos n\omega_1t}{n^2}$.

(1) Since the function is even, only cosines appear and no sines. The only harmonics that appear are those of odd order. This is because the triangle wave has half-wave symmetry $y(t+T/2)=-y(t)$ (for example, $y(0)=A$ and $y(T/2)=A(1-2)=-A$). Since $\cos\bigl(n\omega_1(t+T/2)\bigr)=(-1)^n\cos n\omega_1t$, an even-order component does not change sign under $t\to t+T/2$, and so it cannot be compatible with $y$, which changes sign. This is the same reason that a square wave contains only odd-order harmonics.

(2) With $A=3.0\ \mathrm{V}$, $a_1=\dfrac{8\times3.0}{\pi^2}\approx2.43\ \mathrm{V}$, $a_3=\dfrac{a_1}{9}\approx0.270\ \mathrm{V}$, and $a_5=\dfrac{a_1}{25}\approx0.0973\ \mathrm{V}$.

(3) The amplitudes of the harmonics of a square wave decrease as $1/n$ ($b_3/b_1=1/3$, $b_5/b_1=1/5$), and those of a triangle wave as $1/n^2$ ($a_3/a_1=1/9$, $a_5/a_1=1/25$). The triangle wave is continuous and has only corners, so its harmonics decay faster, it approaches the original waveform with few terms, and there is no Gibbs phenomenon either. A square wave has discontinuities, so its harmonics decay only slowly.

Solution

(1) Substituting $x_0=L/4$ gives $x_0(L-x_0)=\dfrac L4\cdot\dfrac{3L}{4}=\dfrac{3L^2}{16}$, so

$$ C_n=\frac{2hL^2}{n^2\pi^2}\cdot\frac{16}{3L^2}\sin\frac{n\pi}{4}=\frac{32h}{3n^2\pi^2}\sin\frac{n\pi}{4} $$

For $h=2.0\ \mathrm{mm}$, $\dfrac{32h}{3\pi^2}=\dfrac{64}{3\pi^2}\approx2.16\ \mathrm{mm}$, so

$$ \begin{aligned} C_1&=2.16\times\sin\tfrac\pi4\approx1.53\ \mathrm{mm},\qquad C_2=\frac{2.16}{4}\sin\tfrac\pi2\approx0.540\ \mathrm{mm},\qquad C_3=\frac{2.16}{9}\sin\tfrac{3\pi}{4}\approx0.170\ \mathrm{mm},\\ C_4&=\frac{2.16}{16}\sin\pi=0,\qquad C_5=\frac{2.16}{25}\sin\tfrac{5\pi}{4}\approx-0.0611\ \mathrm{mm} \end{aligned} $$

(2) The harmonics with $n=4,8,12,\ldots$ (multiples of 4) vanish. For those $n$ with $\sin(n\pi/4)=0$, the plucked point $x_0=L/4$ coincides with a node of the corresponding normal mode $X_n(x)=\sin(n\pi x/L)$ ($X_n(L/4)=\sin(n\pi/4)=0$). Plucking a string at a node does not excite that normal mode.

(3) If the midpoint is plucked, $C_1\approx1.62\ \mathrm{mm}$, $C_2=0$, $C_3\approx-0.180\ \mathrm{mm}$, $C_4=0$, and $C_5\approx0.0648\ \mathrm{mm}$; all even-order components vanish, and the sound is a clear tone in which the fundamental accounts for almost everything. By contrast, when the point at $L/4$ is plucked, the second harmonic is present with an amplitude of about $0.540/1.53\approx35\%$ of the fundamental (it is 0 for the midpoint), and the third harmonic also remains at a comparable level. Because even-order components are included, the sound is richer in harmonics, and its timbre is sharper and harder than when the midpoint is plucked.

Solution

(1) For the open pipe, $f_n=nv/(2L)$ and $v/(2L)=340/(2\times0.85)=200\ \mathrm{Hz}$, so $f_1,f_2,f_3,f_4=200,\ 400,\ 600,\ 800\ \mathrm{Hz}$. For the closed pipe, $f_n=(2n-1)v/(4L)$ and $v/(4L)=340/(4\times0.85)=100\ \mathrm{Hz}$, so $f_1,f_2,f_3,f_4=100,\ 300,\ 500,\ 700\ \mathrm{Hz}$.

(2) $300\ \mathrm{Hz}=3\times100\ \mathrm{Hz}$ coincides with the $n=2$ natural frequency of the closed pipe, so the closed pipe resonates. The natural frequencies of the open pipe are integer multiples of $200\ \mathrm{Hz}$ ($200,400,\ldots$), so $300\ \mathrm{Hz}$ is not among them.

(3) The fundamental frequency of the open pipe is $v/(2L)=200\ \mathrm{Hz}$. If the fundamental frequency $v/(4L')$ of a closed pipe of length $L'$ equals this, then $L'=v/(4\times200)=0.425\ \mathrm{m}=L/2$. To produce a note of the same pitch, the closed pipe needs only half the length of the open pipe.

(4) At the closed end $x=0$ the displacement has a node, so $X(0)=0$. Substituting into the general solution $X=B\sin kx+C\cos kx$ gives $C=0$, that is, $X=B\sin kx$. At the open end $x=L$ the displacement has an antinode ($\partial y/\partial x=0$), so $X'(L)=Bk\cos kL=0$. Since $B\ne0,\ k\ne0$, we must have $\cos kL=0$, and only the discrete values $kL=\dfrac\pi2,\ \dfrac{3\pi}2,\ \dfrac{5\pi}2,\ldots$, that is, $k_n=\dfrac{(2n-1)\pi}{2L}$ ($n=1,2,\ldots$), are allowed. Then $\lambda_n=2\pi/k_n=4L/(2n-1)$, that is, $L=(2n-1)\lambda_n/4$ ($L$ is an odd multiple of $1/4$ of the wavelength), and we obtain $f_n=vk_n/2\pi=(2n-1)v/(4L)$. It is the requirement that the boundary conditions be satisfied at the two ends simultaneously that makes $k$ discrete.

Solution

(1) Differentiating both sides of $\omega^2=gk+\dfrac{\sigma}{\rho_{\mathrm m}} k^3$ with respect to $k$ gives $2\omega\dfrac{\dd\omega}{\dd k}=g+\dfrac{3\sigma}{\rho_{\mathrm m}}k^2$, so

$$ v_{\mathrm g}=\frac{\dd\omega}{\dd k}=\frac{g+3\sigma k^2/\rho_{\mathrm m}}{2\omega}=\frac{g+3\sigma k^2/\rho_{\mathrm m}}{2\sqrt{gk+\sigma k^3/\rho_{\mathrm m}}} $$

(2) Since $v_{\mathrm p}=\omega/k$, $\dfrac{v_{\mathrm g}}{v_{\mathrm p}}=\dfrac{k\,(g+3\sigma k^2/\rho_{\mathrm m})}{2\omega^2}=\dfrac{g+3s}{2(g+s)}$ (we let $s=\sigma k^2/\rho_{\mathrm m}$ and used $\omega^2=k(g+s)$). In the long-wavelength limit ($s\ll g$) this is $\dfrac{g}{2g}=\dfrac12$, and in the short-wavelength limit ($s\gg g$) it is $\dfrac{3s}{2s}=\dfrac32$. That is, for long waves $v_{\mathrm g}=\tfrac12v_{\mathrm p}$, and for short ripples $v_{\mathrm g}=\tfrac32v_{\mathrm p}$, so the order of the two speeds is reversed.

(3) $\lambda=5.0\ \mathrm{cm}$: $k=2\pi/0.050\approx125.7\ \mathrm{rad/m}$, $\omega^2=9.8\times125.7+\dfrac{0.072}{1.0\times10^3}\times125.7^3\approx1231.5+142.9=1374.4$, and $\omega\approx37.07\ \mathrm{rad/s}$. Hence $v_{\mathrm p}=\omega/k\approx0.295\ \mathrm{m/s}$ and $v_{\mathrm g}=\dfrac{9.8+3\times7.2\times10^{-5}\times125.7^2}{2\times37.07}=\dfrac{9.8+3.41}{74.14}\approx0.178\ \mathrm{m/s}$ ($v_{\mathrm g}/v_{\mathrm p}\approx0.60$).

$\lambda=0.50\ \mathrm{cm}$: $k\approx1256.6\ \mathrm{rad/m}$, $\omega^2=9.8\times1256.6+7.2\times10^{-5}\times1256.6^3\approx1.23\times10^4+1.43\times10^5=1.55\times10^5$, and $\omega\approx394\ \mathrm{rad/s}$. $v_{\mathrm p}\approx0.313\ \mathrm{m/s}$ and $v_{\mathrm g}=\dfrac{9.8+3\times7.2\times10^{-5}\times1256.6^2}{2\times394}=\dfrac{9.8+341}{788}\approx0.445\ \mathrm{m/s}$ ($v_{\mathrm g}/v_{\mathrm p}\approx1.42$; here the group velocity is the faster).

(4) Setting $v_{\mathrm g}=v_{\mathrm p}$ in the formula from (2), $g+3s=2(g+s)$ gives $s=g$, that is, $\sigma k_{\mathrm c}^2/\rho_{\mathrm m}=g$, so

$$ k_{\mathrm c}=\sqrt{\frac{\rho_{\mathrm m} g}{\sigma}}=\sqrt{\frac{1.0\times10^3\times9.8}{0.072}}\approx369\ \mathrm{rad/m},\qquad \lambda_{\mathrm c}=\frac{2\pi}{k_{\mathrm c}}\approx1.70\ \mathrm{cm} $$

At this wavenumber, $\omega^2=k_{\mathrm c}(g+s)=2gk_{\mathrm c}$, so $v_{\mathrm p}=\omega/k_{\mathrm c}=\sqrt{2g/k_{\mathrm c}}\approx0.23\ \mathrm{m/s}$. This is the wavelength at which the phase velocity is minimum (since $v_{\mathrm g}=v_{\mathrm p}+k\,\dd v_{\mathrm p}/\dd k$, the condition $v_{\mathrm g}=v_{\mathrm p}$ is equivalent to $\dd v_{\mathrm p}/\dd k=0$); waves longer than this behave as gravity waves, and shorter ones as capillary waves (surface tension waves).

Solution

(1) From $\omega=\sqrt{gk}$, $\omega'=\dfrac12\sqrt g\,k^{-1/2}$ and $\omega''=-\dfrac14\sqrt g\,k^{-3/2}$. Since $k_0=2\pi/20\approx0.3142\ \mathrm{rad/m}$, we have $k_0^{-3/2}\approx5.68\ \mathrm{m^{3/2}}$ and $\sqrt g\approx3.130\ \mathrm{m^{1/2}/s}$, so

$$ \beta=\omega''(k_0)=-\frac14\times3.130\times5.68\approx-4.44\ \mathrm{m^2/s} $$

(The group velocity is $v_{\mathrm g}=\omega'(k_0)=\tfrac12\sqrt{g/k_0}\approx2.79\ \mathrm{m/s}$.)

(2) The Fourier components of $\psi(x,0)=e^{-x^2/(2\sigma_0^2)}e^{ik_0x}$ (the distribution in $q=k-k_0$) are given by $A(q)=\sigma_0\sqrt{2\pi}\,e^{-\sigma_0^2q^2/2}$. Using $\omega\approx\omega_0+v_{\mathrm g}q+\tfrac12\beta q^2$ and $s=x-v_{\mathrm g}t$, the waveform at time $t$ is

$$ \psi(x,t)=e^{i(k_0x-\omega_0t)}\frac{1}{2\pi}\int A(q)\,e^{iqs-\frac i2\beta q^2t}\,\dd q =e^{i(k_0x-\omega_0t)}\frac{\sigma_0}{\sqrt{2\pi}}\int e^{-\frac12(\sigma_0^2+i\beta t)q^2+iqs}\,\dd q $$

From the Gaussian integral formula ($a=\sigma_0^2+i\beta t$, $\mathrm{Re}\,a=\sigma_0^2\gt0$), we obtain

$$ \psi(x,t)=e^{i(k_0x-\omega_0t)}\,\frac{\sigma_0}{\sqrt{\sigma_0^2+i\beta t}}\exp\!\left[-\frac{s^2}{2(\sigma_0^2+i\beta t)}\right] $$

Taking the real part of $\dfrac{1}{\sigma_0^2+i\beta t}=\dfrac{\sigma_0^2-i\beta t}{\sigma_0^4+\beta^2t^2}$, the envelope is

$$ |\psi|\propto\exp\!\left[-\frac{s^2\sigma_0^2}{2(\sigma_0^4+\beta^2t^2)}\right]=\exp\!\left[-\frac{(x-v_{\mathrm g}t)^2}{2\sigma(t)^2}\right],\qquad \sigma(t)^2=\frac{\sigma_0^4+\beta^2t^2}{\sigma_0^2}=\sigma_0^2\left[1+\left(\frac{\beta t}{\sigma_0^2}\right)^2\right] $$

Hence the envelope remains Gaussian: its center advances as $v_{\mathrm g}t$ while its width broadens to $\sigma(t)=\sigma_0\sqrt{1+(\beta t/\sigma_0^2)^2}$.

(3) $\sigma(t)=\sqrt2\,\sigma_0$ occurs when $(\beta t/\sigma_0^2)^2=1$, that is, at

$$ t=\frac{\sigma_0^2}{|\beta|}=\frac{(50\ \mathrm{m})^2}{4.44\ \mathrm{m^2/s}}\approx5.6\times10^2\ \mathrm{s}\approx9.4\ \text{min} $$

During that time the wave group travels $v_{\mathrm g}t\approx2.79\times563\approx1.6\times10^3\ \mathrm{m}=1.6\ \mathrm{km}$. (Here $\Delta k=1/\sigma_0=0.020\ \mathrm{rad/m}\ll k_0$, so the wave packet is quasi-monochromatic.) Since the group velocity $v_{\mathrm g}=\tfrac12\sqrt{g\lambda/2\pi}$ is larger for longer waves, in a swell generated by a distant storm the waves reach the shore in order, the longest first. (The width of the envelope at time $563\ \mathrm{s}$ is $70.7\ \mathrm{m}\approx\sqrt2\times50\ \mathrm{m}$.)

Solution

(1) The line spacing is $1/T$, and the first zero of the envelope is at $f=1/\tau=1/(2.0\times10^{-6}\ \mathrm{s})=5.0\times10^5\ \mathrm{Hz}=500\ \mathrm{kHz}$ (independent of $T$). The line located there is the one with index $n=f\cdot T=T/\tau$, and its height is 0, so the number of nonzero lines up to the zero is $T/\tau-1$. The amplitude of the fundamental is $a_1=\dfrac{2A}{\pi}\sin\dfrac{\pi\tau}{T}$.

• $T=8.0\ \mathrm{\mu s}$: the spacing is $1/T=125\ \mathrm{kHz}$ and $T/\tau=4$, so there are three nonzero lines, and $a_1=\dfrac{2\times1.0}{\pi}\sin\dfrac\pi4\approx0.450\ \mathrm{V}$.
• $T=80\ \mathrm{\mu s}$: the spacing is $1/T=12.5\ \mathrm{kHz}$ and $T/\tau=40$, so there are 39 nonzero lines, and $a_1=\dfrac2\pi\sin\dfrac\pi{40}\approx0.0500\ \mathrm{V}$.

(2) As $T$ is increased, the line spacing $1/T$ approaches 0, and the number of lines increases without limit. Meanwhile the height of each line approaches 0 in proportion to $\tau/T$, but the shape of the envelope (of the form $\sin(\pi f\tau)/(\pi f\tau)$, with its first zero at $1/\tau$) does not change. As $T\to\infty$ the lines join without gaps into a continuous spectrum with the shape of the envelope. The Fourier series of a periodic pulse train thus turns into the Fourier transform of a single isolated pulse.

(3) (Sample answer) The Fourier series deals with waveforms that have a period and represents them as a sum of sinusoids at the discrete frequencies that are integer multiples of the fundamental frequency, so the spectrum is a line spectrum. The Fourier transform deals with non-periodic waveforms such as a single pulse. It is the limit in which the period is made infinite: the spacing between the lines goes to 0 so that the frequency becomes continuous, and the sum is replaced by an integral. Consequently, the spectrum is continuous. In both cases, the waveform is expressed as a superposition of sinusoids (complex exponentials).

Solution

(1) $\Delta x=c\tau=2.998\times10^8\times100\times10^{-15}\approx3.0\times10^{-5}\ \mathrm{m}=30\ \mathrm{\mu m}$.

(2) From $\Delta t\,\Delta\omega\sim2\pi$, $\Delta f=\Delta\omega/2\pi\sim1/\tau=1/(1.0\times10^{-13}\ \mathrm{s})=1.0\times10^{13}\ \mathrm{Hz}$ ($10\ \mathrm{THz}$). Differentiating $\lambda=c/f$ gives $\Delta\lambda\approx\lambda^2\Delta f/c$, so

$$ \Delta\lambda\approx\frac{(8.0\times10^{-7}\ \mathrm{m})^2\times1.0\times10^{13}\ \mathrm{Hz}}{2.998\times10^8\ \mathrm{m/s}}\approx2.1\times10^{-8}\ \mathrm{m}=21\ \mathrm{nm} $$

(The same value is obtained from the spatial relation $\Delta x\,\Delta k\sim2\pi$: $\Delta\lambda\approx\lambda^2/\Delta x=(8.0\times10^{-7})^2/(3.0\times10^{-5})\approx21\ \mathrm{nm}$.)

(3) The period of the light is $\lambda/c=8.0\times10^{-7}/2.998\times10^8\approx2.7\times10^{-15}\ \mathrm{s}=2.7\ \mathrm{fs}$, so the number of light waves contained in the pulse is $\tau/(\lambda/c)=100/2.67\approx37$.

(4) For $\tau=10\ \mathrm{fs}$, $\Delta x=3.0\ \mathrm{\mu m}$, $\Delta f\sim1/\tau=1.0\times10^{14}\ \mathrm{Hz}$, and $\Delta\lambda\approx\lambda^2\Delta f/c\approx2.1\times10^{-7}\ \mathrm{m}=210\ \mathrm{nm}$ (10 times that in (2)), and the number of waves contained is about $3.7$. The shorter the pulse, the wider the spread in frequency (wavelength); since $\Delta\lambda$ reaches about one quarter of the central wavelength $800\ \mathrm{nm}$, this estimate gives only the order of magnitude.

Solution

The cross-sectional area of a bolt of diameter $d_0=12\ \mathrm{mm}$ is the area of a circle of radius $d_0/2=6\ \mathrm{mm}$:

$$ A_0=\pi\left(\frac{d_0}{2}\right)^2=\pi\times(6\ \mathrm{mm})^2=\pi\times36\ \mathrm{mm^2}\approx113.10\ \mathrm{mm^2} $$

Substituting the load $P=8\ \mathrm{kN}=8\times10^3\ \mathrm N$ into the definition of stress from Chapter 18, $\sigma=P/A$, we obtain

$$ \sigma=\frac{P}{A_0}=\frac{8\times10^3\ \mathrm N}{113.10\ \mathrm{mm^2}}\approx70.7\ \mathrm{N/mm^2}=70.7\ \mathrm{MPa} $$

(we used the fact that $1\ \mathrm{N/mm^2}=1\ \mathrm{MPa}$). Giving the answer to three significant figures, $\sigma\approx70.7\ \mathrm{MPa}$.

Solution

From the definition of Poisson's ratio in Chapter 18, $\nu=\abs{\varepsilon'/\varepsilon}$, and the fact that when the longitudinal strain is positive (tension, $\varepsilon\gt0$) the transverse strain is negative ($\varepsilon'\lt0$), we can write $\varepsilon'=-\nu\varepsilon$. Substituting $\nu=0.30$ and $\varepsilon=2.0\times10^{-4}$, we get

$$ \varepsilon'=-\nu\varepsilon=-0.30\times2.0\times10^{-4}=-6.0\times10^{-5} $$

The magnitude of the transverse strain ($6.0\times10^{-5}$) is about 0.3 times ($=\nu$) the longitudinal strain ($2.0\times10^{-4}$), which is exactly what the definition of Poisson's ratio requires.

Solution

From the definition of shear strain in Chapter 18, $\gamma=\tan\alpha$, we have $\alpha=\arctan\gamma$.

(a) The exact angle is obtained by substituting $\gamma=0.05$:

$$ \alpha_{\text{exact}}=\arctan(0.05)\approx0.049958\ \mathrm{rad}\ (\approx2.862^\circ) $$

(b) Using the small-angle approximation $\alpha\approx\gamma$, we have

$$ \alpha_{\text{approx}}\approx\gamma=0.05\ \mathrm{rad}\ (\approx2.865^\circ) $$

(c) The relative error between the two is

$$ \frac{\abs{\alpha_{\text{exact}}-\alpha_{\text{approx}}}}{\alpha_{\text{exact}}}=\frac{\abs{0.049958-0.050000}}{0.049958}\approx0.083\ \% $$

In Example 18.11 of Chapter 18 ($\gamma=0.10$) the relative error was about $0.33\ \%$, whereas in the present problem ($\gamma=0.05$) it is about $0.083\ \%$, roughly one quarter as large. This can be explained as follows. Since $\tan\alpha=\alpha+\alpha^{3}/3+\cdots$ (Taylor expansion), the relative error of the approximation $\gamma\approx\alpha$ is

$$ \frac{\tan\alpha-\alpha}{\alpha}\approx\frac{\alpha^{2}}{3} $$

When $\alpha$ is halved ($\gamma$ goes from $0.10$ to $0.05$), the relative error becomes $(1/2)^2=1/4$ of its former value, and indeed $0.33\ \%\to0.083\ \%$ agrees exactly with the calculated value. (Note that the absolute error $\tan\alpha-\alpha\approx\alpha^{3}/3$ is proportional to $\alpha^{3}$, so when $\alpha$ is halved it becomes about $1/8$.) In any case, for a small shear strain such as $\gamma=0.05$, the approximation $\alpha\approx\gamma$ can be said to be extremely accurate.

Solution

A metal foam contains a large number of pores (cavities) inside. Definition 18.1 (homogeneous body) in Chapter 18 was the assumption that “the material is a continuous solid with no defects or cavities inside, and every part has the same properties,” so for a metal foam full of pores it is the assumption of a homogeneous body that is most likely to fail. When the size of the pores becomes comparable to the dimensions of the member under consideration (for example, the thickness of a plate), the approximation that “the material has the same properties wherever we take a piece from it” no longer holds, and the positions, sizes, and arrangement of the pores themselves begin to govern the mechanical behavior.

Furthermore, when the shape or distribution of the pores is biased toward a particular direction (for example, a metal foam in which the pores have been stretched and aligned in one direction by the manufacturing process), the assumption of an isotropic body (Definition 18.2) fails at the same time—because the strength and the ease of elongation of the material become different in the direction along which the pores are lined up and in the direction perpendicular to it. Thus the two basic assumptions, homogeneous body and isotropic body, both rest on the premise that “the material can be regarded as a continuum whose structure is sufficiently fine compared with the scale of the problem under consideration,” and for a material with “a non-uniform structure of non-negligible size,” such as pores, both assumptions need to be checked carefully.

Solution

According to Saint-Venant's principle (Law 18.2) of Chapter 18, the disturbance that the details of how an external force is applied produce in the stress becomes small at a distance of about the representative size of the region over which that external force acts. When a concentrated load acts on the end of a flat plate, the “representative size” is the width $b$ of the plate that carries the load.

Therefore, as a general rule of thumb, in a cross section located at least about the plate width $b$ away from the point of application of the load (to be on the safe side, one to two times $b$), the stress distribution can be regarded as nearly uniform. As a numerical example, when the non-uniform stress components of a strip are examined by the theory of elasticity, it is known that the magnitude of the slowest-decaying component decreases, as a function of the distance $x$ from the load point, roughly in proportion to $e^{-4.2x/b}$. Taking its magnitude at the load point to be $1$, this is about $12\ \%$ at $x=0.5b$, about $1.5\ \%$ at $x=b$, and about $0.02\ \%$ at $x=2b$. Likewise, for the circular hole of Example 18.2, if $d$ is the diameter of the hole that produces the disturbance, the increase is about $22\ \%$ at a distance $d$ from the center of the hole, about $4\ \%$ at $2d$, and about $1.5\ \%$ at $3d$, so here too the disturbance had almost vanished at “several times the size of the region of action.” Conversely, within a distance less than the width $b$ of the point of application, the assumption that the stress distribution is uniform (the working assumption usually adopted in mechanics of materials, stated at the end of Law 18.2 in Chapter 18) may fail, and a particularly careful examination (or a detailed analysis of stress concentration such as the one we shall study in Chapter 21) becomes necessary.

Solution

In the first-order approximation formula for the volumetric strain from Chapter 18, $\varepsilon_V\approx\varepsilon_x+\varepsilon_y+\varepsilon_z$ (the simple sum of the longitudinal strains in the three directions), let the strain in the longitudinal direction (the direction of the load) be $\varepsilon_x=\varepsilon$, and let the strains in the two transverse directions perpendicular to it be equal by symmetry, $\varepsilon_y=\varepsilon_z=\varepsilon'$. Then

$$ \varepsilon_V\approx\varepsilon_x+\varepsilon_y+\varepsilon_z=\varepsilon+\varepsilon'+\varepsilon'=\varepsilon+2\varepsilon' $$

and the required formula is obtained. Next, by the definition of Poisson's ratio in Chapter 18, the transverse strain under uniaxial tension ($\varepsilon\gt0$) is $\varepsilon'=-\nu\varepsilon$, so substituting this gives

$$ \varepsilon_V\approx\varepsilon+2(-\nu\varepsilon)=\varepsilon(1-2\nu) $$

an expression in terms of $\varepsilon$ and $\nu$ alone. Substituting $\nu=1/2$ into this expression,

$$ \varepsilon_V\approx\varepsilon\left(1-2\times\frac12\right)=\varepsilon(1-1)=0 $$

and we confirm that the volumetric strain is exactly $0$.

The physical meaning of this result is that “a material with Poisson's ratio $\nu=1/2$ undergoes no change at all in its total volume, no matter in which direction it is pulled or pushed.” Such a material is called incompressible. It is well known that the Poisson's ratio of natural rubber is very close to $\nu\approx0.5$ (the Application in Section 18.4), which means that rubber deforms in such a way that the thinning that occurs when it is pulled is exactly enough to keep the volume constant. Conversely, when a material with $\nu\lt1/2$ (most solids, such as metals and ceramics, fall in this category) is pulled ($\varepsilon\gt0$), we have $1-2\nu\gt0$, so $\varepsilon_V=\varepsilon(1-2\nu)\gt0$; that is, the transverse contraction alone cannot fully cancel the longitudinal elongation, and the volume increases slightly overall. The volume never decreases. For example, for $\nu=0.30$ and $\varepsilon=1.0\times10^{-3}$, the first-order approximation is $\varepsilon(1-2\nu)=4.0\times10^{-4}$, while the exact formula $\varepsilon_V=(1+\varepsilon)(1+\varepsilon')^{2}-1$ (with $\varepsilon'=-3.0\times10^{-4}$) gives $3.995\times10^{-4}$, in good agreement. The bulk modulus $K$ (defined in Chapter 19 as the constant of proportionality between pressure and volumetric strain) is a quantity that expresses this “resistance to volume change,” and it can also be confirmed in the next chapter that the limit $\nu\to1/2$ corresponds to $K\to\infty$ (the volume does not change no matter what pressure is applied).

Solution

(a) The shear plane is a circle of diameter $d=8\ \mathrm{mm}$, so its area is

$$ A=\pi\left(\frac d2\right)^2=\pi\times(4\ \mathrm{mm})^2=16\pi\ \mathrm{mm^2}\approx50.27\ \mathrm{mm^2} $$

Substituting $Q=5\ \mathrm{kN}=5.0\times10^{3}\ \mathrm N$ into the definition of shear stress from Chapter 18, $\tau=Q/A$, we obtain

$$ \tau=\frac QA=\frac{5.0\times10^{3}\ \mathrm N}{50.27\ \mathrm{mm^2}}\approx99.5\ \mathrm{N/mm^2}=99.5\ \mathrm{MPa} $$

This is the shear stress.

(b) The bolt fails when the shear stress reaches the shear strength $\tau_{\max}=300\ \mathrm{MPa}=300\ \mathrm{N/mm^2}$, so rewriting $\tau=Q/A$ as $Q=\tau A$,

$$ Q_{\max}=\tau_{\max}A=300\ \mathrm{N/mm^2}\times50.27\ \mathrm{mm^2}\approx1.508\times10^{4}\ \mathrm N\approx15.1\ \mathrm{kN} $$

The load of $5\ \mathrm{kN}$ in (a) is about $1/3$ of the failure load of $15.1\ \mathrm{kN}$ (consistent with $99.5/300\approx0.33$).

Solution

(a) Since $1\ \mathrm{GPa}=10^{3}\ \mathrm{MPa}=10^{9}\ \mathrm{Pa}$,

$$ 1.5\ \mathrm{GPa}=1.5\times10^{3}\ \mathrm{MPa}=1500\ \mathrm{MPa},\qquad 1.5\ \mathrm{GPa}=1.5\times10^{9}\ \mathrm{Pa} $$

(b) $101.325\ \mathrm{kPa}=101.325\times10^{3}\ \mathrm{Pa}=1.01325\times10^{5}\ \mathrm{Pa}$. Since $1\ \mathrm{MPa}=10^{6}\ \mathrm{Pa}=1\ \mathrm{N/mm^2}$,

$$ 1.01325\times10^{5}\ \mathrm{Pa}=0.101325\times10^{6}\ \mathrm{Pa}=0.101325\ \mathrm{MPa}=0.101325\ \mathrm{N/mm^2} $$

so atmospheric pressure is approximately $0.1\ \mathrm{N/mm^2}$.

(c) Rewriting $\sigma=P/A$ as $P=\sigma A$ and substituting $\sigma=120\ \mathrm{MPa}=120\ \mathrm{N/mm^2}$ and $A=250\ \mathrm{mm^2}$, we obtain

$$ P=\sigma A=120\ \mathrm{N/mm^2}\times250\ \mathrm{mm^2}=3.0\times10^{4}\ \mathrm N=30\ \mathrm{kN} $$

(the unit is $\mathrm{N/mm^2}\times\mathrm{mm^2}=\mathrm N$). Converting to $\mathrm{Pa}$, with $\sigma=1.2\times10^{8}\ \mathrm{Pa}$ and $A=2.5\times10^{-4}\ \mathrm{m^2}$, we still get $P=1.2\times10^{8}\times2.5\times10^{-4}=3.0\times10^{4}\ \mathrm N$, the same result.

Solution

Substitute the cross-sectional area $A_0=5.0\times10^{-5}\ \mathrm{m^2}$, the length $l_0=2.0\ \mathrm{m}$, the load $P=1.0\times10^{4}\ \mathrm N$, and the elongation $\lambda=2.0\ \mathrm{mm}=2.0\times10^{-3}\ \mathrm{m}$ (all units have been converted to SI units) into the equation $E=l_0P/(\lambda A_0)$ of Chapter 19. Then

$$ E=\frac{l_0P}{\lambda A_0}=\frac{2.0\times1.0\times10^{4}}{2.0\times10^{-3}\times5.0\times10^{-5}}=\frac{2.0\times10^{4}}{1.0\times10^{-7}}=2.0\times10^{11}\ \mathrm{Pa}=200\ \mathrm{GPa} $$

This is the value obtained. (The stress is $\sigma=P/A_0=200\ \mathrm{MPa}$ and the strain is $\varepsilon=\lambda/l_0=1.0\times10^{-3}$, and $E=\sigma/\varepsilon$ gives the same value.) This value agrees with the typical value of Young's modulus of steel (Table 19.1 of Chapter 19, $200$–$210\ \mathrm{GPa}$), so the rod is considered to be made of steel.

Solution

Substitute $E=120\ \mathrm{GPa}$ and $\nu=0.35$ into the formulas $G=E/\{2(1+\nu)\}$ and $K=E/\{3(1-2\nu)\}$ of Chapter 19. First we check that $\nu=0.35$ lies within the range $-1\lt\nu\lt1/2$ allowed for an isotropic elastic body (Section 19.3.4)—since $0.35$ is inside this range, the calculation below is physically consistent.

$$ G=\frac{E}{2(1+\nu)}=\frac{120}{2\times(1+0.35)}=\frac{120}{2.70}\approx44.4\ \mathrm{GPa} $$ $$ K=\frac{E}{3(1-2\nu)}=\frac{120}{3\times(1-0.70)}=\frac{120}{3\times0.30}=\frac{120}{0.90}\approx133.3\ \mathrm{GPa} $$

These are the values obtained. The Young's modulus used here is in nearly the same range as the typical Young's modulus of copper (Table 19.1 of Chapter 19, $110$–$128\ \mathrm{GPa}$), so the resulting values of $G,K$ also agree well with the measured values for copper ($G\approx44\ \mathrm{GPa}$ or so, $K\approx130$–$140\ \mathrm{GPa}$ or so).

Solution

Building an analogy with the spring constant. Solving the equation $E=l_0P/(\lambda A_0)$ of Chapter 19 for $\lambda$ gives $\lambda=\dfrac{l_0}{EA_0}P$. This has the same form as the elongation of a spring, $x=F/k$ (that is, $F=kx$), and we see that $l_0/(EA_0)$ corresponds to $1/k$. In other words, we can regard a single rod as a “spring” with the following spring constant:

$$ k=\frac{EA_0}{l_0} $$

Suppose that the two rods have the same cross-sectional area $A_0$ and that each has length $l_0$ (for a total length of $2l_0$ altogether). Then the spring constants of the two rods are

$$ k_1=\frac{E_1A_0}{l_0},\qquad k_2=\frac{E_2A_0}{l_0} $$

respectively.

Using the formula for springs in series. Directly applying the series-combination formula for springs learned in high school physics, $\dfrac{1}{k_{\text{comb}}}=\dfrac{1}{k_1}+\dfrac{1}{k_2}$ (this corresponds to the fact that when two springs are connected in series, the assembly as a whole is subjected to the same load $P$ and stretches by the sum of the individual elongations $\lambda_1,\lambda_2$), we obtain

$$ \frac{1}{k_{\text{comb}}}=\frac{1}{k_1}+\frac{1}{k_2}=\frac{l_0}{E_1A_0}+\frac{l_0}{E_2A_0}=\frac{l_0}{A_0}\left(\frac{1}{E_1}+\frac{1}{E_2}\right)=\frac{l_0}{A_0}\cdot\frac{E_1+E_2}{E_1E_2} $$

Taking the reciprocal of both sides gives

$$ k_{\text{comb}}=\frac{A_0}{l_0}\cdot\frac{E_1E_2}{E_1+E_2} $$

Finding $E_{\text{comb}}$. The total length of the combined rod is $2l_0$, so the Young's modulus after combination follows from the relation $k_{\text{comb}}=E_{\text{comb}}A_0/(2l_0)$ (the same “spring-constant formula for a rod,” applied to a single rod of length $2l_0$):

$$ E_{\text{comb}}=\frac{2l_0}{A_0}k_{\text{comb}}=\frac{2l_0}{A_0}\times\frac{A_0}{l_0}\cdot\frac{E_1E_2}{E_1+E_2}=\frac{2E_1E_2}{E_1+E_2} $$

This is what we obtain. It is exactly the harmonic mean of $E_1,E_2$, namely $\dfrac{2}{1/E_1+1/E_2}$ (putting the fractions in the denominator over a common denominator gives $\dfrac{2}{1/E_1+1/E_2}=\dfrac{2E_1E_2}{E_1+E_2}$). As a numerical example, if $E_1=200\ \mathrm{GPa}$ and $E_2=70\ \mathrm{GPa}$, then

$$ E_{\text{comb}}=\frac{2\times200\times70}{200+70}=\frac{28000}{270}\approx103.7\ \mathrm{GPa} $$

which is smaller than the simple average of the two values, $135\ \mathrm{GPa}$; it is a value pulled toward the softer rod ($70\ \mathrm{GPa}$). The property familiar from springs in series, that the one with the smaller elastic modulus (the softer one) has the stronger influence, appears here as well. This is because, under the same load $P$, the softer rod stretches more.

Solution

At the necked minimum cross section, the true strain is defined as $\varepsilon_{\mathrm t}=\ln(A_0/A)$ (Definition 19.13 of Chapter 19), so the cross-sectional area is $A=A_0e^{-\varepsilon_{\mathrm t}}$. Since the load is $P=\sigma_{\mathrm t}A$, the nominal stress, obtained by dividing the load by the initial cross-sectional area $A_0$, can be written as

$$ \frac{P}{A_0}=\sigma_{\mathrm t}\,e^{-\varepsilon_{\mathrm t}} $$

Differentiating this with respect to $\varepsilon_{\mathrm t}$ (by the product rule) gives

$$ \frac{\dd(P/A_0)}{\dd\varepsilon_{\mathrm t}}=e^{-\varepsilon_{\mathrm t}}\left(\frac{\dd\sigma_{\mathrm t}}{\dd\varepsilon_{\mathrm t}}-\sigma_{\mathrm t}\right) $$

For Hollomon's equation $\sigma_{\mathrm t}=K_{\mathrm H}\varepsilon_{\mathrm t}^{\,n}$ we have $\dd\sigma_{\mathrm t}/\dd\varepsilon_{\mathrm t}=(n/\varepsilon_{\mathrm t})\sigma_{\mathrm t}$, so

$$ \frac{\dd(P/A_0)}{\dd\varepsilon_{\mathrm t}}=e^{-\varepsilon_{\mathrm t}}\sigma_{\mathrm t}\left(\frac{n}{\varepsilon_{\mathrm t}}-1\right) $$

Since $e^{-\varepsilon_{\mathrm t}}\sigma_{\mathrm t}\gt0$, the sign is determined by $n/\varepsilon_{\mathrm t}-1$. It is positive for $\varepsilon_{\mathrm t}\lt n$ ($P/A_0$ increases), $0$ at $\varepsilon_{\mathrm t}=n$ (the maximum value; the tensile strength), and negative for $\varepsilon_{\mathrm t}\gt n$ ($P/A_0$ decreases). On the other hand, the true stress $\sigma_{\mathrm t}$ itself satisfies $\dd\sigma_{\mathrm t}/\dd\varepsilon_{\mathrm t}=(n/\varepsilon_{\mathrm t})\sigma_{\mathrm t}\gt0$, so it keeps increasing with $\varepsilon_{\mathrm t}$.

A qualitative restatement. When the true strain increases slightly by $\dd\varepsilon_{\mathrm t}$, the cross-sectional area $A$ decreases by the fraction $\dd\varepsilon_{\mathrm t}$ ($\dd A/A=-\dd\varepsilon_{\mathrm t}$). Meanwhile, the true stress (the strength due to work hardening) increases by the fraction $(n/\varepsilon_{\mathrm t})\dd\varepsilon_{\mathrm t}$. The load $P=\sigma_{\mathrm t}A$ changes by the fraction given by the difference of these two effects, $(n/\varepsilon_{\mathrm t}-1)\dd\varepsilon_{\mathrm t}$. Work hardening “levels off” as $\varepsilon_{\mathrm t}$ becomes larger (that is, $n/\varepsilon_{\mathrm t}$ becomes smaller), so once $\varepsilon_{\mathrm t}\gt n$, the decrease of the cross-sectional area outweighs the increase of strength due to work hardening, and the load (and therefore $P/A_0$) decreases. This is why, even though the material itself continues to work-harden (the true stress keeps increasing), the load recorded by the testing machine, that is, the nominal stress, goes down once it has passed the tensile strength.

Numerical check ($K_{\mathrm H}=636\ \mathrm{MPa}$, $n=0.166$; these are values obtained by assuming that Hollomon's equation holds unchanged in the necked region). At $\varepsilon_{\mathrm t}=n=0.166$ (the onset of necking), $\sigma_{\mathrm t}\approx472\ \mathrm{MPa}$ and $P/A_0=\sigma_{\mathrm t}e^{-0.166}\approx400\ \mathrm{MPa}$ (the tensile strength). When the true strain of the minimum cross section has advanced to $\varepsilon_{\mathrm t}=0.30$, the true stress increases to $\sigma_{\mathrm t}\approx521\ \mathrm{MPa}$, whereas the nominal stress decreases to $P/A_0=521\times e^{-0.30}\approx386\ \mathrm{MPa}$.

Note that the relation $\sigma_{\mathrm t}=\sigma(1+\varepsilon)$, which holds only while the specimen is elongating uniformly, cannot be used after necking has begun (because the average elongation over the gauge length does not represent the deformation of the minimum cross section). For the explanation after necking, we use $\varepsilon_{\mathrm t}=\ln(A_0/A)$ of the minimum cross section, as above.

Solution

Classifying the tests according to Definition 19.5 of Chapter 19, and adding in a few words the typical property that each test is meant to examine, we obtain the following.

Static load tests (the load is applied slowly, quasi-statically):

• Tensile test: The specimen is pulled, and the mechanical properties in the tensile direction as a whole are examined, such as the elastic modulus (Young's modulus), the yield point and proof stress, the tensile strength, and the elongation and reduction of area.
• Compression test: The specimen is compressed, and the elastic modulus and strength in the compression direction are examined (this is especially important for evaluating the strength of brittle materials such as concrete and ceramics, which are weak in tension but strong in compression).
• Torsion test: A twisting moment is applied to the specimen, and the shear modulus $G$ and the strength in the shear direction (related to the mechanics of torsion treated in Chapter 20) are examined.
• Flexural test: A beam-shaped specimen is bent, and the flexural rigidity and flexural strength are examined (this is often used especially for evaluating the strength of brittle materials from which tensile specimens are hard to make; it is related to the bending of beams in Chapter 20).
• Hardness test: An indenter (a hard sphere or pyramid) is pressed into the surface of the material, and the degree to which the surface resists deformation (hardness) is examined. It is known to have an empirical correlation with other measures of strength, such as the tensile strength.
• Creep test: A constant load (constant stress) is applied for a long time, and the rate of the deformation that proceeds slowly with time (creep deformation) is examined. It is important for evaluating the long-term life of components used at high temperatures (such as turbine blades).

Dynamic load tests (the load is applied suddenly or repeatedly):

• Fatigue test: Even at a stress much smaller than the tensile strength, a material can be brought to fracture by applying the load many times over (fatigue); this test examines the relationship between the stress and the number of repetitions at which this phenomenon occurs.
• Impact test: A large load is applied instantaneously with a hammer or the like, and the ability of the material to withstand the impact and absorb energy while deforming (toughness) and its brittleness at low temperatures (embrittlement) are examined.

Note that the creep test has in common with the other static load tests that the magnitude of the load itself is constant (quasi-static), but it differs in character from tests such as the tensile test, which can be completed with an instantaneous, short-time measurement, in that it targets “deformation that appears only after a long time” (even so, it is clearly distinguished from the dynamic load tests, in which the load is applied suddenly or repeatedly, and is classified as a static load test).

Solution

Prediction: reasoning from the type of bonding. As we saw in “Application: The atomistic origin of elastic moduli” in Section 19.5, Young's modulus reflects the curvature at the bottom of the interatomic potential (the strength of the bonding itself). Considering how each of the four materials resists deformation,

• Rubber: It has a structure in which long chain molecules of hydrocarbons (polymers) are loosely entangled with one another. Most of the deformation when it is pulled does not come from stretching the individual covalent bonds themselves, but from the tangled chains unraveling and extending (a phenomenon called entropic elasticity, which has a statistical-mechanical origin—when rubber is stretched, the “disorder” of the chain configurations decreases, so it tends to return to its original state). Since the bonds themselves are not directly stretched, Young's modulus is extremely small.
• Wood (along the grain): Cellulose molecules are aligned in a fibrous manner, and strong covalent bonds act along the direction of the fibers. When wood is pulled along the fibers, these covalent bonds are stretched directly, so it is much stiffer than rubber, but the bonds are neither as dense nor as uniform as in the steel and glass discussed below (there are also parts where the fibers are connected to one another by weak bonds such as hydrogen bonds), so it has an intermediate Young's modulus.
• Glass (soda-lime glass): Silicate (mainly $\mathrm{SiO_2}$) forms a random network structure dominated by covalent bonds. Because strongly directional covalent bonds extend throughout, it is stiffer than wood. However, the atoms in the network are loosely packed, and the network can also deform through changes in the $\mathrm{Si\text{–}O\text{–}Si}$ angle, so it does not become as stiff as steel.
• Steel: Metallic bonding (bonding in which metal ions are immersed in a “sea” of free electrons), with the atoms densely packed in a crystal lattice. The strength of the bonding itself is not as great as that of strong covalent bonds, but the atoms are arranged regularly at high density and all of the bonds are stretched directly. Consequently its Young's modulus is larger than that of glass, which has a network structure with many gaps.

Therefore, if we assume that Young's modulus is larger the more the “deformation that directly stretches the bonds” dominates, the predicted order is

$$ \text{rubber}\ (E\sim0.001\text{–}0.1\ \mathrm{GPa}) \ \lt\ \text{wood (along the grain)}\ (E\sim10\text{–}16\ \mathrm{GPa}) \ \lt\ \text{glass}\ (E\sim65\text{–}72\ \mathrm{GPa}) \ \lt\ \text{steel}\ (E\sim200\text{–}210\ \mathrm{GPa}) $$

This is the predicted order.

Comparison with literature values. If we line up the values for steel ($200$–$210\ \mathrm{GPa}$) and soda-lime glass ($65$–$72\ \mathrm{GPa}$) listed in Table 19.1 of Chapter 19 together with the typical values found in general engineering-materials data collections—for wood (along the grain, with a range depending on the species), $E\approx10$–$16\ \mathrm{GPa}$ or so, and for rubber, a few $\mathrm{MPa}$ for soft types and about $0.1\ \mathrm{GPa}$ for hard types (with a wide range depending on the degree of vulcanization)—we can confirm that they agree well with the order predicted above (rubber ≪ wood < glass < steel). That the Young's moduli are spread over more than four orders of magnitude (the ratio of steel to rubber is roughly $10^{3}$–$10^{5}$) is an example that shows plainly how great an influence the atomistic differences in the “type and density of the bonds” have on the macroscopic mechanical properties.

Solution

We have $\sigma_x=-40\ \mathrm{MPa}$, $\sigma_y=80\ \mathrm{MPa}$, and $\tau_{xy}=-25\ \mathrm{MPa}$. Computing the mean and the half-difference, we obtain

$$ \frac{\sigma_x+\sigma_y}2=\frac{-40+80}2=20\ \mathrm{MPa},\qquad \frac{\sigma_x-\sigma_y}2=\frac{-40-80}2=-60\ \mathrm{MPa} $$

(1) From the principal stress formula of Chapter 20 (Law 20.2), $\sigma_{1,2}=\dfrac{\sigma_x+\sigma_y}2\pm\sqrt{\left(\dfrac{\sigma_x-\sigma_y}2\right)^2+\tau_{xy}^2}$, the square-root term is $\sqrt{(-60)^2+(-25)^2}=\sqrt{3600+625}=\sqrt{4225}=65\ \mathrm{MPa}$, so

$$ \sigma_1=20+65=85\ \mathrm{MPa},\qquad \sigma_2=20-65=-45\ \mathrm{MPa} $$

The maximum in-plane shear stress, being half the difference of the principal stresses (or the value of the square root itself), is $\tau_{\max}=65\ \mathrm{MPa}$ (since $\sigma_1,\sigma_2$ have opposite signs, this is also the three-dimensional absolute maximum shear stress $(\sigma_1-\sigma_2)/2$).

(2) The center of Mohr's circle is $\left(\dfrac{\sigma_x+\sigma_y}2,0\right)=(20,0)\ \mathrm{MPa}$, and the radius is $R=65\ \mathrm{MPa}$, the same as the value of the square root found above. The points where the circle meets the $\sigma$ axis are $20\pm65$, that is, $85\ \mathrm{MPa}$ on the right and $-45\ \mathrm{MPa}$ on the left, which agree exactly with the $\sigma_1,\sigma_2$ found in (1). The $\tau$ coordinates of the top and bottom of the circle are also $\pm65\ \mathrm{MPa}$, which agree with $\tau_{\max}$. We have thus confirmed that the graphical method (Mohr's circle) and the algebraic method (the eigenvalue computation) arrive at the same answer (confirmation by another method: solving the characteristic equation $\det\begin{pmatrix}-40-\lambda&-25\\-25&80-\lambda\end{pmatrix}=\lambda^2-40\lambda-3825=0$ directly gives $\lambda=(40\pm\sqrt{1600+15300})/2=(40\pm130)/2=85,\,-45$, which agrees.)

Solution

Substituting $E=70\ \mathrm{GPa}$, $\alpha=2.3\times10^{-5}\ \mathrm{K^{-1}}$, and $\Delta T=-30\ \mathrm K$ into the formula $\sigma=-E\alpha\Delta T$ for the thermal stress in a bar whose two ends are fixed to rigid walls (Formula 20.2 of Chapter 20), we obtain

$$ \sigma=-E\alpha\Delta T=-(70\times10^9\ \mathrm{Pa})\times(2.3\times10^{-5}\ \mathrm{K^{-1}})\times(-30\ \mathrm K) $$

Paying attention to the signs in the computation, we have $(-1)\times(-30)=+30$, so

$$ \sigma=70\times10^9\times2.3\times10^{-5}\times30\ \mathrm{Pa}=70\times2.3\times30\times10^4\ \mathrm{Pa}=4830\times10^4\ \mathrm{Pa}=4.83\times10^7\ \mathrm{Pa}\approx48.3\ \mathrm{MPa} $$

Since the sign is positive, this is a tensile stress. This agrees with the physical explanation: when the temperature falls ($\Delta T\lt0$), the bar would naturally contract, but the walls at both ends do not allow it, so a stress in the direction that stretches the bar (tension) arises.

Solution

From the formula $I=\pi d^4/64$ for a circular cross section in Chapter 20, the second moment of area of a circular cross section of diameter $d=60\ \mathrm{mm}=0.060\ \mathrm m$ is

$$ I=\frac{\pi\times(0.060\ \mathrm m)^4}{64}=\frac{\pi\times1.296\times10^{-5}}{64}\ \mathrm{m^4}\approx6.362\times10^{-7}\ \mathrm{m^4} $$

The distance from the neutral axis to the edge is $y_{\max}=d/2=0.030\ \mathrm m$, so from the bending stress formula $\sigma=My/I$,

$$ \sigma_{\max}=\frac{M\,y_{\max}}I=\frac{3.0\times10^3\ \mathrm{N\cdot m}\times0.030\ \mathrm m}{6.362\times10^{-7}\ \mathrm{m^4}}=\frac{90}{6.362\times10^{-7}}\ \mathrm{Pa}\approx1.415\times10^8\ \mathrm{Pa}\approx141.5\ \mathrm{MPa} $$

(unit check: $[\mathrm{N\cdot m\cdot m/m^4}]=[\mathrm{N/m^2}]=[\mathrm{Pa}]$ is correct. Using the section modulus $Z=\pi d^3/32=\pi\times(0.060)^3/32\approx2.121\times10^{-5}\ \mathrm{m^3}$, the same value follows immediately: $\sigma_{\max}=M/Z=3.0\times10^3/2.121\times10^{-5}\approx1.415\times10^8\ \mathrm{Pa}$.) The procedure is the same as in the example with a rectangular cross section (Example 20.8 of Chapter 20); note that only the formula for the second moment of area differs.

Solution

In the interval $L/2\le x\le L$, by Section 20.6.2 of Chapter 20, $M(x)=\dfrac{P(L-x)}2$. We integrate both sides of the differential equation of deflection $EIv''=M(x)$ with respect to $x$. Noting that $\dd(L-x)/\dd x=-1$, and that $\dfrac{\dd}{\dd x}\left[-\dfrac P4(L-x)^2\right]=-\dfrac P4\times2(L-x)\times(-1)=\dfrac P2(L-x)=M(x)$, we see that the result of the first integration is

$$ EIv'=-\frac P4(L-x)^2+C_3 $$

(this can be checked by differentiating the right-hand side with respect to $x$, which indeed returns $M(x)$). Integrating once more in the same way, and using $\dfrac{\dd}{\dd x}\left[\dfrac P{12}(L-x)^3\right]=\dfrac P{12}\times3(L-x)^2\times(-1)=-\dfrac P4(L-x)^2$, we obtain

$$ EIv=\frac P{12}(L-x)^3+C_3x+C_4 $$

We determine the constants of integration $C_3,C_4$ from the following two conditions.

(a) Slope condition: At the interval boundary $x=L/2$, the slope in the right interval must agree with the value at $x=L/2$ of the slope in the left interval, $EIv'=\dfrac P4x^2-\dfrac{PL^2}{16}$, namely $EIv'(L/2)=\dfrac P4\left(\dfrac L2\right)^2-\dfrac{PL^2}{16}=\dfrac{PL^2}{16}-\dfrac{PL^2}{16}=0$ (by symmetry, the bottom of the deflection curve):

$$ -\frac P4\left(\frac L2\right)^2+C_3=0 \quad\Longrightarrow\quad -\frac{PL^2}{16}+C_3=0 \quad\Longrightarrow\quad C_3=\frac{PL^2}{16} $$

(b) Boundary condition at the support: At the right end $x=L$, the deflection at the support must be 0:

$$ EIv(L)=\frac P{12}(L-L)^3+C_3L+C_4=0+\frac{PL^2}{16}\cdot L+C_4=\frac{PL^3}{16}+C_4=0 \quad\Longrightarrow\quad C_4=-\frac{PL^3}{16} $$

Therefore, the deflection curve in the interval $L/2\le x\le L$ is

$$ EIv(x)=\frac P{12}(L-x)^3+\frac{PL^2}{16}x-\frac{PL^3}{16}\qquad\left(\frac L2\le x\le L\right) $$

(expanding with $(L-x)^3=L^3-3L^2x+3Lx^2-x^3$, it can also be put in the form $EIv(x)=\dfrac{PL^3}{12}-\dfrac{PL^2}{4}x+\dfrac{PL}4x^2-\dfrac P{12}x^3+\dfrac{PL^2}{16}x-\dfrac{PL^3}{16}=\dfrac{PL^3}{48}-\dfrac{3PL^2}{16}x+\dfrac{PL}4x^2-\dfrac P{12}x^3$).

(c) Let us use the constants of integration just determined to check that the deflection at $x=L/2$ agrees with the result for the left interval. Substituting $x=L/2$, we obtain

$$ EIv\left(\frac L2\right)=\frac P{12}\left(\frac L2\right)^3+\frac{PL^2}{16}\cdot\frac L2-\frac{PL^3}{16}=\frac{PL^3}{96}+\frac{PL^3}{32}-\frac{PL^3}{16} $$

Putting the terms over a common denominator (96), we obtain

$$ \frac{PL^3}{96}+\frac{3PL^3}{96}-\frac{6PL^3}{96}=\frac{(1+3-6)PL^3}{96}=-\frac{2PL^3}{96}=-\frac{PL^3}{48} $$

That is, $v(L/2)=-\dfrac{PL^3}{48EI}$, which indeed agrees exactly with the value of the maximum deflection $\delta_{\max}=PL^3/(48EI)$ obtained from the left interval in Section 20.6.4, apart from the sign ($v$ is positive upward, so a downward deflection is negative). Although the constants of integration were determined from the two conditions (a) and (b) alone, the continuity of the deflection itself was also satisfied automatically; this confirms that, starting from either the left or the right interval, we arrive without contradiction at the same single deflection curve and the same maximum deflection.

Solution

(1) From the rotational speed $N=1200\ \mathrm{rpm}$, the angular velocity is

$$ \omega=\frac{2\pi N}{60}=\frac{2\pi\times1200}{60}\ \mathrm{rad/s}=2\pi\times20\ \mathrm{rad/s}\approx125.66\ \mathrm{rad/s} $$

From the power formula $P=T\omega$, the required torque is

$$ T=\frac P\omega=\frac{45\times10^3\ \mathrm W}{125.66\ \mathrm{rad/s}}\approx358.1\ \mathrm{N\cdot m} $$

As in Example 20.15 of Chapter 20, we substitute $J=\pi d^4/32$ for a circular cross section and $r=d/2$ into the torsional stress formula $\tau_{\max}=Tr/J$ and simplify, which gives the relation $\tau_{\max}=16T/(\pi d^3)$; substituting $\tau_{\max}=\tau_{\mathrm a}=35\ \mathrm{MPa}$ into this relation and solving for $d$, we obtain

$$ d=\left(\frac{16T}{\pi\tau_{\mathrm a}}\right)^{1/3}=\left(\frac{16\times358.1}{\pi\times35\times10^6}\right)^{1/3}\ \mathrm m=\left(\frac{5730}{1.0996\times10^8}\right)^{1/3}\ \mathrm m\approx\left(5.212\times10^{-5}\right)^{1/3}\ \mathrm m\approx0.03735\ \mathrm m $$

That is, $d\approx37.4\ \mathrm{mm}$. As in Example 20.15, in practice it is customary to choose a standard diameter at least this large (for example $d=40\ \mathrm{mm}$).

(2) The polar second moment of area of a shaft with $d=40\ \mathrm{mm}=0.040\ \mathrm m$ is $J=\dfrac{\pi d^4}{32}=\dfrac{\pi\times(0.040)^4}{32}\ \mathrm{m^4}\approx2.513\times10^{-7}\ \mathrm{m^4}$. The maximum shear stress at the surface is

$$ \tau_{\max}=\frac{16T}{\pi d^3}=\frac{16\times358.1}{\pi\times(0.040)^3}\ \mathrm{Pa}=\frac{5730}{2.011\times10^{-4}}\ \mathrm{Pa}\approx2.85\times10^7\ \mathrm{Pa}\approx28.5\ \mathrm{MPa} $$

which is smaller than the allowable stress of $35\ \mathrm{MPa}$, so there is a margin. For the angle of twist, with $G=79\ \mathrm{GPa}$ and $L=1.5\ \mathrm m$, the formula $\varphi=TL/(GJ)$ gives

$$ \varphi=\frac{358.1\times1.5}{79\times10^9\times2.513\times10^{-7}}\ \mathrm{rad}=\frac{537.2}{1.985\times10^4}\ \mathrm{rad}\approx0.02706\ \mathrm{rad} $$

Converted to degrees, this is $0.02706\times180/\pi\approx1.55°$.

Solution

Substituting the pure shear state $\sigma_x=\sigma_y=0$, $\tau_{xy}=\tau_{\max}$ into the principal stress formula of Chapter 20 (Law 20.2)

$$ \sigma_{1,2}=\frac{\sigma_x+\sigma_y}2\pm\sqrt{\left(\frac{\sigma_x-\sigma_y}2\right)^2+\tau_{xy}^2} $$

we find that, since $\sigma_x=\sigma_y=0$, the mean-value term is 0, and the term $\left(\dfrac{\sigma_x-\sigma_y}2\right)^2$ under the square root is also 0, so

$$ \sigma_{1,2}=0\pm\sqrt{0+\tau_{\max}^2}=\pm\tau_{\max} $$

That is, (1) $\sigma_1=+\tau_{\max}$ (tension) and $\sigma_2=-\tau_{\max}$ (compression).

(2) The orientation of the principal planes is determined by the equation $\tan2\theta_{\mathrm p}=\dfrac{2\tau_{xy}}{\sigma_x-\sigma_y}$ of Chapter 20, but since the denominator $\sigma_x-\sigma_y=0$, this equation refers to the angle for which $\tan2\theta_{\mathrm p}\to\infty$ ($2\theta_{\mathrm p}=90°$). The tangent becomes infinite when the angle is $90°$ (or an angle shifted from it by an integer multiple of that angle), so $2\theta_{\mathrm p}=90°$, that is,

$$ \theta_{\mathrm p}=45° $$

In this case $a=(\sigma_x-\sigma_y)/2=0,\ b=\tau_{\max}$, which corresponds to the sign combination $(\cos2\theta_{\mathrm p},\sin2\theta_{\mathrm p})=(a,b)/r=(0,1)$, so the plane with $\theta_{\mathrm p}=45°$ is the plane of the maximum principal stress $\sigma_1=+\tau_{\max}$. Indeed, substituting $\theta=45°$ ($\cos90°=0,\ \sin90°=1$) into the coordinate transformation formula (Formula 20.1 of Chapter 20) to check, we obtain

$$ \sigma_{x'}=0+0\times0+\tau_{\max}\times1=\tau_{\max},\qquad \tau_{x'y'}=-0\times1+\tau_{\max}\times0=0 $$

so we confirm that on the plane $\theta=45°$ the shear stress is indeed 0 and the normal stress is $\tau_{\max}$ ($=\sigma_1$).

This result is the very basis of the fact stated in Section 20.7, “Application: Principal Stresses in Torsion and the Fracture Surface of a Brittle Material”—that when a shaft with a circular cross section is subjected to torsion, its surface is in a state of pure shear, and a principal stress (the maximum tensile stress) arises in the direction inclined at $45°$ to the axial direction. When a brittle material (such as chalk or cast iron) is twisted, it fractures on the surface perpendicular to this tensile principal stress in the $45°$ direction, that is, on a helical surface inclined at $45°$ to the axis, precisely because of this calculation result.

Solution

We substitute $E=70\,000\ \mathrm{MPa}$, $\nu=0.33$, $\sigma_x=120\ \mathrm{MPa}$, $\sigma_y=-40\ \mathrm{MPa}$, and $\sigma_z=0$ into the generalized Hooke's law of Chapter 20 (Law 20.3).

(1)

$$ \begin{aligned} \varepsilon_x&=\frac{\sigma_x-\nu(\sigma_y+\sigma_z)}E=\frac{120-0.33\times(-40)}{70\,000}=\frac{133.2}{70\,000}\approx1.90\times10^{-3}\\ \varepsilon_y&=\frac{\sigma_y-\nu(\sigma_z+\sigma_x)}E=\frac{-40-0.33\times120}{70\,000}=\frac{-79.6}{70\,000}\approx-1.14\times10^{-3}\\ \varepsilon_z&=\frac{\sigma_z-\nu(\sigma_x+\sigma_y)}E=\frac{0-0.33\times(120-40)}{70\,000}=\frac{-26.4}{70\,000}\approx-3.77\times10^{-4} \end{aligned} $$

Even though $\sigma_z=0$, owing to the Poisson effect of the in-plane stresses a strain $\varepsilon_z=-3.77\times10^{-4}$ (contraction) remains.

(2) The volumetric strain is the sum of the three strains: $\varepsilon_V=1.903\times10^{-3}-1.137\times10^{-3}-0.377\times10^{-3}\approx3.89\times10^{-4}$. Checking with the formula $\varepsilon_V=(1-2\nu)(\sigma_x+\sigma_y+\sigma_z)/E$ of Chapter 20, we get $(1-2\times0.33)\times(120-40)/70\,000=0.34\times80/70\,000\approx3.89\times10^{-4}$, which agrees.

(3) The change in thickness is $\Delta t=\varepsilon_zt=-3.77\times10^{-4}\times5.0\ \mathrm{mm}\approx-1.89\times10^{-3}\ \mathrm{mm}$; that is, the plate becomes thinner by about $1.9\ \mathrm{\mu m}$.

Solution

(1) From the equilibrium of forces, the reaction force at the wall is upward, $R_0=P=2.0\ \mathrm{kN}$. From the equilibrium of moments about the wall, the reaction moment is counterclockwise, $M_0=PL=2.0\times10^3\times1.5\ \mathrm{N\cdot m}=3.0\ \mathrm{kN\cdot m}$.

(2) Taking the fixed end as $x=0$ and counting the forces on the left side of the section, we have $Q(x)=R_0=2.0\ \mathrm{kN}$ (constant over the entire length) and $M(x)=R_0x-M_0=2.0x-3.0\ \mathrm{kN\cdot m}=-2.0\,(1.5-x)\ \mathrm{kN\cdot m}$ ($x$ in $\mathrm m$). We have $M(0)=-3.0\ \mathrm{kN\cdot m}$ and $M(L)=0$, and both satisfy $\dd M/\dd x=2.0=Q$.

(3) The magnitude of the bending moment is largest at the fixed end, $\abs{M}_{\max}=3.0\ \mathrm{kN\cdot m}$. The second moment of area is $I=\dfrac{bh^3}{12}=\dfrac{0.030\times0.060^3}{12}\ \mathrm{m^4}=5.4\times10^{-7}\ \mathrm{m^4}$ and $y_{\max}=h/2=0.030\ \mathrm m$, so

$$ \sigma_{\max}=\frac{\abs{M}_{\max}\,y_{\max}}I=\frac{3.0\times10^3\times0.030}{5.4\times10^{-7}}\ \mathrm{Pa}\approx1.667\times10^8\ \mathrm{Pa}\approx166.7\ \mathrm{MPa} $$

Since $M\lt0$ (the beam bends convex upward), the top surface of the cross section at the fixed end is in tension ($+167\ \mathrm{MPa}$) and the bottom surface of the same cross section is in compression ($-167\ \mathrm{MPa}$). This is smaller than a typical value of the yield stress of steel (about $250\ \mathrm{MPa}$).

(4) From the result $\delta_{\max}=\dfrac{PL^3}{3EI}$ of the derivation in Section 20.6.4 of Chapter 20 (cantilever beam with a concentrated load at the tip), the deflection at the tip is

$$ \delta_{\max}=\frac{2.0\times10^3\times1.5^3}{3\times206\times10^9\times5.4\times10^{-7}}\ \mathrm m=\frac{6750}{3.337\times10^5}\ \mathrm m\approx2.02\times10^{-2}\ \mathrm m\approx20.2\ \mathrm{mm} $$

That is, about $20\ \mathrm{mm}$ downward (about $1/74$ of the length).

Solution

(1) The bending moment between the loading points is $M_{\max}=\dfrac P2\times\dfrac{L-L_{\mathrm i}}2=\dfrac{P(L-L_{\mathrm i})}4$. Substituting $P=960\ \mathrm N$ and $L-L_{\mathrm i}=40-20=20\ \mathrm{mm}=0.020\ \mathrm m$, we obtain

$$ M_{\max}=\frac{960\times0.020}4\ \mathrm{N\cdot m}=4.8\ \mathrm{N\cdot m} $$

The section modulus is $Z=\dfrac{bd^2}6=\dfrac{0.0040\times(0.0030)^2}6\ \mathrm{m^3}=6.0\times10^{-9}\ \mathrm{m^3}$, so the bending strength is

$$ \sigma_{\mathrm f}=\frac{M_{\max}}Z=\frac{4.8}{6.0\times10^{-9}}\ \mathrm{Pa}=8.0\times10^8\ \mathrm{Pa}=800\ \mathrm{MPa} $$

(This also agrees with the formula $\sigma_{\mathrm f}=3P(L-L_{\mathrm i})/(2bd^2)=3\times960\times0.020/(2\times0.0040\times9.0\times10^{-6})=57.6/(7.2\times10^{-8})=8.0\times10^8\ \mathrm{Pa}$ of Chapter 20.)

(2) If the same load $960\ \mathrm N$ is applied in three-point bending with span $L=40\ \mathrm{mm}$, the maximum bending moment is $M_{\max}=PL/4=960\times0.040/4=9.6\ \mathrm{N\cdot m}$, which is twice the $4.8\ \mathrm{N\cdot m}$ of four-point bending (since $L-L_{\mathrm i}=L/2$, $PL/4\div\{P(L/2)/4\}=2$). The maximum bending stress in this case is as large as $9.6/(6.0\times10^{-9})=1.6\times10^9\ \mathrm{Pa}=1600\ \mathrm{MPa}$. In other words, under the same load, in three-point bending a large stress is concentrated at the single point directly beneath the loading point, whereas in four-point bending a stress of half that magnitude acts uniformly over a wide interval between the loading points.

Solution

The polar second moment of area of the hollow shaft is

$$ J=\frac{\pi(D^4-d^4)}{32}=\frac{\pi\times\left\{(0.060)^4-(0.040)^4\right\}}{32}\ \mathrm{m^4}=\frac{\pi\times(1.296\times10^{-5}-2.56\times10^{-6})}{32}\ \mathrm{m^4}=\frac{\pi\times1.04\times10^{-5}}{32}\ \mathrm{m^4}\approx1.021\times10^{-6}\ \mathrm{m^4} $$

The radius of the surface is $D/2=0.030\ \mathrm m$. From $\tau_{\max}=TR/J=\tau_{\mathrm a}$, the allowable torque is

$$ T_{\text{hollow}}=\frac{\tau_{\mathrm a}J}{R}=\frac{40\times10^6\times1.021\times10^{-6}}{0.030}\ \mathrm{N\cdot m}\approx1.361\times10^3\ \mathrm{N\cdot m} $$

Next, consider a solid shaft with the same cross-sectional area. The cross-sectional area of the hollow shaft is $A=\dfrac{\pi(D^2-d^2)}4=\dfrac{\pi(0.060^2-0.040^2)}4\ \mathrm{m^2}=\dfrac{\pi\times2.0\times10^{-3}}4\ \mathrm{m^2}\approx1.571\times10^{-3}\ \mathrm{m^2}$. The diameter $d_{\mathrm s}$ of the solid shaft satisfies $\dfrac{\pi d_{\mathrm s}^2}4=A$, so $d_{\mathrm s}^2=D^2-d^2=2.0\times10^{-3}\ \mathrm{m^2}$, that is, $d_{\mathrm s}=\sqrt{2.0\times10^{-3}}\ \mathrm m\approx0.04472\ \mathrm m=44.7\ \mathrm{mm}$. The polar second moment of area in this case is $J_{\mathrm s}=\dfrac{\pi d_{\mathrm s}^4}{32}=\dfrac{\pi\times(2.0\times10^{-3})^2}{32}\ \mathrm{m^4}\approx3.927\times10^{-7}\ \mathrm{m^4}$ and the radius of the surface is $d_{\mathrm s}/2=0.02236\ \mathrm m$, so the allowable torque is

$$ T_{\text{solid}}=\frac{\tau_{\mathrm a}J_{\mathrm s}}{d_{\mathrm s}/2}=\frac{40\times10^6\times3.927\times10^{-7}}{0.02236}\ \mathrm{N\cdot m}\approx7.03\times10^2\ \mathrm{N\cdot m} $$

Therefore, $T_{\text{hollow}}/T_{\text{solid}}\approx1361/703\approx1.94$, and for the same mass, the hollow shaft can transmit about 1.9 times the torque of the solid shaft. This is because the torsional stress increases with distance from the center, and the material near the center does almost no work, so it is more efficient to move that material outward.

Solution

In Chapter 21 we showed that the radius of curvature at the end of the major axis of an elliptical hole (major axis $2a$, minor axis $2b$) is

$$ \rho=\frac{b^2}{a} $$

and that, using this, Inglis's formula for the stress concentration can be written as

$$ \sigma_{\max}=\sigma\left(1+2\sqrt{\frac a\rho}\right) $$

A circular hole is the special case of an ellipse in which the major and minor axes are equal, that is, $b=a$. In this case the radius of curvature becomes

$$ \rho=\frac{b^2}{a}=\frac{a^2}{a}=a $$

Thus, for a circular hole the radius of curvature $\rho$ is simply equal to the radius $a$ (the radius of curvature of a circle is the radius of that circle itself, so this is also a natural result geometrically). Substituting this into $\sigma_{\max}=\sigma(1+2\sqrt{a/\rho})$ gives

$$ \sigma_{\max}=\sigma\left(1+2\sqrt{\frac a a}\right)=\sigma(1+2\sqrt1)=\sigma(1+2\times1)=3\sigma $$

and we indeed obtain the stress concentration factor of 3. This agrees exactly with the result obtained by substituting $a=b$ directly into Inglis's general formula $\sigma_{\max}=\sigma(1+2a/b)$ (Chapter 21, Example 21.1)—we have confirmed that the route through the radius of curvature and the route that uses the major and minor axes directly arrive at the same conclusion.

Solution

Griffith's criterion (plane stress) derived in Chapter 21 is

$$ \sigma_{\mathrm f}=\sqrt{\frac{2E\gamma_{\mathrm s}}{\pi a}} $$

We solve this for the crack half-length $a$. Squaring both sides gives

$$ \sigma_{\mathrm f}^2=\frac{2E\gamma_{\mathrm s}}{\pi a} $$

Multiplying both sides by $\pi a$ and dividing by $\sigma_{\mathrm f}^2$, we obtain

$$ a=\frac{2E\gamma_{\mathrm s}}{\pi\sigma_{\mathrm f}^2} $$

Here we substitute the values for soda-lime glass, $E=70\ \mathrm{GPa}=70\times10^9\ \mathrm{Pa}$ and $\gamma_{\mathrm s}=0.6\ \mathrm{J/m^2}$, and the measured fracture stress $\sigma_{\mathrm f}=60\ \mathrm{MPa}=60\times10^6\ \mathrm{Pa}$:

$$ a=\frac{2\times70\times10^9\times0.6}{\pi\times(60\times10^6)^2}\ \mathrm m =\frac{8.4\times10^{10}}{\pi\times3.6\times10^{15}}\ \mathrm m =\frac{8.4\times10^{10}}{1.131\times10^{16}}\ \mathrm m \approx7.4\times10^{-6}\ \mathrm m $$

Unit check: the numerator is $[\mathrm{Pa\cdot J/m^2}]=[\mathrm{Pa\cdot Pa\cdot m}]=[\mathrm{Pa^2\cdot m}]$ and the denominator is $[\mathrm{Pa^2}]$, so the quotient is in $[\mathrm m]$, which is correct. Therefore, this specimen is estimated to have contained a crack of half-length $a\approx7.4\ \mu\mathrm m$ (a total length of about $15\ \mu\mathrm m$). This is a typical numerical example showing that even a tiny flaw that is almost invisible to the naked eye can greatly affect the strength of glass.

Solution

The defining equation of the stress intensity factor is

$$ K_{\mathrm I}=Y\sigma\sqrt{\pi a} $$

The condition for a crack to begin to extend unstably is $K_{\mathrm I}=K_{\mathrm{IC}}$, so we take this as the allowable limit and solve for $\sigma$:

$$ \sigma_{\text{allow}}=\frac{K_{\mathrm{IC}}}{Y\sqrt{\pi a}} $$

We substitute $K_{\mathrm{IC}}=30\ \mathrm{MPa\,m^{1/2}}=30\times10^6\ \mathrm{Pa\,m^{1/2}}$ for the aluminum alloy, the flaw depth $a=2\ \mathrm{mm}=2\times10^{-3}\ \mathrm m$, and the geometry factor $Y=1.12$ (surface crack):

$$ \sigma_{\text{allow}}=\frac{30\times10^6}{1.12\times\sqrt{\pi\times2\times10^{-3}}}\ \mathrm{Pa} =\frac{30\times10^6}{1.12\times0.0793}\ \mathrm{Pa} =\frac{30\times10^6}{0.0888}\ \mathrm{Pa} \approx3.38\times10^{8}\ \mathrm{Pa} $$

Therefore, the allowable tensile stress is approximately $338\ \mathrm{MPa}$. Unit check: dividing $K_{\mathrm{IC}}$ ($\mathrm{Pa\cdot m^{1/2}}$) by $\sqrt{\pi a}$ ($\mathrm{m^{1/2}}$) gives $\mathrm{Pa}$, which is correct. Note also that, compared with the representative yield stress of common aluminum alloys (it depends on the type of alloy and the heat treatment, but is roughly 200–300 MPa), this allowable stress of 338 MPa exceeds it, so in practice the member may yield before the crack extends. In design, one must calculate both the allowable stress from fracture mechanics and the allowable stress based on the yield strength, and adopt the smaller one.

Solution

The plane-strain condition of ASTM E399, as we saw in Chapter 21, is

$$ B,\ a,\ (W-a)\ \ge\ 2.5\left(\frac{K_{\mathrm{IC}}}{\sigma_{\mathrm y}}\right)^2 $$

For the thickness $B$, we calculate the minimum thickness required by the right-hand side. Substituting $K_{\mathrm{IC}}=50\ \mathrm{MPa\,m^{1/2}}=50\times10^6\ \mathrm{Pa\,m^{1/2}}$ and $\sigma_{\mathrm y}=900\ \mathrm{MPa}=900\times10^6\ \mathrm{Pa}$, we obtain

$$ B_{\min}=2.5\left(\frac{50\times10^6}{900\times10^6}\right)^2\mathrm m =2.5\times(0.0556)^2\ \mathrm m =2.5\times3.086\times10^{-3}\ \mathrm m \approx7.7\times10^{-3}\ \mathrm m=7.7\ \mathrm{mm} $$

The minimum required thickness is approximately $7.7\ \mathrm{mm}$. By contrast, the thickness of the specimen actually used is $B=6\ \mathrm{mm}$, and since $6\ \mathrm{mm}\lt7.7\ \mathrm{mm}$, this condition is not satisfied. In other words, this specimen is too thin to satisfy the plane-strain condition.

If this condition is not satisfied, the plastic zone at the crack tip (Section 21.6) becomes too large to be neglected relative to the specimen thickness, and the stress state at the tip deviates from the ideal plane strain (triaxial constraint) and approaches plane stress. In a state close to plane stress, plastic deformation occurs more easily, so the apparent fracture toughness is measured as a larger value than the true (plane-strain) $K_{\mathrm{IC}}$. Therefore, the value obtained in this test cannot be used in design as the $K_{\mathrm{IC}}$ intrinsic to the material, and it must be measured again with a thicker specimen.

Solution

Under a constant stress $\sigma$, the energy release rate $\mathcal G(a)=\pi\sigma^2a/E'$ is a straight line through the origin, proportional to the crack half-length $a$ (or to the crack extension $\Delta a$). The theorem of Chapter 21 (the condition for stable and unstable crack growth) states that $\mathcal G=R$ is necessary for the crack to extend even slightly, and that, given this, the relative magnitudes of $\dd\mathcal G/\dd a$ and $\dd R/\dd a$ determine stability or instability.

For a material with a flat R-curve (constant, $\dd R/\dd a=0$) (an ideal brittle material such as glass), as the stress is raised little by little, the straight line of $\mathcal G$ reaches the constant value of $R$ at some stress. At that moment, $\dd\mathcal G/\dd a>0=\dd R/\dd a$ holds, and the instability condition is satisfied immediately. Therefore, the crack shows exactly the behavior of the Griffith theory that we saw in Section 21.2: “nothing happens until then, and the instant a certain stress is reached, it accelerates without warning and fractures.”

For a material with a rising R-curve, at the point where the line of $\mathcal G$ first reaches the curve (the intersection), the slope $\dd R/\dd a$ of the curve is still larger than the slope $\dd\mathcal G/\dd a$ of the line, so the stability condition ($\dd\mathcal G/\dd a\lt\dd R/\dd a$) is satisfied, and the crack stops there for the time being. When the stress is raised further, the slope of the line of $\mathcal G$ also increases (the slope is proportional to $\pi\sigma^2/E'$, so the larger $\sigma$ is, the steeper it becomes), and when the line finally comes to touch the R-curve, the growth becomes unstable beyond that point. In other words, in this material the crack fractures in stages: “each time the stress is raised, it extends stably by a little, and finally, at a certain stress, it goes all at once into unstable fracture.”

Practical advantage: In a material with a rising R-curve, the strength becomes less sensitive to how large the initial crack (flaw) was. Since even a small crack becomes unstable only after it has extended to some extent, the final fracture stress tends to be insensitive to the scatter in the initial crack size. This is an important property for suppressing the scatter in the strength of ceramics (related to the Weibull statistics that we will learn in Chapter 22), and it is one of the reasons why transformation toughening of zirconia and bridging toughening of silicon nitride (Section 21.7) are valued in practical materials.

Solution

As we saw in Chapter 21, the size of the plastic zone at the crack tip is $r_{\mathrm p}=\dfrac{1}{2\pi}(K_{\mathrm I}/\sigma_{\mathrm y})^2$ in Irwin's estimate and $r_{\mathrm D}=\dfrac\pi8(K_{\mathrm I}/\sigma_{\mathrm y})^2$ in the Dugdale model, and both are proportional to $(K_{\mathrm I}/\sigma_{\mathrm y})^2$. Therefore, the plastic zone becomes large (i) when $K_{\mathrm I}$ (that is, the stress or the crack size) is large, or (ii) when the yield stress $\sigma_{\mathrm y}$ is small (a soft material).

Specific example: Consider a member that has a relatively large crack (or notch) and is made of a material such as mild steel (low-carbon steel), whose yield stress is relatively low ($\sigma_{\mathrm y}\sim200$–$300\ \mathrm{MPa}$ or so) and which is highly ductile and can withstand a fairly large $K_{\mathrm I}$ without fracturing. In this case, the size of the plastic zone at the tip can reach the order of a few millimeters to, at times, several tens of millimeters, as calculated for structural steel in Example 21.6 of this chapter. If the plate thickness of the member, or the remaining dimension from the crack to the edge of the member (the ligament), is of the same order as the size of this plastic zone or smaller, the premise of small-scale yielding (that the plastic zone can be regarded as a “small point” relative to the whole member) breaks down. Specifically, this situation applies, for example, to a thin plate-like member (a plate thickness of a few millimeters) made of soft steel and having a relatively large notch or weld defect.

What happens if $K_{\mathrm{IC}}$ is used as it is: In a situation where small-scale yielding does not hold, the stress field around the crack tip no longer follows the elasticity solution given by the equation for the crack-tip K field. The material yields over a wide region around the tip and large-scale plastic deformation occurs (when the whole structure yields, it is called general yielding), so the state at the tip can no longer be represented by the single number $K_{\mathrm I}$. In such a case, if $K_{\mathrm{IC}}$, which is based only on elasticity theory, is used as the basis of the safety factor, the actual fracture behavior (behavior close to ductile fracture, in which the member ruptures after large plastic deformation) cannot be captured correctly, and there is a risk of an excessively conservative evaluation (or, conversely, a non-conservative evaluation that overlooks plastic collapse, which is a different failure mode). In such cases, it is necessary to use methods of elastic-plastic fracture mechanics, such as the J-integral introduced in Section 21.6.4 of this chapter.

Solution

Based on the definitions in Section 21.3.2, we identify the mode by focusing on the direction in which, and the manner in which, the crack surfaces are displaced.

(1) Under a tensile stress normal to the crack plane, the upper and lower faces of the crack are displaced in the direction in which they move apart from each other (the direction that opens the faces). This is a deformation in which the crack opens under tension normal to the crack plane, and it is Mode I (opening mode). The stress intensity factor $K_{\mathrm I}$ and the stress field that we have treated so far in this chapter all represent this situation.

(2) The upper and lower faces on either side of the crack are displaced in opposite directions, parallel to the direction of crack propagation—this is a shear deformation within the crack plane, and it is Mode II (in-plane shear mode). It differs from Mode I in that the crack faces hardly open at all and instead slide within the plane.

(3) In the torsion of a round bar, the faces are displaced out of plane in the direction parallel to the front of an axial crack (the direction of rotation about the crack front as the axis). This is an out-of-plane shear deformation in which the faces slide parallel to the crack front, and it is Mode III (out-of-plane shear mode). As stated in Section 21.3.2, torsional fracture of a round bar is a typical example in which Mode III appears.

In this way, (1)–(3) each involve a different direction of displacement, and Modes I, II, and III can be identified by distinguishing whether the crack faces “open perpendicularly,” “slide within the plane,” or “slide out of the plane.” Note also that in actual members multiple modes often coexist, but as stated in Section 21.3.2, the crack in a brittle material tends to advance in the direction perpendicular to the maximum principal stress, so even under oblique loading it tends to curve naturally toward the direction in which Mode I dominates.

Solution

Consider a material whose S–N curve clearly shows an upper limit of stress amplitude (the fatigue limit $\sigma_{\mathrm w}$) below which failure never occurs however many times the loading is repeated (this is common in body-centered cubic metals such as steel). If the stress amplitude in service is always kept below $\sigma_{\mathrm w}$, then in theory fatigue failure will not occur no matter how many times the loading is repeated. This design policy is called infinite-life design. It is often adopted for components that continue to receive repeated loads of the same amplitude over long periods, such as automobile coil springs and bearings of stationary machines.

In contrast, for a material that shows no clear fatigue limit, such as an aluminum alloy (this is common in face-centered cubic metals), no matter how far the stress amplitude is lowered, failure will in theory eventually occur if the number of cycles is made large enough. For this reason, one first fixes a design life, that is, “how many cycles (or how many years, or tens of thousands of flight hours) the part is to be used safely,” then uses Basquin's law and the Paris law to calculate the allowable stress amplitude corresponding to that life, and keeps the actual working stress at or below it. This design policy is called safe-life design. Typical examples include the airframe of an aircraft (aluminum alloys are widely used there to save weight, and in addition the fatigue limit cannot be relied on); this approach is often combined with an operating practice (damage tolerance design) that assumes cracks are present, checks for their presence and size by periodic inspection, and replaces parts if necessary.

To sum up, infinite-life design is the philosophy of “keep the stress below the fatigue limit and do not worry about the life itself,” whereas safe-life design is the philosophy of “the fatigue limit cannot be used, so calculate the life and manage it.” We see that the design policy to be adopted changes fundamentally depending on whether the material has a fatigue limit.

Solution

The expression for the fatigue life obtained in Chapter 22 by integrating the Paris law (for $n\neq2$) has the form

$$ N_{\mathrm f}=\frac{a_{\mathrm c}^{\,p}-a_0^{\,p}}{p\,C(Y\Delta\sigma)^{n}\pi^{n/2}},\qquad p=1-\frac n2 $$

Suppose $a_0\ll a_{\mathrm c}$ and $n\gt2$ ($p\lt0$). Then $a_{\mathrm c}^{\,p}$ is small compared with $a_0^{\,p}$ and can be neglected, so $N_{\mathrm f}\approx a_0^{\,p}\big/\{|p|\,C(Y\Delta\sigma)^n\pi^{n/2}\}$; that is, approximately $N_{\mathrm f}\propto(\Delta\sigma)^{-n}$ (because $(\Delta\sigma)^n$ appears in the denominator). (Strictly speaking $a_{\mathrm c}$ also changes with the stress, but the effect is small.) Therefore, when the stress range $\Delta\sigma$ is multiplied by $k$, the life becomes about $k^{-n}$ times its original value. Moreover, since the crack growth rate is $\dd a/\dd N=C(\Delta K)^n\propto(\Delta\sigma)^n$, for a crack of the same length the growth rate becomes $k^{n}$ times as large.

For a material with a large $n$ (for example, $n=10$ or so), raising the stress range by only 10% ($k=1.1$) sharply reduces the life to about $1.1^{-10}\approx0.386$ times its original value—less than half—and makes the growth rate about $1.1^{10}\approx2.6$ times as large. Conversely, lowering the stress only slightly extends the life greatly. In other words, a material with a large $n$ has a crack growth rate that is extremely sensitive to stress, so that a small error in estimating the design stress has a large effect on the life prediction; such a material is hard to handle (but, conversely, a long life is easy to obtain as long as the stress is kept sufficiently low).

On the other hand, for a material with a small $n$ ($n=2$–$3$ or so), the change in life for the same 10% change in stress is much more gradual (for $n=3$ the life is $1.1^{-3}\approx0.75$ times its original value and the growth rate is $1.1^3\approx1.33$ times). With such a material, even a modest error in the stress estimate does not throw the life prediction far off, which makes it easy for the designer to handle—but note that whether the absolute life at a given stress is long or short depends not only on $n$ but also on the coefficient $C$, so one cannot simply label a material “good” or “bad” by the size of $n$ alone.

Solution

Miner's rule $D=\sum_i n_i/N_i$ assumes that, whatever the order in which the stress amplitudes are applied, the same combination of $(n_i,N_i)$ gives the same damage $D$. In real materials, however, when a large stress is applied first, local plastic deformation occurs around the crack tip, and after unloading a compressive residual stress may remain in that region. For the small stress amplitudes that follow, this residual stress acts in the direction that hinders the opening and closing of the crack and thus slows crack growth (the crack-growth retardation effect due to overloads). Conversely, if a compressive overload is applied first, the subsequent fatigue life may actually become shorter. Such an effect, which depends on “the order in which the loads were applied,” is called the load sequence effect.

Miner's rule is an approximation that takes no account whatsoever of such load sequence effects and simply adds the contributions of the individual amplitudes linearly. For this reason, for actual load patterns (in particular random loading, in which large and small loads are mixed irregularly), the prediction of Miner's rule can be either an overestimate or an underestimate of the actual life. In the design of important structures (the main structure of an aircraft, piping in a nuclear power plant, and so on), it is essential to use Miner's rule for an initial simple assessment while verifying the design by full-scale fatigue tests that reproduce load patterns close to the real ones, and by more detailed analyses that take load sequence effects into account (for example, a cumulative damage assessment combined with the rainflow counting method, which counts pairs of stress amplitude and number of cycles from an irregular time history of load).

Solution

The equation of the straight line in the Weibull plot is

$$ \ln\ln\frac1{1-P_{\mathrm f}}=m\ln\sigma-m\ln\sigma_0 $$

Its slope is exactly the Weibull modulus $m$. For the line with $m=20$ (steep slope), a small change in the horizontal-axis variable $\ln\sigma$ produces a large change in the vertical axis (and hence in the failure probability $P_{\mathrm f}$)—in other words, the range of stress over which the failure probability goes from nearly 0% to nearly 100% is narrow. This means that the strength values are concentrated very close to the characteristic strength $\sigma_0$ (the scatter is small).

On the other hand, for the line with $m=5$ (gentle slope), the same change in the horizontal axis (stress) produces only a small change in the vertical axis (failure probability). This means that a wider range of stress is needed for the failure probability to go from nearly 0% to nearly 100%, which corresponds to the strength values being widely scattered around $\sigma_0$ (the scatter is large). In short, the steeper the slope of the Weibull plot, the smaller the scatter, and the gentler the slope, the larger the scatter.

A designer who uses a material with a small $m$ (large scatter) in an actual structure must take measures such as the following: (i) carry out strength tests on a statistically sufficient number of specimens to determine $\sigma_0,m$ accurately; (ii) use as the design basis not the average strength but a considerably lower stress corresponding to an acceptable failure probability (for example, 0.1% or 0.01%); and (iii) take into account the size effect and the effective volume seen in Section 22.7, so as to estimate correctly the strength at the actual part size. Moreover, since a small $m$ itself reflects “large scatter in the flaws,” improving the manufacturing process (homogenizing the raw powder, controlling the sintering conditions, and so on) to reduce the scatter of the flaws and thereby increase $m$ is also a fundamental countermeasure.

Solution

In Chapter 22 we showed by integration that the effective volume in the three-point bending test is

$$ V_{\mathrm{eff}}=\frac{V}{2(m+1)^2} $$

(here $V$ is the total volume between the supports). For example, when $m=10$, $V_{\mathrm{eff}}/V=1/\{2\times11^2\}=1/242\approx0.4\%$, so the volume that is “effectively under high stress” in the flexural test is only a tiny fraction of the whole specimen.

However, the size-effect formula of Weibull statistics, $\sigma_1/\sigma_2=(V_2/V_1)^{1/m}$, expressed the fact that “the larger the volume, the higher the probability of containing a flaw of dangerous size, and the lower the strength.” The effective volume in the flexural test is much smaller than the volume when the same specimen is tested in uniform tension (the whole specimen), so the flexural test is, so to speak, the same as measuring the strength with an “effectively very small specimen.” Since the smaller the volume, the higher the measured strength, as the relation

$$ \frac{\sigma_{\text{three-point bending}}}{\sigma_{\text{tensile}}}=\left(\frac{V}{V_{\mathrm{eff}}}\right)^{1/m}=\bigl[2(m+1)^2\bigr]^{1/m} $$

shows, the strength measured by three-point bending is systematically higher than the uniform tensile strength of the same material (about 1.73 times for $m=10$). This is why the “flexural strength” listed in ceramics catalogs is higher than the strength obtained when a uniform tensile stress is actually applied to a part, and using the catalog value as it is for the design of a large part risks overestimating the strength—in design, the strength must be converted by taking into account the difference between the effective volumes of the specimen and of the part.

Solution

Static fatigue of glass is triggered by moisture (humidity) in the atmosphere. The network structure of Si, the main component of glass ($\mathrm{Si\!-\!O\!-\!Si}$ bonds), is hydrolyzed by reaction with water molecules at the stressed crack tip ($\mathrm{Si\!-\!O\!-\!Si}+\mathrm{H_2O}\to2(\mathrm{Si\!-\!OH})$), and the crack grows slowly as the bonds are broken. Therefore, the higher the humidity of the environment, the more readily static fatigue proceeds, and conversely it hardly proceeds at all in a dry environment (in vacuum or in a dry inert gas).

In contrast, delayed fracture of high-strength steel due to hydrogen embrittlement is triggered by hydrogen atoms penetrating into the steel. Hydrogen atoms generated by manufacturing processes such as pickling and electroplating, or by the corrosion reaction itself, penetrate and diffuse into the interstitial sites of the steel lattice, and gather and accumulate especially at places where stress is concentrated and the lattice is distorted, such as dislocations and crack tips. This locally lowers the toughness of the steel and causes fracture after a time delay under constant stress. Therefore, the more an environment exposes the steel to routes by which hydrogen can enter (acidic cleaning solutions, corrosive aqueous solutions, hydrogen gas atmospheres, and so on), the more readily delayed fracture occurs.

Although the environmental factors differ (moisture and humidity in the former case, hydrogen in the latter), both cases have this in common: they can be understood through the mechanism of subcritical crack growth, in which, under a constant sustained stress, chemical species in the environment reach the crack tip and make a crack that would otherwise not advance until the fracture toughness $K_{\mathrm{IC}}$ is reached grow slowly at a stress intensity factor much smaller than $K_{\mathrm{IC}}$. In both cases the growth rate can be approximated by an equation of the same form, $v=AK_{\mathrm I}^n$, and the mathematical treatment is also common to both: the coefficient $A$ depends strongly on the ambient conditions (humidity, hydrogen concentration, and so on).

Solution

(1) In the formula $P_{\mathrm f}=1-\exp\left[-\dfrac V{V_0}\left(\dfrac\sigma{\sigma_0}\right)^m\right]$ of Chapter 22, substitute $V=V_0$, $\sigma=200\ \mathrm{MPa}$, $\sigma_0=300\ \mathrm{MPa}$, and $m=10$. Since $(200/300)^{10}=(2/3)^{10}\approx0.01734$,

$$ P_{\mathrm f}=1-e^{-0.01734}\approx0.0172\quad(\text{about}\ 1.7\%) $$

(2) Setting $P_{\mathrm f}=0.01$ gives $\exp[-(\sigma/\sigma_0)^m]=0.99$. Taking the natural logarithm of both sides, $(\sigma/\sigma_0)^m=-\ln0.99=0.010050$. Therefore,

$$ \sigma=\sigma_0\bigl[-\ln(1-P_{\mathrm f})\bigr]^{1/m}=300\times(0.010050)^{1/10}\ \mathrm{MPa}\approx300\times0.6313\ \mathrm{MPa}\approx189\ \mathrm{MPa} $$

Even for a material with a characteristic strength of $300\ \mathrm{MPa}$, the stress must be held to about 63% of that value to make the failure probability 1%.

(3) To obtain the same failure probability for a part of volume $8V_0$, the exponent $\dfrac V{V_0}\left(\dfrac\sigma{\sigma_0}\right)^m$ must be kept at the same value, so the allowable stress is multiplied by $(V_0/V)^{1/m}=8^{-1/10}\approx0.8123$, giving

$$ \sigma\approx189.4\times0.8123\ \mathrm{MPa}\approx154\ \mathrm{MPa} $$

The larger the volume of a part, the lower the allowable stress (the size effect).

Solution

The surface thermal stress is given by the formula $\sigma=-E\alpha\Delta T/(1-\nu)$ of Chapter 22. For $\Delta T=-80\ \mathrm K$:

Soda-lime glass ($E=70\ \mathrm{GPa}$, $\nu=0.22$, $\alpha=9.0\times10^{-6}\ \mathrm{K^{-1}}$, $\sigma_{\mathrm f}=50\ \mathrm{MPa}$):

$$ \sigma=\frac{70\times10^9\times9.0\times10^{-6}\times80}{1-0.22}\ \mathrm{Pa}=\frac{5.04\times10^7}{0.78}\ \mathrm{Pa}\approx64.6\ \mathrm{MPa} $$

This exceeds the strength of $50\ \mathrm{MPa}$, so the glass breaks. The maximum temperature difference that avoids breakage is $\Delta T_{\mathrm c}=\sigma_{\mathrm f}(1-\nu)/(E\alpha)=3.9\times10^7/6.3\times10^5\ \mathrm K\approx61.9\ \mathrm K$, which is smaller than $80\ \mathrm K$.

Fused quartz ($E=73\ \mathrm{GPa}$, $\nu=0.17$, $\alpha=0.55\times10^{-6}\ \mathrm{K^{-1}}$, $\sigma_{\mathrm f}=70\ \mathrm{MPa}$):

$$ \sigma=\frac{73\times10^9\times0.55\times10^{-6}\times80}{1-0.17}\ \mathrm{Pa}=\frac{3.212\times10^6}{0.83}\ \mathrm{Pa}\approx3.87\ \mathrm{MPa} $$

This is far smaller than the strength of $70\ \mathrm{MPa}$, so the glass does not break. With $\Delta T_{\mathrm c}\approx1447\ \mathrm K$, a temperature difference of $80\ \mathrm K$ is no problem at all.

An ordinary glass cup can crack when boiling water is suddenly poured into it, whereas fused quartz apparatus resists cracking even when moved suddenly from boiling water or a flame into cold water. The difference arises because the coefficients of linear thermal expansion $\alpha$ differ by more than an order of magnitude ($9.0\times10^{-6}$ and $0.55\times10^{-6}\ \mathrm{K^{-1}}$). Note that here we have assumed the most severe case, in which the surface temperature drops instantaneously; the temperature difference at which cracking actually begins is larger than this (see the note on the Biot number in Chapter 22).

Solution

Since this is pulsating loading with $R=0$, $\sigma_{\max}=\Delta\sigma=100\ \mathrm{MPa}$. From the fracture criterion at the maximum stress, $K_{\mathrm{IC}}=Y\sigma_{\max}\sqrt{\pi a_{\mathrm c}}$, the critical crack length is

$$ a_{\mathrm c}=\frac1\pi\left(\frac{K_{\mathrm{IC}}}{Y\sigma_{\max}}\right)^2=\frac1\pi\left(\frac{60}{1\times100}\right)^2\ \mathrm m=\frac{0.36}\pi\ \mathrm m\approx0.1146\ \mathrm m=114.6\ \mathrm{mm} $$

For $n=2$, separation of variables in $\dd a/\dd N=C(Y\Delta\sigma)^2\pi a$ gives $\int a^{-1}\dd a=\ln a$, so

$$ N_{\mathrm f}=\frac{1}{C(Y\Delta\sigma)^2\pi}\ln\frac{a_{\mathrm c}}{a_0}=\frac{1}{1.0\times10^{-11}\times(100)^2\times\pi}\ln\frac{0.1146}{0.001} $$

The denominator is $1.0\times10^{-11}\times10^4\times\pi=3.1416\times10^{-7}$ and $\ln114.6=4.7414$, so

$$ N_{\mathrm f}=\frac{4.7414}{3.1416\times10^{-7}}\approx1.51\times10^{7}\ \text{cycles} $$

Unit check: the unit of $C(Y\Delta\sigma)^2\pi$ is $\mathrm{m/cycle}\cdot(\mathrm{MPa\sqrt m})^{-2}\cdot\mathrm{MPa}^2=1/\mathrm{cycle}$, so $N_{\mathrm f}$ is a number of cycles. Because only the logarithm of the ratio of crack lengths matters, even if the initial crack becomes ten times as long ($10\ \mathrm{mm}$), $\ln(11.46)=2.44$ and the life is only about halved. This differs from a case in which the crack length enters as a power, as in Example 22.3 with $n=3$, in the way the life depends on the size of the initial crack.

Solution

We use $Q=mc\Delta T$ from Formula 23.1 ($m$: mass, $c$: specific heat, $\Delta T$: temperature change). Substituting for the water $m=100\ \mathrm{g}$, specific heat $c=4.2\ \mathrm{J/(g\cdot K)}$, and $\Delta T=60-10=50\ \mathrm{K}$, we obtain

$$ Q=100\times4.2\times50=21000\ \mathrm{J}=2.1\times10^4\ \mathrm{J} $$

This is the quantity of heat. Converting to calories, since $1\ \mathrm{cal}=4.2\ \mathrm{J}$, we get $Q=21000/4.2=5000\ \mathrm{cal}=5.0\ \mathrm{kcal}$ (this is exactly the quantity of heat for $100\ \mathrm{g}\times50\ \mathrm{K}=5000\ \mathrm{g\cdot K}$, from the definition of specific heat, “$1\ \mathrm{cal}$ to raise $1\ \mathrm{g}$ of water by $1\ \mathrm{K}$,” and it can also be checked directly by mental arithmetic).

Solution

Since the temperature is constant, Boyle's law $p_1V_1=p_2V_2$ can be used. Substituting $p_1=1.0\times10^5\ \mathrm{Pa}$, $V_1=2.0\ \mathrm{L}$, and $p_2=2.0\times10^5\ \mathrm{Pa}$ and solving for $V_2$, we obtain

$$ V_2=\frac{p_1V_1}{p_2}=\frac{1.0\times10^5\times2.0}{2.0\times10^5}=1.0\ \mathrm{L} $$

Since the pressure has exactly doubled, the volume has exactly halved—this is perfectly consistent with Figure 23.3 (each time the volume is halved, the pressure doubles).

Solution

In an isochoric process the piston does not move, so no work is done on the gas ($W=0$), and by the first law of thermodynamics $Q=\Delta U$. For a monatomic ideal gas $\Delta U=\dfrac32nR\Delta T$, so substituting $n=1.0\ \mathrm{mol}$, $R=8.31\ \mathrm{J/(mol\cdot K)}$, and $\Delta T=100\ \mathrm{K}$, we obtain

$$ Q=\Delta U=\frac32\times1.0\times8.31\times100=\frac32\times831=1246.5\approx1.2\times10^3\ \mathrm{J} $$

This is the quantity of heat absorbed.

Solution

We start from the first law of thermodynamics $\Delta U=Q-p\Delta V$ (solving the equation for $Q$ gives $Q=\Delta U+p\Delta V$). In an isochoric process the volume does not change ($\Delta V=0$), so all of the heat $Q$ added goes into the increase $\Delta U$ in internal energy: $Q_{\text{isochoric}}=\Delta U$. In an isobaric process, on the other hand, the gas does work $p\Delta V(>0)$ on its surroundings corresponding to the increase in its volume, so part of the added heat is “used up” in this work, and only the remainder goes into raising the internal energy: $Q_{\text{isobaric}}=\Delta U+p\Delta V$.

To achieve the same temperature rise $\Delta T$ (for an ideal gas $\Delta U$ is determined by the temperature change alone, so both processes require the same $\Delta U$), in the isobaric process we must add extra heat equal to the work $p\Delta V(>0)$. That is, $Q_{\text{isobaric}}\gt Q_{\text{isochoric}}$, and through the definition of the molar heat capacity $C_{\ast,\mathrm m}=Q_\ast/(n\Delta T)$ we obtain $C_{p,\mathrm m}\gt C_{V,\mathrm m}$. In fact, when we compute this for a monatomic ideal gas in Section 23.7, the difference is exactly equal to the gas constant $R$ ($C_{p,\mathrm m}-C_{V,\mathrm m}=R$).

Solution

After adiabatic compression, the kinetic energy (temperature) of the gas molecules is larger.

In isothermal compression, the energy corresponding to the work done on the gas from outside escapes as heat to the surroundings (the constant-temperature bath) each time ($Q=W'\lt0$; that is, the gas keeps releasing heat). Since the internal energy of an ideal gas is determined by the temperature alone, $\Delta U=0$, and the temperature does not change either. In adiabatic compression, on the other hand, there is no place at all for heat to escape ($Q=0$), so all of the work done on the gas from outside is converted into an increase in internal energy, $\Delta U=W(\gt0)$, and the speed at which the gas molecules rebound increases, so the temperature rises.

When the gas is compressed by the same amount of volume, in isothermal compression the temperature (and hence the rise in pressure) is held down, whereas in adiabatic compression the increase in internal energy due to the compression accumulates directly as a temperature rise, so the pressure rises more steeply. This is the physical reason why the adiabat has a steeper slope than the isotherm in Figure 23.13 (in terms of equations, the slope at the same point is $-p/V$ for the isotherm and $-\gamma p/V$ ($\gamma\gt1$) for the adiabat, so the adiabat is steeper, as shown in Table 23.1).

Solution

(1) Substituting $\alpha=1.2\times10^{-5}\ \mathrm{K^{-1}}$ into Formula 23.3 ($\beta=3\alpha$, which holds for isotropic materials), we obtain

$$ \beta=3\times1.2\times10^{-5}=3.6\times10^{-5}\ \mathrm{K^{-1}} $$

(2) From the equation for volume expansion $V=V_0(1+\beta t)$, the relative increase in volume is $\Delta V/V_0=\beta\Delta T$ ($V_0$ is, strictly speaking, the volume at $0\ {}^{\circ}\mathrm{C}$, but the difference is only about $\beta\times20\approx7\times10^{-4}$, so it does not affect an answer with two significant figures). Substituting $\Delta T=100\ \mathrm{K}$, we obtain

$$ \frac{\Delta V}{V_0}=3.6\times10^{-5}\times100=3.6\times10^{-3}=0.36\ \% $$

This is the fractional volume change.

Solution

In the convention of this book, $W$ is the work “done on the gas by its surroundings” ($W\gt0$ when the gas is compressed), and $W'$ is the work “done by the gas on its surroundings” ($W'\gt0$ when the gas expands); the two are related by $W'=-W$. The first law of thermodynamics can be written $\Delta U=Q+W$ using $W$, and $\Delta U=Q-W'$ using $W'$. In high school physics as well, there are cases in which one learns $\Delta U=Q+W$ ($W$: work done on the gas) and cases in which one learns $Q=\Delta U+W'$ ($W'$: work done by the gas), and these correspond directly to the way this book writes them.

In contrast, engineering textbooks and the like often adopt the convention of using the same letter $W$ to mean “the work done by the gas” (corresponding to $W'$ in this book). In this convention, the first law of thermodynamics is written $\Delta U=Q-W$ ($W$: work done by the gas).

Relating the two conventions, we obtain the following:

• $W$ in this book (work done on the gas) = $-W$ in the engineering convention (the work done by the gas, with its sign reversed)
• $W'$ in this book (work done by the gas) = $W$ in the engineering convention (work done by the gas, unchanged)

That is, the equation $\Delta U=Q-W'$ in this book expresses exactly the same content as what the engineering convention calls “$\Delta U=Q-W$” ($W$: work done by the gas). Looking only at the letters, this book's $\Delta U=Q+W$ and the engineering $\Delta U=Q-W$ seem to have opposite signs, but this is merely the same physical law written with a different choice of the “direction” of the quantity called work (whether the work done on the gas or the work done by the gas is taken as positive), so there is no contradiction. When you read other textbooks, it is a good idea to get into the habit of first checking which convention the book adopts.

Solution

The molecular weight of Ar, $39.9$, is a dimensionless relative value, and it means that the molar mass (the mass per $1\ \mathrm{mol}$) is $M=39.9\ \mathrm{g/mol}$. Since $1\ \mathrm{mol}=N_{\mathrm A}$ molecules have a total mass of $M$ grams, the mass $m$ of one molecule is found by dividing the molar mass by the Avogadro constant $N_{\mathrm A}$:

$$ m=\frac{M}{N_{\mathrm A}}=\frac{39.9\ \mathrm{g/mol}}{6.022\times10^{23}\ \mathrm{mol^{-1}}}=6.63\times10^{-23}\ \mathrm g=6.63\times10^{-26}\ \mathrm{kg} $$

This is the result (we rounded $39.9/6.022=6.626$ to three significant figures). The order of magnitude of the $m$ obtained in this way (about $10^{-26}\ \mathrm{kg}$) is the same as that of the value used in Example 24.3 of Chapter 24 ($m=5.0\times10^{-26}\ \mathrm{kg}$), and we can confirm that it is a reasonable value.

Solution

Solving the result of the kinetic theory of gases (Formula 24.1 of Chapter 24), $\frac12m\overline{|\bm v|^2}=\frac32\dfrac{R}{N_{\mathrm A}}T$, for $\overline{|\bm v|^2}$ gives $\overline{|\bm v|^2}=\dfrac{3RT}{mN_{\mathrm A}}=\dfrac{3RT}{M}$ ($M=mN_{\mathrm A}$ is the molar mass). At the same temperature $T$, the factor $3RT$ on the right-hand side is a common value independent of the kind of gas, so there is the proportionality $\overline{|\bm v|^2}\propto1/M$. Therefore the ratio of the root-mean-square speeds $\sqrt{\overline{|\bm v|^2}}$ is

$$ \frac{\sqrt{\overline{|\bm v|^2}}_{\mathrm{He}}}{\sqrt{\overline{|\bm v|^2}}_{\mathrm{Ar}}} =\sqrt{\frac{\overline{|\bm v|^2}_{\mathrm{He}}}{\overline{|\bm v|^2}_{\mathrm{Ar}}}} =\sqrt{\frac{3RT/M_{\mathrm{He}}}{3RT/M_{\mathrm{Ar}}}} =\sqrt{\frac{M_{\mathrm{Ar}}}{M_{\mathrm{He}}}} $$

Substituting the molecular weights (the numerical values of the molar masses) $4.00$ and $39.9$, we obtain

$$ \sqrt{\frac{M_{\mathrm{Ar}}}{M_{\mathrm{He}}}}=\sqrt{\frac{39.9}{4.00}}=\sqrt{9.975}\approx3.158 $$

That is, even at the same temperature, the molecules of the lighter He gas fly around about 3.16 times faster than those of Ar gas. This corresponds to the intuition that the lighter a molecule is, the faster it must move in order to keep the same kinetic energy (the same temperature).

Solution

The constant $b$ is a quantity that corrects for the fact that gas molecules are not point masses but have a finite size. The equation of state of an ideal gas assumed that the volume $V$ can be brought arbitrarily close to 0, but in reality the molecules cannot penetrate into one another any further, so no matter how large the pressure is made, the volume cannot become smaller than the volume that the molecules exclude ($b$ per $1\ \mathrm{mol}$). We therefore replace $V$ in the ideal gas equation by the effective volume $V-b$ in which the molecules can move freely. Note that $b$ is not simply $N_{\mathrm A}$ times the volume of the molecules themselves; it is the excluded volume resulting from the fact that the centers of molecules, regarded as spheres of radius $r$, cannot approach each other closer than $2r$, and it corresponds to $b\approx4N_{\mathrm A}\cdot\frac{4\pi}{3}r^3$ (about four times $N_{\mathrm A}$ times the volume of a molecule itself).

The constant $a$ is a quantity that corrects for the weak attractive force (intermolecular force, van der Waals force) acting between gas molecules. A molecule near the wall of the container receives, on average, an inward attractive force from the many molecules inside the container, so the force of its impact when it collides with the wall is slightly weakened, and the actual pressure is lower than the pressure predicted by the ideal gas model. This decrease is the product of the number of molecules striking the wall (proportional to the density $1/V$) and the attractive force that each molecule receives (also proportional to the density), so it is expected to be proportional to the square of the density, $1/V^2$. We therefore use, as a correction term $a/V^2$, the pressure $p+a/V^2$ that the ideal gas model predicts (with no intermolecular force) in place of the measured pressure $p$ ($V$ is the volume per $1\ \mathrm{mol}$).

To summarize, $b$ is the correction to the volume due to “molecular size” and $a$ is the correction to the pressure due to “attraction between molecules”; both are positive constants that express realistic deviations from the ideal gas (see Figure 24.7 of Chapter 24).

Solution

Key point: For $T\gt T_{\mathrm c}$ the isotherm has no point where $\dd p/\dd V=0$; compressing the gas only increases its density continuously, and no boundary separating gas from liquid (a jump in density) appears. Therefore “liquefaction” (the change from gas to liquid) does not occur.

How to read Figure 24.5: An isotherm at low temperature has a trough and a peak, and a real gas behaves by replacing this interval with a horizontal region (the liquid–gas coexistence region) in which only the volume decreases greatly while the pressure hardly changes. The length of this region (its extent in the volume direction) gradually becomes shorter as the temperature is raised, and at the critical temperature (about $31\ °\mathrm C\approx304\ \mathrm K$ for $\mathrm{CO_2}$; for the van der Waals equation in Figure 24.5 as well, $T_{\mathrm c}\approx304\ \mathrm K$) it finally shrinks to a single point (the critical point C).

Check using the equation: The critical point is a degenerate inflection point at which both the first and second derivatives of the graph of $p=f(V)$ vanish (Definition 24.3 of Chapter 24). For the van der Waals equation, $\dfrac{\dd p}{\dd V}=-\dfrac{RT}{(V-b)^2}+\dfrac{2a}{V^3}$, so the condition for $\dfrac{\dd p}{\dd V}\lt0$ (a monotonic decrease at every volume) is

$$ RT\gt g(V)\equiv\frac{2a(V-b)^2}{V^3}\qquad(\text{for all }V\gt b\text{}) $$

Differentiating $g(V)$ gives $g'(V)=\dfrac{2a(V-b)(3b-V)}{V^4}$, and it takes its maximum value $g(3b)=\dfrac{2a\cdot(2b)^2}{27b^3}=\dfrac{8a}{27b}$ at $V=3b$. Therefore, “$\dd p/\dd V\lt0$ for all $V$” is equivalent to $RT\gt\dfrac{8a}{27b}$, that is, $T\gt\dfrac{8a}{27Rb}=T_{\mathrm c}$. For $T\gt T_{\mathrm c}$ the curve is a smooth monotonically decreasing curve with no trough, no peak, and no critical point, and only when the temperature is exactly $T_{\mathrm c}$ does a single point appear at which $\dd p/\dd V=0$, namely at $V=3b=V_{\mathrm c}$.

That the liquid–gas coexistence region (the horizontal part) does not exist at all means that the phenomenon itself, “as the pressure is raised, the gas suddenly changes into liquid at a certain volume,” does not occur. Therefore, for $T\gt T_{\mathrm c}$, no matter how much the pressure of the gas is raised, its volume merely shrinks smoothly and its density rises continuously, without passing through a state of coexistence with the liquid (that is, without producing a clear boundary at which one can say “from here on it is liquid”), and “liquefaction” in the ordinary sense does not occur.

Solution

From Formula 24.2 of Chapter 24 (the internal energy of a monatomic ideal gas, which is obtained from both the kinetic theory of gases and the law of equipartition of energy (Law 24.2)), we have

$$ U=\frac32nRT $$

The molar heat capacity at constant volume is defined as the internal energy per $1\ \mathrm{mol}$ differentiated with respect to temperature (at constant volume):

$$ C_{V,\mathrm m}\equiv\frac1n\left(\frac{\partial U}{\partial T}\right)_V $$

Here $\left(\dfrac{\partial U}{\partial T}\right)_V$ is a partial derivative, meaning that $U$ is differentiated with respect to $T$ alone, regarding $V$ as a constant. Differentiating $U=\frac32nRT$ with respect to $T$ (since $n,R$ are constants, this is the derivative of a linear function of $T$), we obtain

$$ \left(\frac{\partial U}{\partial T}\right)_V=\frac32nR $$

Therefore we obtain

$$ C_{V,\mathrm m}=\frac1n\times\frac32nR=\frac32R $$

Numerically, $C_{V,\mathrm m}=\frac32\times8.31\approx12.47\ \mathrm{J/(mol\,K)}$. This value agrees with the molar heat capacity at constant volume of a monatomic ideal gas obtained in Chapter 27 from the first law of thermodynamics by using the fact that $Q=\Delta U$ in an isochoric process (Chapter 27).

Solution

Solving the van der Waals equation of state $\left(p+\dfrac{a}{V^2}\right)(V-b)=RT$ ($V$ is the volume per $1\ \mathrm{mol}$) for $p$ gives

$$ p=\frac{RT}{V-b}-\frac{a}{V^2} $$

For $V\gg b$, we rewrite $\dfrac1{V-b}=\dfrac1V\cdot\dfrac1{1-b/V}$ and, using the fact that $b/V$ is small, apply the binomial expansion (up to the first-order term) $\dfrac1{1-x}\approx1+x$ ($x=b/V$, $|x|\ll1$), which gives

$$ \frac{1}{V-b}=\frac1V\left(1+\frac bV+\cdots\right)\approx\frac1V+\frac{b}{V^2} $$

Substituting this into the expression for $p$, we obtain

$$ p\approx RT\left(\frac1V+\frac{b}{V^2}\right)-\frac{a}{V^2}=\frac{RT}{V}+\frac{RTb-a}{V^2} $$

Multiplying both sides by $V$, we obtain

$$ pV\approx RT+\frac{RTb-a}{V} $$

The first term $RT$ is the equation of state of an ideal gas itself, and the second term $\dfrac{RTb-a}{V}$ is a correction term that becomes smaller the larger $V$ is (it approaches 0 in proportion to $1/V$). Therefore, the fact that the second term vanishes and $pV\to RT$ as $V\to\infty$ agrees with the result confirmed in Example 24.11, but here we have further shown concretely that the size of the correction for finite $V$ is of order $1/V$. Note also that the sign of this correction term $RTb-a$ (which of $RTb$ and $a$ is larger) determines whether the pressure of a real gas is higher or lower than that of an ideal gas at the same temperature and volume—at high temperature and small $a$ (weak intermolecular force), the volume correction $b$ dominates and $p$ is higher than for an ideal gas, whereas at low temperature and large $a$ (strong intermolecular force), the pressure correction $a$ dominates and $p$ is lower than for an ideal gas.

Solution

From Formula 24.3 of Chapter 24, the internal energy of an ideal gas consisting of molecules with $f$ degrees of freedom is $U=\frac f2nRT$. When the temperature changes by $\Delta T$, the change in internal energy is (since $n,f,R$ are constant)

$$ \Delta U=\frac f2nR\,\Delta T $$

For $\mathrm{O_2}$, $f=5$ (three translational plus two rotational), $n=1\ \mathrm{mol}$, and $\Delta T=350-300=50\ \mathrm K$, so

$$ \Delta U_{\mathrm{O_2}}=\frac52\times1\times8.31\times50=1.04\times10^{3}\ \mathrm J\approx1.04\ \mathrm{kJ} $$

For Ar, $f=3$, so

$$ \Delta U_{\mathrm{Ar}}=\frac32\times1\times8.31\times50=6.23\times10^{2}\ \mathrm J\approx0.623\ \mathrm{kJ} $$

The ratio is $\Delta U_{\mathrm{O_2}}/\Delta U_{\mathrm{Ar}}=5/3\approx1.67$; to warm the gas by the same amount, $\mathrm{O_2}$, which has more degrees of freedom, requires about 1.7 times as much energy. This is because part of the added energy is also distributed to rotational motion.

Solution

Substitute the van der Waals constants of oxygen, $a=0.138\ \mathrm{Pa\,m^6/mol^2}$ and $b=3.19\times10^{-5}\ \mathrm{m^3/mol}$, into the formulas for the critical constants

$$ V_{\mathrm c}=3b,\qquad T_{\mathrm c}=\frac{8a}{27Rb},\qquad p_{\mathrm c}=\frac{a}{27b^2} $$

(Section 25.1.1). Taking $R=8.314\ \mathrm{J/(mol\,K)}$, we obtain

$$ T_{\mathrm c}=\frac{8\times0.138}{27\times8.314\times3.19\times10^{-5}}\ \mathrm K\approx154.2\ \mathrm K $$ $$ p_{\mathrm c}=\frac{0.138}{27\times(3.19\times10^{-5})^2}\ \mathrm{Pa}\approx5.02\times10^{6}\ \mathrm{Pa}\approx49.6\ \mathrm{atm} $$

(The measured values are $T_{\mathrm c}\approx154.6\ \mathrm K$ and $p_{\mathrm c}\approx49.8\ \mathrm{atm}$, but since $a,b$ are values determined so as to fit the critical constants, it is natural that the calculated values nearly agree with them. The small differences come from rounding $a,b$ to three significant figures.) Next, substituting $T=250\ \mathrm K$ and $p=20\ \mathrm{atm}$ into the definitions of the reduced variables, $T_{\mathrm r}=T/T_{\mathrm c},\ p_{\mathrm r}=p/p_{\mathrm c}$, gives

$$ T_{\mathrm r}=\frac{250}{154.2}\approx1.62,\qquad p_{\mathrm r}=\frac{20}{49.6}\approx0.40 $$

The value $T_{\mathrm r}\approx1.62$ is higher than the critical temperature, but it is not as far from the critical point as nitrogen is in Example 25.2 ($T_{\mathrm r}\approx2.38$).

Solution

What the principle of corresponding states (Law 25.1) asserts is that the van der Waals equation of state written in reduced variables

$$ \left(p_{\mathrm r}+\frac{3}{V_{\mathrm r}^2}\right)(3V_{\mathrm r}-1)=8T_{\mathrm r} $$

is just a single equation that does not depend at all on $a,b$ (and hence on the kind of gas). Therefore, if gases A and B are at the same $T_{\mathrm r},p_{\mathrm r}$, the same numbers are substituted into this single equation (a cubic equation), so the resulting real solution for the reduced volume $V_{\mathrm r}$ is also exactly the same for A and B. That is,

$$ V_{\mathrm r}(\mathrm A)=V_{\mathrm r}(\mathrm B) $$

This equality holds (here, if the equation has more than one real solution, we compare the same branch—the gas branch if the state is a gas, the liquid branch if it is a liquid).

Next, for the compression factor, we rewrite the definition $Z=pV_{\mathrm m}/(RT)$ in terms of the reduced variables by substituting $p=p_{\mathrm r}p_{\mathrm c}$, $V_{\mathrm m}=V_{\mathrm r}V_{\mathrm c}=3bV_{\mathrm r}$, and $T=T_{\mathrm r}T_{\mathrm c}$:

$$ Z=\frac{p_{\mathrm r}p_{\mathrm c}\cdot3bV_{\mathrm r}}{R\,T_{\mathrm r}T_{\mathrm c}}=\frac{3p_{\mathrm c}b}{RT_{\mathrm c}}\cdot\frac{p_{\mathrm r}V_{\mathrm r}}{T_{\mathrm r}} $$

Substituting here $p_{\mathrm c}=a/(27b^2)$ and $T_{\mathrm c}=8a/(27Rb)$, we find that $\dfrac{3p_{\mathrm c}b}{RT_{\mathrm c}}=\dfrac{3\times a/(27b^2)\times b}{R\times8a/(27Rb)}=\dfrac{a/(9b)}{8a/(27b)}=\dfrac{27}{9\times8}=\dfrac38$, which is again a universal constant independent of $a,b$. Therefore,

$$ Z=\frac38\cdot\frac{p_{\mathrm r}V_{\mathrm r}}{T_{\mathrm r}} $$

Hence, if gases A and B are at the same $(p_{\mathrm r},T_{\mathrm r})$ (and therefore at the same $V_{\mathrm r}$), then $Z(\mathrm A)=Z(\mathrm B)$ also holds. To summarize, under the principle of corresponding states, different gases in the same reduced state have the same reduced volume and the same compression factor—this is the meaning of “corresponding.” Note that $Z_{\mathrm c}=3/8$ is the value for a van der Waals gas; for real gases the critical compression factor is different, about $0.27$–$0.29$, so the principle of corresponding states holds only approximately for real gases.

Solution

The virial equation of state (Definition 25.3) is

$$ Z=1+\frac{B(T)}{V_{\mathrm m}}+\frac{C}{V_{\mathrm m}^2}+\cdots $$

By definition (Definition 25.4), the Boyle temperature $T_{\mathrm B}$ is the temperature at which this second virial coefficient vanishes, $B(T_{\mathrm B})=0$. So when we substitute $T=T_{\mathrm B}$, the first term that shifts $Z$ away from $1$ (the term proportional to $1/V_{\mathrm m}$, which has the largest effect) vanishes exactly, and only

$$ Z=1+\frac{C}{V_{\mathrm m}^2}+\cdots $$

remains. The $C/V_{\mathrm m}^2$ term is proportional to $1/V_{\mathrm m}^2$, a quantity that tends to zero much faster (at second order) than the $1/V_{\mathrm m}$ term, so even if the molar volume $V_{\mathrm m}$ is not very large (even if the pressure is not very low), the approximation $Z\approx1$ holds over a fairly wide range. On the other hand, at temperatures $T\neq T_{\mathrm B}$ the term proportional to $1/V_{\mathrm m}$ remains, so as $V_{\mathrm m}$ is increased (the pressure is lowered), we approach $Z=1$ only slowly, at $1$st order. Therefore, the Boyle temperature is the special temperature at which a real gas behaves like an ideal gas over the widest range of pressures.

Solution

Solving the definition of the compression factor, $Z=pV_{\mathrm m}/(RT)$, for $V_{\mathrm m}$ gives

$$ V_{\mathrm m}=\frac{ZRT}{p}=Z\times\frac{RT}{p}=Z\,V_{\mathrm m}^{\mathrm{ideal}} $$

(here $V_{\mathrm m}^{\mathrm{ideal}}=RT/p$ is the molar volume of an ideal gas at the same $p,T$). At the same $p,T$, $V_{\mathrm m}^{\mathrm{ideal}}$ is a positive constant, so $V_{\mathrm m}$ is proportional to $Z$. Therefore,

• When $Z\lt1$: $V_{\mathrm m}=Z\,V_{\mathrm m}^{\mathrm{ideal}}\lt V_{\mathrm m}^{\mathrm{ideal}}$; that is, the actual molar volume is smaller than that of the ideal gas.
• When $Z\gt1$: $V_{\mathrm m}=Z\,V_{\mathrm m}^{\mathrm{ideal}}\gt V_{\mathrm m}^{\mathrm{ideal}}$; that is, the actual molar volume is larger than that of the ideal gas.

This agrees with the interpretation in Section 25.2 ($Z\lt1$ means that attraction dominates and the molecules are “drawn together,” so the volume becomes smaller; $Z\gt1$ means that repulsion due to the size of the molecules themselves dominates, so the volume becomes larger).

Solution

In the virial equation of state $Z=1+B/V_{\mathrm m}+C/V_{\mathrm m}^2+\cdots$, the ratio of the third term to the second term is

$$ \frac{C/V_{\mathrm m}^2}{B/V_{\mathrm m}}=\frac{C}{B\,V_{\mathrm m}} $$

Since $B,C$ are constants determined by the gas and its temperature (in order of magnitude, they are the first and second powers, respectively, of a volume of about the size of $b$), this ratio is sufficiently small—that is, the terms from $C/V_{\mathrm m}^2$ onward may be neglected and the first-order approximation $Z\approx1+B/V_{\mathrm m}$ suffices—when

$$ V_{\mathrm m}\gg\left|\frac{C}{B}\right| $$

that is, when the molar volume $V_{\mathrm m}$ is sufficiently large (the gas is sufficiently dilute). In terms of pressure, since $V_{\mathrm m}\approx RT/p$ (an estimate valid when the gas is close to ideal), a large $V_{\mathrm m}$ corresponds to a sufficiently low pressure $p$. To summarize, the higher-order terms of the virial expansion can be neglected in the limit of low pressure and large volume (a dilute gas), and the more we raise the pressure and compress the gas (the smaller the molar volume becomes), the more we must take into account the higher-order terms involving $C,D,\ldots$.

Let us find a concrete rule of thumb for nitrogen. For a van der Waals gas, $B=b-a/(RT)$ and $C=b^2$, so $C/B=b^2/\bigl(b-a/(RT)\bigr)$. At $T=300\ \mathrm K$, $B=3.87\times10^{-5}-0.137/(8.314\times300)\approx-1.62\times10^{-5}\ \mathrm{m^3/mol}$ and $C=(3.87\times10^{-5})^2\approx1.50\times10^{-9}\ \mathrm{m^6/mol^2}$, so

$$ \left|\frac{C}{B}\right|\approx\frac{1.50\times10^{-9}}{1.62\times10^{-5}}\approx9.2\times10^{-5}\ \mathrm{m^3/mol} $$

The condition for the third term to be at most $10\ \%$ of the second term, $\left|C/(BV_{\mathrm m})\right|\le0.1$, is $V_{\mathrm m}\ge9.2\times10^{-4}\ \mathrm{m^3/mol}$. Converting this to a pressure with the ideal-gas value $V_{\mathrm m}\approx RT/p$ gives

$$ p\lesssim\frac{RT}{V_{\mathrm m}}=\frac{8.314\times300}{9.2\times10^{-4}}\ \mathrm{Pa}\approx2.7\times10^{6}\ \mathrm{Pa}\approx27\ \mathrm{atm} $$

That is, for nitrogen at $300\ \mathrm K$, the first-order approximation $Z\approx1+B/V_{\mathrm m}$ suffices at pressures below about $27\ \mathrm{atm}$, and at higher pressures the $C$ term can no longer be neglected.

Solution

Setting $U(r)=4\varepsilon\bigl[(\sigma/r)^{12}-(\sigma/r)^6\bigr]=0$ gives

$$ \left(\frac{\sigma}{r}\right)^{12}=\left(\frac{\sigma}{r}\right)^{6} $$

Let $X\equiv(\sigma/r)^6$. The left-hand side can be written as $(\sigma/r)^{12}=X^2$, so the equation becomes

$$ X^2=X\qquad\Longrightarrow\qquad X^2-X=0\qquad\Longrightarrow\qquad X(X-1)=0 $$

so $X=0$ or $X=1$. The root $X=(\sigma/r)^6=0$ corresponds to $r\to\infty$ (it merely expresses the fact that the potential approaches zero asymptotically far away; it is not a finite distance at which the sign changes), so the solution we want is $X=1$. From $(\sigma/r)^6=1$ we get $\sigma/r=1$, that is,

$$ r=\sigma $$

For $r\lt\sigma$ we have $\sigma/r\gt1$, so $(\sigma/r)^{12}\gt(\sigma/r)^6$ and $U\gt0$ (the potential energy is positive: the molecules are too close together and the energy is high), whereas for $r\gt\sigma$ the opposite holds, $U\lt0$ (the energy is negative, a region of attractive binding)—this is precisely the boundary between the blue and red regions in Figure 25.6.

Next, we examine the direction of the force at this $r=\sigma$. The force is $F=-\dd U/\dd r=\dfrac{24\varepsilon\sigma^6}{r^{7}}\left(\dfrac{2\sigma^6}{r^6}-1\right)$, and at $r=\sigma$ it is $F=\dfrac{24\varepsilon}{\sigma}(2-1)=\dfrac{24\varepsilon}{\sigma}\gt0$, that is, a repulsive force. The force becomes $0$ at $r=2^{1/6}\sigma\approx1.12\,\sigma$ (the distance at which $U$ is minimal), and that is where the force changes from repulsive to attractive. Thus the distance $r=\sigma$ at which the sign of $U$ changes and the distance $r_{\min}=2^{1/6}\sigma$ at which the direction of the force changes are different.

Solution

(1) We substitute the van der Waals constants into $T_{\mathrm B}=a/(Rb)$ and $T_{\mathrm c}=8T_{\mathrm B}/27$. For hydrogen,

$$ T_{\mathrm B}=\frac{0.0248}{8.314\times2.66\times10^{-5}}\ \mathrm K\approx112\ \mathrm K,\qquad T_{\mathrm c}=\frac{8}{27}\times112\ \mathrm K\approx33.2\ \mathrm K $$

and for carbon dioxide,

$$ T_{\mathrm B}=\frac{0.364}{8.314\times4.27\times10^{-5}}\ \mathrm K\approx1025\ \mathrm K,\qquad T_{\mathrm c}=\frac{8}{27}\times1025\ \mathrm K\approx304\ \mathrm K $$

(The values of $T_{\mathrm c}$ agree well with the measured values, $304\ \mathrm K$ for carbon dioxide and $33.2\ \mathrm K$ for hydrogen.)

(2) For hydrogen, $273\ \mathrm K$ is higher than $T_{\mathrm B}\approx112\ \mathrm K$, so $B\gt0$; concretely,

$$ B=b-\frac{a}{RT}=2.66\times10^{-5}-\frac{0.0248}{8.314\times273}\approx2.66\times10^{-5}-1.09\times10^{-5}\approx+1.57\times10^{-5}\ \mathrm{m^3/mol} $$

For carbon dioxide, $273\ \mathrm K$ is far lower than $T_{\mathrm B}\approx1025\ \mathrm K$, so $B\lt0$, and

$$ B=4.27\times10^{-5}-\frac{0.364}{8.314\times273}\approx4.27\times10^{-5}-16.04\times10^{-5}\approx-1.18\times10^{-4}\ \mathrm{m^3/mol} $$

For $p=10\ \mathrm{atm}=1.01325\times10^6\ \mathrm{Pa}$, $V_{\mathrm m}\approx RT/p=8.314\times273/(1.01325\times10^6)\approx2.24\times10^{-3}\ \mathrm{m^3/mol}$, so

$$ Z_{\mathrm{H_2}}\approx1+\frac{1.57\times10^{-5}}{2.24\times10^{-3}}\approx1.007,\qquad Z_{\mathrm{CO_2}}\approx1+\frac{-1.18\times10^{-4}}{2.24\times10^{-3}}\approx0.947 $$

(Solving the van der Waals equation directly gives $1.007$ and $0.945$, so the first-order approximation is in good agreement.)

(3) In Figure 25.5, the $Z$ of hydrogen rises gently above $Z=1$ ($Z\gt1$) because $273\ \mathrm K$ is higher than the $T_{\mathrm B}$ of hydrogen and $B\gt0$ (the effect of molecular size dominates). For carbon dioxide, $273\ \mathrm K$ is far lower than $T_{\mathrm B}$, so $B\lt0$ (attraction dominates) and $Z\lt1$. Because how high $T_{\mathrm B}$ is (the size of $a/b$) differs from gas to gas, the direction in which $Z$ changes can be opposite even at the same temperature.

Solution

(1) The reduced equation of state at $T_{\mathrm r}=0.9$ is $p_{\mathrm r}=8T_{\mathrm r}/(3V_{\mathrm r}-1)-3/V_{\mathrm r}^2$, and its slope is $\dd p_{\mathrm r}/\dd V_{\mathrm r}=-24T_{\mathrm r}/(3V_{\mathrm r}-1)^2+6/V_{\mathrm r}^3$. At $V_{\mathrm r}=1.2$, $3V_{\mathrm r}-1=2.6$, so

$$ p_{\mathrm r}=\frac{7.2}{2.6}-\frac{3}{1.44}\approx2.769-2.083=0.686,\qquad \frac{\dd p_{\mathrm r}}{\dd V_{\mathrm r}}=-\frac{21.6}{6.76}+\frac{6}{1.728}\approx-3.195+3.472=+0.277 $$

At $V_{\mathrm r}=2.0$, $3V_{\mathrm r}-1=5$, so

$$ p_{\mathrm r}=\frac{7.2}{5}-\frac{3}{4}=1.44-0.75=0.690,\qquad \frac{\dd p_{\mathrm r}}{\dd V_{\mathrm r}}=-\frac{21.6}{25}+\frac{6}{8}=-0.864+0.75=-0.114 $$

(2) The state $V_{\mathrm r}=1.2$, where the slope $\dd p_{\mathrm r}/\dd V_{\mathrm r}\gt0$, is one in which the pressure rises when the volume is increased, and it is mechanically unstable. This is also consistent with its lying between the spinodal points $0.719$ and $1.529$ of Example 25.4. On the other hand, $V_{\mathrm r}=2.0$, where the slope is negative, is mechanically stable. However, since $1.529\lt2.0\lt2.349=V_{\mathrm r}^{\mathrm{gas}}$, it lies on the side smaller than the volume $2.349$ of the coexisting gas (the higher-pressure side at the same temperature), so it is not a thermodynamic equilibrium state but is metastable (supersaturated vapor). Only the part with $V_{\mathrm r}\gt2.349$ is a stable gas.

Solution

As an everyday example of path functions, take the distance a car travels and the fuel it consumes. Consider going from point A to point B either by a straight road ($30\ \mathrm{km}$) or by a roundabout road over a mountain ($50\ \mathrm{km}$). The correspondence is as follows.

Thermodynamic term	This example
Initial state and final state	Starting point A and destination B
State function (such as $U$)	Position (elevation). At the moment of arriving at B, the value is the same whichever road was taken
Path (the way the process is carried out)	Which road was taken, and how it was driven
Path function (such as $Q$, $W$)	Distance traveled and amount of fuel consumed. They differ depending on the road: $30\ \mathrm{km}$ or $50\ \mathrm{km}$

Even when the starting point and the end point (the result) are the same, the distance traveled and the fuel consumed differ if the route (the history) differs. Similarly, even to bring about the same change in internal energy $\Delta U$, the values of the heat $Q$ and the work $W'$ done by the system differ if the path differs (Example 26.7).

(Alternative solution: the electricity bill. Even to keep the same room at the same temperature (the initial and final states are the same), the amount of electric energy consumed changes depending on whether the air conditioner is switched on and off frequently or left running. The state corresponds to the room temperature, the path corresponds to the way the air conditioner is operated, and the path function corresponds to the electricity bill.)

Solution

The motion of a diatomic molecule can be divided into three kinds: (a) translational motion of the molecule as a whole (three directions), (b) rotational motion about the two axes perpendicular to the molecular axis, and (c) vibrational motion in which the distance between the two atoms periodically stretches and shrinks. When the energy can be written in terms of quadratic terms (terms in the square of a velocity or a coordinate), each such term counts as one degree of freedom. Translation gives three terms such as $\tfrac12mv_x^2$; rotation gives one term $\tfrac12I\omega^2$ per axis, hence two terms; and vibration gives two terms, the kinetic energy $\tfrac12\mu\dot r^2$ and the elastic energy $\tfrac12k(r-r_0)^2$.

Near room temperature, the typical size of the energy exchanged in collisions between molecules (of the order of the thermal energy $k_{\mathrm B}T$) is much smaller than the energy needed to excite vibration (the energy spacing of the vibrational quantum, determined by the stiffness of the spring of the interatomic bond). Consequently, near room temperature the vibrational motion is hardly excited at all; it is effectively in a “freeze-out” state and takes no part in the exchange of energy. Therefore, near room temperature only 3 translational + 2 rotational = 5 degrees of freedom are effectively used. The temperature that serves as a guide to whether vibration matters (the characteristic temperature) is about $3400\ \mathrm K$ for $\mathrm{N_2}$ and about $2300\ \mathrm K$ for $\mathrm{O_2}$, and room temperature, $300\ \mathrm K$, is far below these.

When the temperature becomes sufficiently high, the collision energy between molecules becomes comparable to the energy needed to excite vibration, and the vibrational motion can also take part in the exchange of energy (the vibration “thaws”). Since vibration has two independent quadratic terms, the kinetic energy and the elastic energy, the degrees of freedom increase by two. As a result, at high temperatures the degrees of freedom become $5+2=7$. (A detailed explanation of the fact that vibration is a quantum-mechanical phenomenon that requires a “quantum” for excitation is something that can be understood after studying quantum mechanics, and goes beyond the scope of this book. Here it is enough to grasp the qualitative picture that “near room temperature vibration is effectively not used, and begins to be used only at high temperatures.”)

Solution

A quasi-static process is, by definition, a process carried out “infinitely slowly.” However, any real operation must be completed in a finite time—expanding a gas over literally infinite time is impossible in practice. In this sense, the quasi-static process is merely an “unrealistic” idealization (one that cannot be realized strictly in reality). That said, “slowly” here means slowly compared with the time it takes for disturbances inside the gas to even out (about $3\times10^{-4}\ \mathrm s$ for sound to travel $10\ \mathrm{cm}$), so an operation such as moving a piston at $1\ \mathrm{cm/s}$ can be regarded as nearly quasi-static.

Nevertheless, the quasi-static process is regarded as “ideal” because, thanks to this idealization, the gas can be regarded as being in thermal equilibrium at every instant during the change (macroscopically uniform, with the pressure and temperature fixed at definite values everywhere). Only an instant at which the gas is in thermal equilibrium can be represented unambiguously as a single point on the $p$-$V$ diagram (Section 26.4). In a quasi-static process, all intermediate states from the starting point to the end point are states of thermal equilibrium, so the whole process can be drawn as a continuous curve on the $p$-$V$ diagram; moreover, the external pressure and the gas pressure balance ($p_{\mathrm{ex}}=p$), so integral calculations such as the work $W'=\int p\,\dd V$ are also well defined. In contrast, in the middle of an abrupt change (an irreversible process), convection and turbulence arise in the gas and destroy the thermal equilibrium, so that “intermediate state” cannot be represented as a single point in the first place—because the pressure and volume take different values at different places.

In other words, the phrase “ideal but unrealistic” means “although it is a limiting idealization that cannot be realized strictly, it is precisely thanks to this idealization that the whole process of change can be treated mathematically in a well-defined way.” All real changes are irreversible to a greater or lesser degree, and the quasi-static process is used as a standard (a yardstick) against which to compare them.

Solution

A bent triatomic molecule (for example $\mathrm{H_2O}$) has its three atoms not lying on one straight line (it is not a linear molecule), so there are three independent directions of axes about which the molecule can rotate while keeping its overall shape (rotations about three mutually perpendicular axes of rotation). Because none of the moments of inertia about the three axes is $0$, three terms of the rotational kinetic energy $\tfrac12I\omega^2$ appear. Therefore, the rotational degrees of freedom number three, and together with the three translational degrees of freedom we obtain $f=3+3=6$ (vibration excluded).

In contrast, a diatomic molecule has a straight line connecting its two atoms (the molecular axis), and rotating the molecule about this molecular axis does not change the shape of the molecule (the arrangement of the atoms) at all—since the two atoms can be regarded as point-like particles (point masses), the moment of inertia about the molecular axis is $0$, and rotation about this axis carries no energy (it is indistinguishable from “nothing happening” even mechanically). For this reason, rotation about the molecular axis is not counted as a degree of freedom. What remains is only rotation about the two directions perpendicular to the molecular axis, and this is why the rotational degrees of freedom of a diatomic molecule are two (not three). A bent molecule has no such “axis about which rotation is indistinguishable,” so rotations in all three directions count as meaningful degrees of freedom, and the rotational degrees of freedom are three.

Linear polyatomic molecules such as $\mathrm{CO_2}$ and $\mathrm{C_2H_2}$, whose atoms lie on a straight line, have a molecular axis just as diatomic molecules do, and the moment of inertia about that axis is $0$, so the only rotations are about the two axes perpendicular to the molecular axis. Therefore, the degrees of freedom of a linear molecule, excluding vibration, are 3 translational + 2 rotational, that is, $f=5$, and not $f=6$.

Solution

The sign of the general expression for work $\dd'W=-p_{\mathrm{ex}}\dd V$ ($W$ is the “work done on the system”) reflects the physical fact that when a gas expands, the gas is the side that gives energy to the surroundings, not the side that receives it.

When the gas expands ($\dd V\gt0$), the gas pushes the piston outward and does work on the surroundings (the other side of the piston)—that is, the gas is giving energy to the surroundings. Since it is giving energy, the work “done on” the gas cannot be positive; it must instead be negative. Indeed, substituting $\dd V\gt0$ into $\dd'W=-p_{\mathrm{ex}}\dd V$ gives $\dd'W\lt0$ (since $p_{\mathrm{ex}}\gt0$), so the signs correspond correctly.

Conversely, when the gas is compressed ($\dd V\lt0$), the surroundings are pouring in energy to compress the gas, so positive work must be being done on the gas; substituting $\dd V\lt0$ into $\dd'W=-p_{\mathrm{ex}}\dd V$ gives $\dd'W\gt0$, and here too the signs correspond correctly.

For the work done by the system, $W'=-W$, the sign is exactly opposite: during expansion ($\dd V\gt0$), $\dd'W'=p_{\mathrm{ex}}\dd V\gt0$ (the gas does positive work on the surroundings), and during compression ($\dd V\lt0$), $\dd'W'\lt0$ (far from doing work on the surroundings, the gas is simply having work done on it). In this way, the minus sign in $\dd'W=-p_{\mathrm{ex}}\dd V$ is the sign needed to reflect correctly, in the quantity $W$ (the work done on the system), the physical direction: “during expansion the gas is the side that gives out energy, and during compression the gas is the side that receives energy.”

In the language of mechanics, the external force (the force due to the external pressure) pushes the gas, together with the piston, inward, and during expansion, in which the piston moves outward, the direction of the force and the direction of the displacement are opposite, so the work done by the external force on the gas is negative. This is the origin of the sign in $\dd'W=(-F)\,\dd x=-p_{\mathrm{ex}}S\,\dd x=-p_{\mathrm{ex}}\dd V$.

Solution

Integrating the infinitesimal change in internal energy $\dd U$ from state $i$ to state $f$ gives, as in Eq. (26.5) of Chapter 26,

$$ \Delta U=\int_i^f\dd U=U_f-U_i $$

so only the two endpoints of the interval of integration (the initial state $i$ and the final state $f$) need to be specified, and the answer ($U_f-U_i$) is determined uniquely, no matter how the path in between is specified. This is the meaning of saying that $\dd U$ is an exact differential—under the integral sign only “from $i$ to $f$” is written, and no information about the path is needed.

On the other hand, when the infinitesimal heat $\dd'Q$ is integrated, as in Eq. (26.6) of Chapter 26,

$$ Q=\int_{i,\,\text{path}}^{f}\dd'Q $$

the information “path” must be specified explicitly under the integral sign. Even between the same initial state $i$ and final state $f$, if the route followed differs, the result of the integral (the amount of heat exchanged) can differ. This is the meaning of saying that $\dd'Q$ is an inexact differential—to determine the result of the integral uniquely, one must specify not only the endpoints but also the path itself in between.

To summarize, the difference is that for an exact differential “the value of the integral is determined by the endpoints alone” (that is, the original quantity is a state function), whereas for an inexact differential “the value of the integral is not determined unless the path in between is also specified” (that is, the original quantity is a path function). The infinitesimal quantity of work $\dd'W$ is also an inexact differential, like heat, so it is written with a prime.

Solution

By the formula of this chapter, the internal energy of a monatomic ideal gas (with $f=3$ degrees of freedom) is $U=\dfrac32nRT$. When the volume is held constant and only the temperature is changed by $\Delta T$, the change in internal energy is obtained by fixing $n$ and taking out only the change in $T$, so we have

$$ \Delta U=\frac{3}{2}nR\,\Delta T $$

Substituting $n=3\ \mathrm{mol}$, $R=8.3145\ \mathrm{J/(mol\cdot K)}$, $\Delta T=80\ \mathrm K$, we obtain

$$ \Delta U=\frac{3}{2}\times3\ \mathrm{mol}\times8.3145\ \mathrm{J/(mol\cdot K)}\times80\ \mathrm K $$

Computing the coefficient first, $\dfrac32\times3=4.5$, so we obtain

$$ \Delta U=4.5\times8.3145\times80\ \mathrm J=4.5\times665.16\ \mathrm J\approx2993\ \mathrm J\approx2.99\ \mathrm{kJ} $$

Checking the units, $[\mathrm{mol}]\times[\mathrm{J/(mol\cdot K)}]\times[\mathrm K]=[\mathrm J]$, which agrees with the unit of internal energy (the joule).

Solution

State $\mathrm A$ is ($1.0\times10^5\ \mathrm{Pa}$, $2.0\times10^{-2}\ \mathrm{m^3}$) and state $\mathrm B$ is ($2.0\times10^5\ \mathrm{Pa}$, $4.0\times10^{-2}\ \mathrm{m^3}$). As found in Example 26.7, since the internal energy is a state function, the change in internal energy is independent of the path: $\Delta U=9.0\times10^3\ \mathrm J$.

Path c is the straight line joining $\mathrm A$ and $\mathrm B$ on the $p$-$V$ diagram. Since the process is quasi-static, $W'$ equals the area under this line (Formula 26.4, $W'=\int p\,\dd V$). The figure under the line is a trapezoid with upper base $p_{\mathrm A}=1.0\times10^5\ \mathrm{Pa}$, lower base $p_{\mathrm B}=2.0\times10^5\ \mathrm{Pa}$, and height $V_{\mathrm B}-V_{\mathrm A}=2.0\times10^{-2}\ \mathrm{m^3}$, so

$$ W'_{\mathrm c}=\frac{p_{\mathrm A}+p_{\mathrm B}}{2}\,(V_{\mathrm B}-V_{\mathrm A})=\frac{1.0\times10^5+2.0\times10^5}{2}\ \mathrm{Pa}\times2.0\times10^{-2}\ \mathrm{m^3}=3.0\times10^3\ \mathrm J $$

(the same value is obtained by integrating the straight line $p=p_{\mathrm A}+\dfrac{p_{\mathrm B}-p_{\mathrm A}}{V_{\mathrm B}-V_{\mathrm A}}(V-V_{\mathrm A})$ from $V_{\mathrm A}$ to $V_{\mathrm B}$). From the first law $\Delta U=Q-W'$, we obtain

$$ Q_{\mathrm c}=\Delta U+W'_{\mathrm c}=9.0\times10^3+3.0\times10^3=1.2\times10^4\ \mathrm J $$

Compare this with path a ($W'=4.0\times10^3\ \mathrm J$, $Q=1.3\times10^4\ \mathrm J$) and path b ($W'=2.0\times10^3\ \mathrm J$, $Q=1.1\times10^4\ \mathrm J$): the $W'$ and $Q$ of path c are exactly intermediate between them. Even if the initial and final states are the same, $W'$ and $Q$ differ depending on the path, and only $\Delta U$ is independent of the path.

Solution

(a) Since the system released heat, $Q=-300\ \mathrm J$; since work was done on it by the surroundings, the work done on the system is $W=+200\ \mathrm J$, and the work done by the system is $W'=-W=-200\ \mathrm J$.

In the formula $\Delta U=Q+W$, which uses the work done on the system,

$$ \Delta U=Q+W=-300+200=-100\ \mathrm J $$

In the formula $\Delta U=Q-W'$, which uses the work done by the system,

$$ \Delta U=Q-W'=-300-(-200)=-100\ \mathrm J $$

Both give $\Delta U=-100\ \mathrm J$, in agreement. The gas received $200\ \mathrm J$ through work but lost $300\ \mathrm J$ as heat, so its internal energy decreases by $100\ \mathrm J$.

(b) $Q=+120\ \mathrm J$, and since the system did $50\ \mathrm J$ of work on the surroundings, the work done by the system is $W'=50\ \mathrm J$. In the formula $\Delta U=Q-W$ of the convention that writes the “work done by the system” as $W$, the $W$ corresponds to $W'$ in this book, so the value to substitute is $50\ \mathrm J$ (a positive value). Then we have

$$ \Delta U=Q-W=120-50=70\ \mathrm J $$

Calculating in the notation of this book as $\Delta U=Q-W'=120-50=70\ \mathrm J$, or using the work done on the system, $W=-50\ \mathrm J$, as $\Delta U=Q+W=120+(-50)=70\ \mathrm J$, gives the same result.

Solution

The criterion for $M\,\dd x+N\,\dd y$ to be an exact differential form is $\partial M/\partial y=\partial N/\partial x$.

(a) Writing $\dd'W=-p\,\dd V=0\cdot\dd p+(-p)\,\dd V$, the coefficient of $\dd p$ is $M=0$ and the coefficient of $\dd V$ is $N=-p$.

$$ \pdiff{M}{V}=\pdiff{0}{V}=0,\qquad\pdiff{N}{p}=\pdiff{(-p)}{p}=-1 $$

The two partial derivatives differ, since $0\ne-1$, so the criterion does not hold. Therefore $\dd'W$ is not an exact differential form (it is an inexact differential).

(b) For a quasi-static change of a monatomic ideal gas, $p=nRT/V$, so

$$ \dd'Q=\frac32nR\,\dd T+p\,\dd V=\frac32nR\,\dd T+\frac{nRT}{V}\,\dd V $$

Regarding $(T,V)$ as the variables, let the coefficient of $\dd T$ be $M=\tfrac32nR$ and the coefficient of $\dd V$ be $N=\dfrac{nRT}{V}$. Then

$$ \pdiff{M}{V}=0,\qquad\pdiff{N}{T}=\frac{nR}{V} $$

The two partial derivatives differ, since $0\ne nR/V$, so the criterion does not hold. Therefore $\dd'Q$ is not an exact differential form (it is an inexact differential).

(c) Dividing $\dd'Q$ by $T$ gives

$$ \frac{\dd'Q}{T}=\frac{3nR}{2T}\,\dd T+\frac{nR}{V}\,\dd V $$

Let $M=\dfrac{3nR}{2T}$ and $N=\dfrac{nR}{V}$. Then

$$ \pdiff{M}{V}=\pdiff{}{V}\left(\frac{3nR}{2T}\right)=0,\qquad\pdiff{N}{T}=\pdiff{}{T}\left(\frac{nR}{V}\right)=0 $$

The two coincide, so the criterion holds. Indeed, we can write $\dfrac{\dd'Q}{T}=\dd\left(\dfrac32nR\ln T+nR\ln V\right)$. The heat $Q$ itself is not a function of state, but dividing by $T$ turns it into the infinitesimal change of a function of state (an exact differential). This $1/T$ is the integrating factor, and it is the starting point of the entropy that we study in Chapter 29.

Solution

The first law (with this book's sign convention, in which the work is that done on the gas) is $\dd U=\dd'Q+\dd'W$, $\dd'W=-p\,\dd V$.

Isochoric process: Since $\dd V=0$, $\dd'W=-p\,\dd V=0$. Therefore $\dd U=\dd'Q$; that is, the heat received goes entirely into increasing the internal energy. Because the volume does not change, the gas does no work of expansion at all.

Isobaric process: As the temperature rises, the gas also expands (for an ideal gas, $pV=nRT$ shows that if $T$ rises at constant $p$, then $V$ increases in proportion). In this case the gas does work $\dd'W'=-\dd'W=p\,\dd V\gt0$ on its surroundings. The heat received, $\dd'Q=\dd U+p\,\dd V$ (the first law rewritten), must supply both the increase $\dd U$ in internal energy and this work of expansion $p\,\dd V$.

When the gas receives the same quantity of heat $\dd'Q$, the increase in internal energy is $\dd U=\dd'Q$ at constant volume and $\dd U=\dd'Q-p\,\dd V\lt\dd'Q$ at constant pressure. At constant pressure it is smaller by the work of expansion $p\,\dd V$. For an ideal gas $\dd U=C_V\,\dd T$, so to obtain the same $\dd T$, the constant-pressure process needs $p\,\dd V$ more heat.

Comparing the heat needed to raise the temperature by the same $\dd T$, the constant-pressure process requires extra heat equal to “the amount spent on the work of expansion.” This is the reason why $C_p=(\partial H/\partial T)_p$ is larger than $C_V=(\partial U/\partial T)_V$ (in general $C_p\ge C_V$; Chapter 27, Formula 27.3). As we saw in Example 27.2 of Chapter 27, for an ideal gas this extra heat is exactly $nR\,\dd T$ (which is precisely Mayer's relation $C_p-C_V=nR$).

Solution

One form of Poisson's formula in Chapter 27 is $T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}$. By the ideal gas law $pV=nRT$ we have $T=pV/(nR)$, so we substitute this into both sides:

$$ \frac{p_1V_1}{nR}\,V_1^{\gamma-1}=\frac{p_2V_2}{nR}\,V_2^{\gamma-1} $$

Canceling the constant $nR$ (which is not $0$) common to both sides, we get

$$ p_1V_1\cdot V_1^{\gamma-1}=p_2V_2\cdot V_2^{\gamma-1} $$

On the left side, $V_1\cdot V_1^{\gamma-1}=V_1^{1+(\gamma-1)}=V_1^\gamma$ (the law of exponents $x^a\cdot x^b=x^{a+b}$), and likewise the right side becomes $V_2^\gamma$, so that

$$ p_1V_1^{\gamma}=p_2V_2^{\gamma} $$

is obtained. This is exactly the other form of the formula. Therefore, as long as we use the ideal gas law in the form $T=pV/(nR)$, the two forms are completely equivalent, and we may start from either one.

Solution

The internal pressure $\pi_T=(\partial U/\partial V)_T$ is the quantity that expresses how much the internal energy changes when only the average spacing between molecules (the volume) is changed without changing the temperature.

In an ideal gas, we assume that the molecules exert no forces on one another at all (one of the basic assumptions of the kinetic theory of gases). In that case, the internal energy of the gas consists solely of the sum of the kinetic energies of the thermal motion of all the molecules. Even if we change the average distance between molecules (the volume), no forces (attractive or repulsive) act between the molecules in the first place, so there is no potential energy that could be stored there from the start. The sum of the kinetic energies is determined only by the temperature (a quantity related to the square of the average speed of the molecules) and does not depend on how far apart the molecules are. Therefore, even if we change $V$ (the volume, and hence the intermolecular distance) while keeping $T$ constant, $U$ does not change at all, and $\pi_T=(\partial U/\partial V)_T=0$.

In a real gas in which attractive forces act between the molecules (for example, the van der Waals gas of Figure 27.5 in Chapter 27), the situation is different. When the molecules attract one another, to increase the volume and pull the molecules apart we must do work against that attraction—this is exactly the same reasoning as the potential energy increasing when we lift an object against the Earth's gravity. To that extent, the potential energy due to the intermolecular forces is added to the internal energy, and $U$ increases as the volume is made larger. Therefore $\pi_T=(\partial U/\partial V)_T\gt0$ (in Figure 27.5 of Chapter 27, we confirmed, with the concrete form $U(V)=-an^2/V$, that $U$ increases monotonically as $V$ is made larger).

Solution

If we replace the Joule–Thomson process (the process of forcing a gas through a porous barrier inside an adiabatic container) by two pistons, then since the whole container is thermally insulated (no heat flows in or out), the net work received by the gas becomes directly the change in internal energy. Combining the work $+p_1V_1$ done by the left piston in pushing the gas in and the work $-p_2V_2$ done in pushing back the right piston, we have $\Delta U=p_1V_1-p_2V_2$, and rearranging this gives $U_1+p_1V_1=U_2+p_2V_2$, that is, $H_1=H_2$ (the enthalpy does not change) (see the derivation in Chapter 27).

In other words, the path of change that actually occurs when “a gas is forced through a porous barrier” is, mathematically, nothing other than a path along which “the pressure $p$ and the temperature $T$ change while the enthalpy $H$ is held constant.” When we want to express how one physical quantity (here the temperature $T$) changes in response to a change in another physical quantity (the pressure $p$), it is natural to take the partial derivative with the quantity actually held constant along that path written as a subscript (the same idea as in $C_V=(\partial U/\partial T)_V$ being a partial derivative along the “path of constant volume”). In the Joule–Thomson process, the “quantity held constant along the path” was the enthalpy, so defining $\mu=(\partial T/\partial p)_H$ as a partial derivative at constant enthalpy is the definition that correctly represents the phenomenon that actually occurs.

Solution

For a gas to be cooled by Joule–Thomson expansion ($\mu\gt0$), the state of the gas, in temperature and pressure, must lie inside the inversion curve in Figure 27.10 of Chapter 27. If it is not inside the inversion curve (that is, it is outside the curve, or in the low-pressure region where the curve does not extend at all), the gas will instead heat up on expansion ($\mu\lt0$).

Looking at Figure 27.10, the sizes of the inversion curves of the three gases are quite different—the inversion curve of nitrogen extends the farthest, that of hydrogen is much smaller, and that of helium is smaller still (the curves become smaller in the order N$_2$, H$_2$, He). The maximum inversion temperature (the high-temperature value as $p\to0$) is, according to measurements, approximately $620\ \mathrm{K}$ for N$_2$, $200\ \mathrm{K}$ for H$_2$, and $40\ \mathrm{K}$ for He (Figure 27.10 is a rough guide based on the van der Waals model, and its values differ, but the trend of decreasing in this order is the same). Room temperature (approximately $300\ \mathrm{K}$) lies inside the inversion curve of nitrogen, but not inside the inversion curves of hydrogen or helium at all.

Therefore, even if we expand helium at room temperature, it is not inside the small curve of helium in the first place (the inside of that curve being, at best, temperatures of about $40\ \mathrm{K}$ or lower), so it is not cooled by the Joule–Thomson effect. So first we make liquid nitrogen ($77\ \mathrm{K}$) by Joule–Thomson expansion of nitrogen at room temperature (since it is inside nitrogen's own wide inversion curve). Next, so that the temperature is below the maximum inversion temperature of hydrogen (approximately $200\ \mathrm{K}$), we precool hydrogen to $77\ \mathrm{K}$ with liquid nitrogen and then let it undergo Joule–Thomson expansion, producing liquid hydrogen ($20\ \mathrm{K}$). Finally, so that the temperature is below the maximum inversion temperature of helium (approximately $40\ \mathrm{K}$), we precool helium to $20\ \mathrm{K}$ with liquid hydrogen and then let it undergo Joule–Thomson expansion, and only then do we obtain liquid helium. The very structure in which the maximum inversion temperature of the gas at each stage decreases in the order N$_2$, H$_2$, He is the reason why helium cannot be liquefied in a single stage and must be precooled in sequence, in a cascade.

Solution

(1) For reversible isothermal expansion, by the equation of Chapter 27 ($W'=nRT\ln(V_2/V_1)$, the same form as in Example 27.1),

$$ W'_{\text{isothermal}}=nRT_1\ln\frac{V_2}{V_1}=1.0\times8.314\times300\times\ln\frac{20}{10} $$

Since $\ln2\approx0.6931$,

$$ W'_{\text{isothermal}}=8.314\times300\times0.6931\approx1728.8\ \mathrm{J}\approx1.73\ \mathrm{kJ} $$

(2) For reversible adiabatic expansion, we first find the temperature after expansion using Poisson's formula $T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}$ of Chapter 27:

$$ T_2=T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}=300\times\left(\frac{10}{20}\right)^{0.4}=300\times0.5^{0.4} $$

Since $0.5^{0.4}=e^{0.4\ln0.5}=e^{0.4\times(-0.6931)}=e^{-0.2773}\approx0.7579$,

$$ T_2\approx300\times0.7579\approx227.4\ \mathrm{K} $$

Since the process is adiabatic, $Q=0$, and therefore (with this book's sign convention) $\Delta U=W$. The work done by the gas on its surroundings is $W'=-W=-\Delta U=-C_V(T_2-T_1)=C_V(T_1-T_2)$. Since this is an ideal gas of diatomic molecules, $C_V=\tfrac52 nR=2.5\times1.0\times8.314=20.785\ \mathrm{J/K}$. Hence,

$$ W'_{\text{adiabatic}}=C_V(T_1-T_2)=20.785\times(300-227.4)=20.785\times72.6\approx1509.9\ \mathrm{J}\approx1.51\ \mathrm{kJ} $$

(3) $W'_{\text{isothermal}}\approx1.73\ \mathrm{kJ}$ is larger than $W'_{\text{adiabatic}}\approx1.51\ \mathrm{kJ}$. As we confirmed in Figure 27.3 of Chapter 27, when we compare the isotherm and the adiabat passing through the same starting point $(V_1,p_1)$, on the side where $V$ increases (the direction of expansion) the adiabat is steeper ($\abs{\dd p/\dd V}=\gamma p/V\gt p/V$) and passes below the isotherm. That is, in isothermal expansion the gas expands while always maintaining a larger pressure for the same change in volume, so the area under the curve in the $p$-$V$ diagram (= the work) is larger. (On the side where $V$ decreases, that is, in the direction of compression, the adiabat is conversely the upper one.) Physically, this corresponds to the following difference: in adiabatic expansion the gas sacrifices its own internal energy (its temperature falls) by the amount of work it does, so the pressure drops more, whereas in isothermal expansion heat is supplied from outside to keep the temperature constant (and correspondingly more work can be extracted). (For the numerical values, see the check below.)

Check: With $n=1.0,\ R=8.314,\ T_1=300,\ V_1=10,\ V_2=20,\ \gamma=1.4$, we have $W'_{\text{isothermal}}=nRT_1\ln(V_2/V_1)\approx1728.85\ \mathrm{J}$, $T_2=T_1(V_1/V_2)^{\gamma-1}\approx227.36\ \mathrm{K}$, and $W'_{\text{adiabatic}}=C_V(T_1-T_2)\approx1509.87\ \mathrm{J}$. Since $1728.85\gt1509.87$, this agrees with the conclusion of (3).

Solution

(1) The bulk modulus $K$ and the isothermal compressibility $\kappa_T$ are reciprocals of each other, $K=1/\kappa_T$, so

$$ \kappa_T=\frac{1}{K}=\frac{1}{1.4\times10^{11}\ \mathrm{Pa}}\approx7.1\times10^{-12}\ \mathrm{Pa^{-1}} $$

(2) We use the general Mayer's relation (Formula 27.3 of Chapter 27) per $1\ \mathrm{mol}$:

$$ C_{p,\mathrm m}-C_{V,\mathrm m}=\frac{\alpha^2TV_{\mathrm m}}{\kappa_T}=\alpha^2TV_{\mathrm m}K $$

Multiplying $\alpha^2=(4.9\times10^{-5})^2=2.401\times10^{-9}\ \mathrm{K^{-2}}$ by $T=300\ \mathrm{K}$ gives $7.203\times10^{-7}\ \mathrm{K^{-1}}$; multiplying by $V_{\mathrm m}=7.1\times10^{-6}\ \mathrm{m^3/mol}$ gives $5.114\times10^{-12}\ \mathrm{m^3/(mol\,K)}$; and multiplying by $K=1.4\times10^{11}\ \mathrm{Pa}$ gives

$$ C_{p,\mathrm m}-C_{V,\mathrm m}\approx0.72\ \mathrm{Pa\,m^3/(mol\,K)}=0.72\ \mathrm{J/(mol\,K)} $$

(3) Relative to $C_{p,\mathrm m}\approx24.5\ \mathrm{J/(mol\,K)}$, the ratio is $0.72/24.5\approx0.03$, that is, about $3\,\%$. A solid hardly expands when heated ($\alpha$ is small) and, moreover, is hard to compress ($K$ is large), so even when it is heated at constant pressure, little of the heat goes into the work of expansion. For this reason we may regard $C_p\approx C_V$. This is in contrast to an ideal gas, for which the difference is as large as $C_p-C_V=R\approx8.3\ \mathrm{J/(mol\,K)}$.

Solution

In the Joule–Thomson process, $\dd H=0$. Setting $\dd H=0$ in the total differential of the enthalpy, $\dd H=-\mu C_p\,\dd p+C_p\,\dd T$, gives $\dd T=\mu\,\dd p$. Regarding $\mu$ as constant and integrating, we get $\Delta T=\mu\,\Delta p$. The change in pressure is $\Delta p=1.0-10.0=-9.0\ \mathrm{atm}$.

(1) Nitrogen: $\Delta T=0.25\ \mathrm{K/atm}\times(-9.0\ \mathrm{atm})=-2.25\ \mathrm{K}\approx-2.3\ \mathrm{K}$. The temperature falls by about $2.3\ \mathrm{K}$, becoming $297.75\ \mathrm{K}$.

(2) Hydrogen: $\Delta T=(-0.03\ \mathrm{K/atm})\times(-9.0\ \mathrm{atm})=+0.27\ \mathrm{K}$. The temperature rises by about $0.3\ \mathrm{K}$, becoming $300.27\ \mathrm{K}$.

(3) The maximum inversion temperature of nitrogen (approximately $620\ \mathrm{K}$ by measurement) is higher than room temperature, $300\ \mathrm{K}$, so room temperature lies inside the inversion curve and $\mu\gt0$ (cooling). The maximum inversion temperature of hydrogen (approximately $200\ \mathrm{K}$) is lower than room temperature, and room temperature lies outside the inversion curve, so $\mu\lt0$ (heating). Thus, for the same change in pressure, nitrogen is cooled while hydrogen is warmed slightly. To cool hydrogen by throttling expansion, it must be precooled in advance to below its maximum inversion temperature.

Solution

The thermal efficiency of a Carnot cycle is $\eta=1-T_1/T_2$, where $T_2$ is the absolute temperature of the hot reservoir and $T_1$ is that of the cold reservoir. Fixing $T_2$ and considering the limit $T_1\to0$ (absolute zero), we obtain

$$ \lim_{T_1\to0}\eta=\lim_{T_1\to0}\left(1-\frac{T_1}{T_2}\right)=1-\frac{0}{T_2}=1 $$

so indeed $\eta\to1$ (an efficiency of 100%).

However, this does not mean that “if only we could prepare a reservoir at absolute zero, we could actually build a heat engine with 100% efficiency.” This limit merely describes the mathematical behavior of the formula when the variable $T_1$ is made to approach 0. According to the third law of thermodynamics (which we study in Chapter 32 of this text), it is physically impossible to reach absolute zero itself by any finite number of operations. Therefore every real cold reservoir has $T_1\gt0$, and $\eta=1-T_1/T_2$ is always strictly less than 1. The relation “$\eta\to1$ as $T_1\to0$” is a limit that shows a tendency: the lower the temperature of the cold reservoir is made, the higher the theoretical upper bound on the efficiency becomes. It does not mean that an efficiency of 100% can actually be achieved.

This is also consistent with the Kelvin–Planck statement in Section 28.1. The value $\eta=1$ corresponds to $Q_1=0$ (all the absorbed heat is converted into work without any heat being discarded to the cold reservoir), and the Kelvin–Planck statement forbids precisely this. The limit $T_1\to0$ does not mean that the cold reservoir disappears; it means that its temperature approaches absolute zero. By the third law, $T_1=0$ cannot be reached, so the limit $\eta\to1$ cannot be reached either, and the two do not contradict each other.

Solution

The four processes of a Carnot cycle must be quasi-static (sufficiently slow) in order to make the cycle a reversible process (a process that can be retraced in the opposite direction without leaving any trace on the surroundings). Being quasi-static is a necessary condition for reversibility; in addition, there must be no dissipation such as friction, and heat must be exchanged with the reservoirs without any temperature difference. If an intermediate process (for example, the adiabatic expansion) is carried out abruptly, nonuniformities (inhomogeneities) of pressure and temperature arise inside the gas, or dissipation such as turbulence and friction occurs, and the same path can no longer be retraced exactly in the opposite direction (the process becomes an irreversible process).

This reversibility was used in an essential way in the proof of Carnot's theorem (Section 28.4). In the proof, in order to compare with an arbitrary heat engine $\mathrm X$, we ran the reversible heat engine $\mathrm R$ (a Carnot cycle) backward, using the work produced by $\mathrm X$ directly to drive $\mathrm R$ as a refrigerator. If $\mathrm R$ contained irreversible processes, the operation of “running it backward” itself would be impossible in the first place (an irreversible process cannot be retraced in the opposite direction with the same amounts of heat and work flowing in and out; even if one forced it to run backward, the amounts of heat and work on the way back would differ from those on the way out, and the relation on which the proof is based—that the magnitudes on the way out and on the way back are the same—would break down). Therefore, if the processes of the Carnot cycle were not quasi-static but irreversible, the part of the proof in which we assumed that “$\mathrm R$ is run backward, with heat and work of the same magnitudes as on the way out flowing in and out” would not hold in the first place, and the whole proof would collapse. Quasi-staticity (reversibility) is an indispensable premise for Carnot's theorem to hold.

Solution

In the forward Carnot cycle, heat $Q_2$ is absorbed from the hot reservoir, heat $Q_1$ ($Q_1\lt Q_2$) is released to the cold reservoir, and the difference is extracted as the work $W'=Q_2-Q_1$ done by the system on the surroundings. When the cycle is run in reverse, all of these flows of heat and work change direction: heat $Q_1$ is absorbed from the cold reservoir, heat $Q_2$ is released to the hot reservoir, and for this purpose work $W_{\mathrm{in}}=Q_2-Q_1$ is supplied from the surroundings.

In other words, the reversed Carnot cycle does exactly what a refrigerator (which removes heat from the cold side, namely its interior) or a heat pump (which delivers heat to the hot side, the inside of a room) does: it “draws heat up from the cold side and discharges it to the hot side.” Unless work $W_{\mathrm{in}}$ is continually supplied from the surroundings, this pumping of heat does not occur—this is why a refrigerator or an air conditioner keeps consuming electricity (work).

This does not contradict the Clausius statement in Section 28.1 (“there is no process in which heat moves from a cold body to a hot body without the supply of work”). What the Clausius statement forbids is only the process in which heat moves spontaneously from cold to hot without the supply of work. In the reversed Carnot cycle, the “other change” of work $W_{\mathrm{in}}$ from the surroundings properly accompanies the transfer, so it is not a forbidden process.

Solution

The Otto cycle treats combustion (the process corresponding to heat absorption from outside) as an isochoric process. This is because in a gasoline engine the compressed air–fuel mixture is ignited in an instant by a spark from the spark plug, and combustion is considered to be complete before the piston has moved appreciably (only the pressure and temperature rise sharply while the volume stays almost constant). The Diesel cycle, on the other hand, treats combustion as an isobaric process. This is because a diesel engine has no spark plug: fuel is injected into air that has been compressed at a very high compression ratio and thereby heated, and combustion continues all the while the fuel is being injected as the piston is pushed out slowly, so the process can be regarded as one in which the volume increases while the pressure is kept almost constant.

This difference corresponds to a difference in the structure of actual engines. Because a gasoline engine (Otto cycle) uses forced ignition by a spark plug, raising the compression ratio too much makes “knocking,” in which the mixture ignites spontaneously before the spark, likely to occur, so there is an upper limit to the compression ratio (the Discussion in Section 28.6). A diesel engine (Diesel cycle), on the other hand, is designed from the outset to use self-ignition by the compression of air, so there is no need to worry about knocking, and a much higher compression ratio than in the Otto cycle ($r=15$–$25$ or so) can be realized. As we confirmed in Example 28.9, the Otto cycle has the higher efficiency when the two are compared at the same compression ratio, but since a diesel engine can have a higher compression ratio itself, it often achieves a better fuel economy (effective efficiency) than a gasoline engine in practice.

Solution

By Carnot's theorem, the efficiency $\eta=1-T_1/T_2$ of a reversible engine operating between the same two reservoirs $T_1,T_2$ is a universal value determined only by the temperatures of the reservoirs, regardless of the working substance (an ideal gas, steam, or any other substance). This is a very strong statement. Conversely, it means that we can redefine “temperature” itself by using the “efficiency,” a quantity that can be measured without relying on the detailed properties of the working substance (such as the form of its equation of state).

Specifically, if the efficiency $\eta$ of a reversible engine operating between two reservoirs could be measured by an experiment (or, for the purposes of the definition, by a thought experiment), then the relation $\eta=1-T_1/T_2$ fixes the ratio of temperatures, $T_1/T_2=1-\eta$. This is entirely different from methods such as the “gas thermometer,” which defines temperature by relying on the properties of a particular substance (for example, the volume expansion of an ideal gas); it is a way of defining temperature that is common to all substances and based only on the efficiency of a heat engine. This is called the thermodynamic temperature scale (also called the absolute temperature scale; it is the original definition of the Kelvin scale; the content is the same as that of the Discussion box in Section 28.4). The fact that the efficiency $\eta=1-T_1/T_2$ of the ideal-gas Carnot cycle calculated in this chapter (where $T$ here is the temperature defined by the equation of state of an ideal gas) coincides with this universal thermodynamic temperature scale confirms that the definition of temperature using an ideal gas does not contradict the thermodynamic definition of temperature, which is independent of the working substance. This idea will serve as the foundation when we study entropy and the second law of thermodynamics more deeply in the following chapters.

Solution

A state quantity is a quantity whose value is determined solely by the present state (for example, the set of pressure, volume, and temperature), regardless of the path by which the substance arrived at that state (the internal energy that we learned about in Section 23.3 is the representative example). The quantity of heat $Q$ is in general not a state quantity in this sense—as we saw in this chapter, even after one full trip around a Carnot cycle we have $Q_2-Q_1\neq0$; “although the system has returned to its original state after one cycle, the net amount of heat that flowed in and out is not zero.” This means that the value of $Q$ depends on which processes (isothermal or adiabatic, along which path) were followed on the way, and this contradicts the definition of a state quantity.

However, as we confirmed in Section 28.5, when we compare $Q_2/T_2$ and $Q_1/T_1$, the quantities obtained by dividing each quantity of heat by the temperature of the reservoir with which it was exchanged, we get $Q_2/T_2-Q_1/T_1=0$, which returns exactly to zero after one cycle. Although $Q$ by itself depends on the path, with the combination $Q/T$ the changes over one cycle exactly cancel to zero—this strongly suggests the possibility that the quantity $S\equiv Q/T$ is, unlike $Q$, a state quantity determined only by the state (not by the path).

What kind of physical property, then, can we expect this new state quantity $S$ (entropy) to represent? The form $Q/T$ seems to express what fraction of the energy exchanged as heat has become a form that can never again be converted into work (heat that has merely been discarded to a low-temperature reservoir). Entropy is said to be a measure of how “disordered” the molecular motion of a system is (we shall study this in detail from Chapter 29 on). For example, if the system is operated reversibly, as in a Carnot cycle, the balance of $Q/T$ comes out exactly zero, but if irreversible processes (friction or abrupt changes) are mixed in along the way, this balance would break down and be biased toward an increase of $S$—this conjecture can be confirmed by learning the precise definition of entropy and its properties in the next chapter (Chapter 29).

Solution

(1) For a hot reservoir at $T_2=800\ \mathrm{K}$ and a cold reservoir at $T_1=300\ \mathrm{K}$, the Carnot efficiency is

$$ \eta_{\mathrm{Carnot}}=1-\frac{T_1}{T_2}=1-\frac{300}{800}=1-0.375=0.625=62.5\,\% $$

(2) Since the actual thermal efficiency is $\eta=0.40$, its ratio to the Carnot efficiency is

$$ \frac{\eta}{\eta_{\mathrm{Carnot}}}=\frac{0.40}{0.625}=0.64 $$

That is, the actual efficiency is about 64% (a little over six-tenths) of the Carnot efficiency (the theoretical upper bound in this temperature range). Irreversibilities such as friction, heat losses, and heat exchange across temperature differences account for the loss of the remainder.

(3) Let the electric output (work per second) be $W'/t=1.0\times10^{9}\ \mathrm{W}=1.0\times10^{9}\ \mathrm{J/s}$. From $\eta=W'/Q_2$, the heat absorbed per second is

$$ \frac{Q_2}{t}=\frac{W'/t}{\eta}=\frac{1.0\times10^{9}\ \mathrm{J/s}}{0.40}=2.5\times10^{9}\ \mathrm{J/s}=2.5\ \mathrm{GW} $$

From $Q_1=Q_2-W'$, the heat discarded per second to the condenser is

$$ \frac{Q_1}{t}=2.5\times10^{9}-1.0\times10^{9}=1.5\times10^{9}\ \mathrm{J/s}=1.5\ \mathrm{GW} $$

Thus $60\,\%$ of the heat obtained from the fuel is discarded into the sea, rivers, or the atmosphere. (For reference: if the efficiency reached the Carnot efficiency of $62.5\,\%$, the heat required for the same electric output would be only $1.0\times10^{9}/0.625=1.6\times10^{9}\ \mathrm{J/s}$.)

Solution

The thermal efficiency of the Otto cycle is $\eta=1-r^{1-\gamma}$. Substituting $r=10$ and $\gamma=1.4$, we obtain

$$ \eta_{\mathrm{Otto}}=1-10^{-0.4}=1-e^{-0.4\ln10}=1-e^{-0.921}\approx1-0.3981=0.602=60.2\,\% $$

The efficiency of a Carnot cycle operating between a maximum temperature of $2000\ \mathrm{K}$ and a minimum temperature of $300\ \mathrm{K}$ is

$$ \eta_{\mathrm{Carnot}}=1-\frac{300}{2000}=1-0.15=0.85=85\,\% $$

so the efficiency of the Otto cycle is the following fraction of the Carnot efficiency:

$$ \frac{0.602}{0.85}\approx0.708 $$

that is, about 71%. The reason the Otto cycle has a lower efficiency than the Carnot cycle is, as explained in the Discussion in Section 28.6, that the heat is absorbed not only from a reservoir at the maximum temperature $T_A$ but while the temperature rises from $T_D$ to $T_A$ (most of the heat is received at lower temperatures).

Solution

The absolute temperatures are $T_1=-18+273.15=255.15\ \mathrm{K}$ (the inside of the freezer, the cold reservoir) and $T_2=27+273.15=300.15\ \mathrm{K}$ (the room, the hot reservoir).

(1) The coefficient of performance (COP) of the reversed Carnot cycle as a refrigerator is

$$ \mathrm{COP}_{\mathrm R}=\frac{T_1}{T_2-T_1}=\frac{255.15}{300.15-255.15}=\frac{255.15}{45.00}\approx5.67 $$

(2) The heat pumped out of the freezer per second is $Q_1/t=300\ \mathrm{J/s}$. From $\mathrm{COP}_{\mathrm R}=Q_1/W_{\mathrm{in}}$, the required power (the minimum electric power) is

$$ \frac{W_{\mathrm{in}}}{t}=\frac{Q_1/t}{\mathrm{COP}_{\mathrm R}}=\frac{300\ \mathrm{J/s}}{5.67}\approx52.9\ \mathrm{J/s}=52.9\ \mathrm{W} $$

(3) From $Q_2=Q_1+W_{\mathrm{in}}$, the heat released into the room per second is

$$ \frac{Q_2}{t}=300+52.9\approx3.53\times10^{2}\ \mathrm{J/s}=353\ \mathrm{W} $$

(Check: from $Q_2/Q_1=T_2/T_1$ we get $300\times300.15/255.15\approx353\ \mathrm{W}$, which agrees.) Because electric power is used to pump out the heat inside the freezer, more heat is released into the room than the heat carried out of the freezer. In an actual freezer, because of irreversibilities, the required electric power is larger than this ideal value.

Solution

Work in the adiabatic processes. Since $A\to B$ is an adiabatic process, $Q_{AB}=0$, and by the first law $\Delta U=Q-W'$ we have $W'_{AB}=-\Delta U_{AB}$. Since the change in the internal energy of an ideal gas is $\Delta U=C_V\Delta T$,

$$ W'_{AB}=-C_V(T_B-T_A)=C_V(T_A-T_B) $$

Similarly, since $C\to D$ is also an adiabatic process,

$$ W'_{CD}=-C_V(T_D-T_C)=C_V(T_C-T_D)\ (\lt0) $$

In the isochoric processes $D\to A$ and $B\to C$, $\dd V=0$, so the work is $0$. Therefore the net work in one cycle is

$$ W'=W'_{AB}+W'_{CD}=C_V(T_A-T_B)-C_V(T_D-T_C)=C_V(T_A-T_D)-C_V(T_B-T_C) $$

Here, as in Example 28.7, $Q_2=C_V(T_A-T_D)$ and $Q_1=C_V(T_B-T_C)$, so $W'=Q_2-Q_1$, which shows that the net work in one cycle equals $Q_2-Q_1$. (Furthermore, if we use $T_B=xT_A$, $T_C=xT_D$, and $x=r^{1-\gamma}$, then $W'=C_V(1-x)(T_A-T_D)=(1-r^{1-\gamma})Q_2$, and $\eta=W'/Q_2=1-r^{1-\gamma}$.)

Numerical values. $C_V=nR/(\gamma-1)=1.0\times8.314/0.4\approx20.8\ \mathrm{J/K}$. Since $x=8^{-0.4}\approx0.4353$,

$$ T_D=\frac{T_C}{x}=T_C\,r^{\gamma-1}=300\times8^{0.4}\approx300\times2.297\approx689\ \mathrm{K},\qquad T_B=xT_A=0.4353\times2000\approx871\ \mathrm{K} $$

Hence,

$$ W'_{AB}=20.8\times(2000-871)\approx2.35\times10^{4}\ \mathrm{J},\qquad W'_{CD}=20.8\times(300-689)\approx-8.09\times10^{3}\ \mathrm{J} $$ $$ W'=W'_{AB}+W'_{CD}\approx(2.35-0.81)\times10^{4}\ \mathrm{J}\approx1.54\times10^{4}\ \mathrm{J} $$

(Check: $Q_2=20.8\times(2000-689)\approx2.72\times10^{4}\ \mathrm{J}$ and $Q_1=20.8\times(871-300)\approx1.19\times10^{4}\ \mathrm{J}$, so $Q_2-Q_1\approx1.54\times10^{4}\ \mathrm{J}$. This also agrees with $W'/Q_2\approx0.565=1-8^{-0.4}$.)

Solution

We have $n=1.0\ \mathrm{mol}$, $T_2=400\ \mathrm{K}$, $T_1=300\ \mathrm{K}$, $V_B/V_A=3.0/1.0=3.0$, $\ln3.0\approx1.0986$, and $R=8.314\ \mathrm{J/(mol\,K)}$. The heats in the isothermal processes are $Q_2=nRT_2\ln(V_B/V_A)$ and $Q_1=nRT_1\ln(V_B/V_A)$, so

$$ Q_2=1.0\times8.314\times400\times1.0986\approx3.65\times10^{3}\ \mathrm{J},\qquad Q_1=1.0\times8.314\times300\times1.0986\approx2.74\times10^{3}\ \mathrm{J} $$

Dividing these by the temperatures of the reservoirs with which the heat was exchanged, we get

$$ \frac{Q_2}{T_2}=\frac{3.65\times10^{3}}{400}\approx9.13\ \mathrm{J/K},\qquad\frac{Q_1}{T_1}=\frac{2.74\times10^{3}}{300}\approx9.13\ \mathrm{J/K} $$

and indeed they are equal (although $Q_2\neq Q_1$, we have $Q_2/T_2=Q_1/T_1$). In terms of formulas, this common value is $Q_2/T_2=nR\ln(V_B/V_A)$ and $Q_1/T_1=nR\ln(V_B/V_A)$, both of which equal $nR\ln(V_B/V_A)=1.0\times8.314\times1.0986\approx9.13\ \mathrm{J/K}$. The temperatures $T_2,T_1$ cancel and disappear, leaving only the volume ratio $V_B/V_A$—this is the numerical evidence that $Q/T$ can be a path-independent quantity (the entropy).

Solution

(1) Substitute $V_2=V_1/3$ into the result of Chapter 29 for an isothermal reversible change, $\Delta S=nR\ln(V_2/V_1)$:

$$ \Delta S_{\text{gas}}=nR\ln\frac{V_1/3}{V_1}=nR\ln\frac{1}{3}=-nR\ln3 $$

($\ln(1/3)=-\ln3\approx-1.0986$.) For $n=1\ \mathrm{mol}$,

$$ \Delta S_{\text{gas}}=-1\times8.314\times1.0986\ \mathrm{J/K}\approx-9.13\ \mathrm{J/K} $$

which is negative. Compression narrows the range of volume in which the gas molecules can move around, from $V_1$ to $V_1/3$, so the disorder of the molecular configurations decreases and the entropy decreases. This is exactly the opposite of what happened in free expansion, in which the volume expands and $\Delta S\gt0$.

(2) For an isothermal change of an ideal gas, $\Delta U=0$, so $Q=W'$, and $Q=nRT\ln(V_2/V_1)$.

$$ Q=1\times8.314\times300\times\ln\frac13\ \mathrm{J}\approx-2.74\times10^3\ \mathrm{J} $$

Since this is negative, the gas releases about $2.74\ \mathrm{kJ}$ of heat to the heat reservoir (consistent with $\Delta S_{\text{gas}}=Q/T=-2740/300\ \mathrm{J/K}\approx-9.13\ \mathrm{J/K}$).

(3) The heat reservoir receives the heat $2.74\ \mathrm{kJ}$ released by the gas at the temperature $T=300\ \mathrm{K}$, so

$$ \Delta S_{\text{reservoir}}=\frac{+2740\ \mathrm{J}}{300\ \mathrm{K}}\approx+9.13\ \mathrm{J/K},\qquad \Delta S_{\text{total}}=-9.13+9.13=0 $$

Thus, looking at the gas alone, the entropy has decreased, but the increase in the reservoir exactly cancels it. Since the change is reversible, the total is $\Delta S_{\text{total}}=0$, which does not violate the law of entropy increase ($\Delta S\ge0$ for the whole). If the same compression were carried out abruptly (irreversibly), the heat received by the reservoir would be more than $2.74\ \mathrm{kJ}$, and $\Delta S_{\text{total}}\gt0$.

Solution

The entropy change of the adiabatic free expansion found in Section 29.8 (volume $V_i\to V_f$, $V_f\gt V_i$) is

$$ \Delta S_{\text{fwd}}=nR\ln\frac{V_f}{V_i}\gt0 $$

The reverse process we now consider is the change in which, within the same adiabatic container, the gas returns spontaneously from volume $V_f$ to $V_i$ ($V_i\lt V_f$). Since entropy is a state quantity (Theorem 29.2), the entropy change of the reversed change is that of the original change with its sign reversed:

$$ \Delta S_{\text{bwd}}=nR\ln\frac{V_i}{V_f}=-nR\ln\frac{V_f}{V_i}=-\Delta S_{\text{fwd}}\lt0 $$

The whole adiabatic container (gas plus container) is an isolated system, and the law of entropy increase (Law 29.1) states that any change that can occur in an isolated system must satisfy $\Delta S\ge0$. In the reverse process $\Delta S_{\text{bwd}}\lt0$ (the entropy decreases), so this law forbids it. Therefore a change in which the expanded gas spontaneously returns to its original volume does not occur—this is the explanation, in terms of entropy, of why adiabatic free expansion is irreversible (does not proceed in the reverse direction). (In principle, the probability that all the gas molecules happen to gather in one half of the container at the same time is not zero, but for a number of molecules of the order of Avogadro's number that probability is astronomically small, and it is not observed in reality. Treating this “smallness of probability” quantitatively is the task of Boltzmann's relation $S=k_{\mathrm{B}}\ln W$ of statistical mechanics, which we study in Part VII.)

Solution

The entropy changes of the two isothermal processes found in Section 29.5.1 are $\Delta S_{AB}=nR\ln(V_B/V_A)$ and $\Delta S_{CD}=nR\ln(V_D/V_C)$. By the property of logarithms $\ln x+\ln y=\ln(xy)$ (High School Mathematics, Chapter 6), their sum can be combined as

$$ \Delta S_{AB}+\Delta S_{CD}=nR\ln\frac{V_B}{V_A}+nR\ln\frac{V_D}{V_C}=nR\ln\left(\frac{V_B}{V_A}\cdot\frac{V_D}{V_C}\right) $$

Here, using $V_B/V_A=V_C/V_D$, which follows from the relation for adiabatic processes (Poisson's formula), the quantity inside the parentheses becomes

$$ \frac{V_B}{V_A}\cdot\frac{V_D}{V_C}=\frac{V_C}{V_D}\cdot\frac{V_D}{V_C}=1 $$

(If we replace $V_B/V_A$ by $V_C/V_D$, which is known to be equal to it, it becomes exactly the reciprocal of the remaining $V_D/V_C$, so their product is exactly $1$.) Therefore,

$$ \Delta S_{AB}+\Delta S_{CD}=nR\ln1=0 $$

so the two cancel exactly, with opposite signs. Physically, the relation $V_B/V_A=V_C/V_D$ means that the volume ratio of the isothermal expansion on the high-temperature side (by what factor the gas expanded) and the reciprocal of the volume ratio of the isothermal compression on the low-temperature side (by what factor it shrank) become exactly the same ratio through the adiabatic processes in between, and this symmetry of the ratio makes the entropy changes of the two isothermal processes equal in magnitude and opposite in sign.

Solution

The Clausius inequality derived in Section 29.7 was, in general,

$$ \dd S\ge\frac{\dd'Q}{T} $$

Only in the case of an adiabatic system ($\dd'Q=0$) does the right-hand side become $0$, giving the simple form $\dd S\ge0$. For a system that exchanges heat with the outside ($\dd'Q\ne0$), the right-hand side $\dd'Q/T$ is not $0$, so in general the sign of $\dd S$ is not determined. In particular, consider the case where the system releases heat ($\dd'Q\lt0$): the right-hand side can be negative, so the inequality $\dd S\ge\dd'Q/T$ does not forbid $\dd S$ from being negative—that is, the entropy of that system alone decreasing.

As a concrete example, consider water freezing in a refrigerator. When water becomes ice, the water molecules line up in a more orderly crystal structure, so the entropy of the water (this system) decreases ($\Delta S_{\text{water}}\lt0$). Looking at the water alone, it seems that “the entropy is decreasing,” but the heat released by the water (the heat of solidification) is carried through the cooling apparatus of the refrigerator on to the outside air, where it increases the entropy of the environment by at least $|\Delta S_{\text{water}}|$. The total entropy summed over the water, the refrigerator, and the outside air satisfies

$$ \Delta S_{\text{total}}=\Delta S_{\text{water}}+\Delta S_{\text{env}}\ge0 $$

so there is no contradiction with Law 29.1 (the total is $=0$ for a reversible change and $\gt0$ for an irreversible one). That is, the law of entropy increase does not hold for every subsystem; it is a law that holds only for the whole that includes all the exchanges with the outside (which can be regarded as an adiabatic system). The entropy of a subsystem alone can be decreased from the outside through the flow of work and heat into and out of it (the refrigerator and the isothermal compression of Exercise 29.1 are examples)—but as the price, the entropy of the outside of the system (the environment) increases by just enough to make up for the system's decrease (exactly enough if the change is reversible, more than enough if it is irreversible).

Solution

A gas thermometer defines temperature on the basis of a property of a specific substance (a gas for which the ideal-gas approximation holds), namely the equation of state of an ideal gas $pV=nRT$ (or the Boyle–Charles law). Therefore, strictly speaking, the temperature value given by a gas thermometer may depend on “which gas was used” and on “how far the deviation from the ideal gas (such as the van der Waals correction) was corrected.”

In contrast, Theorem 29.1 (Carnot's theorem) guarantees that the efficiency $\eta=1-T_1/T_2$ of a reversible engine operating between two heat reservoirs takes exactly the same value whatever the working substance is (an ideal gas, a real gas, or any other substance). Therefore, once the triple point of water is fixed as the reference $T_2=273.16\ \mathrm{K}$, we need only operate a reversible engine (any working substance may be used) between it and a reservoir at an unknown low temperature $T_1$ and measure its efficiency $\eta$, and $T_1$ is determined uniquely from $T_1=(1-\eta)T_2$. The temperature defined by this method is a purely thermodynamic temperature (based only on the second law) that does not depend at all on the choice of the working substance, and it is called the thermodynamic temperature (absolute temperature).

At the triple point, the temperature and pressure at which the three phases ice, water, and water vapor can coexist are fixed to a single state ($273.16\ \mathrm{K}$, about $611.657\ \mathrm{Pa}$), so the same temperature can be reproduced accurately any number of times without adjusting the external pressure. This is its advantage as a reference point. The gas thermometer is widely used because in practice it approximates this thermodynamic temperature well (and becomes more accurate the closer it is brought to the ideal-gas limit), but as a definition in principle, the significance lies in the fact that the method based on Carnot's theorem gives a more fundamental definition of temperature, one that does not rely on the properties of a particular substance.

Solution

Whether an infinitesimal quantity $\dd X$ is an “infinitesimal change of a state quantity (a total differential)” is determined by whether, “when $\dd X$ is integrated along different paths with the initial and final states held fixed, the value is the same for every path.” If it is independent of the path, $X$ is a state quantity determined by the “present state” alone; if it depends on the path, $X$ is not a state quantity. For the two variables $(T,V)$, the condition for $\dd X=M\,\dd T+N\,\dd V$ to be a total differential follows from the fact that the second-order partial derivatives do not depend on the order of differentiation (Schwarz's theorem):

$$ \frac{\partial M}{\partial V}=\frac{\partial N}{\partial T} $$

This is the condition.

For an ideal gas ($n\ \mathrm{mol}$, $C_V$ constant), the infinitesimal quantity of heat in a reversible change is

$$ \dd'Q=C_V\,\dd T+\frac{nRT}{V}\,\dd V $$

Setting $M=C_V$ and $N=nRT/V$, we have

$$ \frac{\partial M}{\partial V}=0,\qquad\frac{\partial N}{\partial T}=\frac{nR}{V}\ne0 $$

which do not agree, so $\dd'Q$ is not a total differential and $Q$ is not a state quantity (this corresponds to the fact that in Example 29.4 the heats absorbed along the two paths differed, $5470\ \mathrm{J}$ and $7199\ \mathrm{J}$).

However, if we consider $\dd'Q/T$, obtained by multiplying this same $\dd'Q$ by $1/T$, we have

$$ \frac{\dd'Q}{T}=\frac{C_V}{T}\dd T+\frac{nR}{V}\dd V,\qquad \frac{\partial}{\partial V}\!\left(\frac{C_V}{T}\right)=0=\frac{\partial}{\partial T}\!\left(\frac{nR}{V}\right) $$

and the condition for a total differential holds. That is, $\dd'Q$ takes a value that changes with the path (it is not a state quantity), but $\dd'Q/T$, obtained by multiplying by $1/T$, becomes independent of the path (it becomes the infinitesimal change $\dd S$ of the state quantity $S$). A factor that, when multiplied by an infinitesimal quantity that is not a total differential, turns it into a total differential is called an integrating factor, and $1/T$ acts as an integrating factor of $\dd'Q$.

This is not a contradiction; it is merely that “the quantity $Q$ itself” and “the quantity obtained by multiplying $Q$ by $1/T$ and integrating ($S$)” are different quantities to begin with. The discussions in Sections 29.1 and 29.5 showed that $\oint\dd'Q/T=0$ holds likewise for a general working substance and a general reversible change. The general theory of integrating factors is studied in University Mathematics, Section 34.4 (Integrating Factors).

Solution

(1) Since the change is isothermal, use the formula $\Delta S=R\ln[(V_2-b)/(V_1-b)]$ of Chapter 29. With $V_1=0.50\times10^{-3}\ \mathrm{m^3}$, $V_2=1.0\times10^{-3}\ \mathrm{m^3}$, and $b=4.27\times10^{-5}\ \mathrm{m^3}$,

$$ \frac{V_2-b}{V_1-b}=\frac{1.0\times10^{-3}-4.27\times10^{-5}}{0.50\times10^{-3}-4.27\times10^{-5}}=\frac{9.573\times10^{-4}}{4.573\times10^{-4}}\approx2.0934 $$ $$ \Delta S=8.314\times\ln2.0934\ \mathrm{J/K}\approx8.314\times0.7388\ \mathrm{J/K}\approx6.14\ \mathrm{J/K} $$

The ideal-gas value is $R\ln2=8.314\times0.6931\ \mathrm{J/K}\approx5.76\ \mathrm{J/K}$, so the deviation is $(6.14-5.76)/5.76\approx0.066$, that is, about $6.6\,\%$ larger.

(2) For $1.0\ \mathrm{L}\to2.0\ \mathrm{L}$ (Example 29.5), $\Delta S\approx5.95\ \mathrm{J/K}$, which was only about $3.2\,\%$ larger than the ideal-gas value. Therefore the effect of $b$ is larger for $0.50\ \mathrm{L}\to1.0\ \mathrm{L}$. The reason is that $b$ is a constant independent of the volume, so the smaller the volume (the larger the ratio $b/V$ of $b$ to $V$), the larger the reduction of the volume $V-b$ in which the gas can actually move around, and the volume ratio $(V_2-b)/(V_1-b)$ deviates from 2. In the limit of sufficiently large volume, $b/V\to0$ and the result approaches that of the ideal gas. Note that the attraction constant $a$ does not appear in $S$, so it has no effect on $\Delta S$ in either case.

Solution

(1) When water of mass $m=1000\ \mathrm{g}$ and specific heat $c=4.19\ \mathrm{J/(g\,K)}$ is heated reversibly, from $\dd'Q_{\text{rev}}=mc\,\dd T$ we obtain

$$ \Delta S=\int_{T_1}^{T_2}\frac{mc\,\dd T}{T}=mc\ln\frac{T_2}{T_1}=1000\times4.19\times\ln\frac{373.15}{293.15}\ \mathrm{J/K}\approx4190\times0.2415\ \mathrm{J/K}\approx1.01\times10^3\ \mathrm{J/K} $$

(2) The final temperature $T_f$ is determined by conservation of heat. The heat required for $100\ \mathrm{g}$ of ice to melt and warm up to $T_f$ is $100\times334+100\times4.19\times(T_f-273.15)$ [J], and the heat released by $1000\ \mathrm{g}$ of water cooling from $293.15\ \mathrm{K}$ to $T_f$ is $1000\times4.19\times(293.15-T_f)$ [J]; since the two are equal,

$$ 33400+419(T_f-273.15)=4190(293.15-T_f)\qquad\therefore\ T_f\approx284.09\ \mathrm{K}\ (\approx10.9\,^{\circ}\mathrm{C}) $$

Since the ice melts completely ($T_f\gt273.15\ \mathrm{K}$), the assumption is not contradicted. Because entropy is a state quantity, we calculate the entropy change by changing each part reversibly.

• Melting of the ice ($334\ \mathrm{J/g}$ at $273.15\ \mathrm{K}$): $\Delta S_1=\dfrac{100\times334}{273.15}\ \mathrm{J/K}\approx122.3\ \mathrm{J/K}$
• Heating of the melted water ($273.15\to284.09\ \mathrm{K}$): $\Delta S_2=419\times\ln\dfrac{284.09}{273.15}\ \mathrm{J/K}\approx419\times0.0393\ \mathrm{J/K}\approx16.4\ \mathrm{J/K}$
• Cooling of the original water ($293.15\to284.09\ \mathrm{K}$): $\Delta S_3=4190\times\ln\dfrac{284.09}{293.15}\ \mathrm{J/K}\approx4190\times(-0.0314)\ \mathrm{J/K}\approx-131.6\ \mathrm{J/K}$

The total is $\Delta S=122.3+16.4-131.6\approx+7.1\ \mathrm{J/K}\gt0$. Since the change takes place inside an adiabatic container, the fact that the total entropy increases shows that this change is irreversible and proceeds spontaneously in this direction (a change in which ice melts into water and heat is transferred proceeds naturally, but the reverse does not).

Solution

(1) The hot body ($T_2=350\ \mathrm{K}$) loses heat $Q=500\ \mathrm{J}$, and the cold body ($T_1=300\ \mathrm{K}$) receives it. Assuming that the temperature of each can be regarded as constant (the heat capacities being large compared with the quantity of heat),

$$ \Delta S_2=\frac{-Q}{T_2}=\frac{-500\ \mathrm{J}}{350\ \mathrm{K}}\approx-1.43\ \mathrm{J/K},\qquad \Delta S_1=\frac{Q}{T_1}=\frac{500\ \mathrm{J}}{300\ \mathrm{K}}\approx+1.67\ \mathrm{J/K} $$ $$ \Delta S_{\text{total}}=\Delta S_2+\Delta S_1\approx-1.4286+1.6667\approx+0.238\ \mathrm{J/K}\ (\gt0) $$

(2) Suppose that the same quantity of heat were transferred in the opposite direction (from the body at $300\ \mathrm{K}$ to the body at $350\ \mathrm{K}$). Every entropy change would then have the opposite sign, so $\Delta S_{\text{total}}\approx-0.238\ \mathrm{J/K}\lt0$. The two bodies taken together can be regarded as an isolated system if they are cut off from the outside by thermal insulation, so this contradicts the law of entropy increase for an isolated system, $\dd S\ge0$. Therefore, heat transfer in this direction does not occur spontaneously.

(3) In general, let $\dd'Q$ be the quantity of heat transferred from the hot body to the cold body. Then the total entropy change is $\dd S_{\text{total}}=\dd'Q\,(1/T_1-1/T_2)$, and since $T_2\gt T_1\gt0$, the quantity in parentheses is positive. For $\dd S_{\text{total}}\ge0$ we need $\dd'Q\ge0$; that is, heat is transferred only from hot to cold. This corresponds exactly to the Clausius statement of the second law of thermodynamics, “Heat does not pass spontaneously from a colder body to a hotter body.” The law of entropy increase is a unified, quantitative formulation of the various statements of the second law (the Clausius statement, the Thomson–Kelvin statement, and others), and when it is applied to the concrete phenomenon of the “direction of heat transfer,” the Clausius statement itself is obtained.

Solution

The vertical axis $C_p/T$ of the $C_p/T$–$T$ graph represents the rate of change of the entropy when the temperature is changed slightly at constant pressure, $(\partial S/\partial T)_p=C_p/T$. We reason in the following three steps, in this order.

(1) Inside each phase (whichever of solid, liquid, or gas), $C_p/T$ describes how the entropy grows as the temperature is raised gradually, and it varies continuously with temperature.

(2) Before and after the melting point $T_{\mathrm{fus}}$ and the boiling point $T_b$, the manner of thermal motion of the molecules (the degrees of freedom) differs from phase to phase, so the value of $C_p$ differs from phase to phase. As a result, the $C_p/T$ graph shows a step connecting the values of the phases on the two sides of each transition point.

(3) At the transition point itself, a finite amount of heat, the heat of fusion or heat of vaporization $\Delta_{\mathrm{trs}}H$, flows in or out while the temperature stays unchanged. Since the heat capacity is “heat ÷ temperature change,” if the temperature change is $0$ while the heat is finite, then $C_p$ is effectively infinite (a $\delta$-function-like divergence), and a thin peak rises in the $C_p/T$ graph. The area of this peak is $\displaystyle\int\frac{C_p}{T}\,\dd T=\frac{\Delta_{\mathrm{trs}}H}{T_{\mathrm{trs}}}=\Delta_{\mathrm{trs}}S$ (Section 30.2). The $C_p/T$ graph in Figure 30.3 draws this peak as a thin line and draws the difference between the values of the phases on the two sides (the step).

In the $S$–$T$ graph, the area under the $C_p/T$ curve gives the increment of $S$, so in the regions of (1), $S$ increases smoothly, and the area of the peak in (3) appears directly as the stepwise jump $\Delta_{\mathrm{trs}}S$ of $S$ at $T_{\mathrm{trs}}$. The “step” in the $C_p/T$ graph represents the difference between the heat capacities of the phases, and the “jump” in the $S$ graph represents the effect of the latent heat itself; the same phase transition thus appears in different forms in the two graphs.

Solution

(1) Heating stage: Since $C_{p,\mathrm{m}}$ is constant, the integral in the equation of Chapter 30 (the entropy change accompanying a temperature change at constant pressure) can be carried out, and

$$ \Delta S_1=\int_{298.15}^{373.15}\frac{C_{p,\mathrm{m}}}{T}\,\dd T=C_{p,\mathrm{m}}\ln\frac{373.15}{298.15}=75.3\times\ln(1.2516)\approx75.3\times0.2244\approx16.9\ \mathrm{J/(K\cdot mol)} $$

(2) Vaporization stage: This is a phase transition at constant temperature ($373.15\ \mathrm{K}$) and constant pressure, so from $\Delta S=\Delta H/T$,

$$ \Delta S_2=\frac{\Delta_{\mathrm{vap}}H}{T_b}=\frac{40660\ \mathrm{J/mol}}{373.15\ \mathrm{K}}\approx109.0\ \mathrm{J/(K\cdot mol)} $$

(3) Total: Since entropy is a state quantity, the change for the whole process is simply the sum of the changes in the two stages:

$$ \Delta S=\Delta S_1+\Delta S_2\approx16.9+109.0\approx125.9\ \mathrm{J/(K\cdot mol)} $$

The vaporization stage is far larger than the heating stage. This is because, once the substance becomes a gas, the molecules can fly about freely and the disorder increases all at once; it is also consistent with Trouton's rule (Section 30.2).

Solution

We retrace the derivation in Section 30.4.1, separating it according to the conditions.

Where the condition “constant volume, no non-expansion work” was used: Since the volume does not change ($\dd V=0$) and there is no non-expansion work, the work done on the system is $\dd'W=-p\,\dd V+\dd'W_{\text{non-expansion}}=0$. From the first law of thermodynamics, $\dd U=\dd'Q+\dd'W$, we get $\dd U=\dd'Q$. Substituting this into the Clausius inequality $\dd S\ge\dd'Q/T$ gives $\dd S\ge\dd U/T$, that is,

$$ \dd U-T\,\dd S\le0 $$

This inequality was derived using only the condition “constant volume, no non-expansion work”; it is a general inequality that does not yet use the condition of constant temperature.

Where the condition “constant temperature” was used: Differentiating the definition of the Helmholtz free energy, $A=U-TS$, gives in general, by the product rule, $\dd A=\dd U-T\,\dd S-S\,\dd T$. If we now add the condition that the temperature is constant ($\dd T=0$), the third term vanishes and $\dd A=\dd U-T\,\dd S$.

Combining the two: The left-hand side of the inequality $\dd U-T\,\dd S\le0$ obtained from “constant volume, no non-expansion work” is, under the condition of “constant temperature,” exactly equal to $\dd A$. Therefore we obtain

$$ \dd A\le0 $$

In other words, $A$ serves as the criterion for spontaneous change only when all three conditions hold simultaneously: constant volume, constant temperature, and no non-expansion work. “Constant volume, no non-expansion work” plays the role of rewriting the Clausius inequality as an inequality in the variables of the system alone, and “constant temperature” plays the role of making $\dd U-T\,\dd S$ coincide with the total differential of $A$ itself.

Solution

(1) In quasi-static (reversible) isothermal compression, $p_{\mathrm{ex}}=p=nRT/V$. With $V_2=V_1/2$, the work done on the system is

$$ W=-\int_{V_1}^{V_1/2}p\,\dd V=-nRT\ln\frac{V_1/2}{V_1}=nRT\ln2=1\times8.314\times300\times0.6931\approx+1729\ \mathrm{J} $$

On the other hand, for an isothermal process of an ideal gas, $\Delta U=0$ and $\Delta S=nR\ln(V_2/V_1)=-nR\ln2$, so

$$ \Delta A=\Delta U-T\Delta S=+nRT\ln2\approx+1729\ \mathrm{J} $$

Since the change is reversible, $W=\Delta A$ (the equality in Law 30.4) holds.

(2) Let the initial pressure be $p_1=nRT/V_1$. Then the final pressure is $p_2=nRT/(V_1/2)=2p_1$, and “compressing all at once” means setting the external pressure to $p_{\mathrm{ex}}=2p_1$ from the start. Since the external pressure is constant, the work done on the system is

$$ W=-p_{\mathrm{ex}}(V_2-V_1)=-2p_1\left(\frac{V_1}{2}-V_1\right)=p_1V_1=nRT=8.314\times300\approx2494\ \mathrm{J} $$

The initial and final states are the same as in (1), so, since $A$ is a state quantity, $\Delta A$ is also the same, $+1729\ \mathrm{J}$, and $W=2494\ \mathrm{J}\ge\Delta A=1729\ \mathrm{J}$ holds. In an irreversible compression, more work is required than in the reversible case (equality holds only for a reversible change).

(3) For an ideal gas $\Delta U=0$, so by the first law $Q=-W=-2494\ \mathrm{J}$; that is, the system releases $2494\ \mathrm{J}$ of heat to the constant-temperature bath. The entropy change of the system is $\Delta S_{\text{sys}}=-nR\ln2\approx-5.76\ \mathrm{J/K}$, the entropy change of the constant-temperature bath (the surroundings) is $2494/300\approx+8.31\ \mathrm{J/K}$, and the total is

$$ \Delta S_{\text{total}}\approx-5.76+8.31\approx+2.55\ \mathrm{J/K}\ (\gt0) $$

Thus we can confirm that the entropy increases in an irreversible change. This also agrees with $(W-\Delta A)/T=(2494-1729)/300\approx2.55\ \mathrm{J/K}$. The extra work added compared with the reversible case corresponds to the amount dissipated as heat into the constant-temperature bath.

Solution

The relation $(\partial G/\partial T)_p=-S$ is a general relation obtained as follows. Substituting the first law for a quasi-static change, $\dd U=T\,\dd S-p\,\dd V$, into the total differential of $G=U+pV-TS$ gives $\dd G=-S\,\dd T+V\,\dd p$ (Section 30.6), and comparing coefficients with the total differential of $G(T,p)$, $\left(\partial G/\partial T\right)_p\dd T+\left(\partial G/\partial p\right)_T\dd p$, yields the relation. It does not depend on the kind of substance, nor on whether the substance is in the solid, liquid, or gas phase. (If we look at the definition $G=H-TS$ only at constant temperature, then $\dd T=0$, and the coefficient of $S$ cannot be read off.)

As in Section 30.3, the entropy $S$ on the right-hand side is expressed as $S(T)=\int_0^T(C_p/T)\,\dd T$ (with $S(0)=0$), and since the heat capacity $C_p\gt0$, we always have $S\gt0$ for $T\gt0$. Therefore $(\partial G/\partial T)_p=-S\lt0$; that is, when the temperature is raised at constant pressure, $G$ necessarily decreases.

If we compare the entropies of gas, liquid, and solid for the same substance at the same temperature, the gas, in which the molecules can fly about freely, has the largest, and the solid, in which the molecules are fixed in a regular lattice, has the smallest ($S_{\text{gas}}\gt S_{\text{liquid}}\gt S_{\text{solid}}$). For example, in the vaporization of water, $S$ increases by as much as $\Delta_{\mathrm{vap}}S\approx109\ \mathrm{J/(K\cdot mol)}$ (Section 30.2). Since the magnitude of $(\partial G/\partial T)_p=-S$ is determined by the magnitude of $S$, the decrease of $G$ with rising temperature is steepest for the gas and gentlest for the solid. The more degrees of freedom of molecular motion a phase has, the more easily it “earns entropy,” and the more $G$ can be lowered as the temperature rises. As a result, the gas is selected as the phase with the lowest $G$ at high temperatures, and the solid at low temperatures.

Solution

(1) The electrical work that can be extracted when the cell is operated reversibly (the maximum non-expansion work) is $nFE^\circ$, and this equals $-\Delta G^\circ$ (Law 30.5). Therefore,

$$ \Delta G^\circ=-nFE^\circ=-2\times96485\ \mathrm{C/mol}\times1.10\ \mathrm{V}\approx-2.12\times10^{5}\ \mathrm{J/mol}=-212\ \mathrm{kJ/mol} $$

(since $\mathrm{C\cdot V}=\mathrm{J}$, the unit is $\mathrm{J/mol}$).

(2) The amount of substance in $1.00\ \mathrm{g}$ of zinc is $1.00/65.38\approx1.53\times10^{-2}\ \mathrm{mol}$, so the maximum electrical work that can be extracted is

$$ -\Delta G^\circ\times1.53\times10^{-2}\ \mathrm{mol}\approx2.123\times10^{5}\times1.53\times10^{-2}\approx3.25\times10^{3}\ \mathrm{J}=3.25\ \mathrm{kJ} $$

In an actual cell, because of irreversibilities such as internal resistance, the work that can be extracted is smaller than this.

Solution

Since the process is at constant pressure and constant temperature (with no non-expansion work), the reaction proceeds spontaneously when $\Delta G\lt0$ (Law 30.3). Assuming that $\Delta H,\Delta S$ do not depend on temperature,

$$ \Delta G=\Delta H-T\Delta S=-50000\ \mathrm{J/mol}-T\times(-100\ \mathrm{J/(K\cdot mol)})=-50000+100\,T\ \ (\mathrm{J/mol}) $$

The condition $\Delta G\lt0$ is $100\,T\lt50000$, that is, $T\lt500\ \mathrm{K}$. At $T=500\ \mathrm{K}$, $\Delta G=0$ and the system is at equilibrium (the forward and reverse reactions balance); for $T\gt500\ \mathrm{K}$, $\Delta G\gt0$ and the reverse reaction becomes spontaneous. Hence the temperature range in which the reaction proceeds spontaneously is $0\lt T\lt500\ \mathrm{K}$. For example, at $T=300\ \mathrm{K}$ we have $\Delta G=-20\ \mathrm{kJ/mol}$, and at $T=600\ \mathrm{K}$ it becomes $+10\ \mathrm{kJ/mol}$.

Looking at the roles of the two terms of $\Delta G=\Delta H-T\Delta S$: the term with $\Delta H\lt0$ (exothermic) always favors the reaction, but when $\Delta S\lt0$ (a reaction in which the arrangement of the molecules loses disorder), $-T\Delta S=+100\,T$ is positive and grows larger at higher temperatures, which disfavors the reaction. At high temperatures the term for the loss of disorder outweighs the benefit of the heat release, so even with $\Delta H\lt0$ the reaction does not proceed. This can also be understood from the viewpoint of Section 30.4 (whether the decrease in the entropy of the system can be compensated by the increase in the entropy of the surroundings, $-\Delta H/T$, caused by the heat $-\Delta H$ released to the surroundings). The higher the temperature, the smaller the increase in the entropy of the surroundings, $-\Delta H/T$, for the same heat received.

Solution

We regard the Gibbs energy (Gibbs free energy) $G=H-TS$ as a function of the pressure $p$ and the temperature $T$. As verified in Section 31.4 of this chapter, its total differential is

$$ \dd G=V\,\dd p-S\,\dd T $$

On the other hand, the general formula for the total differential of $G=G(p,T)$ is

$$ \dd G=\left(\frac{\partial G}{\partial p}\right)_T\dd p+\left(\frac{\partial G}{\partial T}\right)_p\dd T $$

and since $\dd p,\dd T$ are infinitesimal quantities that can be varied independently, comparing the coefficients gives

$$ V=\left(\frac{\partial G}{\partial p}\right)_T,\qquad -S=\left(\frac{\partial G}{\partial T}\right)_p $$

Differentiating the first equation partially with respect to $T$ (holding $p$ constant), we obtain

$$ \left(\frac{\partial V}{\partial T}\right)_p=\frac{\partial^2G}{\partial T\,\partial p} $$

Differentiating the second equation partially with respect to $p$ (holding $T$ constant), we obtain

$$ -\left(\frac{\partial S}{\partial p}\right)_T=\frac{\partial^2G}{\partial p\,\partial T} $$

Since $G$, as a state quantity, is a smooth function (one that can be partially differentiated any number of times), Schwarz's theorem gives $\dfrac{\partial^2G}{\partial T\,\partial p}=\dfrac{\partial^2G}{\partial p\,\partial T}$. Therefore the right-hand sides are equal, and so the left-hand sides are equal as well:

$$ \left(\frac{\partial V}{\partial T}\right)_p=-\left(\frac{\partial S}{\partial p}\right)_T $$

This is Maxwell relation (IV) of Chapter 31, and we have obtained exactly the same result as in the derivation in the main text (Section 31.4).

Solution

We substitute the equation of state $p=\dfrac{nRT}{V-nb}$ into the thermodynamic equation of state of the main text,

$$ \pi_T=\left(\frac{\partial U}{\partial V}\right)_T=T\left(\frac{\partial p}{\partial T}\right)_V-p $$

First, to find $\left(\dfrac{\partial p}{\partial T}\right)_V$ we only need to differentiate partially with respect to $T$ while holding $V,n,b$ constant, so

$$ \left(\frac{\partial p}{\partial T}\right)_V=\frac{nR}{V-nb} $$

Substituting this into the expression for $\pi_T$, we get

$$ \pi_T=T\cdot\frac{nR}{V-nb}-p=\frac{nRT}{V-nb}-\frac{nRT}{V-nb}=0 $$

The two terms on the right-hand side cancel exactly, and we obtain $\pi_T=0$.

This result is exactly the same “zero internal pressure” as for an ideal gas. Physically, since this is a model with $a=0$ (neglecting intermolecular forces), no work is needed to pull the molecules apart, and therefore the internal energy does not change when the volume is changed isothermally. On the other hand, the internal pressure of the van der Waals gas found in Section 31.6 of this chapter, $\pi_T=an^2/V^2$, arises precisely from the attraction term $a$ alone—that is, this exercise shows that the cause of $\pi_T\neq0$ is “intermolecular forces (attraction),” not “the size of the molecules themselves (excluded volume).” In the equation of state, the effect of the excluded volume merely replaces the volume by $V-nb$, so the structure is such that the effect of $b$ cancels out cleanly upon partial differentiation with respect to $T$.

Solution

(a) The entropy of a photon gas is $S=\dfrac43aVT^3$ (here $a$ is the radiation constant, and the entropy is a function of $V,T$). Differentiating this partially with respect to $V$ (holding $T$ constant), we obtain

$$ \left(\frac{\partial S}{\partial V}\right)_T=\frac43aT^3 $$

Substituting this into Maxwell relation (III), $\left(\dfrac{\partial p}{\partial T}\right)_V=\left(\dfrac{\partial S}{\partial V}\right)_T$, gives

$$ \left(\frac{\partial p}{\partial T}\right)_V=\frac43aT^3 $$

Integrating the pressure $p$ with respect to $T$ while holding $V$ constant, we obtain

$$ p=\int\frac43aT^3\,\dd T=\frac43a\cdot\frac{T^4}{4}+f(V)=\frac13aT^4+f(V) $$

($f(V)$ is the constant of integration—but since the integral is taken with $V$ held fixed, it is a “constant” that may depend on $V$ but not on $T$). Now, as a physical boundary condition, at absolute zero $T\to0$ no photons (thermal radiation) exist, so the pressure must also be $0$: $p(V,0)=0$. Substituting this into the equation above gives $0=0+f(V)$, hence $f(V)=0$ (for every $V$). Therefore

$$ p=\frac13aT^4 $$

The fact that the pressure is determined by the temperature alone, independently of the volume $V$ ($\left(\partial p/\partial V\right)_T=0$), is also a property peculiar to the photon gas, in contrast to the ideal gas and the van der Waals gas seen in the main text (in both of which $p$ depends on $V$)—because a photon gas is, in place of “molecules,” a collection of electromagnetic waves with a fixed energy density independent of the size of the box.

(b) Substituting $p=\dfrac13aT^4$ obtained in (a) and $\left(\dfrac{\partial p}{\partial T}\right)_V=\dfrac43aT^3$ into the thermodynamic equation of state (Theorem 31.5), $\left(\dfrac{\partial U}{\partial V}\right)_T=T\left(\dfrac{\partial p}{\partial T}\right)_V-p$, we obtain

$$ \left(\frac{\partial U}{\partial V}\right)_T=T\cdot\frac43aT^3-\frac13aT^4=aT^4 $$

Integrating this with respect to $V$ while holding $T$ constant gives $U=aT^4V+f(T)$ ($f(T)$ is a constant of integration independent of $V$). In a photon gas the photons disappear and $U\to0$ as $V\to0$, so $f(T)=0$, and therefore

$$ U=aVT^4 $$

(energy density $u=U/V=aT^4$. Unit check: $\mathrm{J\,m^{-3}\,K^{-4}}\times\mathrm{m^3}\times\mathrm K^4=\mathrm J$.) As a check, using $S=\dfrac43aVT^3$ and $p=\dfrac13aT^4$ we have $T\,\dd S-p\,\dd V=\dfrac43aT^4\dd V+4aVT^3\dd T-\dfrac13aT^4\dd V=aT^4\dd V+4aVT^3\dd T$, which agrees with $\dd(aVT^4)$.

Solution

We solve the equation of state of an ideal gas, $pV=nRT$, for each variable in turn.

(i) Differentiating $p=nRT/V$ partially with respect to $T$ (holding $V$ constant), we obtain

$$ \left(\frac{\partial p}{\partial T}\right)_V=\frac{nR}{V} $$

(ii) Differentiating $T=pV/(nR)$ partially with respect to $V$ (holding $p$ constant), we obtain

$$ \left(\frac{\partial T}{\partial V}\right)_p=\frac{p}{nR} $$

(iii) Differentiating $V=nRT/p$ partially with respect to $p$ (holding $T$ constant), we obtain

$$ \left(\frac{\partial V}{\partial p}\right)_T=-\frac{nRT}{p^2} $$

Multiplying these three together, we get

$$ \left(\frac{\partial p}{\partial T}\right)_V\left(\frac{\partial T}{\partial V}\right)_p\left(\frac{\partial V}{\partial p}\right)_T =\frac{nR}{V}\times\frac{p}{nR}\times\left(-\frac{nRT}{p^2}\right) =\frac{p}{V}\times\left(-\frac{nRT}{p^2}\right)=-\frac{nRT}{pV} $$

Since the equation of state $pV=nRT$ gives $nRT/(pV)=1$, the product is $-1$. We have thus confirmed that the cyclic relation (triple product rule) $\left(\dfrac{\partial p}{\partial T}\right)_V\left(\dfrac{\partial T}{\partial V}\right)_p\left(\dfrac{\partial V}{\partial p}\right)_T=-1$ indeed holds.

Solution

As we saw in Section 31.8.2 of this chapter, $\left(\dfrac{\partial^2A}{\partial V^2}\right)_T=\dfrac{1}{\kappa_TV}$. If $\kappa_T\lt0$ were to hold over some range of volumes, this second partial derivative would be negative; that is, $A(T,V)$ (as a function of $V$ at constant temperature $T$) would be concave down over that range.

On the graph of a concave-down function, every point in the interior of the region is a candidate for a “local maximum,” not a “local minimum.” However, as we confirmed in Section 31.8.2, a stable equilibrium state at constant volume and constant temperature must be a state in which the Helmholtz energy $A$ is at a local minimum (Law 30.2: the system changes spontaneously in the direction $\dd A\le0$). On an interval where $A(V)$ is concave down, no point can be a local minimum, so the system cannot remain there—with only a slight change of volume, it moves spontaneously in the direction in which $A$ can be lowered still further (toward a neighboring, more stable state).

Specifically, in a region with $\kappa_T\lt0$ we have $\left(\partial p/\partial V\right)_T\gt0$ (the pressure rises when the volume is increased), so a part whose volume has become slightly larger acquires a pressure higher than its surroundings, pushes the surroundings back, and tries to expand further—conversely, a part whose volume has become smaller has its pressure lowered, is pushed by the surroundings, and tries to contract further. Fluctuations in either direction are amplified (positive feedback), so a uniform state cannot be maintained, and in the end the system separates into a part with large volume (low density = gas) and a part with small volume (high density = liquid). Therefore the region with $\kappa_T\lt0$ is not physically realized, and real substances avoid that region and take a state separated into two phases (gas and liquid), that is, gas–liquid coexistence.

Solution

According to Definition 31.1 (natural variables), natural variables are a pair of independent variables $(x,y)$ such that one can write $\dd\Phi=X\,\dd x+Y\,\dd y$ and, moreover, the coefficients $X,Y$ are basic quantities, each being one of $T,S,p,V$ (up to sign). As we derived in Section 31.3 of this chapter, the total differential of the Helmholtz energy is

$$ \dd A=-p\,\dd V-S\,\dd T $$

and the coefficients of $\dd V$ and $\dd T$ are the basic quantities $-p$ and $-S$, respectively. Therefore the natural variables of $A$ are $(T,V)$. In this case $p=-\left(\dfrac{\partial A}{\partial V}\right)_T$ and $S=-\left(\dfrac{\partial A}{\partial T}\right)_V$ are obtained directly as first partial derivatives of $A(T,V)$, and $U=A+TS$, $H=U+pV$, $G=A+pV$ can also be found by differentiation and addition alone, so knowing the single function $A(T,V)$ gives all the thermodynamic information about the system.

By contrast, if we forcibly write $A$ as a function of $(T,p)$, substituting $\dd V=\left(\dfrac{\partial V}{\partial T}\right)_p\dd T+\left(\dfrac{\partial V}{\partial p}\right)_T\dd p$ gives

$$ \dd A=-\Bigl(S+p\left(\frac{\partial V}{\partial T}\right)_p\Bigr)\dd T-p\left(\frac{\partial V}{\partial p}\right)_T\dd p $$

The coefficients $S+p\left(\dfrac{\partial V}{\partial T}\right)_p$ and $-p\left(\dfrac{\partial V}{\partial p}\right)_T$ are complicated quantities different from any of $T,S,p,V$, and $S,p$ cannot be read off directly from the first partial derivatives. In other words, $(T,p)$ are not natural variables of $A$.

When we want a potential whose natural variables are $(T,p)$, we perform the Legendre transformation (Definition 31.2, Table 31.2) that replaces the independent variable $V$ of $A(T,V)$ by $p$, the conjugate variable $-p$ with its sign changed. Subtracting (conjugate variable) $\times$ (variable) from $A$ gives $A-\left(\dfrac{\partial A}{\partial V}\right)_TV=A+pV$, and $A+pV=U-TS+pV=H-TS=G$, that is, the Gibbs energy. Thus the correct answer to the demand “I want to treat $A$ as a function of $(T,p)$” is not to “forcibly make $A$ a function of $(T,p)$,” but to “switch to another state quantity $G$ whose natural variables are $(T,p)$.” Since the Helmholtz energy and the Gibbs energy have different natural variables, both are needed.

Solution

Using Formula 31.5 per $1\ \mathrm{mol}$, we have

$$ C_{p,\mathrm m}-C_{V,\mathrm m}=\frac{\alpha^2TV_{\mathrm m}}{\kappa_T} =\frac{(6.9\times10^{-5}\ \mathrm{K^{-1}})^2(300\ \mathrm K)(1.0\times10^{-5}\ \mathrm{m^3/mol})}{1.4\times10^{-11}\ \mathrm{Pa^{-1}}} $$

The numerator is $(4.76\times10^{-9})(300)(1.0\times10^{-5})=1.43\times10^{-11}$ (in units of $\mathrm{K^{-1}\,m^3\,mol^{-1}}$), and dividing this by $1.4\times10^{-11}\ \mathrm{Pa^{-1}}$ gives

$$ C_{p,\mathrm m}-C_{V,\mathrm m}\approx1.0\ \mathrm{J/(mol\,K)} $$

(Unit check: $\mathrm{K^{-1}\,m^3\,mol^{-1}\,Pa}=\mathrm{K^{-1}\,J\,mol^{-1}}$.) Relative to $C_{p,\mathrm m}\approx24\ \mathrm{J/(mol\,K)}$, the ratio is $1.0/24\approx0.04$, that is, about $4\ \%$. As with copper (about $3\ \%$), the difference between $C_p$ and $C_V$ is small for solids.

Solution

Regarding $\alpha,V,C_p$ in $\left(\dfrac{\partial T}{\partial p}\right)_S=\dfrac{\alpha TV}{C_p}$ as constant, we have $\Delta T\approx\dfrac{\alpha TV_{\mathrm m}}{C_{p,\mathrm m}}\Delta p$. Substituting the numerical values gives

$$ \Delta T=\frac{(5.0\times10^{-5}\ \mathrm{K^{-1}})(300\ \mathrm K)(7.1\times10^{-6}\ \mathrm{m^3/mol})}{24\ \mathrm{J/(mol\,K)}}\times1.0\times10^{8}\ \mathrm{Pa} =\frac{1.07\times10^{-7}}{24}\times10^{8}\ \mathrm K=0.44\ \mathrm K $$

(Unit check: $\mathrm{m^3\,Pa/(J/K)}=\mathrm K$.) The temperature rises by about $0.44\ \mathrm K$.

Next, $C_p\Delta T=24\ \mathrm{J/K}\times0.44\ \mathrm K=1.07\times10^{1}\ \mathrm J\approx11\ \mathrm J$, which equals the heat released during the isothermal compression in Example 31.6, $|Q|=T|\Delta S|=T\alpha V_{\mathrm m}\Delta p\approx11\ \mathrm J$. The reason is as follows. In an adiabatic process the entropy is constant, so the decrease in entropy due to the compression, $-\alpha V\Delta p$ (Formula 31.4), is canceled by the increase in entropy due to the temperature rise, $C_p\Delta T/T$: $C_p\Delta T/T=\alpha V\Delta p$. In other words, “the temperature rise in adiabatic compression multiplied by $C_p$” equals “the heat released in isothermal compression.”

Solution

The third law of thermodynamics (Sommerfeld's formulation) can be explained as follows: “As the temperature of a substance is brought closer to absolute zero, the entropy of that substance (a quantity expressing the disorder of the arrangement and motion of its molecules and atoms) approaches a certain definite value $S_0$. Moreover, this $S_0$ does not depend on how much pressure is applied to the substance, nor on what state of aggregation the substance is in (gas, liquid, solid, or a difference in crystal structure).”

This statement is broader than “the entropy change of a reaction, $\Delta S\to0$” (Nernst's theorem, Law 32.1) in the following respect. Nernst's theorem states only that, for substances A and B in condensed phases that are linked by a reaction, $S_{\mathrm A}$ and $S_{\mathrm B}$ approach the same value as $T\to0$ ($\Delta S=S_{\mathrm B}-S_{\mathrm A}\to0$); it says nothing about what that common value is (whether it may be set equal to $0$), nor about the relation between the entropies of substances that are not linked by a reaction, or of states that differ in pressure or state of aggregation. By contrast, the third law (Sommerfeld's formulation) asserts that the entropy $S$ of each individual substance itself approaches a common value $S_0$ that is independent of pressure and state of aggregation, and this makes it possible to consider an absolute entropy, with $T=0$ as the reference, for any substance and any state.

Solution

The third law (in its stricter form, including Planck's remark) guarantees $S_0=0$ only when the substance is a “pure crystal,” that is, when the atoms or molecules are arranged in exactly one, perfectly regular configuration. In the language of statistical mechanics, this corresponds to the number of microscopic states the substance can take at absolute zero (the thermodynamic weight) being $W=1$ (effectively only one way of arranging the configuration), so that $S=k_{\mathrm B}\ln W=k_{\mathrm B}\ln1=0$.

Glass (an amorphous solid), however, is made by cooling a liquid rapidly, so the disorder of the atomic and molecular arrangement that existed in the liquid state is “frozen in” as it is, without time to crystallize (to rearrange into a regular pattern). That is, even when $T=0$ is reached, many possible patterns ($W\gt1$) of atomic and molecular arrangements still remain, and $S(0)=k_{\mathrm B}\ln W\gt0$ takes a positive value (the residual entropy). This is because glass is not in a “true equilibrium state” in the thermodynamic sense but in a nonequilibrium state in which the arrangement has happened to stall, and it therefore falls outside the “pure crystal” condition that the third law assumes.

Solution

Solving Trouton's rule $\Delta_{\mathrm{vap}}S=\Delta_{\mathrm{vap}}H/T_b\approx85\ \mathrm{J\,K^{-1}mol^{-1}}$ for $T_b$ gives

$$ T_b\approx\frac{\Delta_{\mathrm{vap}}H}{85\ \mathrm{J\,K^{-1}mol^{-1}}}=\frac{8190\ \mathrm{J\,mol^{-1}}}{85\ \mathrm{J\,K^{-1}mol^{-1}}}\approx96.4\ \mathrm K $$

Compared with the measured boiling point of $111.7\ \mathrm K$, the estimate is about $13.7\%$ lower than the measured value. The entropy of vaporization calculated backward from the measured data is

$$ \Delta_{\mathrm{vap}}S=\frac{8190\ \mathrm{J\,mol^{-1}}}{111.7\ \mathrm K}\approx73.3\ \mathrm{J\,K^{-1}mol^{-1}} $$

which is considerably smaller than the Trouton guideline of $85\ \mathrm{J\,K^{-1}mol^{-1}}$. That is, the $\Delta_{\mathrm{vap}}S$ of methane is about $14\ \%$ smaller than the guideline value of $85$ (the amber point in Figure 32.5). Methane is a small, symmetric, nonpolar molecule with no strong intermolecular interactions such as hydrogen bonds; it has a low boiling point ($111.7\ \mathrm K$) and can already move around fairly freely even while still a liquid, so the increase on becoming a gas is presumably smaller. However, the reason for this deviation is not simple, and it is best to take Trouton's rule as “a guideline good to roughly $\pm$ 10–20%.”

Solution

Substituting the ideal-gas approximation $\Delta V\approx V_{\text{vapor}}=RT/p$ into the Clausius–Clapeyron equation $\dd p/\dd T=\Delta H/(T\Delta V)$ gives

$$ \frac{\dd p}{\dd T}=\frac{\Delta_{\mathrm{sub}}H}{T\times(RT/p)}=\frac{p\,\Delta_{\mathrm{sub}}H}{RT^2} $$

Substituting the values at the triple point, $T_{\mathrm{tp}}=273.16\ \mathrm K$, $p_{\mathrm{tp}}=611.657\ \mathrm{Pa}$, $\Delta_{\mathrm{sub}}H=51060\ \mathrm{J\,mol^{-1}}$, and $R=8.314\ \mathrm{J\,mol^{-1}K^{-1}}$, gives

$$ \frac{\dd p}{\dd T}=\frac{611.657\times51060}{8.314\times273.16^2}\approx\frac{3.1233\times10^7}{6.2036\times10^5}\approx50.3\ \mathrm{Pa/K} $$

This value is positive (the same behavior as ordinary substances: raising the pressure also raises the temperature at which sublimation occurs). Compared with the slope of the melting curve found in Example 32.9, $\dd p/\dd T\approx-13.47\ \mathrm{MPa/K}=-1.347\times10^7\ \mathrm{Pa/K}$, we note the following.

• Sign: the sublimation curve has a positive slope and the melting curve a negative one—water (together with gallium, bismuth, silicon, germanium, and others) has a melting curve whose slope is opposite in direction to that of ordinary substances.
• Magnitude: the slope of the sublimation curve ($\sim50\ \mathrm{Pa/K}$) is smaller than the magnitude of the slope of the melting curve ($\sim1.3\times10^7\ \mathrm{Pa/K}$) by a factor of about 270,000 (more than five orders of magnitude). This is because in sublimation the volume change $\Delta V$ of the gas is very large (the denominator is large), whereas in melting there is only a slight volume change between the two condensed phases, solid and liquid (the denominator is very small).

At the triple point, the three curves—the sublimation curve, the melting curve, and the vaporization curve—meet at exactly one point, forming the triple point of the $p$-$T$ phase diagram of water.

Solution

By the third law, the two entropy curves in Figure 32.7, $S(T,B=0)$ (no magnetic field) and $S(T,B=B_1)$ (with a magnetic field), both converge to the same value $S_0$ as $T\to0$. On the other hand, for $T\gt0$ the $B_1$ curve lies below the $B=0$ curve, and the gap between the two approaches $0$ as $T\to0$. One cycle of adiabatic demagnetization (isothermal magnetization followed by adiabatic demagnetization) corresponds to one step of a staircase that goes back and forth between these two curves: by isothermal magnetization we go from the $B=0$ curve to the $B_1$ curve without changing the temperature (lowering the entropy), and by adiabatic demagnetization we go from the $B_1$ curve to the $B=0$ curve without changing the entropy (lowering the temperature).

Suppose, hypothetically, that the two curves approached different values as $T\to0$, for example $S_0(B_1)\lt S_0(0)$. Then the adiabatic-demagnetization line running horizontally from a point with entropy lower than $S_0(0)$ would reach $T=0$ without crossing the $B=0$ curve anywhere. That is, $T=0$ would be reached in a finite number of operations. But the third law requires the two curves to converge to the same value $S_0$ at $T=0$, so a line running horizontally from a point with $T\gt0$ always meets the $B=0$ curve at a temperature $T'\lt T$ that is still higher than $T=0$. Moreover, since the gap between the curves becomes small as $T\to0$, the temperature drop achieved in a single step also becomes small (in the model of Figure 32.7(a), the temperature merely halves with each step). Therefore, no matter how many times the process is repeated, the temperature never exactly coincides with $T=0$, and reaching it would require an infinite number of operations. This is why the third law leads to the unattainability of absolute zero. Note also that since $C_p\to0$ as $T\to0$, even a slight heat leak raises the temperature, so keeping the temperature low is also difficult (Exercise 32.8).

Solution

In vaporization, in which a liquid becomes a gas, the molecules go from a state in which they were immediately adjacent to one another to a state in which they can fly freely about the whole space of the container, so the degrees of freedom of their motion change dramatically. At the boiling point at 1 atm, the volume increases by a factor of roughly $10^2\sim10^3$, and the increase in the entropy of translational motion due to this “expansion of the space in which the molecules can move freely” is of roughly similar magnitude regardless of the substance (for example, for a 300-fold increase in volume, $R\ln300\approx47\ \mathrm{J\,K^{-1}mol^{-1}}$). For that reason, the entropy change accompanying vaporization tends to take very similar values for all liquids. This is why Trouton's rule holds well.

In melting, by contrast, in which a solid becomes a liquid, only the regularity of the positional relationships among the molecules (the periodic arrangement of the crystal) is lost, and the extent of the space in which each molecule can move freely does not change as dramatically as in vaporization. For that reason, the entropy change accompanying melting strongly reflects the individual circumstances of each substance: how regular the original crystal structure was, in what manner the molecules were bonded to one another (the presence or absence of hydrogen bonds, and so on), and whether the disorder of molecular orientation newly increases. As a result, Richards's rule, $\Delta_{\mathrm{fus}}S\approx R$, is adequate for simple substances such as metals, but it deviates greatly for molecular crystals such as ice ($22.0\ \mathrm{J\,K^{-1}mol^{-1}}$) and benzene (about $35\ \mathrm{J\,K^{-1}mol^{-1}}$), and so it does not hold as well as Trouton's rule.

Solution

Looking only at the equation $\Delta G=\Delta H-T\Delta S$, as long as $\Delta S$ does not diverge (stays finite), $T\Delta S\to0$ as $T\to0$, so it is obvious from the equation alone that $\Delta G\to\Delta H$. Before low-temperature research advanced, many researchers took precisely this obvious fact alone to mean that “the behavior as $T\to0$ is understood.”

However, this obvious argument tells us nothing about what specific value $\Delta S$ itself approaches as $T\to0$ (whether it approaches $0$ or stays at some finite value). This is because the fact $\Delta G\to\Delta H$ alone holds equally well whether $\Delta S$ approaches $0$ or approaches some other finite value. What the low-temperature measurements (the measurements that followed Richards's work on electrochemical cells) revealed was a much stronger fact—that when $\Delta G$ is plotted as a function of $T$, the tangent to this curve is horizontal at $T=0$ (the slope $(\partial \Delta G/\partial T)_p$ itself approaches $0$). Since $(\partial \Delta G/\partial T)_p=-\Delta S$, this fact that “the slope approaches $0$” gives precisely the concrete conclusion $\Delta S\to0$, which could not be derived from the earlier obvious argument. It was this nontrivial discovery that later led to Nernst's theorem and the third law of thermodynamics.

Solution

The quantity of heat required to raise the temperature from $T_1$ to $T_2$ is $Q=\displaystyle\int_{T_1}^{T_2}bT^3\,\dd T=\dfrac{b}{4}\left(T_2^4-T_1^4\right)$, so

$$ T_2=\left(T_1^4+\frac{4Q}{b}\right)^{1/4} $$

Substituting $Q=1\ \mu\mathrm J=10^{-6}\ \mathrm J$ and $b=4.817\times10^{-5}\ \mathrm{J\,mol^{-1}K^{-4}}$ gives $4Q/b=4\times10^{-6}/(4.817\times10^{-5})\approx0.0830\ \mathrm{K^4}$.

• $T_1=1\ \mathrm K$: $T_2=(1+0.0830)^{1/4}\approx1.020\ \mathrm K$. The rise is about $0.02\ \mathrm K$ (almost the same as the approximation $\Delta T=Q/C_p=10^{-6}/(4.817\times10^{-5})\approx0.021\ \mathrm K$).
• $T_1=0.1\ \mathrm K$: $T_2=(10^{-4}+0.0830)^{1/4}\approx0.537\ \mathrm K$. With the same $1\ \mu\mathrm J$ of heat, a substance at $0.1\ \mathrm K$ is heated to nearly $0.54\ \mathrm K$, so its temperature rises by more than a factor of five (here $\Delta T=Q/C_p$ cannot be used, because $C_p$ changes rapidly with $T$).

The lower the temperature, the more rapidly $C_p=bT^3$ decreases, and the more the temperature rises even for the same quantity of heat. This is one reason why it is difficult to maintain low temperatures near absolute zero.

Solution

Recall the derivation of the heat conduction equation in Chapter 33 (the heat balance on a small interval $[x,x+\dd x]$). Suppose the lateral surface of the rod is thermally insulated, so that heat flows only in the $x$ direction. By Fourier's law, the quantity of heat flowing into the interval through cross section A (at position $x$) and the quantity of heat flowing out through cross section B (at position $x+\dd x$) during a short time $\dd t$ are given by

$$ \dd'Q_{\mathrm{A}}=\lambda S\left(-\pdiff Tx\right)_{\!\mathrm{A}}\dd t,\qquad \dd'Q_{\mathrm{B}}=\lambda S\left(-\pdiff Tx\right)_{\!\mathrm{B}}\dd t $$

(both quantities are taken to be positive for heat flow in the $+x$ direction). The net quantity of heat received by the interval is the difference

$$ \dd'Q=\dd'Q_{\mathrm{A}}-\dd'Q_{\mathrm{B}}=\lambda S\left[\left(\pdiff Tx\right)_{\!\mathrm{B}}-\left(\pdiff Tx\right)_{\!\mathrm{A}}\right]\dd t=\lambda S\,\pdiff{^2T}{x^2}\,\dd x\,\dd t $$

Here we used $(\partial T/\partial x)_{\mathrm{B}}-(\partial T/\partial x)_{\mathrm{A}}\approx(\partial^2T/\partial x^2)\dd x$ (Eq. ① of Chapter 33). This quantity of heat raises the temperature of the interval by $\dd T=\dd'Q/(c\rho S\,\dd x)$. That is, the larger $\partial^2T/\partial x^2$ (how sharply the graph of the temperature distribution “bends”; the second derivative) is, the larger the net quantity of heat received by the small interval, and therefore the larger the rate of change of temperature $\partial T/\partial t$ as well.

The meaning of the sign is as follows. At a point where $\partial^2T/\partial x^2\gt0$ (the graph is convex downward, that is, the temperature at that point is lower than the average of the temperatures immediately on either side), the slope $\partial T/\partial x$ is larger on the B side than on the A side. That is, $\dd'Q_{\mathrm{A}}\gt\dd'Q_{\mathrm{B}}$: the inflow exceeds the outflow, so heat accumulates on balance and the temperature rises. Conversely, at a point where $\partial^2T/\partial x^2\lt0$ (the graph is convex upward, a “bump” hotter than its surroundings), the outflow exceeds the inflow and heat escapes to the surroundings, so the temperature falls. The larger the absolute value of $\partial^2T/\partial x^2$, the more extreme this “difference from the surroundings” is (a sharper bump or dip), so the imbalance between the heat entering and leaving is also larger, and the temperature changes faster. This is also the reason why, in the Fourier series solution found in the main text (Theorem 33.2 and Theorem 33.3), the higher-order Fourier components that oscillate finely in space (the components with large $n$, for which the absolute value of the coefficient $-n^2\pi^2/a^2$ corresponding to $\partial^2T/\partial x^2$ is large) decay faster.

(The case $T(x)=T_m\sin(\pi x/a)$) Differentiating twice gives

$$ \pdiff Tx=\frac{\pi}{a}T_m\cos\frac{\pi x}{a},\qquad \pdiff{^2T}{x^2}=-\frac{\pi^2}{a^2}T_m\sin\frac{\pi x}{a} $$

At the midpoint $x=a/2$ we have $\sin(\pi/2)=1$, so $\partial^2T/\partial x^2=-\pi^2T_m/a^2\lt0$ (with $T_m\gt0$), and the graph is convex upward (the top of a hill). Hence, by the heat conduction equation, $\partial T/\partial t=\kappa^2\,\partial^2T/\partial x^2\lt0$, and the temperature at the midpoint falls. This expresses the fact that heat escapes from the top of the hill to both sides, which agrees with the $n=1$ mode of the main text decaying with time.

Solution

The heat conduction equation is $\partial T/\partial t=\kappa^2\,\partial^2T/\partial x^2$. A steady state is a state in which the temperature distribution no longer changes with time, so by definition $\partial T/\partial t=0$. Substituting this into the left-hand side of the heat conduction equation, we obtain

$$ 0=\kappa^2\,\pdiff{^2T}{x^2} $$

Now, $\kappa^2=\lambda/(c\rho)$ is built from positive quantities intrinsic to the material (the thermal conductivity, the specific heat, and the density), so $\kappa^2\neq0$. Therefore we can divide both sides by $\kappa^2$, which yields $\partial^2T/\partial x^2=0$.

In a steady state $T$ does not depend on the time $t$, so we may treat it as a function $T=T(x)$ of $x$ alone, and the partial derivative is the same as an ordinary derivative. Integrating both sides of $\dd^2T/\dd x^2=0$ once with respect to $x$ gives $\dd T/\dd x=A$ (a constant of integration), and integrating once more gives

$$ T(x)=Ax+B $$

(where $A,B$ are constants of integration). This is precisely a linear function of $x$ (a straight line with slope $A$ and intercept $B$). That the second derivative vanishes identically means that “the slope of the graph (the first derivative) is constant,” and the only functions with constant slope are precisely the linear functions (straight lines)—this is why the temperature distribution in a steady state is always a straight line.

Solution

The separated equation was $\dfrac{1}{\kappa^2}\dfrac{\dot\Theta(t)}{\Theta(t)}=\dfrac{X''(x)}{X(x)}$ (the equation in Chapter 33). The left-hand side is an expression in $t$ alone ($x$ does not appear at all), and the right-hand side is an expression in $x$ alone ($t$ does not appear at all).

Now fix $x$ at some single value (say $x=x_1$). Since $x_1$ is fixed, the right-hand side $X''(x_1)/X(x_1)$ is now just a number (a constant independent of $t$). Since the equality holds, the left-hand side $\dfrac{1}{\kappa^2}\dfrac{\dot\Theta(t)}{\Theta(t)}$ must also always equal this same number, however we vary $t$. That is, the left-hand side was in fact a constant independent of $t$.

Next, if we fix $t$ at another value ($t=t_1$) and make the same argument, this time the right-hand side $X''(x)/X(x)$ turns out to be a constant independent of $x$.

Combining the two arguments above, both the left-hand side (an expression in $t$ alone) and the right-hand side (an expression in $x$ alone) equal “one and the same number that depends on neither $x$ nor $t$.” This is the logic by which the “constant $\alpha$” always appears in the method of separation of variables. If the left-hand side really changed its value with $t$, the right-hand side (an expression not containing $t$) could not follow it, contradicting the premise that the identity holds for all $x,t$.

Solution

(1) By the formula of Chapter 33 (obtained by dividing $Q=\lambda tS(T_0-T_a)/a$ by $St$), the steady-state heat flux is $q=\lambda\,\Delta T/a$. Substituting $\lambda=403\ \mathrm{W/(m\cdot K)}$, $\Delta T=80\ \mathrm{K}$, and $a=0.10\ \mathrm{m}$, we obtain

$$ q=403\ \mathrm{W/(m\cdot K)}\times\frac{80\ \mathrm{K}}{0.10\ \mathrm{m}}=3.224\times10^{5}\ \mathrm{W/m^2}\approx3.2\times10^{5}\ \mathrm{W/m^2} $$

The cross-sectional area is $S=1.0\ \mathrm{cm^2}=1.0\times10^{-4}\ \mathrm{m^2}$, so the rate of heat flow through the rod (the quantity of heat per unit time) is

$$ P=qS=3.224\times10^{5}\ \mathrm{W/m^2}\times1.0\times10^{-4}\ \mathrm{m^2}=32.24\ \mathrm{W}\approx32\ \mathrm{W} $$

Even in a thin rod, copper conducts heat so well that a temperature difference of only this size drives as much as about $32\ \mathrm{W}$ of heat flow.

(2) Substituting $a=0.10\ \mathrm{m}$, $\kappa^2=1.17\times10^{-4}\ \mathrm{m^2/s}$, and $\pi^2=9.870$ into $\tau_1=a^2/(\pi^2\kappa^2)$, we obtain

$$ \tau_1=\frac{(0.10)^2}{9.870\times1.17\times10^{-4}}=\frac{0.010}{1.155\times10^{-3}}\approx8.7\ \mathrm{s} $$

(the unit is $\mathrm{m^2/(m^2/s)}=\mathrm{s}$). Compared with the rod of Example 33.7 with $a=0.50\ \mathrm{m}$ (about $2.2\times10^{2}\ \mathrm{s}$), the length is now $1/5$ as large, so since $\tau_1\propto a^2$ the time becomes $1/25$ as long, and indeed $217/25\approx8.7$ agrees.

Solution

What is the same: Both start from the boundary condition “zero at both ends,” $X(0)=X(a)=0$ (zero displacement at the fixed ends for a string; zero temperature at both ends for heat conduction). Solving the ordinary differential equation for the spatial part, $X''+\omega^2X=0$ (or an equation of the same form), we obtain $X(x)=A\cos\omega x+B\sin\omega x$. The boundary condition $X(0)=0$ gives $A=0$, and $X(a)=0$ gives $\sin\omega a=0$, that is, $\omega a=n\pi$ ($n=1,2,3,\ldots$); exactly the same quantization condition appears in both cases. In other words, the allowed wave shapes in the spatial direction (the eigenfunctions $\sin(n\pi x/a)$) and the allowed wavenumbers $n\pi/a$ are common to the heat conduction equation and the wave equation—this is because they are a purely mathematical consequence determined only by the form of the boundary conditions.

What differs: The ordinary differential equation for the time part $\Theta(t)$ is different. In the heat conduction equation, by separation of variables $\Theta(t)$ obeys the first-order ordinary differential equation $\dot\Theta=\kappa^2\alpha\Theta$ ($\alpha\lt0$), and the solution decays exponentially, $\Theta(t)=e^{-n^2\pi^2\kappa^2t/a^2}$ (it does not oscillate; it approaches 0 monotonically). On the other hand, for the wave equation of the string, $\partial^2u/\partial t^2=v^2\partial^2u/\partial x^2$, we get a second-order ordinary differential equation $\ddot\Theta=v^2\alpha\Theta$ in $t$, and for $\alpha\lt0$ the solution is $\Theta(t)=C_1\cos(n\pi vt/a)+C_2\sin(n\pi vt/a)$, which oscillates in time as well.

The root cause of this difference lies in the equations themselves: the heat conduction equation is first-order in time ($\partial T/\partial t$), whereas the wave equation is second-order in time ($\partial^2u/\partial t^2$). A first-order time derivative gives the equation of decay or growth, “the rate of change is proportional to the present value” (an exponential function), and a second-order time derivative gives the equation of oscillation, “the acceleration is proportional to the displacement” (trigonometric functions)—this has the same structure as the difference in mechanics between the equation for a resistive force proportional to the velocity (first-order, exponential decay) and the equation for the restoring force of a spring (second-order, simple harmonic motion).

Solution

The time constant is $\tau_n=a^2/(n^2\pi^2\kappa^2)$. Taking the ratio for $n=1$ and $n=3$, the factors $a^2,\pi^2,\kappa^2$ are common to both cases and cancel, so that

$$ \frac{\tau_1}{\tau_3}=\frac{a^2/(1^2\pi^2\kappa^2)}{a^2/(3^2\pi^2\kappa^2)}=\frac{3^2}{1^2}=9 $$

(Numerically as well, computing with $\kappa^2=1.0\times10^{-4}\ \mathrm{m^2/s},\ a=1\ \mathrm{m}$ gives $\tau_1\approx1013\ \mathrm{s}$ and $\tau_3\approx112.6\ \mathrm{s}$, and the ratio agrees with $9.000$.) That is, the $n=3$ mode decays nine times faster than the $n=1$ mode.

What this means is that “fine bumps and dips in a temperature distribution (components with short wavelength, that is, large $n$) disappear much faster than the overall gentle hill (components with long wavelength, that is, small $n$).” Physically, this is because the larger $n$ is, the more rapid the spatial variation is (the shorter the wavelength $2a/n$), and, as we saw in Exercise 33.1, where $\partial^2T/\partial x^2$ (the steepness of the variation) is larger, heat flows in and out more intensely and is evened out faster. As a result, after a sufficiently long time the temperature distribution, no matter how complicated the initial distribution it started from, approaches a smooth shape in which only the lowest-order mode, which decays most slowly ($n=1$, or the steady-state term $T_1(x)$ when the boundary conditions are not 0), survives. This is what we also confirmed numerically in Example 33.6.

Solution

The solution of Theorem 33.3 is $T(x,t)=T_1(x)+T_2(x,t)$, where the steady-state term is $T_1(x)=T_0+\dfrac{T_a-T_0}{a}x$ and the transient term is $T_2(x,t)=-\displaystyle\sum_{k\text{ odd}}\frac{8Aa^2}{k^3\pi^3}e^{-\kappa^2k^2\pi^2t/a^2}\sin\frac{k\pi x}{a}$.

$t\to\infty$: The transient term approaches 0, so the temperature at the midpoint is given by the steady-state term alone:

$$ T_1\!\left(\frac a2\right)=\frac{T_0+T_a}{2}=\frac{40+20}{2}=30\ {}^\circ\mathrm{C} $$

$t=0$: Substituting $x=0.5\ \mathrm{m}$ into the initial condition $T(x,0)=Ax^2+\beta x+T_0$ (with $\beta=(T_a-T_0)/a-Aa=-20-100=-120\ \mathrm{K/m}$), we obtain

$$ T\!\left(\frac a2,0\right)=100\times0.25-120\times0.5+40=25-60+40=5\ {}^\circ\mathrm{C} $$

(This is also the sum $30-25=5\ {}^\circ\mathrm{C}$ of the steady-state term $30\ {}^\circ\mathrm{C}$ and the transient term $T_2(a/2,0)=-Aa^2/4=-25\ \mathrm{K}$.)

$t=1000\ \mathrm{s}$ (the $k=1$ term only): We have $\dfrac{8Aa^2}{\pi^3}=\dfrac{800}{31.006}=25.80\ \mathrm{K}$, $e^{-\kappa^2\pi^2t/a^2}=e^{-\pi^2\times10^{-4}\times1000}=e^{-0.987}=0.3727$, and $\sin(\pi/2)=1$, so

$$ T_2\!\left(\frac a2,1000\right)\approx-25.80\times0.3727=-9.62\ \mathrm{K} $$

Hence

$$ T\!\left(\frac a2,1000\right)\approx30-9.62=20.4\ {}^\circ\mathrm{C} $$

The $k=3$ term is $+25.80/27\times e^{-8.88}\approx1.3\times10^{-4}\ \mathrm{K}$, which is very small and can be neglected. We see that the temperature rises from the initial $5\ {}^\circ\mathrm{C}$ to $20.4\ {}^\circ\mathrm{C}$ after $1000\ \mathrm{s}$ and continues toward the final $30\ {}^\circ\mathrm{C}$ (the midpoint is initially near the bottom of a trough, so heat flows in and its temperature rises; this agrees with the trend of the $t=500\ \mathrm{s}$ curve in Figure 33.5).

Solution

(a) Direct effect on the pressure formula. The normal force on a collision surface perpendicular to the $x$ axis, as we saw in Section 34.3.2, is $F_x/\dd S=2m\displaystyle\int_{v_x\gt0}v_x^2f\,\dd v_x\dd v_y\dd v_z=m\rho\langle v_x^2\rangle$, and it is determined by the distribution of $v_x$ alone (the marginal distribution obtained by integrating out $v_y,v_z$). Therefore, even if there is a correlation between $v_x$ and $v_y$, that alone does not immediately change the normal pressure on a surface perpendicular to the $x$ axis.

(b) Effect on the derivation of the functional form. If independence does not hold, the distribution function can no longer be written as the product $g(v_x)g(v_y)g(v_z)$, and the derivation of Section 34.2 (the product of the values on the axes, $f=a^{-6}f(v_x,0,0)f(0,v_y,0)f(0,0,v_z)$, and the argument that leads from it to the functional equation $G(x)G(y)=G(x+y)$) cannot be used. Then we cannot fix the functional form of $f$ to be the Gaussian form $Ae^{-\alpha(v_x^2+v_y^2+v_z^2)}$, and neither $p=m\rho/(2\alpha)$ of Section 34.3.2 (the form in which $\alpha$ appears) nor Law 34.1 can be derived. (Note that the form $p=m\rho\langle v_x^2\rangle$ itself holds regardless of the form of $f$. What independence is responsible for is not the isotropy of the pressure but the determination of the functional form of $f$. For example, a non-Gaussian distribution determined by the speed alone has $\langle v_xv_y\rangle=0$ even when the components are not independent, and the pressure can be isotropic.)

(c) The contradiction if a correlation really existed. A correlation of the type “the larger the $v_x$ of a molecule, the larger its $v_y$” means $\langle v_xv_y\rangle\ne0$. This means that momentum in the $y$ direction is carried on net through a surface perpendicular to the $x$ axis (a force parallel to the surface, that is, a shear stress, arises). But a shear force is not observed to appear on the wall of a gas at rest. Moreover, if the width of the distribution differed with direction, as in $\langle v_x^2\rangle\ne\langle v_y^2\rangle$, the normal pressure would differ depending on the orientation of the wall, which contradicts the experimental fact that “the pressure of a gas is the same on a wall of any orientation.”

(d) Effect on the verification in Section 34.5.1. In Section 34.5.1 we used the fact that $\langle v_x^2\rangle=\langle v_y^2\rangle=\langle v_z^2\rangle$ holds automatically as a consequence of treating the three directions on an equal footing, and we compared with the equipartition theorem using $\langle v^2\rangle=3\langle v_x^2\rangle$. If independence and isotropy fail, the equality of the three directions is no longer guaranteed, $\langle v^2\rangle=3\langle v_x^2\rangle$ cannot be used, and the correspondence with the equipartition theorem can no longer be asserted immediately.

Solution

The normalization condition is the condition that “the total number of molecules in the container equals the integral of the distribution function over the entire range of positions and velocities,” and it appeared as Eq. (34.1) of Chapter 34, $N=V\displaystyle\int f\,\dd v_x\dd v_y\dd v_z$. Dividing both sides by $V$ gives $\rho=N/V=\displaystyle\int f\,\dd v_x\dd v_y\dd v_z$, and substituting $f=Ae^{-\alpha(v_x^2+v_y^2+v_z^2)}$ into this and using the formula for the Gaussian integral gave the relation Eq. (34.6) of Chapter 34, $\rho=A(\pi/\alpha)^{3/2}$.

If we did not impose this normalization condition, $A$ and $\alpha$ could each independently take any positive value. This is because the functional form $f=Ae^{-\alpha(v_x^2+v_y^2+v_z^2)}$ (the form determined from independence and isotropy) always satisfies the requirements that “the velocity components are independent” and “there is no special direction in space,” regardless of the values of $A,\alpha$. Independence and isotropy alone determine the “form” of the function (that it is Gaussian), but not the two constants, its “height” $A$ and its “spread” $\alpha$. The normalization condition brings in the additional information that “the integral of the distribution function over all of space must coincide exactly with the actual number of molecules $N$ (number density $\rho$),” and thereby gives one relation between $A$ and $\alpha$. However, this single relation alone does not determine the two values $A,\alpha$, and one more independent condition was needed (in the text, $p=m\rho/(2\alpha)$, obtained from the pressure calculation, together with the equation of state of an ideal gas; see Section 34.4).

Incidentally, there is another route that does not go through the pressure calculation. From the energy equipartition theorem of Chapter 24 we know that $\langle v_x^2\rangle=k_{\mathrm B}T/m$, while from the Gaussian distribution we can compute $\langle v_x^2\rangle=1/(2\alpha)$ (Section 34.5.1), so setting the two equal determines $\alpha=m/(2k_{\mathrm B}T)$. In this case the equipartition theorem (an already known result) has been used as the “one more condition.”

Solution

For oxygen molecules $\mathrm{O_2}$ ($M=32.0\times10^{-3}\ \mathrm{kg/mol}$, $\gamma=1.4$), $T=300\ \mathrm K$, and $R=8.314\ \mathrm{J/(mol\,K)}$, the speed of sound is

$$ v_{\text{sound}}=\sqrt{\frac{\gamma RT}{M}}=\sqrt{\frac{1.4\times8.314\times300}{32.0\times10^{-3}}}\approx330\ \mathrm{m/s} $$

The root-mean-square speed is (Formula 34.3 of Chapter 34)

$$ v_{\mathrm{rms}}=\sqrt{\frac{3RT}{M}}=\sqrt{\frac{3\times8.314\times300}{32.0\times10^{-3}}}\approx484\ \mathrm{m/s} $$

Factoring out the part $RT/M$ common to both expressions, we obtain

$$ \frac{v_{\text{sound}}}{v_{\mathrm{rms}}}=\frac{\sqrt{\gamma RT/M}}{\sqrt{3RT/M}}=\sqrt{\frac{\gamma}{3}} $$

and $R,T,M$ all cancel and disappear—that is, this ratio depends on neither the molar mass $M$ nor the temperature $T$, and is determined by the heat capacity ratio $\gamma$ alone. For a diatomic molecule ($\gamma=1.4$) it is $\sqrt{1.4/3}\approx0.683$, in good agreement with the actual $330/484\approx0.682$ (for a monatomic molecule, $\gamma=5/3$ and $\sqrt{5/9}\approx0.745$). Physically, this result shows that the speed of sound is of roughly the same order of magnitude as the speed of the thermal motion of the molecules (the root-mean-square speed)—since sound is, after all, a phenomenon in which collisions between molecules are passed along, the speed of sound cannot greatly exceed the speed of the thermal motion of the molecules.

Solution

By the definition of the expectation value, converting to spherical coordinates (as in Section 34.5.1), we have

$$ \langle v^4\rangle=\left(\frac{m}{2\pi k_{\mathrm B}T}\right)^{3/2}\cdot4\pi\int_0^{\infty}v^4\cdot v^2e^{-\frac{mv^2}{2k_{\mathrm B}T}}\,\dd v =\left(\frac{m}{2\pi k_{\mathrm B}T}\right)^{3/2}\cdot4\pi\int_0^{\infty}v^6e^{-\frac{mv^2}{2k_{\mathrm B}T}}\,\dd v $$

Now let $x=\dfrac{mv^2}{2k_{\mathrm B}T}$. Then $v^2=\dfrac{2k_{\mathrm B}T}{m}x$ and $\dd x=\dfrac{mv}{k_{\mathrm B}T}\dd v$, so $v\,\dd v=\dfrac{k_{\mathrm B}T}{m}\dd x$. Grouping $v^6\dd v=v^5\cdot(v\,\dd v)$ and using $v^5=(v^2)^{5/2}=\left(\dfrac{2k_{\mathrm B}T}{m}x\right)^{5/2}$, we get

$$ \int_0^{\infty}v^6e^{-\frac{mv^2}{2k_{\mathrm B}T}}\,\dd v =\left(\frac{2k_{\mathrm B}T}{m}\right)^{5/2}\frac{k_{\mathrm B}T}{m}\int_0^{\infty}x^{5/2}e^{-x}\,\dd x =4\sqrt2\left(\frac{k_{\mathrm B}T}{m}\right)^{7/2}\Gamma\!\left(\frac72\right) $$

(We used $2^{5/2}=4\sqrt2$.) Here $\Gamma(7/2)=\dfrac52\cdot\dfrac32\cdot\dfrac12\Gamma\!\left(\dfrac12\right)=\dfrac{15}{8}\sqrt\pi$, so

$$ \int_0^{\infty}v^6e^{-\frac{mv^2}{2k_{\mathrm B}T}}\,\dd v=4\sqrt2\cdot\frac{15}{8}\sqrt\pi\left(\frac{k_{\mathrm B}T}{m}\right)^{7/2}=\frac{15\sqrt{2\pi}}{2}\left(\frac{k_{\mathrm B}T}{m}\right)^{7/2} $$

Substituting this and using $\left(\dfrac{m}{2\pi k_{\mathrm B}T}\right)^{3/2}\left(\dfrac{k_{\mathrm B}T}{m}\right)^{7/2}=(2\pi)^{-3/2}\left(\dfrac{k_{\mathrm B}T}{m}\right)^2$, we obtain

$$ \langle v^4\rangle=4\pi\cdot\frac{15\sqrt{2\pi}}{2}\cdot(2\pi)^{-3/2}\left(\frac{k_{\mathrm B}T}{m}\right)^2 =\frac{4\pi\cdot15}{2}\cdot\frac{\sqrt{2\pi}}{(2\pi)^{3/2}}\left(\frac{k_{\mathrm B}T}{m}\right)^2 =30\pi\cdot\frac1{2\pi}\left(\frac{k_{\mathrm B}T}{m}\right)^2=15\left(\frac{k_{\mathrm B}T}{m}\right)^2 $$

(We used $\sqrt{2\pi}/(2\pi)^{3/2}=1/(2\pi)$.) This shows that $\langle v^4\rangle=15(k_{\mathrm B}T/m)^2$. The dimension is $(\mathrm{J/kg})^2=(\mathrm{m^2/s^2})^2=\mathrm{m^4/s^4}$, which agrees with that of $v^4$. Also, since $\langle v^2\rangle=3k_{\mathrm B}T/m$,

$$ \frac{\langle v^4\rangle}{\langle v^2\rangle^2}=\frac{15}{9}=\frac53\approx1.67 $$

If all the molecules had the same speed $v_0$, we would have $\langle v^4\rangle/\langle v^2\rangle^2=1$, so $5/3$ is greater than 1. The more this ratio deviates from 1, the broader the distribution of speeds is (the longer the tail of fast molecules).

Solution

(a) Since $v_{\mathrm p}=\sqrt{2RT/M}$ is proportional to $M^{-1/2}$ at the same temperature,

$$ \frac{v_{\mathrm p}(\mathrm{H_2})}{v_{\mathrm p}(\mathrm{N_2})}=\sqrt{\frac{M_{\mathrm{N_2}}}{M_{\mathrm{H_2}}}}=\sqrt{\frac{28.0}{2.02}}\approx3.72 $$

The most probable speed of $\mathrm{H_2}$ is

$$ v_{\mathrm p}(\mathrm{H_2})=\sqrt{\frac{2\times8.314\times300}{2.02\times10^{-3}}}\approx1.57\times10^3\ \mathrm{m/s} $$

(This agrees with 3.72 times the value $v_{\mathrm p}(\mathrm{N_2})\approx422\ \mathrm{m/s}$ of Example 34.6: $422\times3.72\approx1570$.)

(b) Since $v_{\mathrm p}\propto\sqrt T$, to double $v_{\mathrm p}$ it suffices to double $\sqrt T$, that is, to multiply $T$ by $2^2=4$. We get $T=4\times300\ \mathrm K=1200\ \mathrm K$.

Solution

(a) We have $\langle v_x\rangle=\sqrt{\alpha/\pi}\displaystyle\int_{-\infty}^{\infty}v_xe^{-\alpha v_x^2}\dd v_x$, and the integrand $v_xe^{-\alpha v_x^2}$ is an odd function (it changes sign under $v_x\to-v_x$), so the positive and negative contributions cancel and $\langle v_x\rangle=0$. This corresponds to the fact that the gas as a whole is at rest, and there are equal numbers of molecules flying in the $+x$ direction and in the $-x$ direction.

(b) Since the integrand is even, the mean of $|v_x|$ is twice the one-sided value:

$$ \langle|v_x|\rangle=2\sqrt{\frac\alpha\pi}\int_0^{\infty}v_xe^{-\alpha v_x^2}\dd v_x=2\sqrt{\frac\alpha\pi}\cdot\frac1{2\alpha}=\frac1{\sqrt{\pi\alpha}}=\sqrt{\frac{2k_{\mathrm B}T}{\pi m}} $$

(We used $\int_0^\infty xe^{-\alpha x^2}\dd x=\big[-e^{-\alpha x^2}/(2\alpha)\big]_0^\infty=1/(2\alpha)$ and $\alpha=m/(2k_{\mathrm B}T)$.) On the other hand, since $\langle v_x^2\rangle=k_{\mathrm B}T/m$ (Eq. (34.12) of Chapter 34), the standard deviation is $\sqrt{\langle v_x^2\rangle}=\sqrt{k_{\mathrm B}T/m}$, and the ratio is

$$ \frac{\langle|v_x|\rangle}{\sqrt{\langle v_x^2\rangle}}=\sqrt{\frac2\pi}\approx0.798 $$

The “mean of the absolute value” is smaller than the “square root of the mean square.”

(c) For nitrogen ($M=28.0\times10^{-3}\ \mathrm{kg/mol}$) at $T=300\ \mathrm K$, using $k_{\mathrm B}/m=R/M$ we have

$$ \langle|v_x|\rangle=\sqrt{\frac{2RT}{\pi M}}=\sqrt{\frac{2\times8.314\times300}{\pi\times28.0\times10^{-3}}}\approx238\ \mathrm{m/s} $$

(The standard deviation is $\sqrt{RT/M}\approx298\ \mathrm{m/s}$.) Also, the ratio of $\langle v\rangle=\sqrt{8k_{\mathrm B}T/(\pi m)}$ (Formula 34.2) to the mean absolute value of the velocity component found above is

$$ \frac{\langle v\rangle}{\langle|v_x|\rangle}=\frac{\sqrt{8/\pi}}{\sqrt{2/\pi}}=\sqrt4=2 $$

and numerically, too, $476/238=2$ agrees.

Solution

(a) Computing $v_{\mathrm{rms}}=\sqrt{3RT/M}$ ($T=300\ \mathrm K$, $R=8.314\ \mathrm{J/(mol\,K)}$), we obtain

$$ v_{\mathrm{rms}}(\mathrm{H_2})=\sqrt{\frac{3\times8.314\times300}{2.02\times10^{-3}}}\approx1.92\ \mathrm{km/s},\quad v_{\mathrm{rms}}(\mathrm{He})\approx1.37\ \mathrm{km/s},\quad v_{\mathrm{rms}}(\mathrm{N_2})\approx0.517\ \mathrm{km/s} $$

The ratios to $v_{\mathrm{esc}}=11.2\ \mathrm{km/s}$ are

$$ \frac{v_{\mathrm{esc}}}{v_{\mathrm{rms}}}\approx5.8\ (\mathrm{H_2}),\quad8.2\ (\mathrm{He}),\quad21.7\ (\mathrm{N_2}) $$

The lighter the molecule, the closer $v_{\mathrm{rms}}$ is to the escape velocity.

(b) From Example 34.7, with $u_0=v_{\mathrm{esc}}/v_{\mathrm p}$ ($v_{\mathrm p}=\sqrt{2RT/M}$), the fraction is $\dfrac{2}{\sqrt\pi}u_0e^{-u_0^2}+\operatorname{erfc}(u_0)$. Since $v_{\mathrm p}$ is $1.571,\ 1.117,\ 0.422\ \mathrm{km/s}$ for $\mathrm{H_2}$, $\mathrm{He}$, and $\mathrm{N_2}$, respectively, we have $u_0\approx7.13,\ 10.03,\ 26.5$. When $u_0$ is large, the first term dominates, and the common logarithm can be estimated as $-u_0^2/\ln10+\log_{10}(2u_0/\sqrt\pi)$. For example, for $\mathrm{H_2}$ we have $u_0^2\approx50.8$, so $-50.8/2.303\approx-22.06$, and $\log_{10}(2\times7.13/1.772)\approx\log_{10}8.04\approx0.91$; together these give about $-21.15$, that is, the fraction is about $7\times10^{-22}$. Similarly, it is about $2\times10^{-43}$ for $\mathrm{He}$ and about $5\times10^{-305}$ for $\mathrm{N_2}$.

(c) If the temperature stays at $300\ \mathrm K$, for every gas the fraction of molecules exceeding the escape velocity is extremely small, but the degree of smallness differs by many orders of magnitude. Compared with about $10^{-21}$ for $\mathrm{H_2}$, $\mathrm{N_2}$ is effectively zero (around $10^{-304}$). Since the fraction has the form $e^{-u_0^2}$, that is, $e^{-Mv_{\mathrm{esc}}^2/(2RT)}$, it is exponentially sensitive to the mass $M$, so the lighter the molecule, the vastly greater the number of fast molecules exceeding the escape velocity. In the actual upper atmosphere the temperature is much higher (the larger $T$ is, the smaller $u_0$ becomes), and in addition many billions of years have passed, so this small fraction accumulates, and the lighter the gas, the more easily it escapes into space.

Solution

(a) Using the same reasoning as in Section 34.3.2, the number of molecules with velocities in the range $(v_x,v_y,v_z)\sim(v_x+\dd v_x,\ldots)$ that collide with a wall of area $\dd S$ during $\Delta t$ is the number of molecules in a column of base area $\dd S$ and height $v_x\Delta t$, namely $v_x\Delta t\,\dd S\,f\,\dd v_x\dd v_y\dd v_z$ ($v_x\gt0$). Dividing this by $\Delta t\,\dd S$ and summing over the entire range of velocities, the number of collisions per unit time and unit area, $\Phi$, is

$$ \Phi=\int_0^{\infty}\!\!\int_{-\infty}^{\infty}\!\!\int_{-\infty}^{\infty}v_xf\,\dd v_x\dd v_y\dd v_z =\rho\left(\frac\alpha\pi\right)^{3/2}\cdot\frac{1}{2\alpha}\cdot\frac\pi\alpha=\rho\,\frac{\alpha^{3/2}}{\pi^{3/2}}\cdot\frac{\pi}{2\alpha^2}=\frac{\rho}{2\sqrt{\pi\alpha}} $$

(we used $f=\rho(\alpha/\pi)^{3/2}e^{-\alpha v^2}$ and $\int_0^\infty v_xe^{-\alpha v_x^2}\dd v_x=1/(2\alpha)$, and, for the integrals over $v_y,v_z$, $\sqrt{\pi/\alpha}$ for each). Since $\alpha=m/(2k_{\mathrm B}T)$ gives $\dfrac{1}{\sqrt{\pi\alpha}}=\sqrt{\dfrac{2k_{\mathrm B}T}{\pi m}}$, we have

$$ \Phi=\frac\rho2\sqrt{\frac{2k_{\mathrm B}T}{\pi m}}=\rho\sqrt{\frac{k_{\mathrm B}T}{2\pi m}} $$

On the other hand, $\dfrac14\langle v\rangle=\dfrac14\sqrt{\dfrac{8k_{\mathrm B}T}{\pi m}}=\sqrt{\dfrac{8k_{\mathrm B}T}{16\pi m}}=\sqrt{\dfrac{k_{\mathrm B}T}{2\pi m}}$, so $\Phi=\tfrac14\rho\langle v\rangle$ has been shown.

(b) From Example 34.4, $\rho=p/(k_{\mathrm B}T)=2.41\times10^{25}\ \mathrm{m^{-3}}$, and from Example 34.6, $\langle v\rangle\approx476\ \mathrm{m/s}$. Since $S=1.0\ \mathrm{mm^2}=1.0\times10^{-6}\ \mathrm{m^2}$, the number of molecules leaving per second is

$$ \Phi S=\frac14\times2.41\times10^{25}\times476\times1.0\times10^{-6}\approx2.9\times10^{21}\ \text{molecules/s} $$

(In terms of the amount of substance, this is $2.9\times10^{21}/6.02\times10^{23}\approx4.8\times10^{-3}\ \mathrm{mol/s}$.)

(c) For the same $p,T$, $\rho=p/(k_{\mathrm B}T)$ is the same regardless of the kind of gas, so the ratio of the numbers of molecules effusing equals the ratio of $\langle v\rangle$, and since $\langle v\rangle\propto M^{-1/2}$,

$$ \frac{\Phi_{\mathrm{He}}}{\Phi_{\mathrm{N_2}}}=\sqrt{\frac{M_{\mathrm{N_2}}}{M_{\mathrm{He}}}}=\sqrt{\frac{28.0}{4.00}}=\sqrt7\approx2.65 $$

This result—that the rate of effusion of a gas through a small hole is inversely proportional to the square root of the molar mass—is Graham's law (the law of effusion).

Solution

Using a large cube. If we attach eight cubes (side $a$), each having the point charge $Q$ at a vertex, around the charge, we obtain a large cube of side $2a$ centered on the charge. The surface of the large cube is a closed surface, and the only charge inside is $Q$, so by Gauss's law (integral form) the flux through the entire surface is

$$ \iint_{\text{large cube}}\bm E\cdot\bm n\,\dd S=\frac{Q}{\varepsilon_0} $$

The large cube has a symmetry centered on the charge (interchanges of faces, rotations, and reflections), so the fluxes through the six faces are all equal, each being $\dfrac{Q}{6\varepsilon_0}$. Furthermore, one face (a square of side $2a$) is divided by two lines through the midpoints of its edges into four small squares of side $a$. The four small squares are interchanged with one another by reflections in the two planes that pass through the center of the face and are parallel to its edges, so the flux through each of them is also the same, namely

$$ \frac{1}{4}\times\frac{Q}{6\varepsilon_0}=\frac{Q}{24\varepsilon_0} $$

per small square (side $a$).

(a) The surface of the large cube consists of 24 such small squares ($=6\times4$). The three faces of the original cube (side $a$, with the charge at one vertex) that do not contain the charge coincide exactly with three of the small squares on the surface of the large cube. Therefore, the flux through one of these faces is $\dfrac{Q}{24\varepsilon_0}$. Numerically,

$$ \frac{Q}{24\varepsilon_0}=\frac{3.0\times10^{-9}\ \mathrm C}{24\times8.854\times10^{-12}\ \mathrm{C^2/(N\,m^2)}}\approx14.1\ \mathrm{N\,m^2/C} $$

(the unit is $\mathrm{N\,m^2/C}=\mathrm{V\,m}$). The three faces have a symmetry under which they are interchanged with one another around the three edges passing through a vertex of the cube, so the value is the same for each face.

(b) The three faces that contain the charge lie in planes passing through the charge (the vertex). A line drawn from the charge to a point on such a face lies within the face, so the electric field of the point charge, $\bm E\parallel\bm r$, is parallel to the face everywhere on it. The normal vector $\bm n$ of the face is perpendicular to the face, so $\bm E\cdot\bm n=0$, and the flux is $0$.

(c) The flux through the entire surface of the cube is the sum of the three faces in (a) and the three faces in (b):

$$ 3\times\frac{Q}{24\varepsilon_0}+3\times0=\frac{Q}{8\varepsilon_0}\approx42.4\ \mathrm{N\,m^2/C} $$

This also agrees with the self-evident result that each of the eight cubes accounts for $\dfrac18$ of the total flux $Q/\varepsilon_0$. The side length $a$ does not appear in the result. Therefore, the flux does not change even if $a$ is doubled. This is due to a property of Gauss's law: the electric field of a point charge weakens in proportion to $1/r^2$, while the area of a surface grows in proportion to $r^2$, so the two effects cancel (the flux through a surface is determined only by the solid angle that the surface subtends as seen from the charge).

Solution

The divergence of $\bm F=(x^2y,\ y^2z,\ z^2x)$ is obtained by taking the partial derivative of each component with respect to its own variable and adding the results (as in the definitions in Section 35.3 and in University Mathematics, Section 19.1). We have $\pdiff{}{x}(x^2y)=2xy$ ($y$ is treated as a constant), $\pdiff{}{y}(y^2z)=2yz$ ($z$ is treated as a constant), and $\pdiff{}{z}(z^2x)=2zx$ ($x$ is treated as a constant), so

$$ \operatorname{div}\bm F=2xy+2yz+2zx $$

The gradient of $\varphi=x^2+y^2+z^2$ is the vector formed by arranging the three partial derivatives with respect to each variable, so from $\pdiff\varphi x=2x,\ \pdiff\varphi y=2y,\ \pdiff\varphi z=2z$ we get

$$ \grad\varphi=(2x,\ 2y,\ 2z) $$

(Since $\varphi=\abs{\bm r}^2$, this can also be written $\grad\varphi=2\bm r$.)

Next, for the position vector field $\bm f=(x,y,z)$ (its divergence, computed according to Definition 19.2 as the three-dimensional version of Illustration 35.1, is $\operatorname{div}\bm f=\pdiff xx+\pdiff yy+\pdiff zz=1+1+1=3$), we compute directly the surface integral over the entire surface of a sphere of radius $R$. At each point on the sphere, $\bm f=(x,y,z)$ is the position vector of that point itself; its magnitude equals the radius $R$ of the sphere, and its direction is outward from the center of the sphere—that is, exactly the same direction as the outward normal $\bm n$ of the sphere. Therefore, at every point on the sphere $\bm f\cdot\bm n=\abs{\bm f}\abs{\bm n}\cos0°=R\times1=R$ (constant), and since the surface area of the sphere is $4\pi R^2$,

$$ \iint_S\bm f\cdot\bm n\,\dd S=R\times4\pi R^2=4\pi R^3 $$

On the other hand, the right-hand side of the divergence theorem is the product of the divergence $\operatorname{div}\bm f=3$ (constant) and the volume of the sphere, $\frac43\pi R^3$, so

$$ \iiint_V(\operatorname{div}\bm f)\,\dd V=3\times\frac43\pi R^3=4\pi R^3 $$

Thus the surface integral (left-hand side) and the volume integral (right-hand side) both equal $4\pi R^3$ and agree—we have confirmed that Gauss's divergence theorem (University Mathematics, Section 20.5) also holds in this concrete example.

Solution

The equation of motion is $m\ddot x=qE$. The right-hand side $qE$ is constant, so the acceleration $\ddot x=qE/m\equiv a$ is also constant. Integrating this once with respect to time (with constant of integration $C_1$), we get

$$ \dot x(t)=at+C_1 $$

Substituting the initial condition $\dot x(0)=v_0$ gives $\dot x(0)=C_1=v_0$, so $\dot x(t)=at+v_0$. Integrating once more with respect to time (with constant of integration $C_2$), we get

$$ x(t)=\frac12at^2+v_0t+C_2 $$

Substituting the initial condition $x(0)=x_0$ gives $x(0)=C_2=x_0$, so the correct expression is

$$ x(t)=x_0+v_0t+\frac12at^2=x_0+v_0t+\frac12\cdot\frac{qE}{m}\,t^2 $$

The incorrect expression $x(t)=x_0+v_0t+\dfrac{qE}{m}t^2$ lacks the coefficient $\dfrac12$; the second of the two integrations (when the velocity expression $\dot x=at+v_0$ is integrated once more, the term $\frac12at^2$ arises from the $at$ term) has not been carried out correctly. (This is the same material as in Sections 1.4 and 35.6. Note that differentiating the correct expression twice with respect to $t$ gives back $\ddot x=a=qE/m$, in agreement with the equation of motion. The incorrect expression gives $\ddot x=2qE/m$, which does not match.)

Solution

Substituting $q=Q=1\ \mathrm C$ and $r=1\ \mathrm m$ into Coulomb's law (Law 35.1 of Chapter 35), we obtain

$$ \abs{\bm F}=\frac{1}{4\pi\varepsilon_0}\cdot\frac{(1\ \mathrm C)(1\ \mathrm C)}{(1\ \mathrm m)^2}=k\times1\approx8.988\times10^{9}\ \mathrm N $$

This is an enormous force of about 9 billion newtons. The mass $M$ of an object whose weight $Mg$ equals this force is, from $F=Mg$,

$$ M=\frac{F}{g}=\frac{8.988\times10^{9}\ \mathrm N}{9.8\ \mathrm{m/s^2}}\approx9.2\times10^{8}\ \mathrm{kg} $$

This is the same magnitude as the gravitational force acting on an object of about 920,000 metric tons (the weight of about three very large crude-oil tankers of the 300,000-deadweight-ton class). The fact that a charge of just $1\ \mathrm C$, an amount that does not feel especially large in everyday life, produces such an enormous force is consistent with the conclusion “the Coulomb force is far stronger than universal gravitation” seen in Example 35.1 of Section 35.1—in practice, gathering as much as $1\ \mathrm C$ of charge at a single point is itself extremely difficult, because of the strong repulsion between charges of the same sign.

Solution

The electric field lines start radially from the positive charge $+q_0$, curve along the direction of the electric field on the way, and enter the negative charge $-q_0$. On the line segment joining the two charges, the electric field line is a straight line going from $+q_0$ to $-q_0$, and on both sides of it, the electric field lines leaving $+q_0$ draw arcs and come back to $-q_0$. Some electric field lines also head to places far from the two charges; these curve around the charge on the opposite side while gradually spreading out radially (the general shape is shown in the figure below; it is the pattern of electric field lines of an electric dipole, the same shape as that of the magnetic field lines that emerge from the N pole of a magnet and return to the S pole).

Next, let us examine the electric field on the perpendicular bisecting plane of the segment joining the two charges (the set of points equidistant from the two charges). Let the separation of the two charges be $2a$, with $+q_0$ placed at $(-a,0)$ and $-q_0$ at $(a,0)$. At a point $\mathrm P=(0,s)$ on the bisecting plane (at distance $s$ from the midpoint), the distance from each charge is the same, $\sqrt{a^2+s^2}$, so the two electric fields have equal magnitude, $E_1=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_0}{a^2+s^2}$. As for direction, the field of $+q_0$ points from $+q_0$ toward $\mathrm P$, and the field of $-q_0$ points from $\mathrm P$ toward $-q_0$, so in components,

$$ \bm E_+=E_1\cdot\frac{(a,\ s)}{\sqrt{a^2+s^2}},\qquad \bm E_-=E_1\cdot\frac{(a,\ -s)}{\sqrt{a^2+s^2}} $$

The components of the two fields perpendicular to the direction joining the two charges (the $y$ components, i.e., the components parallel to the bisecting plane) are $\pm E_1s/\sqrt{a^2+s^2}$, which are opposite in direction and equal in magnitude, so they cancel each other. On the other hand, the components along the direction joining the two charges (the $x$ components) are both $+E_1a/\sqrt{a^2+s^2}$, and they reinforce each other in the same direction (from the $+q_0$ side to the $-q_0$ side). Therefore,

$$ \bm E=\bm E_++\bm E_-=\left(\frac{2E_1a}{\sqrt{a^2+s^2}},\ 0\right)=\left(\frac{q_0\,a}{2\pi\varepsilon_0\,(a^2+s^2)^{3/2}},\ 0\right) $$

Thus, on the perpendicular bisecting plane the electric field is perpendicular to the bisecting plane (parallel to the line joining the two charges, pointing from the $+q_0$ side to the $-q_0$ side), and the electric field lines cross this plane at right angles. (For example, with $q_0=1\ \mathrm{nC}$, $a=1\ \mathrm m$, and $s=1.5\ \mathrm m$, the components of the field of each charge are $(1.53,\ \pm2.30)\ \mathrm{V/m}$ and the total is $(3.07,\ 0)\ \mathrm{V/m}$, which confirms numerically that the $y$ components cancel.)

Solution

The integral form of Gauss's law $\displaystyle\iint_S\bm E\cdot\bm n\,\dd S=Q_{\text{enc}}/\varepsilon_0$ is the statement that “for a closed surface $S$ of any shape, the net flux of the electric field through $S$ is determined by the total amount of charge inside $S$.” The differential form $\operatorname{div}\bm E=\rho/\varepsilon_0$, on the other hand, is a local statement (one that focuses on a single point only): “at each point in space, the rate at which the electric field emanates from the immediate neighborhood of that point (per unit volume) is determined by the charge density at that point”—at first glance, one is a statement about information on “the whole surface” and the other about information at “a single point,” so the scales are completely different.

What connects the two is Gauss's divergence theorem $\displaystyle\iint_S\bm f\cdot\bm n\,\dd S=\iiint_V(\operatorname{div}\bm f)\,\dd V$ (University Mathematics, Section 20.5). The content of this theorem rests on a “cancellation” argument (Section 35.3 and the sketch of the proof in University Mathematics, Section 20.5): “If the region $V$ is divided into countless tiny volumes, then on an internal face shared by adjacent tiny volumes, the outward flux as seen from one volume and the outward flux as seen from its neighbor have exactly opposite signs and equal magnitudes (since we are merely viewing the same face from opposite sides), so when everything is added up, all the internal faces cancel in pairs and vanish. What survives is only the part on the outer boundary surface $S$, which is not shared with any tiny volume.”

Applying this theorem to the left-hand side of the integral form gives $\displaystyle\iiint_V(\operatorname{div}\bm E)\,\dd V=Q_{\text{enc}}/\varepsilon_0=\dfrac1{\varepsilon_0}\displaystyle\iiint_V\rho\,\dd V$, an equality between volume integrals. This equality holds no matter how small $V$ is made and wherever it is taken. In particular, if we shrink $V$ indefinitely around a single point in space, the value of the volume integral approaches (provided the integrand is continuous at that point) the value of the integrand at that point multiplied by the volume. Dividing both sides by that tiny volume and taking the limit, the volume-integral sign drops out, and we obtain the differential-form equation $\operatorname{div}\bm E=\rho/\varepsilon_0$, which concerns a single point only. Conversely, if the differential form holds (at every point in the region), then integrating it over the whole volume $V$ and using the divergence theorem from right to left (that is, using the same equality in the reverse direction), we obtain the integral form. Thus the relation is this: “the sum over all of space of the relations for tiny volumes is the integral form, and the integral form collapsed to a single point is the differential form.” Because the divergence theorem ties this “adding up” operation and this “collapsing to a point” operation together with neither excess nor deficiency, the two forms are mathematically completely equivalent.

Solution

By Coulomb's law, the electric field around a point charge $Q$ placed at the origin points in the radial direction and has magnitude

$$ E(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2} $$

By the definition of potential difference (Section 36.1 of the main text), the electric potential at a point $A$ at distance $r$ from the origin, taking infinity as the reference ($\varphi_\infty=0$), is $\varphi(r)=-\displaystyle\int_\infty^A\bm E\cdot\dd\bm r$. If we choose as the integration path a straight line from infinity to $A$ (in the radial direction), the electric field and the direction of the path are aligned, so the inner product (dot product) becomes a mere product. Writing the distance of a point on the path from the origin as the integration variable $r'$ (so that $r'=r$ at the point $A$), we have

$$ \varphi(r)=-\int_{r'=\infty}^{r'=r}E(r')\,\dd r'=-\int_\infty^r\frac{1}{4\pi\varepsilon_0}\frac{Q}{r'^2}\,\dd r' $$

Using the fact that an antiderivative of $1/r'^2$ is $-1/r'$ (one need only check that differentiating it returns $\dd(-1/r')/\dd r'=1/r'^2$), we obtain

$$ \varphi(r)=-\frac{1}{4\pi\varepsilon_0}\left[-\frac{Q}{r'}\right]_\infty^r=-\frac{1}{4\pi\varepsilon_0}\left(-\frac{Q}{r}-\lim_{c\to\infty}\left(-\frac{Q}{c}\right)\right) $$

The limit inside the parentheses is $\lim_{c\to\infty}(-Q/c)=0$ (the denominator diverges to infinity, so the fraction as a whole approaches 0), and therefore

$$ \varphi(r)=-\frac{1}{4\pi\varepsilon_0}\left(-\frac{Q}{r}\right)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r} $$

which agrees with Formula 36.2 of the main text.

Solution

Viewed individually, a plane of surface density $+\sigma$ and a plane of surface density $-\sigma$ each produce an electric field of magnitude $\sigma/(2\varepsilon_0)$ (Example 36.4 of the main text, the electric field of an infinite plane, $E=\sigma/(2\varepsilon_0)$). The field points outward on both sides of the $+\sigma$ plane (away from it), and inward from both sides toward the $-\sigma$ plane (toward it). Now place the $+\sigma$ plane on the left and the $-\sigma$ plane on the right, and take rightward as positive.

(1) The region between the two planes: This region is on the right side of the left $+\sigma$ plane (between the two planes), so the field produced by $+\sigma$ points rightward, with magnitude $\sigma/(2\varepsilon_0)$. It is on the left side of the right $-\sigma$ plane (again between the two planes), so the field produced by $-\sigma$ also points rightward (toward $-\sigma$), with magnitude $\sigma/(2\varepsilon_0)$. Both fields point in the same (rightward) direction, so they add:

$$ E_{\text{between}}=\frac{\sigma}{2\varepsilon_0}+\frac{\sigma}{2\varepsilon_0}=\frac{\sigma}{\varepsilon_0}\quad(\text{rightward, pointing from + to −}) $$

(2) The regions outside the two planes: For example, on the left (farther left than the $+\sigma$ plane), the field produced by $+\sigma$ points leftward (outward), with magnitude $\sigma/(2\varepsilon_0)$. On the other hand, the field produced by $-\sigma$ points, as seen from there, toward $-\sigma$, that is, rightward, with magnitude $\sigma/(2\varepsilon_0)$. The magnitudes are equal and the directions opposite, so they cancel:

$$ E_{\text{outside}}=\frac{\sigma}{2\varepsilon_0}-\frac{\sigma}{2\varepsilon_0}=0 $$

For the outer region on the right, the same argument with left and right interchanged gives $0$ as well. In summary, between two parallel plane charges of opposite sign there is a uniform electric field $\sigma/\varepsilon_0$ (directed from + to −), and outside them there is no electric field—this is precisely the superposition argument for the electric field of a parallel-plate capacitor used in Section 36.6 of the main text.

Solution

(1) Series connection: When two capacitors $C_1,C_2$ are connected in a single line, the two plates joined by the wire in between (the negative plate of $C_1$ and the positive plate of $C_2$) can be regarded as a single conductor isolated from the outside. Since this isolated conductor initially carried charge $0$, the sum of the charges produced on it by electrostatic induction must always be $0$. Therefore, if $+Q$ is placed on $C_1$, then $-Q$ is induced on the connected conductor, and the positive plate of $C_2$, which pairs with it, receives $+Q$—that is, the capacitors $C_1,C_2$ carry a common charge $Q$. The total potential difference is the sum of the potential differences of the two capacitors (the potential difference of the battery is split exactly into the two potential differences along the way):

$$ V=V_1+V_2=\frac{Q}{C_1}+\frac{Q}{C_2}=Q\left(\frac1{C_1}+\frac1{C_2}\right) $$

From the definition of the equivalent capacitance $C=Q/V$ (the same form as Definition 36.4 of the main text),

$$ \frac1C=\frac VQ=\frac1{C_1}+\frac1{C_2} $$

(2) Parallel connection: When two capacitors are connected in parallel, both are connected directly to the same two wires (the + terminal and the − terminal of the battery), so the potential difference $V$ is common to both. The total charge is the sum of the charges stored in the two capacitors:

$$ Q=Q_1+Q_2=C_1V+C_2V=(C_1+C_2)V $$

Therefore, the equivalent capacitance is

$$ C=\frac QV=C_1+C_2 $$

Correspondence with resistors: The rules for combining resistances, which we will learn in Section 37.5, are $R=R_1+R_2$ for a series connection (a simple sum) and $1/R=1/R_1+1/R_2$ for a parallel connection (the reciprocal of the sum of the reciprocals). For capacitors it is exactly the opposite: for a series connection we take the reciprocal of the sum of the reciprocals, and for a parallel connection we take a simple sum. This is because the quantity that is common has been interchanged. In a series connection of resistors, the current $I$ is common and the voltage is divided. In a series connection of capacitors, the charge $Q$ is common and the voltage is divided ($V_1:V_2=\dfrac1{C_1}:\dfrac1{C_2}$). In a parallel connection, the voltage $V$ is common in both cases. The difference in the common quantity is the reason why the forms of the formulas (sum and reciprocal sum) are interchanged.

Solution

We carry out the same procedure as in Examples 36.7 and 36.8 of the main text, now with $\rho=\rho_0(1-r/a)$.

Electric field: Taking a sphere of radius $r$ as the Gaussian surface, for $0\le r\le a$ the enclosed charge is (using the integral of the angular part, $\displaystyle\int_0^{2\pi}\!\!\int_0^\pi\sin\theta\,\dd\theta\,\dd\phi=4\pi$)

$$ Q(r)=4\pi\int_0^r\rho_0\left(1-\frac{r'}{a}\right)r'^2\,\dd r'=4\pi\rho_0\left[\frac{r'^3}{3}-\frac{r'^4}{4a}\right]_0^r=4\pi\rho_0\left(\frac{r^3}{3}-\frac{r^4}{4a}\right) $$

From Gauss's law $E(r)\cdot4\pi r^2=Q(r)/\varepsilon_0$,

$$ E(r)=\frac{Q(r)}{4\pi\varepsilon_0r^2}=\frac{\rho_0}{\varepsilon_0}\left(\frac r3-\frac{r^2}{4a}\right)\qquad(0\le r\le a) $$

The total charge is $Q=Q(a)=4\pi\rho_0a^3\left(\dfrac13-\dfrac14\right)=\dfrac{\pi}{3}\rho_0a^3$. For $r\gt a$, the total charge inside is $Q$, so

$$ E(r)=\frac{Q}{4\pi\varepsilon_0r^2}=\frac{\rho_0a^3}{12\varepsilon_0}\cdot\frac1{r^2}\qquad(r\gt a) $$

Position of the maximum electric field: Differentiating the interior expression with respect to $r$ and setting the result equal to $0$, we obtain $\dfrac{\dd E}{\dd r}=\dfrac{\rho_0}{\varepsilon_0}\left(\dfrac13-\dfrac{r}{2a}\right)=0$, so $r=\dfrac{2a}{3}$. At this point,

$$ E_{\max}=\frac{\rho_0}{\varepsilon_0}\left(\frac{2a}{9}-\frac{(2a/3)^2}{4a}\right)=\frac{\rho_0}{\varepsilon_0}\left(\frac{2a}{9}-\frac{a}{9}\right)=\frac{\rho_0a}{9\varepsilon_0} $$

The value at the surface is $E(a)=\dfrac{\rho_0a}{12\varepsilon_0}$, which is smaller than this. Since the density becomes smaller as the surface is approached, the maximum of the electric field occurs not at the surface but in the interior (at $r=2a/3$).

Electric potential: For $r\gt a$,

$$ \varphi(r)=\int_r^\infty\frac{\rho_0a^3}{12\varepsilon_0}\frac1{r'^2}\,\dd r'=\frac{\rho_0a^3}{12\varepsilon_0}\left[-\frac1{r'}\right]_r^\infty=\frac{\rho_0a^3}{12\varepsilon_0\,r}\qquad(r\gt a) $$

For $0\le r\le a$, we split the integration interval into $[r,a]$ (the interior expression) and $[a,\infty)$ (the exterior expression). Since the exterior potential at $r=a$ is $\varphi(a)=\dfrac{\rho_0a^2}{12\varepsilon_0}$,

$$ \varphi(r)=\int_r^a\frac{\rho_0}{\varepsilon_0}\left(\frac{r'}3-\frac{r'^2}{4a}\right)\dd r'+\frac{\rho_0a^2}{12\varepsilon_0}=\frac{\rho_0}{\varepsilon_0}\left[\frac{r'^2}{6}-\frac{r'^3}{12a}\right]_r^a+\frac{\rho_0a^2}{12\varepsilon_0} $$ $$ =\frac{\rho_0}{\varepsilon_0}\left(\frac{a^2}{12}-\frac{r^2}{6}+\frac{r^3}{12a}\right)+\frac{\rho_0a^2}{12\varepsilon_0}=\frac{\rho_0}{12\varepsilon_0a}\left(2a^3-2ar^2+r^3\right)\qquad(0\le r\le a) $$

Comparison with a point charge: Substituting $\rho_0a^3=\dfrac{3Q}{\pi}$ into the exterior potential gives $\varphi(r)=\dfrac{1}{12\varepsilon_0r}\cdot\dfrac{3Q}{\pi}=\dfrac{1}{4\pi\varepsilon_0}\dfrac Qr$, which agrees with the potential of a point charge with the total charge $Q$ concentrated at the center (Formula 36.2 of the main text). Also, at $r=a$ the interior and exterior expressions for $E$ and $\varphi$ both give the same values ($\dfrac{\rho_0a}{12\varepsilon_0}$ and $\dfrac{\rho_0a^2}{12\varepsilon_0}$), so they join continuously.

Solution

The charge $Q$ is held constant (the dielectric is inserted with the capacitor disconnected from the external power supply), so from $C=Q/V$ in Definition 36.4 we have $V=Q/C$. Let $C_0$ be the capacitance before the dielectric is inserted, and $V_0=Q/C_0$ the potential difference. Inserting the dielectric makes the capacitance $C=\varepsilon_rC_0$, so

$$ V=\frac{Q}{C}=\frac{Q}{\varepsilon_rC_0}=\frac{V_0}{\varepsilon_r} $$

(1) Therefore, the potential difference becomes $1/\varepsilon_r$ times its value before the dielectric was inserted (since $\varepsilon_r\gt1$, the potential difference decreases).

(2) For the electrostatic energy, we use the form $U=Q^2/(2C)$ (Formula 36.5 of the main text), which is convenient when $Q$ is constant. Before the insertion, $U_0=Q^2/(2C_0)$, and after the insertion,

$$ U=\frac{Q^2}{2C}=\frac{Q^2}{2\varepsilon_rC_0}=\frac{U_0}{\varepsilon_r} $$

Therefore, the electrostatic energy also becomes $1/\varepsilon_r$ times its value before the dielectric was inserted. Only the capacitance increases, while the charge $Q$ (and hence the attraction between the charges on the plates) is unchanged, so both the potential difference and the energy needed to store the same charge become smaller. This is the effect of inserting a dielectric—it helps to understand it with a picture similar to the way the water level drops when the same amount of water is poured into a wider container. The detailed mechanism by which a dielectric increases the capacitance (the weakening of the electric field by polarization) will be studied in Chapter 41.

Solution

The electric field on the axis due to the charge on a disk, found in Example 36.15 of the main text, was, for $x\gt0$,

$$ E(x)=\frac{\sigma}{2\varepsilon_0}\left(1-\frac{x}{\sqrt{a^2+x^2}}\right) $$

Taking $x\to+0$ (the limit of approaching the surface of the disk very closely from the positive side), the numerator of the second term satisfies $x\to0$, so

$$ \lim_{x\to+0}E(x)=\frac{\sigma}{2\varepsilon_0}\left(1-\frac{0}{\sqrt{a^2+0}}\right)=\frac{\sigma}{2\varepsilon_0}(1-0)=\frac{\sigma}{2\varepsilon_0} $$

which agrees exactly with the result $E=\sigma/(2\varepsilon_0)$ of Example 36.4 of the main text (the electric field of an infinitely wide charged plane).

The physical reason can be explained as follows—when the observation point is very close to the surface of the disk (the distance $x$ from the surface being sufficiently small compared with the radius $a$ of the disk), the rim of the disk (its outline of radius $a$) appears to be “very far away” within the range visible from the observation point. That is, even for a disk of finite size, if we look only at the immediate neighborhood of the observation point, it looks as if it were an infinitely wide plane charge, and the electric field also agrees with the infinite-plane result in accordance with this approximation (away from the disk, once $x$ becomes about $a$ or more, the effect of the rim can no longer be neglected, and the field deviates from the infinite-plane result).

The opposite side ($x\lt0$): By symmetry about the plane of the disk, at the point mirrored across the plane the magnitude of the electric field is the same, $\sigma/(2\varepsilon_0)$, and its direction is away from the disk (the $-x$ direction). Therefore, the electric field jumps discontinuously as it crosses the plane of the disk, from $-\sigma/(2\varepsilon_0)$ to $+\sigma/(2\varepsilon_0)$, by a difference of $\sigma/\varepsilon_0$ (the same holds for the infinite plane). Note that the value exactly at the center of the disk, $x=0$, is not determined, and what we found above is a one-sided limit.

Solution

From Poisson's equation $\nabla^2\varphi=-\rho/\varepsilon_0$ we have $\rho=-\varepsilon_0\nabla^2\varphi$, and since the system is spherically symmetric we use $\nabla^2\varphi=\dfrac1{r^2}\dfrac{\dd}{\dd r}\left(r^2\dfrac{\dd\varphi}{\dd r}\right)$. First, differentiating $\varphi=V_0\exp(-r^2/a^2)$ with respect to $r$ (differentiation of a composite function), we obtain

$$ \frac{\dd\varphi}{\dd r}=V_0\cdot\left(-\frac{2r}{a^2}\right)\exp(-r^2/a^2)=-\frac{2r}{a^2}\varphi $$

Multiply by $r^2$ and differentiate once more with respect to $r$ (using the product rule):

$$ \frac{\dd}{\dd r}\left(r^2\frac{\dd\varphi}{\dd r}\right)=\frac{\dd}{\dd r}\left(-\frac{2r^3}{a^2}\varphi\right)=-\frac{6r^2}{a^2}\varphi-\frac{2r^3}{a^2}\cdot\left(-\frac{2r}{a^2}\right)\varphi=\left(-\frac{6r^2}{a^2}+\frac{4r^4}{a^4}\right)\varphi $$

Dividing by $r^2$ gives $\nabla^2\varphi=\left(-\dfrac{6}{a^2}+\dfrac{4r^2}{a^4}\right)\varphi$, so

$$ \rho(r)=-\varepsilon_0\nabla^2\varphi=\varepsilon_0V_0\left(\frac{6}{a^2}-\frac{4r^2}{a^4}\right)\exp(-r^2/a^2) $$

(unit check: $\varepsilon_0V_0/a^2$ has units $\dfrac{\mathrm{F}}{\mathrm m}\cdot\dfrac{\mathrm V}{\mathrm{m^2}}=\dfrac{\mathrm C}{\mathrm{m^3}}$, which agrees with the unit of charge density).

Total charge: Since the system is spherically symmetric, we take $\dd V=4\pi r^2\,\dd r$, so that

$$ Q_{\text{total}}=4\pi\int_0^\infty\rho\,r^2\,\dd r=4\pi\varepsilon_0V_0\left[\frac{6}{a^2}\int_0^\infty r^2\exp(-r^2/a^2)\dd r-\frac{4}{a^4}\int_0^\infty r^4\exp(-r^2/a^2)\dd r\right] $$

Substituting the values of the integrals given in the hint, we have $\dfrac{6}{a^2}\cdot\dfrac{\sqrt\pi}{4}a^3=\dfrac{3\sqrt\pi}{2}a$ and $\dfrac{4}{a^4}\cdot\dfrac{3\sqrt\pi}{8}a^5=\dfrac{3\sqrt\pi}{2}a$, so

$$ Q_{\text{total}}=4\pi\varepsilon_0V_0\left(\frac{3\sqrt\pi}{2}a-\frac{3\sqrt\pi}{2}a\right)=0 $$

Physically, $\rho(r)$ is positive near the center ($r\lt\sqrt{3/2}\,a$) and negative outside it, and the distribution as a whole is electrically neutral. The reason the potential approaches $0$ at large distances in a Gaussian fashion (far faster than $1/r$) is that, seen from far away, the total charge is $0$ (no electric field of the $1/r^2$ form remains).

Solution

(1) The right-hand side of $I=\dd Q/\dd t$ is “the electric charge divided by the time,” so its unit is

$$ \left[\frac{\dd Q}{\dd t}\right]=\frac{[\mathrm C]}{[\mathrm s]}=\mathrm{C/s} $$

On the other hand, the right-hand side of $I=envS$ is the product of $e$ (electric charge, unit $\mathrm C$), $n$ (number per $1\ \mathrm m^3$, unit $\mathrm{1/m^3}$), $v$ (speed, unit $\mathrm{m/s}$), and $S$ (area, unit $\mathrm{m^2}$), so multiplying the units together one by one gives

$$ [envS]=\mathrm C\times\frac{1}{\mathrm m^3}\times\frac{\mathrm m}{\mathrm s}\times\mathrm m^2 =\mathrm C\times\frac{\mathrm m^3}{\mathrm m^3}\times\frac{1}{\mathrm s} =\mathrm{C/s} $$

(Counting the powers of $\mathrm m$: the numerator has $\mathrm m^1\times\mathrm m^2=\mathrm m^3$ and the denominator has $\mathrm m^3$, so they cancel exactly.) Both have the unit $\mathrm{C/s}=\mathrm A$ (ampere, Definition 37.1), which confirms that the two ways of expressing current represent quantities of the same dimension. Note also that, for the third expression $I=\iint_S\bm j\cdot\bm n\,\dd S$, since the unit of the current density $\bm j$ is $[\mathrm{A/m^2}]$ (Definition 37.2 in Section 37.2), multiplying by an area in $\mathrm m^2$ gives $\mathrm A$, which is again consistent.

(2) The cross-sectional area for a diameter of $d=0.50\ \mathrm{mm}$ is

$$ S=\pi\left(\frac d2\right)^2=\pi\times(0.25\times10^{-3}\ \mathrm m)^2\approx1.96\times10^{-7}\ \mathrm{m^2} $$

Because the current is perpendicular to the cross section and uniform, the magnitude of the current density is

$$ j=\frac IS=\frac{0.50\ \mathrm A}{1.96\times10^{-7}\ \mathrm{m^2}}\approx2.5\times10^{6}\ \mathrm{A/m^2} $$

From $j=env$, the average velocity is

$$ v=\frac{j}{en}=\frac{2.55\times10^{6}}{(1.602\times10^{-19})\times(8.49\times10^{28})}\approx1.9\times10^{-4}\ \mathrm{m/s} $$

This is faster than in the thick copper wire of Example 37.1 ($7.4\times10^{-5}\ \mathrm{m/s}$), but it is still only about $0.2\ \mathrm{mm/s}$. The thinner the wire, the faster the electrons must move to carry the same current.

Solution

(1) Let the area of the cross section perpendicular to the axis be $S=1.0\ \mathrm{mm^2}$. When the normal $\bm n$ of the cut makes an angle $\theta=30^\circ$ with the axis (the direction of $\bm j$), the area of the cut is $S'=S/\cos\theta$ (the width of the wire does not change, and the cut becomes longer because it is slanted). Therefore,

$$ S'=\frac{1.0\ \mathrm{mm^2}}{\cos30^\circ}=\frac{1.0}{0.8660}\ \mathrm{mm^2}\approx1.15\ \mathrm{mm^2} $$

(2) On the cut, $\bm j\cdot\bm n=j\cos\theta=(2.0\times10^{6})\times0.8660\approx1.73\times10^{6}\ \mathrm{A/m^2}$, and since $\bm j$ is uniform over the whole cut, the surface integral reduces to a simple product:

$$ I'=\iint_{S'}\bm j\cdot\bm n\,\dd S=(j\cos\theta)\,S'=(1.73\times10^{6})\times(1.155\times10^{-6})\approx2.0\ \mathrm A $$

On the other hand, the current through the perpendicular cross section is $I=jS=(2.0\times10^{6})\times(1.0\times10^{-6})=2.0\ \mathrm A$, so $I'=I$, and the two agree.

(3) For a general $\theta$,

$$ I'=(j\cos\theta)\,S'=(j\cos\theta)\,\frac{S}{\cos\theta}=jS=I $$

so the $\cos\theta$ factors cancel, and the current is the same regardless of the inclination $\theta$. This is a check, in the simplest case, of the derivation in Section 37.2 of the main text (using $\nabla\cdot\bm j=0$ and the divergence theorem) that the currents through two cross sections of a wire are equal.

Solution

In the Drude model of Section 37.3 of the main text, the resistivity is expressed as

$$ \rho=\frac{m_{\mathrm e}}{ne^2\tau} $$

($\tau$: the relaxation time, the average time interval between scatterings of a free electron by positive ions and the like). When the temperature of a metal rises, the positive ions undergo more violent thermal vibration about their lattice sites. From the free electron's point of view, the more strongly the ions vibrate, the higher the probability that the free electron is scattered—both the distance it can travel and the time it can go between one collision and the next become shorter. Therefore, as the temperature rises, the relaxation time $\tau$ becomes shorter.

Note that $\tau$ is in the denominator of the expression for $\rho$: if $\tau$ becomes shorter (assuming that the other quantities $m_{\mathrm e},n,e$ do not change), $\rho$ becomes larger. This is why the resistivity of a metal usually increases with temperature.

This explanation implicitly relies on the assumption that “the free-electron density $n$ is almost independent of temperature.” If $n$ also changed greatly with temperature, whether $\rho=m_{\mathrm e}/(ne^2\tau)$ increases or decreases would be decided by the balance between the changes in both $n$ and $\tau$, and we could no longer simply say that “$\tau$ decreases, so $\rho$ increases.” In a metal, the free electrons originate from the conduction electrons (valence electrons) released by each atom, and their total number (= the free-electron density $n$) is hardly affected by the arrangement of the atoms or the intensity of their thermal vibration—this is because the atoms merely vibrate while staying in place, and whether or not they release electrons does not change greatly with temperature. In contrast, in a semiconductor, the free electrons (conduction-band electrons) are newly created by thermal excitation, and as the temperature rises $n$ itself increases sharply, in an exponential manner, so this assumption does not hold, and as a result the resistivity often decreases with temperature.

Quantitative part: substituting $\alpha=4\times10^{-3}\ \mathrm{K^{-1}}$ and $T-T_0=100-20=80\ \mathrm K$ into $\rho(T)=\rho_0[1+\alpha(T-T_0)]$ gives $\rho(100\ ^\circ\mathrm C)=\rho_0\times(1+4\times10^{-3}\times80)=1.32\,\rho_0$. If the shape of the wire ($l,S$) does not change, the resistance is also multiplied by $1.32$, so

$$ R(100\ ^\circ\mathrm C)=0.168\ \Omega\times1.32\approx0.222\ \Omega $$

Also, if $n$ is constant, then $\rho\propto1/\tau$, so $\tau(100\ ^\circ\mathrm C)/\tau(20\ ^\circ\mathrm C)=1/1.32\approx0.76$. Using $\tau=2.49\times10^{-14}\ \mathrm s$ from Example 37.6, at $100\ ^\circ\mathrm C$ it becomes about $1.9\times10^{-14}\ \mathrm s$.

Solution

(a) Series connection. By Formula 37.2, the equivalent resistance is $R=R_1+R_2=6.0+3.0=9.0\ \Omega$. Since the connection is in series, the same current flows through each of $R_1,R_2$, and the total current flowing out of the source equals it:

$$ I=I_1=I_2=\frac{V}{R}=\frac{12}{9.0}=\frac43\ \mathrm A\approx1.33\ \mathrm A $$

The power dissipated in each resistor is, by Formula 37.1 ($P=I^2R$),

$$ P_1=I^2R_1=\left(\frac43\right)^2\times6.0\approx10.7\ \mathrm W,\qquad P_2=I^2R_2=\left(\frac43\right)^2\times3.0\approx5.33\ \mathrm W $$

The total power supplied by the source is $P=VI=12\times\dfrac43=16\ \mathrm W$, which agrees with $P_1+P_2=10.7+5.33=16\ \mathrm W$. (Using $P=V^2/R=144/9.0=16\ \mathrm W$ gives the same result.)

(b) Parallel connection. Since the connection is in parallel, the same voltage $V=12\ \mathrm V$ is applied across each of $R_1,R_2$ (Section 37.5.2 of the main text). By Ohm's law,

$$ I_1=\frac{V}{R_1}=\frac{12}{6.0}=2.0\ \mathrm A,\qquad I_2=\frac{V}{R_2}=\frac{12}{3.0}=4.0\ \mathrm A $$

The power dissipated in each resistor is, by Formula 37.1 ($P=V^2/R$),

$$ P_1=\frac{V^2}{R_1}=\frac{12^2}{6.0}=24\ \mathrm W,\qquad P_2=\frac{V^2}{R_2}=\frac{12^2}{3.0}=48\ \mathrm W $$

Next, by Formula 37.3 (the equivalent resistance of a parallel connection, in its two-term closed form),

$$ R=\frac{R_1R_2}{R_1+R_2}=\frac{6.0\times3.0}{6.0+3.0}=\frac{18}{9.0}=2.0\ \Omega $$

The total current flowing out of the source is $I=V/R=12/2.0=6.0\ \mathrm A$ (this also agrees with $I_1+I_2=2.0+4.0=6.0\ \mathrm A$, and is a concrete example of Kirchhoff's first law—Law 37.3 of the main text). The total power supplied by the source is, by Formula 37.1 ($P=IV$),

$$ P=IV=6.0\times12=72\ \mathrm W $$

On the other hand, $P_1+P_2=24+48=72\ \mathrm W$, which confirms that $P=P_1+P_2$ holds.

In both connections, the energy supplied by the source equals the simple sum of the energy dissipated in the individual resistors (the energy conservation law). The formulas for the equivalent resistance were derived from the sum of voltages in the series case and the sum of currents in the parallel case (Kirchhoff's laws), and this check shows that, as a result, they are also free of contradiction in the energy balance. For the same voltage, the parallel connection ($R=2.0\ \Omega$) dissipates more power ($72\ \mathrm W$ versus $16\ \mathrm W$) than the series connection ($R=9.0\ \Omega$).

Solution

We use the same setup and the same choice of current directions as Example 37.9 of the main text (in every battery, the upper side is the positive terminal). Let $I_1$ be the current flowing into node N from the left branch, $I_3$ the current flowing into it from the right branch, and $I_2$ the current flowing out of node N through the central branch ($R_2$). By Kirchhoff's first law (Law 37.3) of Chapter 37,

$$ I_1+I_3=I_2 \qquad\cdots\ \text{(a)} $$

Applying the second law (Law 37.4) to the left loop (containing $\mathcal E_1,R_1,R_2$), starting from the grounded point, we get $+\mathcal E_1-I_1R_1-I_2R_2-\mathcal E_2=0$. Here $\mathcal E_2=0\ \mathrm V$ (the central branch has no battery, only the resistor $R_2$), so

$$ \mathcal E_1-\mathcal E_2=I_1R_1+I_2R_2 \qquad\Longrightarrow\qquad \mathcal E_1=I_1R_1+I_2R_2 \qquad\cdots\ \text{(b)} $$

In the same way for the right loop (containing $\mathcal E_3,R_3,R_2$),

$$ \mathcal E_3-\mathcal E_2=I_3R_3+I_2R_2 \qquad\Longrightarrow\qquad \mathcal E_3=I_3R_3+I_2R_2 \qquad\cdots\ \text{(c)} $$

Substituting the numerical values ($R_1=4\ \Omega,R_2=2\ \Omega,R_3=6\ \Omega,\mathcal E_1=8\ \mathrm V,\mathcal E_3=5\ \mathrm V$), (b) becomes $8=4I_1+2I_2$ and (c) becomes $5=6I_3+2I_2$. Solving the three simultaneous linear equations (a), (b), (c) in three unknowns gives

$$ I_1=\frac{27}{22}\approx1.227\ \mathrm A,\qquad I_2=\frac{17}{11}\approx1.545\ \mathrm A,\qquad I_3=\frac{7}{22}\approx0.318\ \mathrm A $$

Since $I_1,I_2,I_3$ are all positive, the directions assumed at the outset ($I_1,I_3$ flowing into node N, and $I_2$ flowing out of node N) are the actual current directions. As a check, substituting into (a) gives $I_1+I_3=27/22+7/22=34/22=17/11=I_2$, which indeed holds.

Solution

Consider an infinitesimal rectangular box whose edges, of lengths $\Delta x,\Delta y,\Delta z$, are parallel to the coordinate axes. Let one of its vertices be $(x,y,z)$, and focus on the faces perpendicular to the $x$ direction (the two faces parallel to the $y$-$z$ plane). The current flowing in through the face on the smaller-$x$ side (area $\Delta y\Delta z$) is, in terms of the $x$ component $j_x(x,y,z,t)$ of the current density at that face,

$$ j_x(x,y,z,t)\,\Delta y\Delta z $$

The current flowing out through the face on the $x+\Delta x$ side is, because $\Delta x$ is an infinitesimal quantity, approximated by the first-order Taylor expansion (University Mathematics, the first-order approximation of a function of several variables) as

$$ j_x(x+\Delta x,y,z,t)\,\Delta y\Delta z\approx\left[j_x(x,y,z,t)+\pdiff{j_x}{x}\Delta x\right]\Delta y\Delta z $$

Therefore, the “net outflow” (outflow minus inflow) when we look only at the two faces in the $x$ direction is

$$ \left[j_x(x,y,z,t)+\pdiff{j_x}{x}\Delta x\right]\Delta y\Delta z-j_x(x,y,z,t)\,\Delta y\Delta z=\pdiff{j_x}{x}\,\Delta x\Delta y\Delta z $$

Exactly the same holds for the $y$ and $z$ directions, and the net outflows are $\pdiff{j_y}{y}\Delta x\Delta y\Delta z$ and $\pdiff{j_z}{z}\Delta x\Delta y\Delta z$, respectively. The total net outflow through all six faces of the box is the sum over these three directions,

$$ \left(\pdiff{j_x}{x}+\pdiff{j_y}{y}+\pdiff{j_z}{z}\right)\Delta x\Delta y\Delta z=(\nabla\cdot\bm j)\,\Delta x\Delta y\Delta z $$

($\nabla\cdot\bm j=\partial j_x/\partial x+\partial j_y/\partial y+\partial j_z/\partial z$ is precisely the definition of the divergence that we learned in University Mathematics, Section 19.1). On the other hand, as in the argument of the main text, this net outflow equals the rate of decrease of the charge $Q=\rho\,\Delta x\Delta y\Delta z$ inside the box ($\rho$ can be regarded as almost constant within the box because the volume is infinitesimal):

$$ (\nabla\cdot\bm j)\,\Delta x\Delta y\Delta z=-\frac{\dd Q}{\dd t}=-\pdiff{\rho}{t}\,\Delta x\Delta y\Delta z $$

Dividing both sides by the common volume $\Delta x\Delta y\Delta z$ gives

$$ \nabla\cdot\bm j=-\pdiff{\rho}{t} $$

and we obtain, from the infinitesimal box (a local argument), exactly the same result as the continuity equation (Law 37.1) that was derived in Section 37.2 of the main text using the divergence theorem (a global argument for a closed surface). In fact, Gauss's divergence theorem is precisely “the relation that holds for each infinitesimal box, summed over the entire region,” and the argument of this problem is itself the skeleton of the proof of the divergence theorem (see “Physical Meaning of Divergence” in University Mathematics, Section 19.2).

Solution

By Definition 37.5 of the main text, the time constant is

$$ \tau_{\mathrm{RC}}=RC=(2.0\times10^{3}\ \Omega)\times(100\times10^{-6}\ \mathrm F)=0.20\ \mathrm s $$

The voltage of the capacitor during charging is obtained by dividing the main text's expression for charging, $Q(t)=C\mathcal E(1-\exp(-t/\tau_{\mathrm{RC}}))$, by $C$:

$$ V_C(t)=\mathcal E\left(1-\exp(-t/\tau_{\mathrm{RC}})\right) $$

Substituting $t=0.10\ \mathrm s$ and $\tau_{\mathrm{RC}}=0.20\ \mathrm s$, we have $t/\tau_{\mathrm{RC}}=0.10/0.20=0.50$, so

$$ V_C(0.10)=5.0\times\left(1-\exp(-0.50)\right)=5.0\times(1-0.6065)=5.0\times0.3935\approx1.97\ \mathrm V $$

Next, for discharge from the fully charged state (voltage $\mathcal E$) through the resistor alone, dividing the main text's expression for discharging, $Q(t)=Q_0\exp(-t/\tau_{\mathrm{RC}})$, by $C$ gives

$$ V_C(t)=\mathcal E\,\exp(-t/\tau_{\mathrm{RC}}) $$

The time $t_{1/2}$ at which the voltage drops to half is found by setting $V_C(t_{1/2})=\mathcal E/2$ and solving

$$ \frac12=\exp(-t_{1/2}/\tau_{\mathrm{RC}}) \qquad\Longrightarrow\qquad t_{1/2}=\tau_{\mathrm{RC}}\ln2 $$

(we took the natural logarithm of both sides and used $\ln(1/2)=-\ln2$). Substituting the numerical values,

$$ t_{1/2}=0.20\times\ln2\approx0.20\times0.6931\approx0.139\ \mathrm s $$

The relation $t_{1/2}=\tau_{\mathrm{RC}}\ln2$ comes from the same structure as the half-life of radioactive decay: “after each equal interval of time, the remaining quantity decreases by the same fraction.”

Solution

(1) The heat generated per unit volume per unit time is $p=\bm j\cdot\bm E=\rho j^2$ (Section 37.4 of the main text):

$$ p=\rho j^2=(1.68\times10^{-8}\ \Omega\cdot\mathrm m)\times(1.0\times10^{6}\ \mathrm{A/m^2})^2=1.68\times10^{4}\ \mathrm{W/m^3} $$

(unit check: $\Omega\cdot\mathrm m\cdot\mathrm{A^2/m^4}=(\mathrm{V/A})\,\mathrm A^2/\mathrm m^3=\mathrm{W/m^3}$ is correct).

(2) The heat capacity of copper per unit volume is (density) × (specific heat) $=(8.96\times10^{3}\ \mathrm{kg/m^3})\times(385\ \mathrm{J/(kg\cdot K)})\approx3.45\times10^{6}\ \mathrm{J/(m^3\cdot K)}$. In the adiabatic case, all of the heat generation $p$ is used for the temperature rise, so

$$ \frac{\dd T}{\dd t}=\frac{p}{3.45\times10^{6}}=\frac{1.68\times10^{4}}{3.45\times10^{6}}\approx4.9\times10^{-3}\ \mathrm{K/s} $$

and over $1$ minute ($60\ \mathrm s$), $\Delta T\approx0.29\ \mathrm K$.

(3) In a real wire, heat escapes from the surface to the surrounding air (heat dissipation), so the dissipation increases as the temperature rises, and the temperature settles at the value at which heat generation and heat dissipation balance. Since the heat generation is proportional to the volume of the wire and the heat dissipation to its surface area, a thicker wire cools less easily even at the same current density. This is why the allowable current is determined by the thickness of the wire and the heat-resistance temperature of its insulation.

Solution

(1) Since the internal resistance $r$ and the external resistance $R$ are in series, the total resistance is $R+r$, and by Kirchhoff's second law $\mathcal E=IR+Ir$, that is,

$$ I=\frac{\mathcal E}{R+r} $$

The terminal voltage is obtained by subtracting the voltage drop $Ir$ across the battery's internal resistance:

$$ V=\mathcal E-Ir=\mathcal E-\frac{\mathcal E r}{R+r}=\frac{R}{R+r}\,\mathcal E $$

(This equals the voltage $IR$ across the external resistance $R$, and is also the voltage-divider formula $V=R\mathcal E/(R+r)$.) The more current is drawn, the further the terminal voltage falls below the electromotive force (emf).

(2) The power dissipated in the external resistance $R$ is

$$ P_R=I^2R=\frac{\mathcal E^2R}{(R+r)^2} $$

Differentiating with respect to $R$,

$$ \frac{\dd P_R}{\dd R}=\mathcal E^2\,\frac{(R+r)^2-2R(R+r)}{(R+r)^4}=\mathcal E^2\,\frac{r-R}{(R+r)^3} $$

This is positive for $R\lt r$ and negative for $R\gt r$, so $P_R$ is maximum at $R=r$, and the maximum value is

$$ P_{\max}=\frac{\mathcal E^2r}{(2r)^2}=\frac{\mathcal E^2}{4r} $$

(When the load resistance is matched to the internal resistance of the battery, the largest power can be extracted. In that case half of the power becomes heat in the internal resistance $r$, so the efficiency is $50\%$.)

(3) For $\mathcal E=1.5\ \mathrm V$ and $r=0.50\ \Omega$: with $R=0.50\ \Omega$, $I=1.5/(0.50+0.50)=1.5\ \mathrm A$ and $P_R=I^2R=1.5^2\times0.50=1.125\ \mathrm W\approx1.1\ \mathrm W$ (which agrees with $\mathcal E^2/(4r)=2.25/2.0=1.125\ \mathrm W$). With $R=2.0\ \Omega$, $I=1.5/2.5=0.60\ \mathrm A$ and $P_R=0.60^2\times2.0=0.72\ \mathrm W$. The power is larger for $R=r$.

Solution

The Biot–Savart law is $\dd\bm H=\dfrac{1}{4\pi r^2}\left(I\dd\bm\ell\times\hat{\bm r}\right)$. Since the current element points in the $+x$ direction and the direction to the observation point is the $+y$ direction, we can write $I\dd\bm\ell=I\dd\ell\,\bm e_x$ and $\hat{\bm r}=\bm e_y$. For the cross product of unit vectors, in a right-handed system (a coordinate system in which the $x,y,z$ axes are taken along the directions of the thumb, index finger, and middle finger of the right hand, in this order) we have $\bm e_x\times\bm e_y=\bm e_z$. Therefore,

$$ \dd\bm H=\frac{I\dd\ell}{4\pi r^2}\,\bm e_x\times\bm e_y=\frac{I\dd\ell}{4\pi r^2}\,\bm e_z $$

which shows that $\dd\bm H$ points in the $+z$ direction. Checking with the right hand: when the fingers of the right hand are curled from the direction of $\dd\bm\ell$ ($+x$) toward the direction of $\hat{\bm r}$ ($+y$), the thumb points in the $+z$ direction (turning a right-handed screw from the $x$ axis toward the $y$ axis advances the screw in the positive direction of the $z$ axis). Also, since $\dd\bm\ell\perp\hat{\bm r}$ ($\theta=90^\circ$, $\sin\theta=1$), the magnitude is $\dd H=\dfrac{I\dd\ell}{4\pi r^2}$, and $\dd\bm H$ is perpendicular to both $\dd\bm\ell$ and $\bm r$.

To draw the figure, proceed as follows. Take the $x$ axis rightward, the $y$ axis upward, and the $z$ axis pointing out of the page toward the viewer (a right-handed system). From the current element at the origin, draw $\dd\bm\ell$ as an arrow pointing right ($+x$) and $\hat{\bm r}$, directed toward the observation point, as an arrow pointing up ($+y$); at the observation point, draw $\dd\bm H$ as an arrow pointing out of the page toward the viewer ($+z$) (the symbol $\odot$, a dot enclosed in a circle). That the thumb points toward the viewer when the fingers of the right hand are curled from rightward to upward corresponds to the relationship among these three arrows.

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Solution

The foot of the perpendicular dropped from the center of the square onto each side is the midpoint of that side. The distance from the center to that foot is $r=L/2$, and the two ends of each side are at $\pm L/2$ from the foot, so the half-length is $a=L/2$. Substituting $a=r=L/2$ into Formula 38.2 of Chapter 38 (the magnetic field of a wire of finite length, $H=\dfrac{I}{2\pi r}\sin\left(\tan^{-1}\dfrac ar\right)$), the magnitude of the magnetic field produced by one side is

$$ H_1=\frac{I}{2\pi\cdot\frac L2}\sin\left(\tan^{-1}1\right)=\frac{I}{\pi L}\sin\frac\pi4=\frac{I}{\pi L}\cdot\frac{\sqrt2}{2}=\frac{\sqrt2\,I}{2\pi L} $$

(here $\tan^{-1}1=\pi/4$ and $\sin(\pi/4)=\sqrt2/2$). When the current flows counterclockwise, by the Biot–Savart law (the right-hand rule), the magnetic field produced at the center by the current element on any of the four sides is perpendicular to the page and in the same direction (out of the page toward the viewer; as in the case of a circular current, for each current element the direction of the cross product of the vector from the side to the center and the direction of the current is the same for all sides). Therefore, the four contributions, which all have the same direction, can simply be added, and

$$ H=4H_1=4\times\frac{\sqrt2\,I}{2\pi L}=\frac{2\sqrt2\,I}{\pi L}\approx0.900\,\frac{I}{L} $$

As a check, the magnetic field at the center of a circular current inscribed in the square (radius $r=L/2$) is $\dfrac{I}{2r}=\dfrac{I}{L}$, and the value for the square is slightly weaker than that (about $90\ \%$), which is consistent with the fact that the corners add portions that are farther from the center.

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Solution

Take a coordinate $x$ along the wire, with the endpoint at the origin O and the wire on the $x\ge0$ side. The observation point is a point P at distance $r$ from the endpoint, measured perpendicular to the wire. For this point the endpoint is the “foot of the perpendicular,” and with the same geometry as in Section 38.3 (let $\theta$ be the angle that the line from P to the current element makes with the perpendicular; then $x=r\tan\theta$, the distance from the current element to P is $r/\cos\theta$, and the angle between the direction of the current and $\bm r$ is $\theta+\pi/2$), the magnitude of the magnetic field produced by a current element is

$$ \dd H=\frac{I\cos^3\theta}{4\pi r^2}\,\dd x=\frac{I}{4\pi r}\cos\theta\,\dd\theta $$

(we substituted $\dd x=\dfrac{r}{\cos^2\theta}\dd\theta$). Since the wire extends from $x=0$ to $x=\infty$, the angle $\theta$ satisfies $\theta=0$ when $x=0$ and $\theta\to\pi/2$ as $x\to\infty$. Therefore,

$$ H=\int_0^{\pi/2}\frac{I}{4\pi r}\cos\theta\,\dd\theta=\frac{I}{4\pi r}\Bigl[\sin\theta\Bigr]_0^{\pi/2}=\frac{I}{4\pi r}\,(1-0)=\frac{I}{4\pi r} $$

This is exactly half of the magnetic field $\dfrac{I}{2\pi r}$ of an infinitely long wire at the same distance $r$. Thinking in terms of symmetry, an infinitely long wire can be regarded as two semi-infinite wires joined at their endpoints, each producing a magnetic field of the same magnitude and the same direction, so this is a natural result. Note also that at a point on the extension of the wire, $\dd\bm\ell\parallel\bm r$ ($\sin\theta=0$), so the magnetic field is $0$.

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Solution

Let $R$ be the radius of the solenoid, and in cylindrical coordinates centered on the axis suppose that the magnetic field inside ($s\lt R$) is parallel to the axis with magnitude $H(s)$. Consider a rectangular path consisting of two sides of length $\ell$ parallel to the axis, at distances $s_1,s_2$ from the axis (both less than $R$), joined by two sides perpendicular to the axis (in the radial direction). This rectangle fits entirely inside the solenoid and encloses no winding at all, so $I_{\text{enc}}=0$.

On the two sides perpendicular to the axis, the magnetic field (parallel to the axis) and the path (radial) are orthogonal, so $\bm H\cdot\dd\bm\ell=0$ and the contribution is $0$. On the two sides parallel to the axis, going once around the rectangle, one side is traversed in the same direction as the magnetic field and the other in the opposite direction, so

$$ \oint_C\bm H\cdot\dd\bm\ell=H(s_1)\,\ell-H(s_2)\,\ell=I_{\text{enc}}=0\qquad\therefore\ H(s_1)=H(s_2) $$

Since $s_1,s_2$ are arbitrary values inside the solenoid, the magnetic field inside is uniform, independent of the distance from the axis.

Next, we find the magnitude of the magnetic field inside by means of a different path. Of the two sides of the rectangle parallel to the axis, take one (of length $\ell$, traversed in the same direction as the magnetic field) at a distance $s_1=R/2$ inside, and the other (traversed in the direction opposite to the magnetic field) at a distance $s_2\gt R$ outside the solenoid. The magnetic field outside is $0$ (the Discussion in Section 38.4), and the contributions of the two sides perpendicular to the axis are $0$, so $\oint_C\bm H\cdot\dd\bm\ell=H\ell$. The current enclosed by this path is the current $n\ell I$ flowing in the $n\ell$ wires in the interval of length $\ell$. By Ampère's law, $H\ell=n\ell I$, that is, $H=nI$, which is the same result as in the main text (Figure 38.11). Since $\ell$ cancels from both sides, the result does not change however the length $\ell$ of the rectangle is chosen. Also, the same result is obtained wherever the inner side is placed inside the solenoid, that is, at any distance $s_1$ (because the magnetic field inside is uniform, as shown above).

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Solution

Consider the unit of the right-hand side of the straight-current formula $H=\dfrac{I}{2\pi r}$. Since $2\pi$ is a dimensionless number, the unit is $[I]/[r]=\mathrm{A}/\mathrm{m}$. In the solenoid formula $H=nI$, $n$ is the “number of turns per unit length,” and since a number of turns is a (dimensionless) number, $[n]=\mathrm{1/m}$; multiplying this by $[I]=\mathrm A$ gives $[nI]=\dfrac{1}{\mathrm m}\cdot\mathrm A=\mathrm{A/m}$. The right-hand sides of both formulas are in $\mathrm{A/m}$, which agrees with the unit of the magnetic field $H$ ($[\mathrm{N/Wb}]=[\mathrm{A/m}]$, derived from Definition 38.2 and confirmed by an equation in Chapter 38).

Next, we examine the unit of $B=\mu_0H$. Since $[\mu_0]=\mathrm{N/A^2}$ and $[H]=\mathrm{A/m}$,

$$ [B]=\frac{\mathrm N}{\mathrm A^2}\cdot\frac{\mathrm A}{\mathrm m}=\frac{\mathrm N}{\mathrm A\,\mathrm m} $$

On the other hand, using $[\mathrm{Wb}]=\dfrac{\mathrm N\,\mathrm m}{\mathrm A}$ (an equation in Chapter 38), we obtain

$$ \frac{[\mathrm{Wb}]}{\mathrm m^2}=\frac{\mathrm N\,\mathrm m}{\mathrm A\,\mathrm m^2}=\frac{\mathrm N}{\mathrm A\,\mathrm m} $$

and hence $[\mathrm{N/(A\,m)}]=[\mathrm{Wb/m^2}]=[\mathrm T]$ is verified.

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Solution

(1) The magnetic flux density at a point on the central axis of a single circular coil ($N$ turns, current $I$), at axial distance $u$ from the center of the coil, is given by the following, obtained by multiplying Formula 38.5 ($H(z)=\dfrac{Ia^2}{2(a^2+z^2)^{3/2}}$) by $N$ and setting $B=\mu_0H$:

$$ B_1(u)=\frac{\mu_0NIa^2}{2\,(a^2+u^2)^{3/2}}=\frac{\mu_0NIa^2}{2}\,f(u),\qquad f(u)=(a^2+u^2)^{-3/2} $$

Placing the centers of the two coils at $z=\mp d/2$, the magnetic flux density at the point $z$ on the axis is the sum of the contributions of the two coils (which have the same direction), namely

$$ B(z)=\frac{\mu_0NIa^2}{2}\left[f\!\left(z-\tfrac d2\right)+f\!\left(z+\tfrac d2\right)\right] $$

The midpoint is $z=0$, and for $d=a$, $f(\pm a/2)=\left(a^2+\dfrac{a^2}{4}\right)^{-3/2}=\left(\dfrac54\right)^{-3/2}a^{-3}$, so that

$$ B(0)=\frac{\mu_0NIa^2}{2}\times2\times\left(\frac45\right)^{3/2}\frac{1}{a^3}=\left(\frac45\right)^{3/2}\frac{\mu_0NI}{a}=\frac{8}{5\sqrt5}\,\frac{\mu_0NI}{a}\approx0.716\,\frac{\mu_0NI}{a} $$

This is the value of the magnetic flux density at the midpoint.

(2) Since $f(u)$ is an even function, $f'(u)$ is an odd function and $f''(u)$ is an even function. Differentiating $B(z)$ with respect to $z$ and setting $z=0$ gives

$$ B'(0)=\frac{\mu_0NIa^2}{2}\left[f'\!\left(-\tfrac d2\right)+f'\!\left(\tfrac d2\right)\right]=0 $$

(the two terms cancel because the derivative is an odd function). The second derivative is

$$ B''(0)=\frac{\mu_0NIa^2}{2}\left[f''\!\left(-\tfrac d2\right)+f''\!\left(\tfrac d2\right)\right]=\mu_0NIa^2\,f''\!\left(\tfrac d2\right) $$

Differentiating $f'(u)=-3u(a^2+u^2)^{-5/2}$ once more with the product rule gives

$$ f''(u)=-3(a^2+u^2)^{-5/2}+15u^2(a^2+u^2)^{-7/2}=\frac{-3(a^2+u^2)+15u^2}{(a^2+u^2)^{7/2}}=\frac{3(4u^2-a^2)}{(a^2+u^2)^{7/2}} $$

Substituting $u=d/2$, we have $4u^2=d^2$, so

$$ B''(0)=\mu_0NIa^2\cdot\frac{3(d^2-a^2)}{\left(a^2+\frac{d^2}{4}\right)^{7/2}} $$

This becomes $0$ if and only if $d^2=a^2$, that is, $d=a$ ($d\gt0$). Since $B(z)$ is an even function of $z$ (the two coils are symmetric about the origin), the expansion has the form $B(z)=B(0)+\dfrac12B''(0)z^2+\dfrac1{24}B''''(0)z^4+\cdots$. When $d=a$, we have $B''(0)=0$, so the $z^2$ term also vanishes, and the $z^4$ term is the first correction. For this reason the magnetic field is very uniform near the midpoint.

(3) For $a=0.15\ \mathrm m$, $N=130$, and $I=1.0\ \mathrm A$, we get

$$ B(0)=\left(\frac45\right)^{3/2}\times\frac{(1.2566\times10^{-6})\times130\times1.0}{0.15}\approx0.7155\times1.089\times10^{-3}\approx7.79\times10^{-4}\ \mathrm T=0.78\ \mathrm{mT} $$

The magnetic flux density at the center of a single coil of the same kind, obtained by multiplying Formula 38.3 by $N$, is $B=\mu_0\dfrac{NI}{2a}=\dfrac{(1.2566\times10^{-6})\times130\times1.0}{2\times0.15}\approx5.45\times10^{-4}\ \mathrm T=0.54\ \mathrm{mT}$, so the Helmholtz coil produces a magnetic field about $1.43$ times ($=2\times(4/5)^{3/2}$) that of a single coil, nearly uniform over a wide region.

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Solution

The current flows in the axial direction and the system has rotational symmetry about the axis, so the magnetic field points tangentially to circles centered on the axis, and its magnitude is determined only by the distance $r$ from the axis. Taking as the path a circle of radius $r$ centered on the axis, we have $\oint_C\bm H\cdot\dd\bm\ell=2\pi r\,H(r)=I_{\text{enc}}$. We find $I_{\text{enc}}$ for each region.

(i) $r\lt a$: The current in the inner conductor is uniform over its cross section, so the current passing through a disk of radius $r$ is $I_{\text{enc}}=I\dfrac{\pi r^2}{\pi a^2}=I\dfrac{r^2}{a^2}$, and therefore $H=\dfrac{Ir}{2\pi a^2}$.

(ii) $a\lt r\lt b$: The path encloses only the entire current $I$ of the inner conductor, so $H=\dfrac{I}{2\pi r}$.

(iii) $b\lt r\lt c$: In addition to the current $I$ of the inner conductor, the path encloses the part of the outer conductor (current $-I$, uniform over the cross-sectional area $\pi(c^2-b^2)$) that lies inside radius $r$, whose current is $-I\dfrac{\pi(r^2-b^2)}{\pi(c^2-b^2)}$. Therefore,

$$ I_{\text{enc}}=I-I\frac{r^2-b^2}{c^2-b^2}=I\frac{c^2-r^2}{c^2-b^2}\qquad\therefore\ H=\frac{I}{2\pi r}\cdot\frac{c^2-r^2}{c^2-b^2} $$

(iv) $r\gt c$: The path encloses the entire current $I$ of the inner conductor and the entire current $-I$ of the outer conductor, so $I_{\text{enc}}=I-I=0$, and therefore $H=0$.

Continuity at each boundary can be verified from the fact that at $r=a$ both (i) and (ii) give $\dfrac{I}{2\pi a}$, at $r=b$ both (ii) and (iii) give $\dfrac{I}{2\pi b}$, and at $r=c$ both (iii) and (iv) give $0$. The numerical values are as follows.

$r=0.50\ \mathrm{mm}$: $H=\dfrac{5.0\times0.50\times10^{-3}}{2\pi\times(1.0\times10^{-3})^2}\approx3.98\times10^{2}\ \mathrm{A/m}$. $r=2.0\ \mathrm{mm}$: $H=\dfrac{5.0}{2\pi\times2.0\times10^{-3}}\approx3.98\times10^{2}\ \mathrm{A/m}$. $r=3.5\ \mathrm{mm}$: $I_{\text{enc}}=5.0\times\dfrac{16-12.25}{16-9}\approx2.68\ \mathrm A$, $H=\dfrac{2.68}{2\pi\times3.5\times10^{-3}}\approx1.22\times10^{2}\ \mathrm{A/m}$. $r=5.0\ \mathrm{mm}$: $H=0$.

The reason the magnetic field is $0$ outside the cable is that the currents in the inner conductor (the outgoing current) and in the outer conductor (the return current) are equal in magnitude and opposite in direction, so the net current enclosed by any circle outside the cable is $0$. For this reason, a coaxial cable does not leak a magnetic field to the outside and is also not easily affected by external magnetic fields (noise), and it is used for television antenna lines and high-frequency signal lines.

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Solution

(1) The time for the electron to go once around the circular orbit is $T=\dfrac{2\pi a_0}{v}=\dfrac{2\pi\times5.29\times10^{-11}}{2.19\times10^{6}}\approx1.52\times10^{-16}\ \mathrm s$. During this time the charge of magnitude $e$ passes a given point on the orbit once, so the magnitude of the current is

$$ I=\frac eT=\frac{ev}{2\pi a_0}=\frac{(1.602\times10^{-19})\times(2.19\times10^{6})}{2\pi\times5.29\times10^{-11}}\approx1.06\times10^{-3}\ \mathrm A $$

This is about $1.06\ \mathrm{mA}$ (since the electron's charge is negative, the direction of the current is opposite to the electron's motion).

(2) By Definition 38.5, $m=I\times\pi a_0^2$. Substituting $I=\dfrac{ev}{2\pi a_0}$ from (1), we obtain

$$ m=\frac{ev}{2\pi a_0}\cdot\pi a_0^2=\frac{e v a_0}{2}=\frac{(1.602\times10^{-19})\times(2.19\times10^{6})\times(5.29\times10^{-11})}{2}\approx9.28\times10^{-24}\ \mathrm{A\,m^2} $$

This is nearly equal to the Bohr magneton $9.27\times10^{-24}\ \mathrm{A\,m^2}$ ($\mathrm{J/T}$) (the difference is about $0.1\ \%$, which is within the range of the significant figures of the given $v$ and $a_0$). We see that the fundamental unit of the magnetic properties of atoms can be estimated from the value obtained by regarding the orbital motion of the electron as a microscopic circular current.

(3) By Formula 38.3, the magnetic field at the center of the circle (the position of the nucleus) is

$$ H=\frac{I}{2a_0}=\frac{1.056\times10^{-3}}{2\times5.29\times10^{-11}}\approx9.98\times10^{6}\ \mathrm{A/m} $$

and the magnetic flux density is $B=\mu_0H\approx(1.2566\times10^{-6})\times(9.98\times10^{6})\approx12.5\ \mathrm T$. This is a very strong magnetic field, exceeding that of the powerful electromagnets used in laboratories (a few T), and it shows that the orbital motion of the electron alone produces such a magnetic field at the position of the nucleus.

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Solution

The formulas for the radius and period of the circular motion of a charged particle that enters perpendicular to a uniform magnetic field (Formula 39.2 of Chapter 39) are

$$ r=\frac{mv}{qB}, \qquad T=\frac{2\pi m}{qB} $$

We substitute the electron mass $m_{\mathrm e}=9.109\times10^{-31}\ \mathrm{kg}$, the magnitude of its charge $e=1.602\times10^{-19}\ \mathrm C$, the speed $v=3.0\times10^{6}\ \mathrm{m/s}$, and the magnetic flux density $B=1.0\times10^{-3}\ \mathrm T$.

The radius is

$$ r=\frac{m_{\mathrm e}v}{eB}=\frac{9.109\times10^{-31}\times3.0\times10^{6}}{1.602\times10^{-19}\times1.0\times10^{-3}}\ \mathrm m=\frac{2.733\times10^{-24}}{1.602\times10^{-22}}\ \mathrm m\approx1.71\times10^{-2}\ \mathrm m\approx1.7\ \mathrm{cm} $$

The period is

$$ T=\frac{2\pi m_{\mathrm e}}{eB}=\frac{2\pi\times9.109\times10^{-31}}{1.602\times10^{-19}\times1.0\times10^{-3}}\ \mathrm s\approx3.57\times10^{-8}\ \mathrm s\approx36\ \mathrm{ns} $$

The mass of a proton is about 1800 times that of an electron (compare with the proton mass $1.673\times10^{-27}\ \mathrm{kg}$ in Example 39.5), so if a proton circled at the same speed in the same magnetic flux density, both its radius and its period would be about 1800 times larger. We see that, being light, the electron revolves with a small radius and a short period.

Solution

We take rightward as the positive direction of the $x$ axis, upward as the positive direction of the $y$ axis, and out of the page as the positive direction of the $z$ axis (a right-handed system, $\bm e_x\times\bm e_y=\bm e_z$). As in the hint, we take the direction of the current (the direction of the wire) as the positive $x$ direction and the magnetic field as the positive $y$ direction: $\bm\ell=\ell\,\bm e_x$, $\bm B=B\,\bm e_y$.

By the component calculation of the cross product (University Mathematics, Section 15.4) and $\bm e_x\times\bm e_y=\bm e_z$,

$$ \bm F=I\bm\ell\times\bm B=I\ell B\,(\bm e_x\times\bm e_y)=I\ell B\,\bm e_z $$

That is, the force points in the $+z$ direction (out of the page) and has magnitude $I\ell B$.

Next, we check this with Fleming's left-hand rule. Spread the thumb, index finger, and middle finger of the left hand at right angles to one another, and align the middle finger with the direction of the current ($+x$) and the index finger with the direction of the magnetic field ($+y$). The three fingers of the left hand then form a left-handed system. Let $\bm p$ be the direction of the thumb, $\bm q$ that of the index finger, and $\bm r$ that of the middle finger. For a right-handed system we would have $\bm p\times\bm q=\bm r$, but for a left-handed system $\bm p\times\bm q=-\bm r$. Here $\bm q=\bm e_y,\ \bm r=\bm e_x$, so the thumb must satisfy $\bm p\times\bm e_y=-\bm e_x$. Of the two candidates $\bm p=\pm\bm e_z$, the answer is $\bm p=+\bm e_z$, for which $\bm e_z\times\bm e_y=-\bm e_x$ holds, so the thumb points in the $+z$ direction. This agrees with the result of the cross product, $\bm F\parallel+\bm e_z$.

Let us also check another configuration. When the current points in the $+x$ direction and the magnetic field points into the page (the $-z$ direction, $\bm B=-B\bm e_z$), then since $\bm e_x\times\bm e_z=-\bm e_y$,

$$ \bm F=I\ell\,\bm e_x\times(-B\bm e_z)=-I\ell B\,(\bm e_x\times\bm e_z)=+I\ell B\,\bm e_y $$

so the force points upward. With the left-hand rule, $\bm q=-\bm e_z,\ \bm r=\bm e_x$ gives $\bm p\times(-\bm e_z)=-\bm e_x$, that is, $\bm p\times\bm e_z=\bm e_x$. The vector $\bm p$ satisfying this is the direction of the thumb, and since $\bm e_y\times\bm e_z=\bm e_x$ we get $\bm p=+\bm e_y$ (upward), which again agrees with the result of the cross product.

In fact, in general, when $\bm\ell\perp\bm B$ the force $\bm F=I\bm\ell\times\bm B$ satisfies $\bm F\times\bm B=-B^2\,I\bm\ell$ (since $(\bm\ell\times\bm B)\times\bm B=-\ell B^2\,\bm e_\ell$, where $\bm e_\ell$ is the unit vector in the direction of $\bm\ell$), so the triple $(\bm F,\bm B,\bm\ell)$ always stands in the same relationship as the (thumb, index finger, middle finger) of the left hand. Fleming's left-hand rule is a geometric device devised so that the direction of the force can be found from the shape of the fingers alone, without knowing the operation called the cross product, and it essentially expresses the same physical law.

Solution

When a charged particle (charge $q$, mass $m$) is acted on by forces from both an electric field $\bm E$ and a magnetic field $\bm B$ (perpendicular to each other), its equation of motion is

$$ m\frac{\dd\bm v}{\dd t}=q\bm E+q\bm v\times\bm B $$

We choose coordinates such that $\bm B=B\bm e_z$ and $\bm E=E\bm e_x$ ($\bm E\perp\bm B$). We compute the drift velocity $\bm v_{\mathrm d}=\bm E\times\bm B/B^2$ in components. Since $\bm e_x\times\bm e_z=-\bm e_y$,

$$ \bm E\times\bm B=EB\,(\bm e_x\times\bm e_z)=-EB\,\bm e_y,\qquad \bm v_{\mathrm d}=\frac{-EB\,\bm e_y}{B^2}=-\frac{E}{B}\,\bm e_y $$

Next, to find the Lorentz force that $\bm v_{\mathrm d}$ experiences from the magnetic field, we compute $\bm v_{\mathrm d}\times\bm B$; since $\bm e_y\times\bm e_z=\bm e_x$,

$$ \bm v_{\mathrm d}\times\bm B=-\frac{E}{B}\,B\,(\bm e_y\times\bm e_z)=-E\,\bm e_x=-\bm E $$

This is what we obtain (the same result holds for $\bm E\perp\bm B$ in a general orientation: using the vector triple product formula $\bm a\times(\bm b\times\bm c)=\bm b(\bm a\cdot\bm c)-\bm c(\bm a\cdot\bm b)$, we have $(\bm E\times\bm B)\times\bm B=-\bm B\times(\bm E\times\bm B)=-[\bm E(\bm B\cdot\bm B)-\bm B(\bm B\cdot\bm E)]=-B^2\bm E$ (we used $\bm B\cdot\bm E=0$), so that $\bm v_{\mathrm d}\times\bm B=-\bm E$). Therefore,

$$ q\bm E+q\bm v_{\mathrm d}\times\bm B=q\bm E-q\bm E=\bm 0 $$

Thus, for the component moving with the drift velocity $\bm v_{\mathrm d}$, the Coulomb force due to the electric field and the Lorentz force that this motion itself experiences from the magnetic field exactly cancel, and no net force acts. This does not depend on the sign of $q$ or on $m$.

We therefore split the velocity as $\bm v=\bm v_{\mathrm d}+\bm v'$ and substitute it into the equation of motion. Since $\bm v_{\mathrm d}$ is constant, $\dd\bm v_{\mathrm d}/\dd t=\bm 0$, and

$$ m\frac{\dd\bm v'}{\dd t}=q\bm E+q(\bm v_{\mathrm d}+\bm v')\times\bm B=\bigl(q\bm E+q\bm v_{\mathrm d}\times\bm B\bigr)+q\bm v'\times\bm B=q\bm v'\times\bm B $$

This is the equation of the circular motion (cyclotron motion) due to the Lorentz force that we learned in this chapter, in the same form as in the absence of an electric field.

Therefore, the motion of the charged particle is a combination of “a straight drift at the constant velocity $\bm v_{\mathrm d}=\bm E\times\bm B/B^2$” and “an ordinary circular motion superimposed on it.” When only an electric field is present, the particle keeps being acted on by a force and its speed increases steadily, but when a magnetic field is present at the same time, the part due to the electric force is transformed into a “sideways flow (drift)” at a constant velocity, whose speed does not change. This is why, under crossed electric and magnetic fields, a particle does not keep being accelerated but instead flows along at a constant velocity.

Solution

Consider a charged particle moving slowly, in a helical motion, through a region where the strength of the magnetic field varies from place to place (we assume that the magnetic field changes so gradually that its change during one revolution of the particle is almost negligible). Let the velocity component perpendicular to the magnetic field be $v_\perp=v\sin\theta$ and the parallel component be $v_\parallel=v\cos\theta$. Then the radius of the circular motion is $r=mv_\perp/(qB)$, and the magnetic flux through the surface enclosed by the circle traced out in one revolution is

$$ \Phi=\pi r^2B=\pi\left(\frac{mv_\perp}{qB}\right)^2B=\frac{\pi m^2}{q^2}\cdot\frac{v_\perp^2}{B} $$

When the magnetic field changes slowly, the magnetic flux $\Phi$ enclosed by the circle that the particle goes around is kept approximately constant. This is a fact known as the adiabatic invariant; it can be understood as follows: when the magnetic field changes, electromagnetic induction (Chapter 40) produces an electric field along the circular orbit, which changes the speed of the particle and keeps the magnetic flux nearly constant. This is an advanced topic beyond the scope of this chapter, so here we accept it as a fact and use it. Since $\pi m^2/q^2$ is a constant, $\Phi$ being constant means that $v_\perp^2/B$ is constant, that is,

$$ \frac{v_\perp^2}{B}=\text{constant}\qquad\left(\text{equivalently, the magnetic moment}\ \mu=\frac{mv_\perp^2}{2B}\ \text{is constant}\right) $$

Therefore, as the particle enters a region where the magnetic field $B$ becomes larger, the perpendicular component of its speed increases according to $v_\perp^2\propto B$. On the other hand, since the Lorentz force does no work (Section 39.6), the speed $v=\sqrt{v_\perp^2+v_\parallel^2}$ of the particle stays constant throughout the motion. Since $v_\perp$ increases while $v$ as a whole does not change, the component parallel to the magnetic field, $v_\parallel^2=v^2-v_\perp^2$, decreases by the corresponding amount.

As the particle moves into regions of ever stronger magnetic field, $v_\perp$ keeps increasing and $v_\parallel$ keeps decreasing. When it reaches a sufficiently strong field, we finally have $v_\parallel=0$ (the instant at which the particle no longer advances at all along the direction of the field). This point is called the mirror point. After $v_\parallel=0$, $v_\parallel$ changes sign and becomes negative, and the particle is pushed back in the direction from which it came. This closely resembles a ball climbing a slope: at the point where it has used up its kinetic energy its velocity becomes zero, and it rolls back down the slope and returns—the variation in the strength of the magnetic field plays a role like that of the “slope.” This phenomenon is called the magnetic mirror effect. It is at work in the phenomenon in which charged particles approaching the polar regions (where the magnetic field is strong) along the Earth's magnetic field lines are reflected and confined in the magnetosphere (the Van Allen belts), and it is used as the principle of magnetic confinement devices for nuclear fusion (mirror machines).

However, a particle with a large $v_\parallel$ (a particle whose angle $\theta$ with the magnetic field is small), even though $v_\perp^2/B$ is constant, still has $v_\parallel\gt0$ when it reaches the maximum magnetic field near the pole, and so it is not reflected but precipitates into the atmosphere (the loss cone). It is the particles that precipitate in this way that make the aurora of Section 39.7 glow.

Solution

The expression for the Hall voltage derived in Chapter 39 is

$$ V_{\mathrm H}=\frac{IB}{nqt} $$

Solving this for the carrier density $n$ gives

$$ n=\frac{IB}{qV_{\mathrm H}t} $$

Since the carriers are electrons (n-type), $q=-e$ and $V_{\mathrm H}$ is also negative, but here we use the magnitudes $|q|=e=1.602\times10^{-19}\ \mathrm C$ and $|V_{\mathrm H}|=1.0\times10^{-3}\ \mathrm V$. Substituting the values $I=2.0\times10^{-3}\ \mathrm A$, $B=0.80\ \mathrm T$, and $t=0.50\times10^{-3}\ \mathrm m$ gives

$$ n=\frac{(2.0\times10^{-3})\times0.80}{(1.602\times10^{-19})\times(1.0\times10^{-3})\times(0.50\times10^{-3})}\ \mathrm{m^{-3}} $$

The numerator is $1.6\times10^{-3}$ and the denominator is $1.602\times10^{-19}\times1.0\times10^{-3}\times0.50\times10^{-3}=8.01\times10^{-26}$, so

$$ n\approx\frac{1.6\times10^{-3}}{8.01\times10^{-26}}\ \mathrm{m^{-3}}\approx2.0\times10^{22}\ \mathrm{m^{-3}} $$

This is smaller by about six orders of magnitude than the free-electron density of a typical metal ($\sim10^{28}$–$10^{29}\ \mathrm{m^{-3}}$), and about 20 times the value for the semiconductor treated in Example 39.10 ($n=1.0\times10^{21}\ \mathrm{m^{-3}}$). It is the value for a rather heavily doped n-type semiconductor, close to the upper end of the carrier densities of doped semiconductors ($10^{20}$–$10^{22}\ \mathrm{m^{-3}}$) described in the Application box of Chapter 39.

Solution

Sample answer (about 200 characters in Japanese):

The Ampère force is a force acting on the whole visible object, the wire, but its true nature is the Lorentz force that each individual free electron moving inside the wire experiences from the magnetic field. Even though the force on each electron is tiny, the wire contains more than $10^{22}$ electrons, and all of them are acted on by forces in the same direction, so the forces simply add up to a force of visible size. A single microscopic law, accumulated, appears as a macroscopic law.

(Key points of the solution: as long as your answer touches on three points—that each individual Lorentz force is tiny, that the number of free electrons is enormous, and that the forces can simply be added because their directions are aligned—you may rephrase it in your own words.)

Solution

(a) The Hall voltage was defined as “the potential of the $-y$ face minus the potential of the $+y$ face.” $V_{\mathrm H}=-25\ \mathrm{mV}\lt0$ means that the potential of the $-y$ face is lower than that of the $+y$ face, that is, that the $-y$ face is charged negatively. When the current is in the $+x$ direction and the magnetic field is in the $+z$ direction, the carriers are pushed toward the $-y$ side and accumulate there whether they are holes or electrons (derivation (1) in Section 39.8), so the charge of the carriers accumulated on the $-y$ face is negative. Therefore the carriers are electrons, and this semiconductor is n-type (Table 39.1).

(b) The charge of the carriers is $q=-e$, and solving the equation $V_{\mathrm H}=IB/(nqt)$ for $n$ gives, in magnitude,

$$ n=\frac{IB}{e\,|V_{\mathrm H}|\,t}=\frac{5.0\times10^{-3}\times0.10}{1.602\times10^{-19}\times25\times10^{-3}\times0.20\times10^{-3}}\ \mathrm{m^{-3}}=\frac{5.0\times10^{-4}}{8.01\times10^{-25}}\ \mathrm{m^{-3}}\approx6.2\times10^{20}\ \mathrm{m^{-3}} $$

The Hall coefficient, with its sign, is

$$ R_{\mathrm H}=\frac{1}{nq}=-\frac{1}{ne}=-\frac{1}{6.24\times10^{20}\times1.602\times10^{-19}}\ \mathrm{m^3/C}\approx-1.0\times10^{-2}\ \mathrm{m^3/C} $$

As a check, $V_{\mathrm H}=R_{\mathrm H}IB/t=-1.0\times10^{-2}\times5.0\times10^{-3}\times0.10/(0.20\times10^{-3})\ \mathrm V=-25\ \mathrm{mV}$, which agrees with the measured value.

Solution

(a) Solving the equation $F/\ell=\mu_0I_1I_2/(2\pi a)$ for $\mu_0$ gives $\mu_0=2\pi a\,(F/\ell)/(I_1I_2)$. Substituting the values of the old definition, $I_1=I_2=1\ \mathrm A,\ a=1\ \mathrm m,\ F/\ell=2\times10^{-7}\ \mathrm{N/m}$, we get

$$ \mu_0=\frac{2\pi\times1\ \mathrm m\times2\times10^{-7}\ \mathrm{N/m}}{1\ \mathrm A\times1\ \mathrm A}=4\pi\times10^{-7}\ \mathrm{N/A^2} $$

(The unit is $\mathrm{m\cdot(N/m)/A^2}=\mathrm{N/A^2}$.)

(b) Since $4\pi\times10^{-7}=1.256\,637\,061\,44\times10^{-6}\ \mathrm{N/A^2}$, the relative deviation is

$$ \frac{\mu_0-4\pi\times10^{-7}}{4\pi\times10^{-7}}=\frac{(1.256\,637\,062\,12-1.256\,637\,061\,44)\times10^{-6}}{1.256\,637\,061\,44\times10^{-6}}\approx\frac{6.8\times10^{-10}}{1.2566}\approx5.4\times10^{-10} $$

This is larger than the relative uncertainty of the measured value (about $1.5\times10^{-10}$), but extremely small.

(c) Since $F\propto\mu_0$, the force also becomes larger by a relative amount of about $5.4\times10^{-10}$, giving $2\times10^{-7}\times(1+5.4\times10^{-10})\ \mathrm{N/m}\approx2.000\,000\,001\times10^{-7}\ \mathrm{N/m}$. To nine significant figures it remains $2\times10^{-7}\ \mathrm{N/m}$, so there is practically no difference between the old definition and the present value.

Solution

Since the magnet is moved from left to right with its N pole in front, we consider three stages according to the position of the magnet.

(a) While the magnet is approaching the coil. The magnetic field lines emerging from the N pole pass through the coil toward the right, and this flux increases as the magnet approaches. By Lenz's law, the current flows in the direction that produces a leftward magnetic field canceling the increase of the rightward flux. By the right-hand rule, this is counterclockwise as seen from the side of the magnet (the left side) (Table 40.1, (i)).

(b) When the magnet is at the center of the coil. The rightward flux through the coil is at its maximum, and its rate of change $\dd\Phi/\dd t$ is $0$ (the instant at which it changes from increasing to decreasing). By Faraday's law $V=-N\,\dd\Phi/\dd t$, the induced electromotive force (emf) is $0$ at this moment, and the induced current is also $0$. The current reverses its direction from counterclockwise to clockwise, passing through $0$.

(c) While the magnet passes through and moves away. The rear end of the magnet (the S pole side) is leaving the coil. At this time the flux through the coil is still rightward (the field lines outside the magnet go out from the N pole and enter the S pole, so to the left of the S pole they point rightward, toward the magnet), but it decreases as the magnet moves away. By Lenz's law, the current flows in the direction that produces a rightward magnetic field to make up for the decrease of the rightward flux, and this is clockwise as seen from the side of the magnet (the same as (ii) in Table 40.1).

To summarize: when the magnet approaches, the current is counterclockwise; when it passes through and moves away, it is clockwise; and at the center of the coil the current is $0$. The way the flux $\Phi(t)$ changes as “increase → maximum → decrease” (the change of sign of $\dd\Phi/\dd t$) appears directly as the reversal of the direction of the current.

Solution

Model answer (about 200 characters in the Japanese original). The divergence $\operatorname{div}$ is an operator that measures outflow (sources), and the curl $\operatorname{rot}$ is an operator that measures the strength of vortices;$\operatorname{div}\bm D=\rho$ expresses that electric charges are the sources of the electric field, and $\operatorname{rot}\bm E=-\partial\bm B/\partial t$ expresses that the electric field does net work along a closed curve (it circulates in vortices). On the right-hand side, whereas the former has a source, the charge density, the latter has no intrinsic source corresponding to charge, and instead the time variation of the magnetic field appears. The source of the vortices of the electric field is not a new kind of charge but the time variation of the magnetic field.

Supplement. In a vacuum $\bm D=\varepsilon_0\bm E$, so $\operatorname{div}\bm D=\rho$ has the same content as $\operatorname{div}\bm E=\rho/\varepsilon_0$. Also, if magnetic monopoles (magnetic charges) existed, a term corresponding to the flow of magnetic charge would have to be added to the right-hand side of the equation for $\operatorname{rot}\bm E$ as well, but no such thing has been observed. In an electrostatic field $\operatorname{rot}\bm E=\bm 0$, so vortices of the electric field appear only when the magnetic field varies in time.

Solution

From the Lorentz force expression $F=qvB$ we have $B=F/(qv)$, and using $q=It$ (charge = current × time), the unit of magnetic flux density is

$$ 1\ \mathrm T=\frac{\mathrm N}{\mathrm C\cdot\mathrm{m/s}}=\frac{\mathrm N}{\mathrm A\,\mathrm s\cdot\mathrm{m/s}}=\frac{\mathrm N}{\mathrm A\,\mathrm m} $$

Since $1\ \mathrm N=\mathrm{kg\,m/s^2}$,

$$ 1\ \mathrm T=\frac{\mathrm{kg\,m\,s^{-2}}}{\mathrm A\,\mathrm m}=\mathrm{kg\,s^{-2}\,A^{-1}} $$

Therefore, since $\ell$ is in $\mathrm m$ and $v$ is in $\mathrm{m/s}$, the unit of $B\ell v$ is

$$ [B][\ell][v]=(\mathrm{kg\,s^{-2}\,A^{-1}})\cdot\mathrm m\cdot(\mathrm{m\,s^{-1}})=\mathrm{kg\,m^2\,s^{-3}\,A^{-1}} $$

On the other hand, the unit of voltage, $\mathrm V$, is “power ÷ current”; the unit of power, $\mathrm W$, is “work ÷ time”; and the unit of work (energy), $\mathrm J$, is $\mathrm{kg\,m^2/s^2}$ (from the unit of the kinetic energy $\frac12mv^2$). Hence

$$ 1\ \mathrm V=\frac{1\ \mathrm W}{1\ \mathrm A}=\frac{1\ \mathrm{J/s}}{1\ \mathrm A}=\frac{\mathrm{kg\,m^2\,s^{-2}}}{\mathrm s\cdot\mathrm A}=\mathrm{kg\,m^2\,s^{-3}\,A^{-1}} $$

which is exactly the same as the unit of $B\ell v$. Therefore the right-hand side of $V=B\ell v$ indeed has the unit of voltage (this check starts only from the Lorentz force and the definition of electric power, without relying on the unit in Faraday's law, $\mathrm{Wb}=\mathrm{V\,s}$, so it is an independent check).

Solution

(1) In the steady state the current in the coil is no longer changing ($\dd I/\dd t=0$), so the voltage drop across the coil is zero, and the entire emf of the battery appears across the resistor. By Ohm's law,

$$ I_\infty=\frac{V_0}{R}=\frac{20}{100}=0.20\ \mathrm A $$

(2) Using the formula $U_L=\frac12LI^2$ of the main text, the energy stored in the coil is

$$ U_L=\frac12LI_\infty^2=\frac12\times0.50\times0.20^2=\frac12\times0.50\times0.040=0.010\ \mathrm J $$

(3) After the battery is disconnected, the decay of the current follows the equation of the main text ($I(t)=I_0\exp(-t/\tau)$, $\tau=L/R$). Here $I_0=I_\infty=0.20\ \mathrm A$, and the time constant is

$$ \tau=\frac{L}{R}=\frac{0.50}{100}=5.0\times10^{-3}\ \mathrm s $$

To find the time $t_{1/2}$ at which the current becomes half of its steady value, $I_0/2$, it suffices to solve $I_0\exp(-t_{1/2}/\tau)=I_0/2$. Dividing both sides by $I_0$ gives $\exp(-t_{1/2}/\tau)=1/2$; taking the natural logarithm of both sides gives $-t_{1/2}/\tau=\ln(1/2)=-\ln2$, and therefore

$$ t_{1/2}=\tau\ln2\approx5.0\times10^{-3}\times0.6931\approx3.47\times10^{-3}\ \mathrm s\approx3.5\ \mathrm{ms} $$

This is the required time.

Solution

By the formula of the main text, the angular frequency of an LC circuit is $\omega=1/\sqrt{LC}$ and the period is $T=2\pi/\omega=2\pi\sqrt{LC}$. If the capacitance is changed as $C\to4C$ without changing the self-inductance $L$, the new period is

$$ T'=2\pi\sqrt{L\cdot4C}=2\pi\sqrt{4}\sqrt{LC}=2\times2\pi\sqrt{LC}=2T $$

In other words, quadrupling the capacitance doubles the period (and halves the angular frequency). Concretely, for the original period $T=1.0\times10^{-3}\ \mathrm s$, the new period is $T'=2.0\times10^{-3}\ \mathrm s$.

By the equation of the main text, the amplitude of the current is $I_{\max}=Q_0\omega$. Since the initial charge $Q_0$ of the capacitor is unchanged, the amplitude is proportional to $\omega'=\omega/2$ (from $C\to4C$, $\omega=1/\sqrt{LC}\to1/\sqrt{4LC}=\omega/2$), so

$$ \frac{I_{\max}'}{I_{\max}}=\frac{Q_0\omega'}{Q_0\omega}=\frac{\omega'}{\omega}=\frac12 $$

This shows that the amplitude of the current becomes $\dfrac12$ times its original value. We see that when the capacitance is increased (the voltage is lower for the same charge), the oscillation becomes slower and the amplitude of the current also becomes smaller.

Solution

Suppose that Lenz's law were reversed—that is, that the induced current flowed in the direction that reinforces the change of the flux. For example, when the N pole of a magnet is brought close to the coil, the flux through the coil is increasing to begin with. If Lenz's law were reversed, the induced current would produce a magnetic field in the direction that pushes this increase still further. Such a coil (an electromagnet) would exert on the magnet a force in the direction that attracts it. Then the magnet would be drawn toward the coil and accelerate on its own, without any external force; the closer it got to the coil, the larger the rate of change of the flux would become, the stronger the induced current would become, and the stronger the attractive force would become—both the kinetic energy of the magnet and the Joule heat that the induced current generates in the resistance would keep increasing without limit, even though no energy is supplied from outside. This clearly violates the law of conservation of energy and is physically impossible. Therefore the induced current must be in the direction that cancels the change of the flux (the direction in which it exerts a force that pushes the magnet away)—this is the reason for the direction in Lenz's law.

The same holds for electromagnetic braking (eddy-current braking). When a conductor moves in a magnetic field, it functions as a “brake” precisely because the force produced by the eddy currents always acts in the direction that opposes the motion of the conductor. If the direction of the force were reversed—the direction that reinforces the motion—the conductor, instead of being braked, would keep accelerating on its own although no energy is supplied from outside, which would also violate the law of conservation of energy. The very fact that electromagnetic brakes observed in practice always act as braking (deceleration) is one piece of evidence that the direction given by Lenz's law is physically correct.

Solution

(1) The angular velocity is $\omega=2\pi f=2\pi\times60\approx376.99\ \mathrm{rad/s}$. By Formula 40.6, the maximum value (amplitude) is

$$ V_0=NBA\omega=50\times0.30\times0.040\times376.99\ \mathrm V=0.60\times376.99\ \mathrm V\approx226.2\ \mathrm V\approx2.3\times10^{2}\ \mathrm V $$

By Formula 40.7, the effective value is

$$ V_{\mathrm{eff}}=\frac{V_0}{\sqrt2}=\frac{226.2}{1.4142}\ \mathrm V\approx159.94\ \mathrm V\approx1.6\times10^{2}\ \mathrm V $$

(2) For the same coil and magnetic field, $V_0=NBA\cdot2\pi f$ is proportional to the frequency $f$, and $V_{\mathrm{eff}}$ is also proportional to $f$. To make the effective value $100\ \mathrm V$, the frequency should be multiplied by $100/159.94\approx0.6252$, so

$$ f'=60\times\frac{100}{159.94}\ \mathrm{Hz}\approx37.5\ \mathrm{Hz} $$

(To find it directly, solve $V_{\mathrm{eff}}=NBA\cdot2\pi f/\sqrt2=100\ \mathrm V$ for $f$: $f=\dfrac{100\sqrt2}{2\pi\times50\times0.30\times0.040}\ \mathrm{Hz}\approx37.5\ \mathrm{Hz}$.)

Solution

(1) The free charges in the plate experience a force $\bm v\times\bm B$ per unit charge (the Lorentz force of Section 40.4). Since $\bm v\perp\bm B$, its magnitude is $vB$. In Ohm's law $\bm j=\sigma\bm E_{\mathrm{eff}}$, if we use this force in place of the electric field, the current density is

$$ \bm j=\sigma\,\bm v\times\bm B,\qquad j=\sigma vB $$

The unit is $\mathrm{S/m}\cdot\mathrm{m/s}\cdot\mathrm T=\mathrm{A/V\cdot(1/s)\cdot V\,s/m^2}=\mathrm{A/m^2}$, which agrees with the unit of current density.

(2) The magnetic force per unit volume (the density of the Ampère force) is $\bm f=\bm j\times\bm B$. For $\bm v\perp\bm B$, using the relation $(\bm v\times\bm B)\times\bm B=-B^2\bm v$ (from $\bm a\times(\bm b\times\bm c)=\bm b(\bm a\cdot\bm c)-\bm c(\bm a\cdot\bm b)$ we get $(\bm v\times\bm B)\times\bm B=-\bm B\times(\bm v\times\bm B)=-[\bm v(\bm B\cdot\bm B)-\bm B(\bm B\cdot\bm v)]=-B^2\bm v$), we obtain

$$ \bm f=\sigma(\bm v\times\bm B)\times\bm B=-\sigma B^2\,\bm v $$

Its magnitude is $f=\sigma B^2v$, which is proportional to the speed $v$, and its direction is opposite to the velocity (the direction that opposes the motion). This means that the proportionality constant $\gamma$ in $F=\gamma v$ of Example 40.1 is a quantity proportional to $\sigma B^2$. The chain of causation—“the greater the speed, the greater the induced emf ($vB$), the greater the current, and the greater the force that this current experiences from the magnetic field”—produces a braking force proportional to the first power of $v$.

(3) Substituting the numerical values,

$$ f=\sigma B^2v=3.5\times10^{7}\times0.50^2\times2.0\ \mathrm{N/m^3}=1.75\times10^{7}\ \mathrm{N/m^3} $$

The magnitude of the force acting on a portion of volume $1.0\ \mathrm{cm^3}=1.0\times10^{-6}\ \mathrm{m^3}$ is

$$ F=f\times1.0\times10^{-6}\ \mathrm{m^3}=17.5\ \mathrm N\approx18\ \mathrm N $$

Also, the current density is $j=\sigma vB=3.5\times10^{7}\times2.0\times0.50\ \mathrm{A/m^2}=3.5\times10^{7}\ \mathrm{A/m^2}$, and the power dissipated as Joule heat in this portion is

$$ \frac{j^2}{\sigma}\times1.0\times10^{-6}\ \mathrm{m^3}=\frac{(3.5\times10^{7})^2}{3.5\times10^{7}}\times10^{-6}\ \mathrm W=35\ \mathrm W $$

This agrees with the power of the braking force,$Fv=17.5\times2.0\ \mathrm W=35\ \mathrm W$, which confirms that the kinetic energy is entirely converted into Joule heat. Note that this is a simple model that ignores such matters as the need for the current to close over the whole plate, so it serves as a guide for estimating an upper bound on the actual value.

Solution

(a) Case where the charge $Q$ is held constant (the battery is disconnected before the dielectric is inserted). Let $C_0$ be the capacitance before insertion and $C=\varepsilon_{\mathrm{r}}C_0$ the capacitance after insertion (Law 41.3 of Chapter 41). Since the charge $Q$ does not change, using the electrostatic energy $U=Q^2/(2C)$ gives

$$ U_i=\frac{Q^2}{2C_0},\qquad U_f=\frac{Q^2}{2C}=\frac{Q^2}{2\varepsilon_{\mathrm{r}}C_0}=\frac{U_i}{\varepsilon_{\mathrm{r}}} $$

Thus, if $\varepsilon_{\mathrm{r}}\gt1$, then $U_f\lt U_i$; that is, the energy decreases. Since there is no battery, the entire change in the capacitor's energy equals the work done by the hand (the external force), so

$$ W_{\text{hand}}=U_f-U_i=-\Bigl(1-\frac1{\varepsilon_{\mathrm{r}}}\Bigr)U_i\lt0 $$

That the work done by the hand is negative means that the dielectric experiences a force in the direction of being pulled in between the electrodes, so the hand actually has to hold the dielectric back (do negative work) to keep it from rushing in suddenly. Numerical example: for $C_0=88.5\ \mathrm{pF}$, $V=100\ \mathrm V$ ($Q=C_0V=8.85\ \mathrm{nC}$), and $\varepsilon_{\mathrm{r}}=5$, we have $U_i=\frac12C_0V^2=4.43\times10^{-7}\ \mathrm J$, $U_f=U_i/5=8.85\times10^{-8}\ \mathrm J$, and $W_{\text{hand}}=-3.54\times10^{-7}\ \mathrm J$.

(b) Case where the voltage $V$ is held constant (the dielectric is inserted with the battery still connected). Since the voltage $V$ does not change, using $U=\frac12CV^2$ gives

$$ U_i=\frac12C_0V^2,\qquad U_f=\frac12CV^2=\frac12\varepsilon_{\mathrm{r}}C_0V^2=\varepsilon_{\mathrm{r}}U_i $$

Here the energy stored in the capacitor instead increases: $\Delta U_{\text{cap}}=U_f-U_i=\frac12(\varepsilon_{\mathrm{r}}-1)C_0V^2$. In this case the battery sends an additional charge $\Delta Q=(C-C_0)V=(\varepsilon_{\mathrm{r}}-1)C_0V$ into the capacitor, so the battery does work $W_{\text{battery}}=V\Delta Q=(\varepsilon_{\mathrm{r}}-1)C_0V^2$. This is exactly twice $\Delta U_{\text{cap}}$. From conservation of energy (the increase in the capacitor's energy $=$ the work done by the hand $+$ the work done by the battery),

$$ W_{\text{hand}}=\Delta U_{\text{cap}}-W_{\text{battery}}=\frac12(\varepsilon_{\mathrm{r}}-1)C_0V^2-(\varepsilon_{\mathrm{r}}-1)C_0V^2=-\frac12(\varepsilon_{\mathrm{r}}-1)C_0V^2\lt0 $$

Numerical example: $\Delta U_{\text{cap}}=1.77\times10^{-6}\ \mathrm J$, $W_{\text{battery}}=3.54\times10^{-6}\ \mathrm J$, and $W_{\text{hand}}=-1.77\times10^{-6}\ \mathrm J$. Half of the energy $W_{\text{battery}}$ supplied by the battery becomes the increase in the electrostatic energy of the capacitor, and the other half goes into the work done on the dielectric by the force that pulls it in (which becomes the kinetic energy of the dielectric if the hand does not hold it back, or heat if there is friction).

In both cases $W_{\text{hand}}\lt0$, and the conclusion is the same: “the dielectric is pulled in between the electrodes, and the hand does work on the side of holding it back.” This pulling-in force arises not because the dielectric is attracted to the electrodes, but because the electric field at the edges of the electrodes (the fringe field), which lies near the rims of the electrodes and has a component parallel to the plate surfaces, polarizes the dielectric and pulls it toward the region of stronger field (between the electrodes). Once the dielectric has been fully inserted, there is no such force.

Solution

Take a rectangular path $C$ in a plane containing the axis of the solenoid—one side (of length $L$) runs through the interior of the solenoid parallel to the axis, the opposite side runs outside the solenoid (where the field can be regarded as $\bm 0$ because the solenoid is sufficiently long), and the remaining two sides are taken perpendicular to the axis of the solenoid (perpendicular to the interior field $\bm H$). Then the contribution to the line integral $\displaystyle\oint_C\bm H\cdot\dd\bm l$ comes only from the one side passing through the interior (the outer side has $H\approx0$, and the two perpendicular sides are orthogonal to $\bm H$ and so do not contribute). If the interior side has length $L$ and the magnetic field strength there is $H$ (along the axis, assumed constant), then

$$ \oint_C\bm H\cdot\dd\bm l=HL $$

On the other hand, the free current that this path $C$ “picks up” is the free current threading the windings of the solenoid. With $n$ turns per unit length and current $I$ in the wire, the number of turns contained in a section of length $L$ is $nL$, so

$$ I_{\mathrm{f},\text{enc}}=nL\times I=nLI $$

By Ampère's law (integral form) in Law 41.6, $\displaystyle\oint_C\bm H\cdot\dd\bm l=I_{\mathrm{f},\text{enc}}$, we have

$$ HL=nLI\qquad\therefore\quad H=nI $$

This result contains no trace of the permeability $\mu$ of the iron core (or the relative permeability $\mu_{\mathrm{r}}$). This is because Law 41.6 sees only the free current, and the effect of the response of the magnetic material (the magnetization current) has already been absorbed into the definition of $\bm H$ (thanks to the symmetry of an infinitely long solenoid, or of a solenoid long enough that end effects can be neglected, $H$ is determined from the integral form alone). Therefore $H=nI$ does not change whether or not there is an iron core. The magnetic flux density follows from the constitutive equation $\bm B=\mu_0\mu_{\mathrm{r}}\bm H$ of a linear magnetic material (Law 41.4):

$$ B=\mu_0\mu_{\mathrm{r}}H=\mu_0\mu_{\mathrm{r}}nI $$

This agrees completely with the result of Example 41.7 (the same $H,B$ are obtained regardless of how the path length $L$ is chosen).

Solution

(a) In a dielectric, for each of electronic polarization, ionic polarization, and orientational polarization (Section 41.1), the dipoles align parallel to the electric field $\bm E$ (the induced dipole is parallel to $\bm E$, in the orientation that minimizes $U=-\bm p\cdot\bm E$). The electric field produced inside a material polarized in this way points from the positive charges to the negative charges, that is, in the direction opposite to $\bm P$ (inside a uniformly polarized flat plate such as that in a capacitor it is $-\bm P/\varepsilon_0$, the depolarizing field of Section 41.1.3). Hence polarization acts in the direction of weakening the field for a fixed free charge, and the static relative permittivity is always $\varepsilon_{\mathrm{r}}\geq1$ (at high frequencies, near a resonance, $\varepsilon_{\mathrm{r}}\lt1$ is possible, but this is not a static property). For magnetic materials, the situation stands in contrast. The magnetic field produced by a magnetic dipole (circular current) points in the same direction as $\bm m$ inside the loop ($\bm B=\mu_0\bm M$ inside a uniformly magnetized cylinder), so when permanent moments align parallel to $\bm H$, as in paramagnetism and ferromagnetism, $\bm B=\mu_0(\bm H+\bm M)$ is strengthened and $\mu_{\mathrm{r}}\gt1$. Moreover, magnetic materials exhibit diamagnetism, which has no counterpart on the electric side. This arises because the magnetic moment induced in the orbital motion of the electrons when a magnetic field is applied appears, by Lenz's law, in the direction that weakens the original field ($\bm M$ opposite to $\bm H$), and this is the reason why $\mu_{\mathrm{r}}\lt1$ (on the electric side, there is no such thing as an induced dipole pointing opposite to $\bm E$). In ferromagnetism, because the exchange interaction (Section 41.2.3) produces an alignment so strong that $\bm M$ is larger than $\bm H$ by orders of magnitude, $\mu_{\mathrm{r}}\gg1$.

(b) A ferroelectric (Section 41.1.4) has a spontaneous polarization $P_{\mathrm{s}}$ even without an external electric field, owing to the asymmetry of its crystal structure, and its $P$–$E$ relation is nonlinear and shows hysteresis (a history-dependent phenomenon in which the polarization does not reverse immediately when the electric field is reversed). A ferromagnet (Section 41.2.4) likewise has spontaneous magnetization in its magnetic domains even without an external magnetic field, and its $M$–$H$ (or $B$–$H$) relation is nonlinear and shows hysteresis (Law 41.5). Both share the two properties of “having spontaneous order (spontaneous polarization, spontaneous magnetization)” and “the response not following immediately when the external field is reversed (history, hysteresis),” and neither can be described by a single constant such as the relative permittivity or the relative permeability. This similarity originates from the fact that in both there is, inside the system, “an interaction by which neighboring units (dipoles, magnetic moments) tend to align in the same direction” (the exchange interaction in a ferromagnet; in a ferroelectric, the interaction within the crystal and its coupling to the displacement of the lattice), and a phase transition to a state with spontaneous order occurs below the Curie temperature.

Solution

(a) Solving the definition $\bm D=\varepsilon_0\bm E+\bm P$ for $\bm P$ gives

$$ \bm P=\bm D-\varepsilon_0\bm E $$

Solving for $\bm E$ gives

$$ \bm E=\frac{\bm D-\bm P}{\varepsilon_0} $$

(b) Solving the definition $\bm H=\bm B/\mu_0-\bm M$ (that is, $\bm B=\mu_0(\bm H+\bm M)$) for $\bm M$ gives

$$ \bm M=\frac{\bm B}{\mu_0}-\bm H $$

Solving for $\bm H$ gives (the original definition itself)

$$ \bm H=\frac{\bm B}{\mu_0}-\bm M $$

(c) In a linear medium, substituting $\bm P=\varepsilon_0\chi_{\mathrm{e}}\bm E$ into $\bm D=\varepsilon_0\bm E+\bm P$ gives

$$ \bm D=\varepsilon_0\bm E+\varepsilon_0\chi_{\mathrm{e}}\bm E=\varepsilon_0(1+\chi_{\mathrm{e}})\bm E\equiv\varepsilon\bm E,\qquad\varepsilon=\varepsilon_0(1+\chi_{\mathrm{e}}) $$

Similarly, substituting $\bm M=\chi_{\mathrm{m}}\bm H$ into $\bm B=\mu_0(\bm H+\bm M)$ gives

$$ \bm B=\mu_0(\bm H+\chi_{\mathrm{m}}\bm H)=\mu_0(1+\chi_{\mathrm{m}})\bm H\equiv\mu\bm H,\qquad\mu=\mu_0(1+\chi_{\mathrm{m}}) $$

Both relations are thus obtained.

Solution

(a) Comparing the orders of magnitude of the susceptibility, the susceptibility of ordinary diamagnetic materials such as copper and bismuth is of the order of $\chi_{\mathrm{m}}\sim-10^{-5}$–$-10^{-4}$, and they weaken the external magnetic field only very slightly (see Figure 41.9 in this chapter). In contrast, the Meissner effect of a superconductor makes the magnetic flux density inside completely $\bm B=\bm 0$. From $\bm B=\mu_0(\bm H+\bm M)=\bm 0$, this corresponds to $\bm M=-\bm H$, that is, $\chi_{\mathrm{m}}=-1$, which is larger in magnitude than that of ordinary diamagnetic materials by four to five orders of magnitude ($10^4$–$10^5$ times), a response that is stronger by orders of magnitude.

(b) As seen in Section 41.2.3, ordinary diamagnetism is an effect (analogous to electromagnetic induction) in which the orbital motion of the electrons within individual atoms and molecules is slightly altered by the magnetic field; it is merely a weak induced effect, far from completely excluding the magnetic field. The Meissner effect in a superconductor, on the other hand, is an essentially macroscopic quantum effect that arises when a phase-coherent macroscopic quantum state (a condensate of Cooper pairs) forms throughout the material and a supercurrent reflecting that macroscopic quantum state (a persistent current without decay) flows near the surface of the material. The magnetic field created by this supercurrent exactly cancels the external magnetic field and makes the internal $\bm B$ completely $\bm 0$. That is, whereas ordinary diamagnetism is a microscopic and slight effect, “the orbital motion of electrons within individual atoms,” the Meissner effect is a phenomenon entirely different in origin, “a macroscopic supercurrent threading the entire material” (the detailed mechanism of superconductivity itself is beyond the scope of this book).

(c) Even a “perfect conductor” with electrical resistance $0$ keeps its internal $\bm B$ at $\bm 0$ if it is cooled before being placed in a magnetic field (a state with $\bm B=\bm 0$) and the field is applied afterward: a current is induced (by Lenz's law) in the direction that cancels the change in flux and continues to flow without decay. However, when it is cooled with the field already applied, there is no change in flux in a perfect conductor, so the internal $\bm B$ stays at its value from before the cooling (the flux is trapped). A superconductor, in contrast, even when cooled with the field applied, expels the flux from its interior and makes $\bm B=\bm 0$ once the temperature falls below the critical temperature. That is, the Meissner effect (flux expulsion) cannot be explained by the resistance being $0$ (a perfect conductor) alone; it is a property of the superconducting state itself. Note that the $\bm B=\bm 0$ in this problem refers to a type-I superconductor when the magnetic field is smaller than the critical field. In a type-II superconductor, between the lower critical field and the upper critical field, the flux partially penetrates as thin vortex filaments (quantized flux), and $\chi_{\mathrm{m}}=-1$ does not hold.

Solution

(a) The capacitance of one layer is $C_1=\varepsilon_0\varepsilon_{\mathrm{r}}S/d$, so

$$ C_1=(8.854\times10^{-12}\ \mathrm{F/m})\times3000\times\frac{1.0\times10^{-6}\ \mathrm{m^2}}{2.0\times10^{-6}\ \mathrm m}\approx1.33\times10^{-8}\ \mathrm F=13.3\ \mathrm{nF} $$

(b) In a parallel connection the capacitances add, so for 100 layers $C=100\,C_1\approx1.33\times10^{-6}\ \mathrm F=1.33\ \mu\mathrm F$. Even for the same volume, stacking hundreds of thin, high-permittivity layers yields a large capacitance.

(c) $E=V/d=(5\ \mathrm V)/(2.0\times10^{-6}\ \mathrm m)=2.5\times10^{6}\ \mathrm{V/m}$, and $D=\varepsilon_0\varepsilon_{\mathrm{r}}E=(8.854\times10^{-12})(3000)(2.5\times10^{6})\approx6.6\times10^{-2}\ \mathrm{C/m^2}$. This $E$ is about $83\%$ of the dielectric breakdown field of air (about $3\times10^{6}\ \mathrm{V/m}$), a strength at which air would be on the verge of discharge, but the breakdown field of ceramics is far higher than that of air, so there is no problem in using it. When the layers are made thinner, the electric field becomes stronger even at the same voltage ($E=V/d$), so the voltage that can be withstood is determined by the thickness of the layers and the breakdown field of the material.

(d) $V=E_{\text{breakdown}}d=(3\times10^{6}\ \mathrm{V/m})(1\times10^{-3}\ \mathrm m)=3\times10^{3}\ \mathrm V=3\ \mathrm{kV}$.

Solution

(a) Apply Ampère's law (Law 41.6) to the path going once around inside the core (of length $2\pi r$). The free current threading this path is $NI$ (the winding passes through it $N$ times), so

$$ H\cdot2\pi r=NI\qquad\therefore\quad H=\frac{NI}{2\pi r}=\frac{500\times0.10}{2\pi\times0.050}\ \mathrm{A/m}\approx1.59\times10^{2}\ \mathrm{A/m} $$

The magnetic flux density is $B=\mu_0\mu_{\mathrm{r}}H=(1.2566\times10^{-6})(2000)(159.2)\approx0.40\ \mathrm T$ (without the iron core, $\mu_0H\approx0.20\ \mathrm{mT}$). The magnetization is $M=(\mu_{\mathrm{r}}-1)H=1999\times159.2\approx3.18\times10^{5}\ \mathrm{A/m}$. Check: $\mu_0(H+M)=\mu_0\times3.18\times10^{5}\approx0.40\ \mathrm T=B$.

(b) The hysteresis loop area of $300\ \mathrm{J/m^3}$ is the thermal energy lost per unit volume in one cycle (one repetition of magnetization). The volume of the core is $V=m/\rho=(1.0\ \mathrm{kg})/(7.65\times10^3\ \mathrm{kg/m^3})=1.31\times10^{-4}\ \mathrm{m^3}$, and the energy lost in one cycle is $300\times1.31\times10^{-4}\approx3.9\times10^{-2}\ \mathrm J$. At $50\ \mathrm{Hz}$ there are 50 cycles per second, so the heating power is

$$ P=(3.9\times10^{-2}\ \mathrm J)\times50\ \mathrm{s^{-1}}\approx1.96\ \mathrm W\approx2.0\ \mathrm W $$

That is about $2\ \mathrm W$ per $1\ \mathrm{kg}$, and this accumulates into non-negligible heat generation in large transformers. This is the reason for using soft magnetic materials (such as silicon steel), whose hysteresis loops have a small area.

Solution

(a) From $\chi_{\mathrm{m}}=C/T$, $\chi_{\mathrm{m}}(77\ \mathrm K)/\chi_{\mathrm{m}}(300\ \mathrm K)=300/77\approx3.90$. Cooling to the temperature of liquid nitrogen increases the susceptibility by a factor of about four.

(b) $\mu_{\mathrm B}B/(k_{\mathrm B}T)=\dfrac{(9.274\times10^{-24}\ \mathrm{J/T})(1.0\ \mathrm T)}{(1.380649\times10^{-23}\ \mathrm{J/K})(300\ \mathrm K)}\approx2.24\times10^{-3}$. Since this ratio is much smaller than $1$, $\tanh x\approx x$ can be used, giving

$$ M\approx N\mu_{\mathrm B}\cdot\frac{\mu_{\mathrm B}B}{k_{\mathrm B}T}=(5.0\times10^{27})(9.274\times10^{-24})(2.24\times10^{-3})\approx1.04\times10^{2}\ \mathrm{A/m} $$

The susceptibility is, from $B\approx\mu_0H$, $\chi_{\mathrm{m}}=M/H\approx\mu_0M/B=(1.2566\times10^{-6})(1.04\times10^{2})/1.0\approx1.3\times10^{-4}$. This also agrees with $C/T=0.039/300\approx1.3\times10^{-4}$, obtained using the Curie constant $C=\mu_0N\mu_{\mathrm B}^2/k_{\mathrm B}\approx0.039\ \mathrm K$.

(c) The number density of iron atoms is $N=\rho N_{\mathrm A}/M_{\mathrm{mol}}=(7.87\times10^3\ \mathrm{kg/m^3})(6.022\times10^{23}\ \mathrm{mol^{-1}})/(0.05585\ \mathrm{kg/mol})\approx8.49\times10^{28}\ \mathrm{m^{-3}}$. The saturation magnetization when all the moments are aligned is $M_{\mathrm s}=N\times2.2\mu_{\mathrm B}\approx1.73\times10^{6}\ \mathrm{A/m}$, giving $\mu_0M_{\mathrm s}\approx2.2\ \mathrm T$, which is in good agreement with the saturation magnetic flux density of iron (about $2.1$–$2.2\ \mathrm T$).

Solution

(1) The contradiction when the displacement current is not included. Ampère's law without the displacement current is $\oint_C\bm H\cdot\dd\bm\ell=I_{\text{through}}$ ($I_{\text{through}}$ is the total conduction current passing through a surface bounded by $C$). Consider two surfaces that have the same closed curve $C$ (a loop going once around the wire) as their boundary.

• (a) The flat surface $S_1$ cutting straight across the wire: the current flowing in the wire passes through it directly, so $I_{\text{through}}=I=1.5\ \mathrm A$. Therefore $\oint_C\bm H\cdot\dd\bm\ell=1.5\ \mathrm A$.
• (b) The surface $S_2$ passing between the plates: there is no wire between the plates and no movement of charge (conduction current), so $I_{\text{through}}=0$. Therefore $\oint_C\bm H\cdot\dd\bm\ell=0$.

Although the left-hand side $\oint_C\bm H\cdot\dd\bm\ell$ is the same, its value comes out as either $1.5\ \mathrm A$ or $0$ depending on the choice of surface, which is a contradiction. For example, for the magnetic field on a circle $C$ at distance $r$ from the wire, using $S_1$ gives $\oint_C\bm H\cdot\dd\bm\ell=I$, which agrees with the actual field $H=I/(2\pi r)$. Using $S_2$, however, makes the right-hand side $0$, which contradicts the actual field (the magnetic field itself does not change with the choice of surface; rather, the evaluation of the right-hand side of the law disagrees between the two surfaces).

(2) When the displacement current is included. Since the capacitor is charged with a constant current $I$, the charge on the plates at time $t$ (with $t=0$ being the instant when the charge is $0$) is $Q=It$. Neglecting edge effects, the electric field between the plates is uniform, with $E=\sigma/\varepsilon_0=Q/(\varepsilon_0A)=It/(\varepsilon_0A)$. Therefore we have

$$ \pdiff Et=\frac{I}{\varepsilon_0A}=\frac{1.5}{(8.854\times10^{-12})(1.0\times10^{-2})}\approx1.69\times10^{13}\ \mathrm{V/(m\cdot s)} $$

The displacement current density is $j_{\mathrm d}=\pdiff Dt=\varepsilon_0\pdiff Et=I/A=1.5/(1.0\times10^{-2})=1.5\times10^2\ \mathrm{A/m^2}$ (the units agree: $\mathrm{F/m}\cdot\mathrm{V/(m\,s)}=\mathrm{C/(m^2\,s)}=\mathrm{A/m^2}$). Since the surface $S_2$ covers the whole plate area $A$ and is pierced by the electric field, the displacement current through $S_2$ is

$$ I_{\mathrm d}=j_{\mathrm d}A=1.5\times10^2\times1.0\times10^{-2}=1.5\ \mathrm A $$

Using the Ampère–Maxwell law $\oint_C\bm H\cdot\dd\bm\ell=\iint_S\left(\bm j+\pdiff{\bm D}{t}\right)\cdot\dd\bm S$, we find:

• $S_1$: conduction current $1.5\ \mathrm A$; since this surface does not cut through the plates, the displacement current can be neglected; total $1.5\ \mathrm A$.
• $S_2$: conduction current $0$, displacement current $1.5\ \mathrm A$; total $1.5\ \mathrm A$.

Thus, whichever surface is chosen, the right-hand side takes the same value, $1.5\ \mathrm A$, and the contradiction is resolved.

Solution

Let $\bm f=[f,g,h]$. By the definition of the curl,

$$ \rot\bm f=\left[\pdiff hy-\pdiff gz,\ \pdiff fz-\pdiff hx,\ \pdiff gx-\pdiff fy\right] $$

For this $\rot\bm f$, the $y$ component of its curl is obtained by shifting the indices one step cyclically in the definition of the curl ($x\to y\to z\to x$):

$$ \bigl[\rot(\rot\bm f)\bigr]_y=\pdiff{}{z}(\rot\bm f)_x-\pdiff{}{x}(\rot\bm f)_z $$

Substituting $(\rot\bm f)_x,(\rot\bm f)_z$ gives

$$ \bigl[\rot(\rot\bm f)\bigr]_y=\pdiff{}{z}\left(\pdiff hy-\pdiff gz\right)-\pdiff{}{x}\left(\pdiff gx-\pdiff fy\right) =\frac{\partial^2h}{\partial z\partial y}-\frac{\partial^2g}{\partial z^2}-\frac{\partial^2g}{\partial x^2}+\frac{\partial^2f}{\partial x\partial y} $$

The order of partial differentiation can be exchanged (Schwarz's theorem), so we can write $\dfrac{\partial^2h}{\partial z\partial y}=\dfrac{\partial^2h}{\partial y\partial z}$ and $\dfrac{\partial^2f}{\partial x\partial y}=\dfrac{\partial^2f}{\partial y\partial x}$. Now adding and subtracting $\dfrac{\partial^2g}{\partial y^2}$ (that is, adding $0$), we obtain

$$ \bigl[\rot(\rot\bm f)\bigr]_y=\underbrace{\left(\frac{\partial^2f}{\partial y\partial x}+\frac{\partial^2g}{\partial y^2}+\frac{\partial^2h}{\partial y\partial z}\right)}_{=\ \pdiff{}{y}\left(\pdiff fx+\pdiff gy+\pdiff hz\right)} -\underbrace{\left(\frac{\partial^2g}{\partial x^2}+\frac{\partial^2g}{\partial y^2}+\frac{\partial^2g}{\partial z^2}\right)}_{=\ \nabla^2g} $$

The first part is $\pdiff{}{y}(\operatorname{div}\bm f)=\bigl[\grad(\operatorname{div}\bm f)\bigr]_y$, and the second part is $\nabla^2g=\bigl[\nabla^2\bm f\bigr]_y$. Hence

$$ \bigl[\rot(\rot\bm f)\bigr]_y=\bigl[\grad(\operatorname{div}\bm f)\bigr]_y-\bigl[\nabla^2\bm f\bigr]_y $$

holds. The $z$ component can be computed in the same way, and the identity $\rot(\rot\bm f)=\grad(\operatorname{div}\bm f)-\nabla^2\bm f$ has been verified for all three components.

Supplement (checking types). Since the left-hand side $\rot(\rot\bm f)$ is a vector, every term on the right-hand side must also be a vector. $\grad(\operatorname{div}\bm f)$ is “the gradient of the scalar $\operatorname{div}\bm f$,” so it is a vector, and one cannot add $\operatorname{div}\bm f$ itself (a scalar). In the $y$ component, the expression $\pdiff{}{y}(\operatorname{div}\bm f)$ that appeared midway, that is, the operation of “partial differentiation with respect to $y$,” corresponds to the $y$ component of the gradient.

Solution

(1) The names of the laws and their starting points.

• ① Integral form $\oint_S\bm D\cdot\dd\bm S=Q_{\text{enc}}$: Gauss's law (Gauss's law for the electric field). In Chapter 35 it was derived starting from Coulomb's law, using the divergence theorem.
• ② Integral form $\oint_S\bm B\cdot\dd\bm S=0$: Gauss's law for the magnetic field (conservation of magnetic flux). In Chapter 38 it was derived using the divergence theorem, from the fact that magnetic field lines have no ends anywhere and always close on themselves.
• ③ Integral form $\oint_C\bm E\cdot\dd\bm\ell=-\dfrac{\dd}{\dd t}\iint_S\bm B\cdot\dd\bm S$: Faraday's law of induction. In Chapter 40 it was derived from the relation between the time variation of magnetic flux and the induced electromotive force (emf), as found in experiments on electromagnetic induction.
• ④ Integral form $\oint_C\bm H\cdot\dd\bm\ell=\iint_S\left(\bm j+\pdiff{\bm D}{t}\right)\cdot\dd\bm S$: Ampère–Maxwell law. This is Ampère's law of Chapter 38 (an electric current produces a magnetic field), with the displacement current $\pdiff{\bm D}{t}$ added in Section 42.1 of this chapter for consistency with conservation of charge.

(2) Why ② means that “magnetic monopoles do not exist.” Equation ① has the form $\operatorname{div}\bm D=\rho$, and the charge density $\rho$ on the right-hand side represents the sources (positive charge) and sinks (negative charge) of the electric field lines. Objects carrying only positive charge and objects carrying only negative charge actually exist, so there are places where $\rho\neq0$, and electric field lines emerge from them. On the other hand, ② can be written $\operatorname{div}\bm B=0$, so the “magnetic charge density” corresponding to $\rho$ in ① is $0$ everywhere. This means that there are no isolated magnetic poles (magnetic monopoles) to serve as sources or sinks of magnetic field lines, and magnetic field lines always form closed loops (field lines leaving the north pole always enter the south pole). This also agrees with the fact (Example 42.3) that when a bar magnet is broken in half, no piece with only a north pole or only a south pole is produced; each piece becomes a smaller magnet having a north pole and a south pole again. The asymmetry between electricity and magnetism—that electric charge exists as single particles (positive and negative charges) while nothing of the kind has been found for magnetism—appears as the difference between the right-hand sides of ① and ② ($\rho$ versus $0$).

Solution

We have $\bm E=[0,E_y,0]$ and $E_y=E_0\cos(kx-\omega t)$, where $E_y$ is a function of $x,t$ only. Computing the curl component by component,

$$ \rot\bm E=\left[\pdiff{E_z}{y}-\pdiff{E_y}{z},\ \pdiff{E_x}{z}-\pdiff{E_z}{x},\ \pdiff{E_y}{x}-\pdiff{E_x}{y}\right]=\left[0,\ 0,\ \pdiff{E_y}{x}\right] $$

($E_x=E_z=0$, and since $E_y$ does not contain $y,z$, $\pdiff{E_y}{z}=0$). Since $\pdiff{E_y}{x}=-kE_0\sin(kx-\omega t)$, we get $\rot\bm E=[0,0,-kE_0\sin(kx-\omega t)]$, and this time only the $z$ component is not $0$. From Faraday's law $\rot\bm E=-\pdiff{\bm B}{t}$, letting $\bm B=[0,0,B_z]$, we have

$$ -\pdiff{B_z}{t}=-kE_0\sin(kx-\omega t)\qquad\therefore\ \pdiff{B_z}{t}=kE_0\sin(kx-\omega t) $$

We integrate with respect to $t$. Since $\displaystyle\int\sin(kx-\omega t)\,\dd t=\frac1\omega\cos(kx-\omega t)+C$ (setting $u=kx-\omega t$ gives $\dd u=-\omega\,\dd t$), and assuming there is no extra constant magnetic field apart from the wave, so that $C=0$, we obtain

$$ B_z=\frac k\omega E_0\cos(kx-\omega t) $$

Substituting into the wave equation $\pdiff{^2E_y}{x^2}=\dfrac1{c^2}\pdiff{^2E_y}{t^2}$ gives $-k^2E_0\cos(\cdots)=-\dfrac{\omega^2}{c^2}E_0\cos(\cdots)$, so $\omega=ck$, that is, $k/\omega=1/c$. Therefore we obtain

$$ \bm B=\left[0,\ 0,\ \frac{E_0}{c}\cos(kx-\omega t)\right] $$

This is the magnetic field.

• $\bm E$ points in the $y$ direction, $\bm B$ in the $z$ direction, and the propagation direction $\bm k$ in the $x$ direction, so they are mutually perpendicular ($\bm E\perp\bm B\perp\bm k$).
• $E_y=E_0\cos(kx-\omega t)$ and $B_z=\dfrac{E_0}{c}\cos(kx-\omega t)$ are in phase, and the ratio of their magnitudes is $E/B=c$, that is, $E=cB$.
• $\bm E\times\bm B=[0,E_y,0]\times[0,0,B_z]=[E_yB_z,\ 0,\ 0]=\left[\dfrac{E_0^2}{c}\cos^2(kx-\omega t),\ 0,\ 0\right]$ ($\bm e_y\times\bm e_z=\bm e_x$), which always points in the $+x$ direction, that is, it coincides with the direction in which the wave travels.

Solution

For the problem on the square region ($L=1$, $u=0$ on the four sides) with initial velocity $0$, the general solution, as derived in Section 42.7, is

$$ u(x,y,t)=\sum_{n=1}^\infty\sum_{m=1}^\infty D_{mn}\sin m\pi x\sin n\pi y\cos\bigl(c\pi\sqrt{m^2+n^2}\,t\bigr) $$

Substituting $t=0$ gives $\displaystyle\sum_{m,n}D_{mn}\sin m\pi x\sin n\pi y=\sin\pi x\sin2\pi y$. The functions $\sin m\pi x\sin n\pi y$ ($m,n=1,2,\ldots$) on the left-hand side are mutually orthogonal (University Mathematics, Chapter 27), and the right-hand side is itself the term with $m=1,n=2$. Determining the coefficients from orthogonality,

$$ D_{mn}=4\int_0^1\!\!\int_0^1\sin\pi x\,\sin2\pi y\,\sin m\pi x\,\sin n\pi y\,\dd x\,\dd y =4\left(\int_0^1\sin\pi x\sin m\pi x\,\dd x\right)\left(\int_0^1\sin2\pi y\sin n\pi y\,\dd y\right) $$

where $\displaystyle\int_0^1\sin p\pi x\sin q\pi x\,\dd x=\tfrac12\delta_{pq}$ ($p,q$ are positive integers), so $D_{mn}=4\cdot\tfrac12\delta_{m1}\cdot\tfrac12\delta_{n2}=\delta_{m1}\delta_{n2}$. That is, $D_{12}=1$, and all the other $D_{mn}$ are $0$. Therefore,

$$ u(x,y,t)=\sin\pi x\,\sin2\pi y\,\cos\bigl(\sqrt5\,\pi c\,t\bigr) $$

($\omega_{12}=c\pi\sqrt{1^2+2^2}=\sqrt5\,\pi c$. Here we work in units where $L=1$. For general $L$, $\omega_{12}=\sqrt5\,\pi c/L$, and the solution is $\sin\dfrac{\pi x}{L}\sin\dfrac{2\pi y}{L}\cos\left(\dfrac{\sqrt5\,\pi c}{L}t\right)$.)

Check. Wave equation: $\pdiff{^2u}{t^2}=-5\pi^2c^2u$ and $\pdiff{^2u}{x^2}+\pdiff{^2u}{y^2}=(-\pi^2-4\pi^2)u=-5\pi^2u$, so $\pdiff{^2u}{t^2}=c^2\left(\pdiff{^2u}{x^2}+\pdiff{^2u}{y^2}\right)$ holds. Boundary conditions: at $x=0,1$, $\sin\pi x=0$; at $y=0,1$, $\sin2\pi y=0$. Initial conditions: $u(x,y,0)=\sin\pi x\sin2\pi y$ and $\pdiff ut(x,y,0)=-\sqrt5\pi c\sin\pi x\sin2\pi y\sin0=0$. All are satisfied. This solution does not change its spatial shape; the whole pattern merely stretches and shrinks with angular frequency $\sqrt5\,\pi c$ (period $2/(\sqrt5\,c)$ when $L=1$), so it is a standing wave (a single normal mode).

Solution

The energy density is $u=\dfrac12\varepsilon_0E^2+\dfrac1{2\mu_0}B^2$. For a plane electromagnetic wave, $E=cB$, so substituting $B=E/c$ into the second term gives

$$ \frac{1}{2\mu_0}B^2=\frac{1}{2\mu_0}\cdot\frac{E^2}{c^2}=\frac{1}{2\mu_0c^2}E^2 $$

Since $c^2=\dfrac1{\varepsilon_0\mu_0}$, we have $\dfrac1{\mu_0c^2}=\varepsilon_0$, so the second term is equal to $\dfrac12\varepsilon_0E^2$, which has the same magnitude as the first term (the energy density of the electric field, $u_E$); that is, $u_B=u_E$. Therefore

$$ u=\frac12\varepsilon_0E^2+\frac12\varepsilon_0E^2=\varepsilon_0E^2 $$

This is twice the expression for the electric field alone ($u_E=\tfrac12\varepsilon_0E^2$), and it is confirmed that the electric field and the magnetic field each carry half of the energy.

Supplement. In this case the magnitude of the Poynting vector is $\abs{\bm S}=\dfrac{EB}{\mu_0}=\dfrac{E^2}{\mu_0c}=\varepsilon_0cE^2=u\,c$. That is, the relation “energy density $\times$ speed $c$ = energy flux density” holds, which expresses that energy is carried along with the electromagnetic wave at speed $c$. For time averages, $\langle u\rangle=\tfrac12\varepsilon_0E_0^2$ and $\langle S\rangle=\tfrac12\varepsilon_0cE_0^2=c\langle u\rangle$.

Solution

(a) The radiated power $P$ passes isotropically through a sphere of radius $r$ (area $4\pi r^2$) centered on the source. The time average $\langle S\rangle$ of the magnitude of the Poynting vector is the energy passing through unit area per unit time, so

$$ \langle S\rangle=\frac{P}{4\pi r^2}=\frac{100\ \mathrm W}{4\pi\times(2.0\ \mathrm m)^2}=\frac{100}{50.27}\ \mathrm{W/m^2}\approx1.99\ \mathrm{W/m^2} $$

(b) Solving the plane-wave intensity formula $\langle S\rangle=\tfrac12\varepsilon_0cE_0^2$ for $E_0$, we obtain

$$ E_0=\sqrt{\frac{2\langle S\rangle}{\varepsilon_0c}}=\sqrt{\frac{2\times1.989}{(8.854\times10^{-12})(2.998\times10^8)}}=\sqrt{\frac{3.979}{2.654\times10^{-3}}}\ \mathrm{V/m}=\sqrt{1.499\times10^{3}}\ \mathrm{V/m}\approx38.7\ \mathrm{V/m} $$

(c) $B_0=E_0/c=\dfrac{38.7}{2.998\times10^8}\ \mathrm T\approx1.29\times10^{-7}\ \mathrm T$ (about $0.13\ \mu\mathrm T$).

Since the intensity is proportional to $1/r^2$, the amplitudes $E_0,B_0$ decrease in inverse proportion to the distance $r$. Note that the position $2\ \mathrm m$ from the source is taken to be sufficiently far compared with the size of the source, so this is an approximation in which the spherical wave is regarded locally as a plane wave.

Solution

(a) In matter, $\varepsilon_0,\mu_0$ are replaced by $\varepsilon=\varepsilon_{\mathrm r}\varepsilon_0$ and $\mu=\mu_{\mathrm r}\mu_0$, so the speed of the wave is $v=1/\sqrt{\varepsilon\mu}=c/\sqrt{\varepsilon_{\mathrm r}\mu_{\mathrm r}}=c/n$. The refractive index is

$$ n=\sqrt{\varepsilon_{\mathrm r}\mu_{\mathrm r}}=\sqrt{2.25\times1}=1.5 $$

and the speed is

$$ v=\frac cn=\frac{2.998\times10^8\ \mathrm{m/s}}{1.5}\approx2.00\times10^8\ \mathrm{m/s} $$

(b) In vacuum, $c=f\lambda_0$, so

$$ f=\frac{c}{\lambda_0}=\frac{2.998\times10^8\ \mathrm{m/s}}{600\times10^{-9}\ \mathrm m}\approx5.0\times10^{14}\ \mathrm{Hz} $$

(c) The frequency does not change in the glass either (the tangential component of the electric field must be continuous at the boundary, and the field varies in time in the same way on both sides of the boundary, so the frequency is the same before and after the boundary). From the wave relation $v=f\lambda$,

$$ \lambda=\frac vf=\frac{c/n}{c/\lambda_0}=\frac{\lambda_0}{n}=\frac{600\ \mathrm{nm}}{1.5}=400\ \mathrm{nm} $$

That is, in glass the wavelength is $1/n$ times that in vacuum (since the speed becomes $1/n$ of its vacuum value while the frequency does not change, the wavelength becomes shorter).

(Note: The relative permittivity of real glass depends on the frequency of the light (dispersion). Here we used the value $\varepsilon_{\mathrm r}=2.25$, given as the value at the frequency of this light.)

Solution

(1) The definition of the generalized momentum is $p_i=\partial T/\partial\dot q_i$. Differentiate $T=\tfrac12m\left(\dot r^2+r^2\dot\theta^2+r^2\sin^2\theta\,\dot\varphi^2\right)$ partially with respect to each of $\dot r,\dot\theta,\dot\varphi$. In partial differentiation, every variable other than the one being differentiated (including $r,\theta,\varphi$ themselves and the other velocities) is regarded as a constant. The only term in $T$ that contains $\dot r$ is $\tfrac12m\dot r^2$, so

$$ p_r=\pdiff{T}{\dot r}=m\dot r $$

Similarly, the only term containing $\dot\theta$ is $\tfrac12mr^2\dot\theta^2$, and the only term containing $\dot\varphi$ is $\tfrac12mr^2\sin^2\theta\,\dot\varphi^2$, so

$$ p_\theta=\pdiff{T}{\dot\theta}=mr^2\dot\theta,\qquad p_\varphi=\pdiff{T}{\dot\varphi}=mr^2\sin^2\theta\,\dot\varphi $$

As for the units, noting that angles (radians) are dimensionless, $p_r=m\dot r$ has units $\mathrm{kg\cdot m/s}$ (ordinary momentum), $p_\theta=mr^2\dot\theta$ has units $\mathrm{kg\cdot m^2/s}$ (the same dimension as angular momentum), and $p_\varphi=m(r\sin\theta)^2\dot\varphi$ has units $\mathrm{kg\cdot m^2/s}$. $p_\varphi$ is none other than the angular momentum $L_z$ about the $z$ axis for motion around a circle of radius $r\sin\theta$ (the distance from the $z$ axis) at angular velocity $\dot\varphi$.

(2) Substitute $m=2\ \mathrm{kg},\ r=3\ \mathrm m,\ \theta=\pi/3,\ \dot r=1\ \mathrm{m/s},\ \dot\theta=0.5\ \mathrm{s^{-1}},\ \dot\varphi=2\ \mathrm{s^{-1}}$. We have $\sin^2(\pi/3)=(\sqrt3/2)^2=3/4$.

$$ p_r=2\times1=2\ \mathrm{kg\,m/s} $$ $$ p_\theta=2\times3^2\times0.5=2\times9\times0.5=9\ \mathrm{kg\,m^2/s} $$ $$ p_\varphi=2\times3^2\times\frac34\times2=2\times9\times0.75\times2=27\ \mathrm{kg\,m^2/s} $$

Solution

When all the forces are conservative, the generalized force is $Q_i=-\partial U/\partial q_i$. Since $U=U(r)$ does not contain $\theta,\varphi$, partial differentiation with respect to $\theta,\varphi$ gives 0:

$$ Q_r=-\pdiff{U}{r}=-\diff{U}{r},\qquad Q_\theta=-\pdiff{U}{\theta}=0,\qquad Q_\varphi=-\pdiff{U}{\varphi}=0 $$

As a check, let us also compute directly from the definition $Q_i=\sum_jF_j\,\partial x_j/\partial q_i$. The central force can be written $\bm F=-\dfrac{\dd U}{\dd r}\,\bm e_r$ (where $\bm e_r=(\sin\theta\cos\varphi,\ \sin\theta\sin\varphi,\ \cos\theta)$ is the unit vector in the radial direction). Differentiating $\bm x=r\bm e_r$ partially with respect to $\theta$ gives $\partial\bm x/\partial\theta=r\,\partial\bm e_r/\partial\theta$, which is perpendicular to $\bm e_r$ (since $\bm e_r$ is a unit vector of constant magnitude, its derivative is orthogonal to $\bm e_r$). Hence $Q_\theta=\bm F\cdot\partial\bm x/\partial\theta\propto\bm e_r\cdot\partial\bm e_r/\partial\theta=0$. The same holds for $\varphi$, so $Q_\varphi=0$. On the other hand, since $\partial\bm x/\partial r=\bm e_r$, we get $Q_r=\bm F\cdot\bm e_r=-\dd U/\dd r$, in agreement with the result above.

Physical meaning: $Q_r$ is the component of the force in the radial direction, and $Q_\theta,Q_\varphi$ are the torques (moments of force) in the respective angular directions (the same as $Q_\theta=rF_\theta$ for plane polar coordinates in Example 43.4). A central force points toward the origin, so it has no moment about the origin—that is the meaning of $Q_\theta=Q_\varphi=0$. For example, since $T$ does not contain $\varphi$, we have $\dot p_\varphi=Q_\varphi+\partial T/\partial\varphi=0$, and $p_\varphi=mr^2\sin^2\theta\,\dot\varphi$ (the angular momentum about the $z$ axis) is conserved. This is a restatement, in generalized coordinates, of what we learned in Chapters 2 and 8, that “angular momentum is conserved under a central force.”

Solution

(1) Substituting $\mu=0$ into the expression for the velocity, the terms containing $\mu mg/k$ vanish, and

$$ \dot x(t)=v_0e^{-\frac km t} $$

Since $k\gt0,\ m\gt0$, the quantity $-kt/m$ is negative, and $e^{-kt/m}\to0$ as $t\to\infty$. Therefore

$$ \lim_{t\to\infty}\dot x(t)=0 $$

(the terminal velocity is 0). Here is a physical reason why this is plausible: when there is no friction, the only horizontal force acting on the particle (point mass) is the air resistance $-k\dot x$, which always opposes the motion, so the velocity keeps decreasing. On the other hand, the smaller the velocity becomes, the smaller the resistance is, so the decrease gradually slows down, and the velocity approaches 0 exponentially (with time constant $m/k$). Since there is no other force to keep the particle moving, the final velocity can never be a value other than 0. Note also that the stopping time in the presence of friction, $t_{\text{stop}}=\dfrac mk\ln\!\left(1+\dfrac{kv_0}{\mu mg}\right)$, diverges to $\infty$ as $\mu\to0$. Without friction, the particle never comes to a complete stop in a finite time.

(2) Substituting $\mu=0$ into the expression for the position, the two terms containing $\mu$ vanish, and

$$ x(t)=\frac mkv_0\left(1-e^{-\frac km t}\right) $$

Since $e^{-kt/m}\to0$ as $t\to\infty$, the distance traveled before the particle stops is

$$ \lim_{t\to\infty}x(t)=\frac{mv_0}{k} $$

For $m=1\ \mathrm{kg},\ v_0=5\ \mathrm{m/s},\ k=0.5\ \mathrm{kg/s}$, this is $\dfrac{1\times5}{0.5}=10\ \mathrm m$. The units also work out: $\mathrm{kg\cdot(m/s)/(kg/s)}=\mathrm m$, which agrees with the dimension of length.

Solution

The fact that the equation of motion in the $\theta$ direction gave $\ddot\theta=0$ is due to the form $Q_\theta'=-mgR\cos\theta$, in which the external force $F_\theta=-mg\cos\theta$ cancels the tangential component of gravity, and it does not depend on the initial conditions. Therefore $\dot\theta$ is constant and remains $\dot\theta=\omega=\sqrt{g/R}$. Integrating this once gives $\theta(t)=\omega t+C$ ($C$ is the constant of integration). From the initial condition $\theta(0)=\theta_0$ we get $C=\theta_0$, so

$$ \theta(t)=\theta_0+\sqrt{\frac gR}\,t $$

The position is obtained by substituting into $x=R\cos\theta,\ y=R(1-\sin\theta)$:

$$ x=R\cos\!\left(\theta_0+\sqrt{\frac gR}\,t\right),\qquad y=R\left[1-\sin\!\left(\theta_0+\sqrt{\frac gR}\,t\right)\right] $$

(These expressions can be used while $\theta$ is in the range $0\le\theta\le\pi$, that is, while the particle moves from one rim, through the bottom, to the other rim.)

For the normal force, substituting $\dot\theta^2=g/R$ into the $r$-direction equation $N=mR\dot\theta^2+mg\sin\theta$ gives $N=mg(1+\sin\theta)$. This expression is a function of $\theta$ alone and does not contain the initial phase $\theta_0$. Therefore the expression for $N$ as a function of $\theta$ does not change. (However, the value of $\theta$ at a given time $t$ depends on $\theta_0$, so if $N$ is written as a function of time, $N(t)=mg\left[1+\sin(\theta_0+\omega t)\right]$, which depends on $\theta_0$.)

If the angular velocity is taken to be a general (constant) $\omega$, then $\ddot\theta=0$ is unchanged, and the $r$-direction equation becomes

$$ N=mR\omega^2+mg\sin\theta $$

If $\omega=\sqrt{g/R}$, then $mR\omega^2=mg$ and $N=mg(1+\sin\theta)$; if $\omega=0$ (the case in which an external force holds the particle at the same position on the surface), then

$$ N=mg\sin\theta $$

which agrees with the normal force that a stationary particle receives from the surface (the balance with the component $mg\sin\theta$ of gravity perpendicular to the surface).

Solution

(1) When $q=x$, the coordinates of the particle are $x_1=x,\ x_2=y$, and $\partial x/\partial x=1,\ \partial y/\partial x=0$, so from the definition $Q_i=\sum_jF_j\,\partial x_j/\partial q_i$,

$$ Q_x=F_x\cdot1+F_y\cdot0=F_x $$

The unit is $\mathrm N$, the same as that of a force.

(2) When $q=\theta$ (the rotation angle about the origin), keeping the distance $r$ from the origin constant, we have $x=r\cos\theta,\ y=r\sin\theta$. Since $\partial x/\partial\theta=-r\sin\theta=-y,\ \partial y/\partial\theta=r\cos\theta=x$,

$$ Q_\theta=F_x\pdiff{x}{\theta}+F_y\pdiff{y}{\theta}=-yF_x+xF_y $$

This is the moment of the force about the origin (the torque). The unit is $\mathrm{m\cdot N}=\mathrm{N\,m}$ (since an angle is dimensionless, the unit of $\partial x/\partial\theta$ is the unit of length, $\mathrm m$).

Reason for the difference: The generalized force $Q_i$ is defined so that the infinitesimal work is $\dd W=\sum_iQ_i\,\dd q_i$. The unit of work is always the unit of energy, $\mathrm J=\mathrm{N\,m}$, so the unit of $Q_i$ is “$\mathrm J$ divided by (the unit of $q_i$).” If $q_i=x$ (a length, unit $\mathrm m$), the unit of $Q_x$ is $\mathrm J/\mathrm m=\mathrm N$; if $q_i=\theta$ (an angle, unit $\mathrm{rad}$, dimensionless), the unit of $Q_\theta$ is $\mathrm J/\mathrm{rad}=\mathrm{N\,m}$. Depending on whether the generalized coordinate is a length or an angle, the unit of $\partial x_j/\partial q_i$ differs, and so the unit of the generalized force differs as well. Similarly, the generalized momentum is the quantity for which $p_i\dot q_i$ has the dimension of energy, and if $q_i$ is an angle, $p_i$ is an angular momentum ($\mathrm{kg\,m^2/s}$).

Solution

Take the pivot as the origin and the vertically downward direction as the positive direction of the $y$ axis; then the position of the bob is $(x,y)=\ell(\sin\theta,\ \cos\theta)$ (Figure 43.1). The only conservative force is gravity, and the kinetic energy is $T=\tfrac12m\ell^2\dot\theta^2$ and the potential energy is $U=-mg\ell\cos\theta$ (Example 43.6), so

$$ L=T-U=\frac12m\ell^2\dot\theta^2+mg\ell\cos\theta $$

From $\partial L/\partial\dot\theta=m\ell^2\dot\theta$ we get $\dfrac{\dd}{\dd t}\!\left(\pdiff{L}{\dot\theta}\right)=m\ell^2\ddot\theta$, and $\partial L/\partial\theta=-mg\ell\sin\theta$.

The nonconservative force is the air resistance, which is proportional to the speed $\ell\dot\theta$ of the bob and acts in the direction opposing the motion. The unit vector in the tangential direction in which $\theta$ increases is $\bm e_t=\pdiff{(x,y)}{\theta}\Big/\ell=(\cos\theta,\ -\sin\theta)$, so the resistive force is

$$ \bm F'=-c\ell\dot\theta\,\bm e_t $$

(If $\dot\theta\gt0$, the bob is moving in the direction in which $\theta$ increases, so the force acts in the direction of $-\bm e_t$. The expression has the same form and is also correct for $\dot\theta\lt0$.) As in the definition (Section 43.3), the generalized force is obtained by multiplying the force by $\partial(x,y)/\partial\theta=\ell\bm e_t$:

$$ Q_\theta'=\bm F'\cdot\pdiff{(x,y)}{\theta}=-c\ell\dot\theta\,\bm e_t\cdot\ell\bm e_t=-c\ell^2\dot\theta $$

(here $\bm e_t\cdot\bm e_t=1$). This is the magnitude of the moment about the pivot, namely the magnitude $c\ell\dot\theta$ of the resistive force multiplied by the arm length $\ell$. By Law 43.2,

$$ m\ell^2\ddot\theta-\left(-mg\ell\sin\theta\right)=-c\ell^2\dot\theta\quad\Longrightarrow\quad m\ell^2\ddot\theta+c\ell^2\dot\theta+mg\ell\sin\theta=0 $$

Dividing both sides by $m\ell^2$ gives

$$ \ddot\theta+\frac cm\dot\theta+\frac g\ell\sin\theta=0 $$

Unit check: since $F=c\,v$, the unit of $c$ is $\mathrm{N/(m/s)}=\mathrm{kg/s}$, so the unit of $c/m$ is $\mathrm{s^{-1}}$ and the unit of $g/\ell$ is $\mathrm{s^{-2}}$; every term is thus consistent in $\mathrm{s^{-2}}$ (angular acceleration). The tension of the string never appeared in the calculation in the first place.

When $\theta$ is small and $\sin\theta\approx\theta$ can be assumed, the equation becomes $\ddot\theta+(c/m)\dot\theta+(g/\ell)\theta=0$, which has the same form as $\ddot x+(n/m)\dot x+(k/m)x=0$, obtained by dividing the damped-oscillation equation $m\ddot x+n\dot x+kx=0$ of Section 7.3 by $m$. The discriminant of the characteristic equation is $(c/m)^2-4g/\ell$, and for $c\lt2m\sqrt{g/\ell}$ the motion is a damped oscillation.

Solution

Compute the time derivative of $x=\rho\cos\varphi$ with the product rule and the chain rule ($\dd(\cos\varphi)/\dd t=-\dot\varphi\sin\varphi$). Since $\rho,\varphi$ are both functions of $t$,

$$ \dot x=\dot\rho\cos\varphi+\rho\,(-\dot\varphi\sin\varphi)=\dot\rho\cos\varphi-\rho\dot\varphi\sin\varphi $$

Now differentiate this partially with respect to $\dot\varphi$. Since $\rho,\varphi,\dot\rho$ are variables independent of $\dot\varphi$, they may be treated as constants. The term $\dot\rho\cos\varphi$ does not contain $\dot\varphi$, so it gives 0, and $-\rho\dot\varphi\sin\varphi$ is linear in $\dot\varphi$, so

$$ \pdiff{\dot x}{\dot\varphi}=-\rho\sin\varphi $$

On the other hand, differentiating $x=\rho\cos\varphi$ partially with respect to $\varphi$ gives

$$ \pdiff{x}{\varphi}=-\rho\sin\varphi $$

and the two are equal. That is, $\partial\dot x/\partial\dot\varphi=\partial x/\partial\varphi$ (Eq. ② of the main text) has been verified.

The other combinations are similar. We have $\partial\dot x/\partial\dot\rho=\cos\varphi=\partial x/\partial\rho$. Also, for $y=\rho\sin\varphi$, we have $\dot y=\dot\rho\sin\varphi+\rho\dot\varphi\cos\varphi$, and $\partial\dot y/\partial\dot\varphi=\rho\cos\varphi=\partial y/\partial\varphi$ and $\partial\dot y/\partial\dot\rho=\sin\varphi=\partial y/\partial\rho$. For the $z$ coordinate, $z$ itself is a generalized coordinate, so $\partial\dot z/\partial\dot z=1=\partial z/\partial z$.

Why does it hold? It holds because the equation $\dot x=\dfrac{\partial x}{\partial\rho}\dot\rho+\dfrac{\partial x}{\partial\varphi}\dot\varphi+\dfrac{\partial x}{\partial z}\dot z$ (derived in Section 43.2, the expression of $\dot\alpha_j$ in terms of the velocities of the generalized coordinates) is linear in $\dot\rho,\dot\varphi,\dot z$ and its coefficients are exactly $\partial x/\partial\rho,\ \partial x/\partial\varphi,\ \partial x/\partial z$.

Solution

(1) Let $\phi=\pi/2-\theta$; then $\sin\theta=\cos\phi$ and $\dot\theta=-\dot\phi$. The kinetic energy is $T=\tfrac12mR^2\dot\theta^2=\tfrac12mR^2\dot\phi^2$ and the potential energy is $U=mgR(1-\sin\theta)=mgR(1-\cos\phi)$, so in the absence of external forces ($Q_\phi'=0$),

$$ L=\frac12mR^2\dot\phi^2-mgR\,(1-\cos\phi) $$

From $\partial L/\partial\dot\phi=mR^2\dot\phi$ we get $\dfrac{\dd}{\dd t}\!\left(\pdiff{L}{\dot\phi}\right)=mR^2\ddot\phi$, and $\partial L/\partial\phi=-mgR\sin\phi$. Lagrange's equation of motion $\dfrac{\dd}{\dd t}\!\left(\pdiff{L}{\dot\phi}\right)-\pdiff{L}{\phi}=0$ becomes

$$ mR^2\ddot\phi+mgR\sin\phi=0\qquad\therefore\ \ddot\phi=-\frac gR\sin\phi $$

This has the same form as the equation of motion of a simple pendulum of length $R$ (Example 43.6(4)); the cross section of the bowl is a circle of radius $R$, so the particle moves along a circular arc of radius $R$. It also agrees with $\ddot\theta=(g/R)\cos\theta$ of Example 43.12, since $\ddot\theta=-\ddot\phi$ and $\sin\phi=\cos\theta$.

(2) When $\phi$ is small, $\sin\phi\approx\phi$, so $\ddot\phi=-(g/R)\phi$, which is simple harmonic motion with angular frequency $\omega=\sqrt{g/R}$. The period is

$$ T_0=\frac{2\pi}{\omega}=2\pi\sqrt{\frac Rg} $$

which is the same as the period of small oscillations of a simple pendulum of length $R$. For $R=0.50\ \mathrm m$, $T_0=2\pi\sqrt{0.50/9.8}\approx1.42\ \mathrm s$, and $1/4$ of this is $0.355\ \mathrm s$.

(3) The rim is at $\phi=\pi/2$, and since the particle is released gently, $\dot\phi=0$. By conservation of mechanical energy (the normal force does no work), for any $\phi$ we have

$$ \frac12mR^2\dot\phi^2+mgR(1-\cos\phi)=mgR(1-\cos\tfrac\pi2)=mgR $$

Rearranging gives $\tfrac12mR^2\dot\phi^2=mgR\cos\phi$, that is, $\dot\phi^2=\dfrac{2g}{R}\cos\phi$. Since $\phi$ decreases from the rim toward the bottom, $\dot\phi=-\sqrt{2g\cos\phi/R}$, and separating variables and integrating from $\phi=\pi/2$ to $0$, the time required is

$$ t=\int_{\pi/2}^{0}\frac{\dd\phi}{-\sqrt{2g\cos\phi/R}}=\sqrt{\frac{R}{2g}}\int_0^{\pi/2}\frac{\dd\phi}{\sqrt{\cos\phi}} $$

The integrand of the integral on the right-hand side diverges at $\phi=\pi/2$, but the integral itself is finite, and numerical integration gives $\approx2.622$ (it can be written as $\sqrt2\,K(1/\sqrt2)$ using the complete elliptic integral of the first kind). Therefore, for $R=0.50\ \mathrm m,\ g=9.8\ \mathrm{m/s^2}$,

$$ t=\sqrt{\frac{0.50}{2\times9.8}}\times2.622=0.1597\ldots\ \mathrm s\times2.622\approx0.419\ \mathrm s $$

This is about 18% longer than the time $0.355\ \mathrm s$ for the case in which the angular velocity is kept constant by an external force (Example 43.11(1)). The time from the rim to the bottom without external force does not coincide with $1/4$ of the period of small oscillations in (2) ($0.355\ \mathrm s$). This is because, when the amplitude is large ($\phi_0=\pi/2$), the period of a simple pendulum becomes longer than the period of small oscillations. On the other hand, the time of the motion kept at $\dot\theta=\sqrt{g/R}$ by the external force is $0.355\ \mathrm s$ merely as a result of choosing the angular velocity to be the angular frequency of small oscillations.

Solution

Let $x$ be the displacement of the particle (point mass) and $s$ the elongation of the first spring (the one on the wall side). Then the elongation of the second spring is $x-s$. The kinetic energy is only the $\frac12m\dot x^2$ of the particle (the connection point has zero mass), and the potential energy is the sum of the elastic energies of the two springs, so the Lagrangian is

$$ L=\frac12m\dot x^2-\frac12k_1s^2-\frac12k_2(x-s)^2 $$

First we write down Lagrange's equation of motion for the coordinate $s$. Since $L$ does not contain $\dot s$, $\partial L/\partial\dot s=0$; moreover $\partial L/\partial s=-k_1s+k_2(x-s)$ (differentiating $-\frac12k_2(x-s)^2$ with respect to $s$ gives $+k_2(x-s)$), so

$$ \diff{}{t}\left(\pdiff{L}{\dot s}\right)-\pdiff{L}{s}=0-\big[-k_1s+k_2(x-s)\big]=0 \qquad\therefore\ k_1s=k_2(x-s) $$

This expresses the balance of forces at the massless connection point (the force with which the first spring pulls equals the force with which the second spring pulls). Solving it for $s$ gives

$$ s=\frac{k_2}{k_1+k_2}\,x,\qquad x-s=\frac{k_1}{k_1+k_2}\,x $$

Next, the equation for the coordinate $x$: since $\partial L/\partial\dot x=m\dot x$ and $\partial L/\partial x=-k_2(x-s)$,

$$ \diff{}{t}(m\dot x)+k_2(x-s)=0 \qquad\therefore\ m\ddot x+k_2\cdot\frac{k_1}{k_1+k_2}\,x=0 $$

Hence

$$ m\ddot x+\frac{k_1k_2}{k_1+k_2}\,x=0 $$

which has the form of the equation of motion of simple harmonic motion, $m\ddot x=-k_{\text{eq}}x$. Therefore

$$ k_{\text{eq}}=\frac{k_1k_2}{k_1+k_2}\ \left(\text{that is,}\ \frac1{k_{\text{eq}}}=\frac1{k_1}+\frac1{k_2}\right),\qquad \omega=\sqrt{\frac{k_{\text{eq}}}{m}}=\sqrt{\frac{k_1k_2}{m(k_1+k_2)}} $$

These are the results (the high school physics formula for the “equivalent spring constant of springs in series” has come out of Lagrange's equations of motion). Note also that eliminating $s$ by substituting the above relation into $L$ gives $L=\frac12m\dot x^2-\frac12\dfrac{k_1k_2}{k_1+k_2}x^2$, which leads to the same result.

Numerical values: $k_{\text{eq}}=\dfrac{30\times60}{30+60}\ \mathrm{N/m}=20\ \mathrm{N/m}$, $\omega=\sqrt{\dfrac{20\ \mathrm{N/m}}{0.20\ \mathrm{kg}}}=\sqrt{100}\ \mathrm{s^{-1}}=10\ \mathrm{rad/s}$, and the period is $T_0=2\pi/\omega=0.628\ \mathrm s$.

Solution

(1) In rectangular coordinates, $T=\frac12m(\dot x^2+\dot y^2)$, and the potential energy is $U=-mgy$ (since $y$ is positive downward). If we write the Lagrangian without taking the constraint into account, we have

$$ L=\frac12m(\dot x^2+\dot y^2)+mgy $$

If we apply Lagrange's equations of motion to $x,y$ as they stand, we obtain $m\ddot x=0,\ m\ddot y=mg$, that is, unconstrained free fall. This is because the condition $x^2+y^2=\ell^2$, that the length of the rod is constant, has not been included; to incorporate it, we use the method of Lagrange multipliers of Section 44.7 (Example 44.11).

(2) In polar coordinates, $T=\frac12m(\dot r^2+r^2\dot\theta^2)$ and $U=-mgr\cos\theta$ (the particle is lower than the fixed point by $r\cos\theta$), so

$$ L=\frac12m(\dot r^2+r^2\dot\theta^2)+mgr\cos\theta $$

The partial derivatives are $\partial L/\partial\dot r=m\dot r$, $\partial L/\partial r=mr\dot\theta^2+mg\cos\theta$, $\partial L/\partial\dot\theta=mr^2\dot\theta$, and $\partial L/\partial\theta=-mgr\sin\theta$. The tension $N$ acts in the radial direction and pulls the particle toward the origin (the direction in which $r$ decreases), so $Q_r'=-N$; it has no component in the $\theta$ direction, so $Q_\theta'=0$. From Lagrange's equations of motion with nonconservative forces,

\begin{align*} r\text{-direction:}&\quad m\ddot r-mr\dot\theta^2-mg\cos\theta=-N\\ \theta\text{-direction:}&\quad \diff{}{t}(mr^2\dot\theta)+mgr\sin\theta=0 \end{align*}

Substituting the constraint $r=\ell$ ($\dot r=\ddot r=0$), the $r$-direction equation becomes $-m\ell\dot\theta^2-mg\cos\theta=-N$, that is,

$$ N=m\ell\dot\theta^2+mg\cos\theta $$

and the $\theta$-direction equation becomes $m\ell^2\ddot\theta+mg\ell\sin\theta=0$, that is, $\ddot\theta=-(g/\ell)\sin\theta$. Both agree with the results of Section 44.2.3 and Example 44.11 ($\ddot\theta=-(g/\ell)\sin\theta$, $N=mg\cos\theta+m\ell\dot\theta^2$).

(3) For $x=\ell\sin\theta,\ y=\ell\cos\theta$, we have $\dot x=\ell\dot\theta\cos\theta,\ \dot y=-\ell\dot\theta\sin\theta$, so $\dot x^2+\dot y^2=\ell^2\dot\theta^2(\cos^2\theta+\sin^2\theta)=\ell^2\dot\theta^2$. Also, $mgy=mg\ell\cos\theta=mg\ell-mg\ell(1-\cos\theta)$, so

$$ L=\frac12m\ell^2\dot\theta^2+mg\ell\cos\theta =\left[\frac12m\ell^2\dot\theta^2-mg\ell(1-\cos\theta)\right]+mg\ell $$

which differs from the $L$ of Section 44.2.3 only by the constant $mg\ell$. A constant vanishes upon differentiation with respect to $\theta$ or $\dot\theta$, so it does not affect the equation of motion, and the same $\ddot\theta=-(g/\ell)\sin\theta$ is obtained.

Even when we change the choice of coordinates (only $\theta$, with the constraint built in from the start; $(r,\theta)$, keeping the constraint and adding the generalized force $Q_r'=-N$ of the tension; or $(x,y)$, treating the constraint with a multiplier), the procedure of “forming $L$ and substituting it into an equation of the same form” does not change, and we arrive at the same equation of motion—this is advantage (III) of Section 44.1.

Solution

Differentiate $E=\frac12m\big(\dot r^2+r^2\dot\theta^2\big)-\dfrac{GMm}{r}$ with respect to time. By the product rule and the rule for differentiating a composite function,

$$ \diff{E}{t}=m\dot r\ddot r+mr\dot r\dot\theta^2+mr^2\dot\theta\ddot\theta+\frac{GMm}{r^2}\dot r $$

(here we used $\dfrac{\dd}{\dd t}\left(\dfrac12mr^2\dot\theta^2\right)=mr\dot r\dot\theta^2+mr^2\dot\theta\ddot\theta$ and $\dfrac{\dd}{\dd t}\left(-\dfrac{GMm}{r}\right)=+\dfrac{GMm}{r^2}\dot r$). Regrouping this into the terms containing $\dot r$ and the terms containing $\dot\theta$, we obtain

$$ \diff{E}{t}=m\dot r\left(\ddot r-r\dot\theta^2+\frac{GM}{r^2}\right)+mr\dot\theta\left(2\dot r\dot\theta+r\ddot\theta\right) $$

(expanding the right-hand side gives $m\dot r\ddot r-mr\dot r\dot\theta^2+\frac{GMm}{r^2}\dot r+2mr\dot r\dot\theta^2+mr^2\dot\theta\ddot\theta$, and since $-mr\dot r\dot\theta^2+2mr\dot r\dot\theta^2=mr\dot r\dot\theta^2$, this agrees with the expression above). By the equation of motion $m(\ddot r-r\dot\theta^2)=-GMm/r^2$, the first parenthesis is $0$, and by the equation of motion $m(2\dot r\dot\theta+r\ddot\theta)=0$, the second parenthesis is also $0$. Therefore $\dd E/\dd t=0$, that is, $E$ is constant.

For the circular orbit $r=R$ ($\dot r=0$), using $\dot\theta=\Omega$ and $\Omega^2=GM/R^3$ (Example 44.5), the kinetic energy is $T=\frac12mR^2\Omega^2=\frac12mR^2\cdot\dfrac{GM}{R^3}=\dfrac{GMm}{2R}$ and the potential energy is $U=-\dfrac{GMm}{R}$, so

$$ E=T+U=\frac{GMm}{2R}-\frac{GMm}{R}=-\frac{GMm}{2R} $$

$E$ is negative, which expresses that the planet is bound to the Sun. Also, the relation $E=-T=U/2$ holds.

Solution

Take the partial derivatives of $L=\frac12m\big(\dot r^2+r^2\dot\theta^2\big)-U(r)$ treating $r,\dot r,\theta,\dot\theta$ as independent variables. The only term containing $\dot r$ is $\frac12m\dot r^2$; the terms containing $r$ are $\frac12mr^2\dot\theta^2$ and $-U(r)$; the only term containing $\dot\theta$ is $\frac12mr^2\dot\theta^2$; and there is no term in $L$ containing $\theta$ itself ($U$ is a function of $r$ only). Hence

$$ \pdiff{L}{\dot r}=m\dot r,\qquad \pdiff{L}{r}=mr\dot\theta^2-\diff{U}{r},\qquad \pdiff{L}{\dot\theta}=mr^2\dot\theta,\qquad \pdiff{L}{\theta}=0 $$

Lagrange's equations of motion are

\begin{align*} r\text{-direction:}&\quad \diff{}{t}(m\dot r)-\left(mr\dot\theta^2-\diff{U}{r}\right)=0 \ \Longrightarrow\ m\big(\ddot r-r\dot\theta^2\big)=-\diff{U}{r}\\ \theta\text{-direction:}&\quad \diff{}{t}\big(mr^2\dot\theta\big)-0=0 \end{align*}

The right-hand side $-\dd U/\dd r$ of the $r$-direction equation is exactly the radial component $F_r$ of the central force (if $U=-GMm/r$, then $-\dd U/\dd r=-GMm/r^2$, which agrees with the equation of Section 44.3). The $\theta$-direction equation, regardless of the specific form of $U(r)$, means

$$ \diff{}{t}\big(mr^2\dot\theta\big)=0\qquad\therefore\ p_\theta=mr^2\dot\theta=\text{constant} $$

The reason angular momentum is conserved for any central force is that $L$ does not depend on $\theta$ itself (that is, $L$ does not change when the coordinate system is rotated).

Solution

If we carry out the same calculation as in Section 44.5 without imposing the fixed-endpoint condition, the boundary term produced by integration by parts remains, and

$$ \delta S=\left[\pdiff{L}{\dot q}\,\delta q\right]_{t_1}^{t_2}+\int_{t_1}^{t_2}\left(\pdiff{L}{q}-\diff{}{t}\left(\pdiff{L}{\dot q}\right)\right)\delta q\,\dd t $$

The first term is the difference, between the two endpoints, of the product of the generalized momentum $p=\partial L/\partial\dot q$ and the displacement $\delta q$ at the endpoint.

Take the free particle ($L=\frac12m\dot x^2$) as an example. The actual motion is uniform rectilinear motion $x=v_0t$ ($v_0\ne0$), and since $\ddot x=0$, the integrand of the second term is 0. As the displacement, choose for example $\delta x(t)=ct$ ($c$ is a small constant); this is 0 at $t=0$, but at $t=t_2$ it is $ct_2\ne0$ (the final point is not fixed). Then

$$ \delta S=\int_0^{t_2}m\dot x\,\diff{(\delta x)}{t}\,\dd t=\int_0^{t_2}mv_0c\,\dd t=mv_0c\,t_2=\Big[m\dot x\,\delta x\Big]_0^{t_2} $$

so only the boundary term $p\,\delta x$ remains, and it does not become $0$. That is, if we do not fix the position of the final point and demand “$\delta S=0$ for any displacement $\delta x(t_2)$,” we must have $p(t_2)=mv_0=0$ (the momentum is 0 at the final point). This contradicts the motion of a particle that is actually moving ($v_0\ne0$). Therefore, Hamilton's principle that “the actual motion is stationary” must be stated within the family of paths with fixed endpoints.

Conversely, if we restrict ourselves to the family with fixed endpoints, the boundary term vanishes, the conclusion that “the coefficient of the integrand is identically 0” (the fundamental lemma of the calculus of variations) can be used, and the Euler–Lagrange equation is obtained. Physically, fixing the endpoints corresponds to the given data of “where the particle was at times $t_1,t_2$.” Incidentally, the boundary term $p\,\delta q$ expresses how much the action changes when the position of the final point is slightly changed, and this quantity plays an important role in the Hamilton–Jacobi equation (Section 46.5).

Solution

A conservative force is a force for which the work done by the force in moving between two points A and B is determined only by the positions of the two endpoints, independent of the path. Only in that case is the potential energy $U(\bm r)$, defined as the work from a reference point, uniquely determined, and we can write $\bm F=-\nabla U$ (Section 4.4).

The kinetic friction force has magnitude $\mu mg$, and its direction is always opposite to the direction of motion. Therefore, the work done by kinetic friction, with $s$ the total length of the path, is $W=-\mu mg\,s$, which is proportional to the length of the path. Putting in numbers, $\mu mg=0.30\times1.0\ \mathrm{kg}\times9.8\ \mathrm{m/s^2}=2.94\ \mathrm N$, and

\begin{align*} \text{(a) Straight line (}s=2.0\ \mathrm m\text{):}&\quad W=-2.94\ \mathrm N\times2.0\ \mathrm m=-5.88\ \mathrm J\\ \text{(b) Semicircle (radius }1.0\ \mathrm m\text{, so }s=\pi\times1.0\ \mathrm m=3.14\ \mathrm m\text{):}&\quad W=-2.94\ \mathrm N\times3.14\ \mathrm m=-9.24\ \mathrm J \end{align*}

Even for the same movement from A to B the work differs, so the “work from A to B” is not determined by the positions of the endpoints alone. Hence a potential energy as a function of position alone cannot be defined, and friction is a nonconservative force. Moreover, the direction of the frictional force is determined not by position but by the direction of motion (even at the same position, motion in the opposite direction gives a force in the opposite direction), so it is not a function $\bm F(\bm r)$ of position alone in the first place. For this reason, Lagrange's equations of motion cannot be used as they stand in the form with $L=T-U$, and friction must be added on the right-hand side as a generalized force $Q_i'$ (Section 43.5).

Solution

(1) Under the endpoint conditions $x(0)=x(\tau)=0$, the only solution of the equation of motion $\ddot x=-\omega^2x$ when $\omega\tau\ne n\pi$ is $x=0$ (in the general solution $x=C_1\cos\omega t+C_2\sin\omega t$, $x(0)=0$ gives $C_1=0$, and $x(\tau)=0$ gives $C_2\sin\omega\tau=0$; since $\sin\omega\tau\ne0$, we get $C_2=0$). The action of $x=0$ is $S=0$.

The displaced path $x_\varepsilon=\varepsilon\sin(\pi t/\tau)$ satisfies $x_\varepsilon(0)=x_\varepsilon(\tau)=0$. Since $\dot x_\varepsilon=\varepsilon(\pi/\tau)\cos(\pi t/\tau)$, the action is

$$ S(\varepsilon)=\int_0^\tau\left[\frac12m\varepsilon^2\frac{\pi^2}{\tau^2}\cos^2\frac{\pi t}{\tau}-\frac12m\omega^2\varepsilon^2\sin^2\frac{\pi t}{\tau}\right]\dd t $$

Using $\int_0^\tau\cos^2(\pi t/\tau)\,\dd t=\int_0^\tau\sin^2(\pi t/\tau)\,\dd t=\tau/2$ (by the half-angle formula $\cos^2\phi=(1+\cos2\phi)/2$ and so on), we obtain

$$ S(\varepsilon)=\frac12m\varepsilon^2\frac{\pi^2}{\tau^2}\cdot\frac\tau2-\frac12m\omega^2\varepsilon^2\cdot\frac\tau2 =\frac{m\varepsilon^2\tau}{4}\left[\left(\frac\pi\tau\right)^2-\omega^2\right] $$

Note also that there is no term of first order in $\varepsilon$ ($x=0$ is a stationary point).

(2) For $\varepsilon\ne0$ we have $m\varepsilon^2\tau/4>0$, so the sign of $S(\varepsilon)$ is determined by the sign of $(\pi/\tau)^2-\omega^2$.

• When $\omega\tau<\pi$ ($\tau$ is shorter than the half period $\pi/\omega$), $S(\varepsilon)>0=S(0)$, so $x=0$ minimizes the action (for displacements of this form). For example, for $m=1\ \mathrm{kg}$, $\tau=1\ \mathrm s$, $\omega=2\ \mathrm{s^{-1}}$, $\varepsilon=0.1\ \mathrm m$, we have $S=\dfrac{0.01\times1}{4}(\pi^2-4)=0.0147\ \mathrm{J\,s}>0$.

• When $\omega\tau>\pi$ ($\tau$ is longer than the half period), $S(\varepsilon)<0=S(0)$, so there exists a path with a smaller action than $x=0$. Therefore $x=0$ is not a minimum. For example, for $\omega=4\ \mathrm{s^{-1}}$, $S=\dfrac{0.01}{4}(\pi^2-16)=-0.0153\ \mathrm{J\,s}<0$. The same calculation shows that the action of the $n$th-order displacement $\varepsilon\sin(n\pi t/\tau)$ is $\dfrac{m\varepsilon^2\tau}{4}\big[(n\pi/\tau)^2-\omega^2\big]$, so if $\pi<\omega\tau<2\pi$, the action decreases for the $n=1$ displacement and increases for displacements with $n\ge2$. That is, $x=0$ is a saddle point, neither a local minimum nor a local maximum.

• When $\omega\tau=\pi$, the coefficient is 0 and the second-order change is also 0 (a time interval of a half period marks the boundary at which the action ceases to be a minimum).

From the above, we see that the essence of Hamilton's principle is “stationary” ($\delta S=0$), not “minimum,” and that there are examples in which, when the time interval is long, the action ceases to be a minimum.

Solution

Since $F=y\sqrt{1+y'^2}$ (the constant $2\pi$ does not affect the stationarity condition, so it has been dropped) does not contain $x$ explicitly, the Beltrami identity $F-y'\partial F/\partial y'=$ constant can be used. Since $\partial F/\partial y'=y\,y'/\sqrt{1+y'^2}$,

$$ F-y'\pdiff{F}{y'}=y\sqrt{1+y'^2}-\frac{y\,y'^2}{\sqrt{1+y'^2}} =\frac{y\big[(1+y'^2)-y'^2\big]}{\sqrt{1+y'^2}}=\frac{y}{\sqrt{1+y'^2}}=C $$

(where $C>0$ is a constant of integration). Squaring both sides and rearranging gives $y^2=C^2(1+y'^2)$, that is, $y'^2=y^2/C^2-1$, so

$$ y'=\sqrt{\frac{y^2}{C^2}-1}\qquad(y\ge C) $$

(we consider the side with $y'\ge0$). Separating variables gives $\dfrac{\dd y}{\sqrt{y^2/C^2-1}}=\dd x$. If we set $u=y/C$, the left-hand side becomes $C\dfrac{\dd u}{\sqrt{u^2-1}}$; using $\displaystyle\int\frac{\dd u}{\sqrt{u^2-1}}=\operatorname{arcosh}u$ (University Mathematics, Section 2.6), we obtain

$$ C\operatorname{arcosh}\frac yC=x-x_0\qquad\therefore\ y=C\cosh\frac{x-x_0}{C} $$

(here $x_0$ is a constant of integration).

Check: when $y=C\cosh\frac{x-x_0}{C}$, we have $y'=\sinh\frac{x-x_0}{C}$, so $\sqrt{1+y'^2}=\cosh\frac{x-x_0}{C}$ and $y/\sqrt{1+y'^2}=C$, which indeed satisfies the identity above. Also, $y'^2=\sinh^2=\cosh^2-1=y^2/C^2-1$. The surface obtained by rotating this curve about the axis is the shape of the soap film (the catenoid). The constants $C,x_0$ are determined by two conditions, the radii and the positions of the two end rings. For two rings of the same radius $R$, if the separation exceeds about $1.33R$, no solution of this form exists, and the film breaks.

Solution

(1) Take the partial derivatives of $L=\frac12m(\dot r^2+r^2\dot\theta^2)+GMm/r$ treating $r,\dot r,\theta,\dot\theta$ as independent variables.

Quantity in abstract form	Explicit form
$\partial L/\partial\dot r$, and then $\diff{}{t}(\partial L/\partial\dot r)$	$m\dot r$, and then $m\ddot r$
$\partial L/\partial r$	$mr\dot\theta^2-GMm/r^2$
$\partial L/\partial\dot\theta$, and then $\diff{}{t}(\partial L/\partial\dot\theta)$	$mr^2\dot\theta$, and then $m(2r\dot r\dot\theta+r^2\ddot\theta)$ (product rule)
$\partial L/\partial\theta$	$0$ ($\theta$ itself does not appear in $L$)

(Differentiating $GMm/r=GMm\,r^{-1}$ with respect to $r$ gives $-GMm\,r^{-2}$.) Substituting these, the $r$-direction equation is $m\ddot r-mr\dot\theta^2+GMm/r^2=0$ and the $\theta$-direction equation is $m(2r\dot r\dot\theta+r^2\ddot\theta)=0$, which agree with the results of Section 44.3.

(2) Rewrite the $r$-direction equation as $m\ddot r=mr\dot\theta^2-GMm/r^2$. The left-hand side $m\ddot r=\dot p_r$ is the time rate of change of the radial momentum. The first term on the right, $mr\dot\theta^2=\partial T/\partial r$, is a term corresponding to the centrifugal force that appears because we describe the motion in polar coordinates; it points outward (positive). The second term, $-GMm/r^2=Q_r=-\partial U/\partial r$, is the force of universal gravitation, which acts toward the Sun (the direction in which $r$ decreases, negative). This is a concrete example of the fact that in general we can write $\dot p_i=Q_i+\partial T/\partial q_i$.

(3) For a circular orbit, $\dot r=\ddot r=0$, so from the $r$-direction equation $-mr\Omega^2+GMm/r^2=0$, that is, $\Omega^2=GM/r^3$. Putting in numbers, $GM=6.674\times10^{-11}\times1.989\times10^{30}=1.3275\times10^{20}\ \mathrm{m^3/s^2}$ and $r^3=(1.496\times10^{11})^3=3.348\times10^{33}\ \mathrm{m^3}$, so

$$ \Omega=\sqrt{\frac{1.3275\times10^{20}}{3.348\times10^{33}}}\ \mathrm{s^{-1}}=1.99\times10^{-7}\ \mathrm{rad/s},\qquad P=\frac{2\pi}{\Omega}=3.16\times10^7\ \mathrm s $$

Indeed, $3.16\times10^7\ \mathrm s\div(86400\ \mathrm{s/day})=365$ days, which is one year.

Solution

(1) The method using Snell's law applies, in turn, a local rule at each boundary (“determine the next direction from the angle of incidence at the present point”) and builds up the path of the light little by little from the starting point. Fermat's principle is a global formulation: fix the starting point and the end point in advance, and choose the path for which the travel time along the whole path is stationary. As we saw in Section 44.6.3, $n\sin\theta=$ constant is derived from Fermat's principle by the calculus of variations, so the two formulations are mathematically different expressions of the same content.

(2) Once the initial position and velocity are given, the actual motion is uniquely determined, all the way into the future, by the equation of motion at each instant alone. No such function as “knowing the final point” is needed there. Hamilton's principle sounds as if it were “choosing the final point and minimizing the action” because, as the conditions that specify the motion, it chooses to give “the positions of the starting point and the final point” instead of “the initial position and velocity.” It is merely specifying the same motion again by different conditions. Furthermore, the stationary value is not necessarily a minimum (Exercise 44.7), and when the time interval is long it is not even a minimum, so the teleological interpretation that “nature aims at the minimum” does not hold as it stands. In modern physics, $\delta S=0$ is understood as an alternative expression of the same content as the local equation of motion, written as a condition on the whole path.

Solution

Differentiating $L(\dot q)=\tfrac12 m\dot q^2-c\dot q$ with respect to $\dot q$, the generalized momentum is

$$ p=\pdiff{L}{\dot q}=m\dot q-c $$

Solving this for $\dot q$ gives $\dot q=(p+c)/m$. Substituting this into the Legendre transformation $H=p\dot q-L$, we obtain

$$ H=p\cdot\frac{p+c}{m}-\left[\frac12 m\left(\frac{p+c}{m}\right)^2-c\cdot\frac{p+c}{m}\right] $$

Simplifying inside the brackets, using $\tfrac12 m(p+c)^2/m^2=(p+c)^2/(2m)$, we have

$$ H=\frac{p(p+c)}{m}-\frac{(p+c)^2}{2m}+\frac{c(p+c)}{m} =\frac{(p+c)(p+c)}{m}-\frac{(p+c)^2}{2m} =\frac{(p+c)^2}{m}-\frac{(p+c)^2}{2m} $$

(In the middle step we combined the terms as $p(p+c)+c(p+c)=(p+c)(p+c)=(p+c)^2$). Therefore,

$$ H(p)=\frac{(p+c)^2}{2m} $$

Setting $c=0$, we can confirm that we recover the same form as Example 45.1 ($H=p^2/(2m)$, with $a\to m$). The linear term $-c\dot q$ has the effect of shifting the origin of momentum by $p\to p+c$.

Solution

This exercise is practice in reproducing the derivation in Section 45.2 by yourself. We restate only the key points.

Differentiate the defining equation directly. Differentiating $H=\sum_i p_i\dot q_i-L(q,\dot q,t)$, the product rule gives

$$ \dd H=\sum_i\left(\dot q_i\,\dd p_i+p_i\,\dd\dot q_i\right)-\dd L $$

On the other hand, the total differential of $L$ is $\dd L=\sum_i(\partial L/\partial q_i)\dd q_i+\sum_i(\partial L/\partial\dot q_i)\dd\dot q_i+(\partial L/\partial t)\dd t=\sum_i(\partial L/\partial q_i)\dd q_i+\sum_i p_i\,\dd\dot q_i+(\partial L/\partial t)\dd t$ (we used the definition of the generalized momentum, $p_i=\partial L/\partial\dot q_i$), so substituting this gives

$$ \dd H=\sum_i\dot q_i\,\dd p_i-\sum_i\pdiff{L}{q_i}\dd q_i-\pdiff{L}{t}\dd t $$

and the terms $\sum_i p_i\,\dd\dot q_i$ cancel exactly (this is the essential effect of the Legendre transformation).

Use the fact that $H$ is a function of $(q,p,t)$. On the other hand, from the formula for the total differential of a function of several variables, the total differential of $H(q,p,t)$ can also be written as

$$ \dd H=\sum_i\pdiff{H}{q_i}\dd q_i+\sum_i\pdiff{H}{p_i}\dd p_i+\pdiff{H}{t}\dd t $$

Comparing the coefficients of the independent infinitesimal quantities $\dd q_i,\dd p_i,\dd t$ in the two expressions, we obtain the identities of the Legendre transformation

$$ \pdiff{H}{p_i}=\dot q_i,\qquad \pdiff{H}{q_i}=-\pdiff{L}{q_i},\qquad \pdiff{H}{t}=-\pdiff{L}{t} $$

Finally, from Lagrange's equation $\dd(\partial L/\partial\dot q_i)/\dd t=\partial L/\partial q_i$ and the definition of the generalized momentum, we have $\partial L/\partial q_i=\dot p_i$, so the second equation becomes $\dot p_i=-\partial H/\partial q_i$. In this way, Hamilton's canonical equations

$$ \dot q_i=\pdiff{H}{p_i},\qquad \dot p_i=-\pdiff{H}{q_i} $$

are derived. Note that Lagrange's equation (the equation of motion) is used only once, at the very end; the identities before that hold independently of the equation of motion.

As an example with explicit time dependence, take the forced oscillation $L=\tfrac12 m\dot q^2-\tfrac12 kq^2+qF_0\cos\omega t$. Then $p=m\dot q$ and $H=p^2/(2m)+\tfrac12kq^2-qF_0\cos\omega t$, and $\partial H/\partial t=qF_0\omega\sin\omega t=-\partial L/\partial t$ indeed holds.

Solution

The general solution of a single second-order linear ordinary differential equation (for example, $\ddot x+\omega^2x=0$) contains exactly two arbitrary constants ($x=C_1\cos\omega t+C_2\sin\omega t$). This reflects the fact that “finding a function that satisfies a second-order differential equation corresponds to integrating twice, and each integration produces one arbitrary constant.” In the Lagrangian formalism with $n$ generalized coordinates, there is one second-order equation for each coordinate, $n$ equations in all, so the total number of arbitrary constants is $2n$—in terms of initial conditions, these are the initial position $q_i(0)$ and the initial velocity $\dot q_i(0)$ of each coordinate, $2n$ in total.

On the other hand, the general solution of a single first-order linear ordinary differential equation (for example, $\dot y+ay=0$) contains only one arbitrary constant (because a single integration suffices). In the Hamiltonian formalism, there is one first-order equation for each of the $2n$ variables $q_i$ and $p_i$ taken together, $2n$ equations in all, so the total number of arbitrary constants is again $2n$—in terms of initial conditions, these are the initial values $q_i(0),p_i(0)$ of the variables, $2n$ in total.

Therefore, “$n$ second-order equations (a total of $2n$ constants)” and “$2n$ first-order equations (a total of $2n$ constants)” are exactly the same in the amount of information needed to determine the motion uniquely, and the physics that can be described does not change whichever form we use. The only difference is in how that information is carried: whether it is “distributed two to each of the $n$ coordinates” or “one to each of the $2n$ variables” (that is, configuration space or phase space).

Solution

The kinetic energy is $T=\tfrac12 m(\dot x^2+\dot y^2)$ and the potential energy is $U=\tfrac12 k(x^2+y^2)$, and the constraints do not depend explicitly on time, so $H=T+U$ holds (Theorem 45.2). Since the generalized momenta are $p_x=m\dot x,\ p_y=m\dot y$, we can rewrite $T=(p_x^2+p_y^2)/(2m)$, and

$$ H=\frac{p_x^2}{2m}+\frac{p_y^2}{2m}+\frac12 k(x^2+y^2) $$

The canonical equations, written for each of $x,y$ separately, are

$$ \dot x=\pdiff{H}{p_x}=\frac{p_x}{m},\qquad \dot p_x=-\pdiff{H}{x}=-kx, \qquad \dot y=\pdiff{H}{p_y}=\frac{p_y}{m},\qquad \dot p_y=-\pdiff{H}{y}=-ky $$

four equations in all. Combining the two equations for $x$ gives $\ddot x=-(k/m)x$, and combining the two equations for $y$ gives $\ddot y=-(k/m)y$; $y$ does not appear at all in the equation for $x$, and $x$ does not appear at all in the equation for $y$. That is, the Hamiltonian $H$ of the two-dimensional isotropic harmonic oscillator can be decomposed into the sum $H=H_x+H_y$ of the one-dimensional harmonic oscillator in the $x$ direction, $H_x=p_x^2/(2m)+\tfrac12 kx^2$, and the one-dimensional harmonic oscillator in the $y$ direction, $H_y=p_y^2/(2m)+\tfrac12 ky^2$, and the two directions oscillate independently of each other (with exactly the same mathematics as in Example 45.2).

Solution

When the potential energy is a function $U(r)$ of the radial coordinate only (containing no angle $\theta$—this is precisely the condition for a “central force”), the kinetic energy is exactly the same as in the case of universal gravitation, $T=\tfrac12 m(\dot r^2+r^2\dot\theta^2)$, so the Hamiltonian is

$$ H=\frac{p_r^2}{2m}+\frac{p_\theta^2}{2mr^2}+U(r) $$

(here $p_r=m\dot r,\ p_\theta=mr^2\dot\theta$, the same as in Example 45.7). Looking at this $H$, we see that, regardless of the specific form of $U(r)$, $\theta$ does not appear explicitly anywhere:

$$ \pdiff{H}{\theta}=0 $$

Therefore, from the canonical equations, $\dot p_\theta=-\partial H/\partial\theta=0$, that is, $p_\theta=mr^2\dot\theta$ is constant. This is a proof, in the Hamiltonian formalism, that conservation of angular momentum follows solely from the geometric property that “a central force is always directed toward the origin”—whether $U(r)$ is the universal-gravitation form $-GMm/r$, the Hooke's-law form $\tfrac12 kr^2$, or any other function of $r$ alone, the argument is unchanged. For reference, let us also find the radial equation of motion for a general $U(r)$: combining $\dot p_r=-\partial H/\partial r=p_\theta^2/(mr^3)-U'(r)$ with $\dot r=p_r/m$, we obtain

$$ \ddot r=r\dot\theta^2-\frac{U'(r)}{m} $$

(When $U(r)=-GMm/r$, we have $U'(r)=GMm/r^2$, which agrees with the result of Example 45.7, $\ddot r-r\dot\theta^2=-GM/r^2$.)

Solution

(1) Substituting $\ell=0.5\ \mathrm{m}$, $m=0.2\ \mathrm{kg}$, and $g=9.8\ \mathrm{m/s^2}$, the energy of the separatrix is

$$ E_{\mathrm{sep}}=2mg\ell=2\times0.2\times9.8\times0.5=1.96\ \mathrm{J}\approx2.0\ \mathrm{J} $$

(2) When the maximum swing angle is $\theta_0=150^\circ$, the energy is, from $\cos150^\circ=-\cos30^\circ=-\sqrt3/2\approx-0.8660$,

$$ E=mg\ell(1-\cos\theta_0)=0.2\times9.8\times0.5\times(1-(-0.8660)) =0.98\times1.8660\approx1.8\ \mathrm{J} $$

Since $E\approx1.8\ \mathrm{J}$ is smaller than $E_{\mathrm{sep}}\approx2.0\ \mathrm{J}$ ($E\lt E_{\mathrm{sep}}$), by the classification in Section 45.5.2 this motion is a libration. The bob swings back and forth over the range $\pm150^\circ$ centered on the lowest point, without reaching the top ($\theta=\pm180^\circ$) (since the bob is on the end of a rod, it can be lifted to $150^\circ$ and released gently). Note that as $\theta_0$ is brought closer to $180^\circ$, $E$ approaches $E_{\mathrm{sep}}$, and at $\theta_0=180^\circ$ (the limit in which the bob comes to rest exactly at the top) we have exactly $E=E_{\mathrm{sep}}$.

(3) For $\theta_0=60^\circ$ ($\cos60^\circ=0.5$),

$$ E=mg\ell(1-\cos60^\circ)=0.2\times9.8\times0.5\times0.5=0.49\ \mathrm{J} $$

At the lowest point $\theta=0$ we have $U=0$, so $H=p_\theta^2/(2m\ell^2)=E$. Since $m\ell^2=0.2\times0.5^2=0.050\ \mathrm{kg\,m^2}$,

$$ p_\theta=\sqrt{2m\ell^2E}=\sqrt{2\times0.050\times0.49}=\sqrt{0.049}\approx0.22\ \mathrm{kg\,m^2/s} $$

(At this point the angular velocity is $\dot\theta=p_\theta/(m\ell^2)\approx4.4\ \mathrm{rad/s}$ and the speed of the bob is $\ell\dot\theta\approx2.2\ \mathrm{m/s}$, which agrees with $\tfrac12m(\ell\dot\theta)^2\approx0.49\ \mathrm J$.)

Solution

Differentiating the Lagrangian $L=\frac12m(\dot x^2+\dot y^2)-U(x)$ partially with respect to $y$, we note that $U(x)$ does not contain $y$, and the kinetic-energy term contains only $\dot y$ (not $y$ itself), so

$$ \pdiff{L}{y}=0 $$

This is precisely the condition for a cyclic coordinate in Definition 46.1 (this chapter), so $y$ is a cyclic coordinate. By Theorem 46.1, the conjugate momentum

$$ p_y=\pdiff{L}{\dot y}=m\dot y $$

is constant in time, that is, conserved (this can also be verified directly: substituting $\pdiff{L}{y}=0$ into the Lagrange equation for the $y$ direction, $\dd(m\dot y)/\dd t-\pdiff{L}{y}=0$, gives $\dd(m\dot y)/\dd t=0$). This is a natural result, corresponding to the fact that $U$ depends only on $x$ and no force acts in the $y$ direction (the system is uniform in the $y$ direction).

Solution

$T=\frac12\sum_{i,j}m_{ij}(q)\dot q_i\dot q_j$ ($m_{ij}=m_{ji}$) is a homogeneous function of degree 2 in the $\dot q_i$. In general, for a homogeneous function $f(\dot q)$ of degree $n$ (a function such that $f\to\lambda^nf$ when all the $\dot q_i$ are multiplied by $\lambda$), Euler's theorem on homogeneous functions, $\sum_i\dot q_i\pdiff{f}{\dot q_i}=nf$, holds. Indeed, differentiating $T$ partially with respect to $\dot q_k$ (noting that $m_{ij}=m_{ji}$) gives

$$ \pdiff{T}{\dot q_k}=\sum_im_{ik}\dot q_i $$

so that

$$ \sum_k\dot q_k\pdiff{T}{\dot q_k}=\sum_{k,i}m_{ik}\dot q_i\dot q_k=2T $$

(the last equality is just the definition $T=\frac12\sum_{i,k}m_{ik}\dot q_i\dot q_k$). This is Euler's theorem for the case $n=2$. Since $U=U(q)$ does not depend on $\dot q$, we have $\pdiff{U}{\dot q_i}=0$; hence, for $L=T-U$,

$$ h=\sum_i\dot q_i\pdiff{L}{\dot q_i}-L=\sum_i\dot q_i\pdiff{T}{\dot q_i}-(T-U)=2T-(T-U)=T+U $$

and $h=T+U$ has been shown (a calculation with a concrete symmetric matrix $m_{ij}$ for two degrees of freedom gives the same result).

Solution

The standard Lagrangian of a double pendulum of length $\ell$ and mass $m$ (deflection angles $\theta_1,\theta_2$) is

$$ L=m\ell^2\dot\theta_1^2+\frac12m\ell^2\dot\theta_2^2+m\ell^2\dot\theta_1\dot\theta_2\cos(\theta_1-\theta_2)+2mg\ell\cos\theta_1+mg\ell\cos\theta_2 $$

Differentiating this partially with respect to each of $\theta_1,\theta_2$ gives

$$ \pdiff{L}{\theta_1}=-2mg\ell\sin\theta_1-m\ell^2\dot\theta_1\dot\theta_2\sin(\theta_1-\theta_2),\qquad \pdiff{L}{\theta_2}=-mg\ell\sin\theta_2+m\ell^2\dot\theta_1\dot\theta_2\sin(\theta_1-\theta_2) $$

and neither is zero in general. For example, substituting $\theta_1=\theta_2=\pi/2,\ \dot\theta_1=\dot\theta_2=0$ gives $\pdiff{L}{\theta_1}=-2mg\ell\neq0$ ($\pdiff{L}{\theta_1}$ vanishes only for special values such as $\theta_1=\theta_2=0$ with zero angular velocities; it is not identically zero). Therefore neither of $\theta_1,\theta_2$ is a cyclic coordinate. In planetary motion (Section 46.1 of this chapter), $\theta$ became a cyclic coordinate because universal gravitation does not depend on the azimuthal angle $\theta$ (there is rotational symmetry); in the double pendulum, however, gravity points in a particular direction, vertically downward, so the deflection angles $\theta_1,\theta_2$ themselves appear in the potential energy, and there is no rotational symmetry. This example teaches the lesson that, in general, there are not necessarily as many cyclic coordinates as degrees of freedom; the number is determined by what symmetries the system has. Incidentally, for a three-dimensional spherical pendulum (polar angle $\theta$, azimuthal angle $\varphi$), gravity is symmetric under rotations about the vertical axis, so the Lagrangian $\frac12m\ell^2(\dot\theta^2+\sin^2\theta\,\dot\varphi^2)+mg\ell\cos\theta$ does not contain $\varphi$, and $\varphi$ becomes a cyclic coordinate (the corresponding conserved quantity $p_\varphi=m\ell^2\sin^2\theta\,\dot\varphi$ is the angular momentum about the vertical axis).

Solution

As a concrete example, consider a spring pendulum whose spring constant varies periodically with time, $U(x,t)=\frac12k_0\bigl(1+\cos\omega_{\rm d}t\bigr)x^2$ ($k_0,\omega_{\rm d}$ are constants; this represents, for example, a situation in which the stiffness of the spring fluctuates slowly because of temperature changes and the like). The Lagrangian is $L=\frac12m\dot x^2-U(x,t)$, and differentiating it partially with respect to the time $t$ gives

$$ \pdiff{L}{t}=-\pdiff{U}{t}=\frac12k_0\omega_{\rm d}x^2\sin\omega_{\rm d}t $$

This is in general not zero (when $x\neq0$ and $\sin\omega_{\rm d}t\neq0$). Therefore, by Theorem 46.2 of this chapter (the equation $\dd h/\dd t=-\pdiff{L}{t}$), the energy function $h=\frac12m\dot x^2+U(x,t)$ is not conserved:

$$ \diff{h}{t}=-\pdiff{L}{t}=-\frac12k_0\omega_{\rm d}x^2\sin\omega_{\rm d}t\neq0 $$

Physically, the operation of externally varying the stiffness of the spring does work on the pendulum or takes energy from it, so it is natural that energy is not conserved when we look at the pendulum alone (of course, energy is conserved for the whole system including the external operator). This confirms, with a different concrete example, the same kind of phenomenon as in Example 46.5 of this chapter (the simple pendulum whose support is shaken).

Solution

We start from the definition of the Poisson bracket (the equation $\{A,B\}=\sum_i(\partial_{q_i}A\,\partial_{p_i}B-\partial_{p_i}A\,\partial_{q_i}B)$ of this chapter). Writing out $\{B,A\}$ according to the same definition, we have

$$ \{B,A\}=\sum_i\left(\pdiff{B}{q_i}\pdiff{A}{p_i}-\pdiff{B}{p_i}\pdiff{A}{q_i}\right) $$

Now, looking closely at each term on the right-hand side, we see that the products of the same two factors that appear in the definition of $\{A,B\}$ simply appear with their order changed (the value is unchanged, because the order of multiplication can be exchanged): $\pdiff{B}{q_i}\pdiff{A}{p_i}=\pdiff{A}{p_i}\pdiff{B}{q_i}$, $\pdiff{B}{p_i}\pdiff{A}{q_i}=\pdiff{A}{q_i}\pdiff{B}{p_i}$. Therefore,

$$ \{B,A\}=\sum_i\left(\pdiff{A}{p_i}\pdiff{B}{q_i}-\pdiff{A}{q_i}\pdiff{B}{p_i}\right)=-\sum_i\left(\pdiff{A}{q_i}\pdiff{B}{p_i}-\pdiff{A}{p_i}\pdiff{B}{q_i}\right)=-\{A,B\} $$

and $\{A,B\}=-\{B,A\}$ has been shown directly from the definition (letting $A=B$ gives $\{A,A\}=-\{A,A\}$, that is, $2\{A,A\}=0$, so $\{A,A\}=0$ follows at the same time).

Solution

Suppose that $A(q,p)$ does not contain the time $t$ explicitly. Taking the total differential of $A(q(t),p(t))$ with respect to the time $t$ by the differentiation rule for composite functions (the chain rule), we obtain

$$ \diff{A}{t}=\sum_i\left(\pdiff{A}{q_i}\dot q_i+\pdiff{A}{p_i}\dot p_i\right) $$

Substituting the canonical equations $\dot q_i=\pdiff{H}{p_i},\ \dot p_i=-\pdiff{H}{q_i}$ here, we obtain

$$ \diff{A}{t}=\sum_i\left(\pdiff{A}{q_i}\pdiff{H}{p_i}-\pdiff{A}{p_i}\pdiff{H}{q_i}\right) $$

The right-hand side is nothing other than the definition of the Poisson bracket (with $B=H$):

$$ \diff{A}{t}=\{A,H\} $$

This is Theorem 46.6 of this chapter, derived on our own from the beginning for the case in which $A$ does not contain $t$ explicitly ($\partial A/\partial t=0$) (for example, in the concrete case $A=qp$ with $H$ taken to be the Hamiltonian of simple harmonic motion, $\dd A/\dd t$ calculated by the chain rule and the Poisson bracket $\{A,H\}$ both equal $p^2/m-kq^2$, and they agree).

Solution

First we compute $\{L_z,H\}$ from Definition 46.3. The partial derivatives of $L_z=xp_y-yp_x$ are

$$ \pdiff{L_z}{x}=p_y,\qquad\pdiff{L_z}{y}=-p_x,\qquad\pdiff{L_z}{p_x}=-y,\qquad\pdiff{L_z}{p_y}=x,\qquad\pdiff{L_z}{z}=\pdiff{L_z}{p_z}=0 $$

Setting $(q_1,q_2,q_3)=(x,y,z)$ and $(p_1,p_2,p_3)=(p_x,p_y,p_z)$ and substituting into Definition 46.3, we obtain

$$ \{L_z,H\}=\left(\pdiff{L_z}{x}\pdiff{H}{p_x}+\pdiff{L_z}{y}\pdiff{H}{p_y}\right)-\left(\pdiff{L_z}{p_x}\pdiff{H}{x}+\pdiff{L_z}{p_y}\pdiff{H}{y}\right) =p_y\pdiff{H}{p_x}-p_x\pdiff{H}{p_y}+y\pdiff{H}{x}-x\pdiff{H}{y} $$

that is,

$$ \{L_z,H\}=y\pdiff{H}{x}-x\pdiff{H}{y}+p_y\pdiff{H}{p_x}-p_x\pdiff{H}{p_y} $$

The meaning of this expression becomes clear if we calculate the change in $H$ under an infinitesimal rotation about the $z$ axis, $x\to x-\varepsilon y,\ y\to y+\varepsilon x,\ p_x\to p_x-\varepsilon p_y,\ p_y\to p_y+\varepsilon p_x$ (positions and momenta rotate in the same way). By the chain rule,

$$ \delta H=\varepsilon\left(-y\pdiff{H}{x}+x\pdiff{H}{y}-p_y\pdiff{H}{p_x}+p_x\pdiff{H}{p_y}\right)=-\varepsilon\{L_z,H\} $$

Therefore,

$H$ is invariant under rotations about the $z$ axis ($\delta H=0$, rotational symmetry) $\iff$ $\{L_z,H\}=0$ ($L_z$ is a conserved quantity)

and this equivalence holds in general, not only for $H$ of a particular form. Let us check it with concrete examples.

(i) For a central force, $H=\dfrac{p_x^2+p_y^2+p_z^2}{2m}+U(r)$, $r=\sqrt{x^2+y^2+z^2}$, we have $\pdiff{H}{x}=U'(r)\dfrac{x}{r},\ \pdiff{H}{y}=U'(r)\dfrac{y}{r},\ \pdiff{H}{p_x}=\dfrac{p_x}{m},\ \pdiff{H}{p_y}=\dfrac{p_y}{m}$, so

$$ \{L_z,H\}=yU'\frac{x}{r}-xU'\frac{y}{r}+p_y\frac{p_x}{m}-p_x\frac{p_y}{m}=0 $$

A similar calculation gives $\{L_x,H\}=\{L_y,H\}=0$ as well, so for a central force all three components of the angular momentum are conserved.

(ii) Even when $H$ contains a mixed term such as $xp_x+yp_y$, this quantity is invariant under rotations, so $\{L_z,H\}=0$ (from $\pdiff{H}{x}=p_x,\ \pdiff{H}{y}=p_y,\ \pdiff{H}{p_x}=x,\ \pdiff{H}{p_y}=y$ we get $yp_x-xp_y+p_yx-p_xy=0$). A “function of $x^2+y^2$ and $p_x^2+p_y^2$ only” is merely one example of a rotation-invariant $H$, and a rotation-invariant $H$ is not limited to this form.

(iii) If $cx$ ($c$ is a constant; this is the potential corresponding to a uniform force in the $x$ direction) is added to $H$, then $c$ is added to $\pdiff{H}{x}$, so $\{L_z,H\}=cy\neq0$. The rotational symmetry is broken, and $L_z$ is not conserved: it changes at the rate $\dd L_z/\dd t=\{L_z,H\}=cy$ (this agrees with the moment $xF_y-yF_x=cy$ of the force $-c\,\bm e_x$).

Solution

We compute the criterion $\{Q,P\}=1$ directly from the definition. From $Q=q$ we get $\pdiff{Q}{q}=1,\ \pdiff{Q}{p}=0$. From $P=p+f(q)$ we get $\pdiff{P}{q}=f'(q)$ (the ordinary derivative of $f$ with respect to $q$) and $\pdiff{P}{p}=1$. Substituting these,

$$ \{Q,P\}=\pdiff{Q}{q}\pdiff{P}{p}-\pdiff{Q}{p}\pdiff{P}{q}=1\times1-0\times f'(q)=1 $$

so that, whatever (differentiable) function $f(q)$ is, $\{Q,P\}=1$ always holds. Therefore $Q=q,\ P=p+f(q)$ is always a canonical transformation. This corresponds to taking $Q=q$ ($Q'=1$) and setting $g(q)=f(q)$ in the general form of the point transformation, $P=p/Q'(q)+g(q)$ (Section 46.4.1). This transformation adds to the momentum a “shift determined by the coordinate alone,” and it is another typical way of constructing canonical transformations, paired with the point transformation seen in Example 46.9 of this chapter (a transformation that relabels only the coordinates). (Physically, it has the same mathematical structure as the fact that the canonical momentum $p=m\dot q+eA(q)$ of a charged particle (charge $e$) in an electromagnetic field ($A$ is the vector potential) has the form of the mechanical momentum $m\dot q$ plus a term determined by the position alone—we do not go into this within the scope of the electromagnetism in this book, but it is one example showing that the idea of canonical transformations appears in such situations.)

Solution

(1) From $U_{\rm eff}'(r)=-\dfrac{p_\theta^2}{mr^3}+\dfrac{GMm}{r^2}=0$, the radius of the circular orbit is

$$ r_0=\frac{p_\theta^2}{GMm^2} $$

The value there follows from $\dfrac{p_\theta^2}{2mr_0^2}=\dfrac{G^2M^2m^3}{2p_\theta^2}$ and $\dfrac{GMm}{r_0}=\dfrac{G^2M^2m^3}{p_\theta^2}$:

$$ E_{\min}=U_{\rm eff}(r_0)=\frac{G^2M^2m^3}{2p_\theta^2}-\frac{G^2M^2m^3}{p_\theta^2}=-\frac{G^2M^2m^3}{2p_\theta^2}=-\frac{GMm}{2r_0} $$

(the second derivative $U_{\rm eff}''(r_0)=GMm/r_0^3\gt0$ shows that this is a minimum point).

(2) The graph has the shape of a valley: as $r\to0$ the centrifugal term $p_\theta^2/(2mr^2)\to+\infty$; as $r\to\infty$, $U_{\rm eff}\to0$ (it approaches from the negative side because $-GMm/r$ dominates); and at $r=r_0$ it takes the minimum value $E_{\min}\lt0$ (Figure 8.5). Since the radial motion obeys $E=\frac12m\dot r^2+U_{\rm eff}(r)$, the condition $\dot r^2\ge0$ means that the particle can move only in the range $E\ge U_{\rm eff}(r)$, and the intersections of the horizontal line at $E$ with the curve are the turning points. Solving $E=U_{\rm eff}(r)$ as the quadratic equation in $u=1/r$, $\frac{p_\theta^2}{2m}u^2-GMmu-E=0$, we obtain

$$ u=\frac{m}{p_\theta^2}\left[GMm\pm\sqrt{G^2M^2m^2+\frac{2Ep_\theta^2}{m}}\right] $$

as the two roots.

• (a) $E=E_{\min}$: the quantity under the square root is 0, and there is only one turning point, $r=r_0$. A circular orbit that stays at $r=r_0$.
• (b) $E_{\min}\lt E\lt0$: the quantity under the square root is positive, and both roots for $u$ are positive, so there are two turning points of $r$, $r_{\min}\lt r_0\lt r_{\max}$ (the perihelion and the aphelion). An elliptical orbit (bound orbit) confined to the range $r_{\min}\le r\le r_{\max}$.
• (c) $E=0$: the roots are $u=2GMm^2/p_\theta^2$ (a positive root) and $u=0$, so there is only one turning point, $r_{\min}=p_\theta^2/(2GMm^2)=r_0/2$, and the particle can go out to $r\to\infty$. A parabolic orbit (the boundary between bound and unbound).
• (d) $E\gt0$: there is only one positive root $u$ (the other root is negative), so there is only one turning point, $r_{\min}\lt r_0/2$, and the particle can go out to $r\to\infty$. A hyperbolic orbit (unbound orbit).

This corresponds to the fact that, in terms of the eccentricity $\varepsilon=\sqrt{1+E/|E_{\min}|}$ (a rewriting of the relation $E=-\frac{GMm}{2r_0}(1-\varepsilon^2)$ in Section 8.4), we have $\varepsilon=0$ (circle), $0\lt\varepsilon\lt1$ (ellipse), $\varepsilon=1$ (parabola), and $\varepsilon\gt1$ (hyperbola).

Solution

(1) Substituting $F=qQ$ into the equations $p=\pdiff{F}{q},\ P=-\pdiff{F}{Q}$, we obtain

$$ p=\pdiff{(qQ)}{q}=Q,\qquad P=-\pdiff{(qQ)}{Q}=-q $$

that is, $Q=p,\ P=-q$ (since $\dfrac{\partial^2F}{\partial q\,\partial Q}=1\neq0$, the equations can be solved for $Q$).

(2) From $\pdiff{Q}{q}=0,\ \pdiff{Q}{p}=1,\ \pdiff{P}{q}=-1,\ \pdiff{P}{p}=0$,

$$ \{Q,P\}=\pdiff{Q}{q}\pdiff{P}{p}-\pdiff{Q}{p}\pdiff{P}{q}=0\times0-1\times(-1)=1 $$

holds.

(3) Since $F$ does not contain $t$, $K=H$ (in the equation $K=H+\pdiff{F}{t}$, $\pdiff{F}{t}=0$). Substituting $q=-P,\ p=Q$ into $H=\dfrac{p^2}{2m}+\dfrac12kq^2$ gives

$$ K(Q,P)=\frac{Q^2}{2m}+\frac12kP^2 $$

The new canonical equations are $\dot Q=\pdiff{K}{P}=kP$ and $\dot P=-\pdiff{K}{Q}=-\dfrac{Q}{m}$. Returning to the original variables, $\dot Q=\dot p=-kq=kP$ (which agrees with the original equation of motion $\dot p=-kq$) and $\dot P=-\dot q=-\dfrac{p}{m}=-\dfrac{Q}{m}$ (which agrees with the original $\dot q=p/m$), so there is no contradiction. Even when the roles are exchanged, the equations of motion keep the same form (the canonical equations).

Solution

(1) Differentiating $\rho=\rho(H(q,p))$ partially, the chain rule gives $\pdiff{\rho}{q_i}=\rho'(H)\pdiff{H}{q_i}$ and $\pdiff{\rho}{p_i}=\rho'(H)\pdiff{H}{p_i}$. Substituting these into the definition of $\{\rho,H\}$, we obtain

$$ \{\rho,H\}=\sum_i\left(\pdiff{\rho}{q_i}\pdiff{H}{p_i}-\pdiff{\rho}{p_i}\pdiff{H}{q_i}\right)=\rho'(H)\sum_i\left(\pdiff{H}{q_i}\pdiff{H}{p_i}-\pdiff{H}{p_i}\pdiff{H}{q_i}\right)=0 $$

(the two terms in the parentheses are the same product and cancel). Therefore, by the Liouville equation, $\pdiff{\rho}{t}=-\{\rho,H\}=0$; that is, $\rho$ does not depend on time.

(2) In the phase space of simple harmonic motion, $H=p^2/(2m)+\frac12kx^2$, the curve $H=E$ (constant) is an ellipse centered at the origin. Since $\rho=\exp(-\beta H)$ is a function of $H$, it takes the same value at every point on each ellipse, so its value is fixed ellipse by ellipse (the farther out the ellipse, the larger $H$ and the smaller $\rho$). By the canonical equations, a point in phase space goes round and round on its own ellipse while $H$ stays constant. Since the points merely move along the same ellipse, the density on the ellipse does not change, and the shape of the distribution of $\rho$ (the shading in concentric ellipses) does not change as time passes. This is the reason why $\rho=\rho(H)$ is a stationary distribution.

Solution

By linearity, $\{L_x,\abs{\bm L}^2\}=\{L_x,L_x^2\}+\{L_x,L_y^2\}+\{L_x,L_z^2\}$. Leibniz's rule for the second argument follows from antisymmetry and Leibniz's rule of Theorem 46.4: $\{C,AB\}=-\{AB,C\}=-A\{B,C\}-\{A,C\}B=A\{C,B\}+\{C,A\}B$. Using this with $C=L_x$, $A=B=L_j$, we obtain $\{L_x,L_j^2\}=2L_j\{L_x,L_j\}$ (the factors $L_j$ in the product commute with each other). Therefore,

$$ \{L_x,L_x^2\}=2L_x\{L_x,L_x\}=0 $$ $$ \{L_x,L_y^2\}=2L_y\{L_x,L_y\}=2L_yL_z $$ $$ \{L_x,L_z^2\}=2L_z\{L_x,L_z\}=2L_z\times(-L_y)=-2L_zL_y $$

(we used $\{L_x,L_z\}=-\{L_z,L_x\}=-L_y$). Adding the three gives $0+2L_yL_z-2L_zL_y=0$; that is, $\{L_x,\abs{\bm L}^2\}=0$ has been shown. By the cyclic permutation $x\to y\to z\to x$, $\{L_y,\abs{\bm L}^2\}=\{L_z,\abs{\bm L}^2\}=0$ also holds.

Meaning: $\abs{\bm L}^2$ commutes with each of $L_x,L_y,L_z$ (the Poisson bracket is 0). Therefore, as a set whose members have vanishing brackets with one another, we can, for example, choose $\abs{\bm L}^2$ and $L_z$ simultaneously. In quantum mechanics, because $[\hat{\bm L}^2,\hat L_z]=0$, $\hat{\bm L}^2$ and $\hat L_z$ can simultaneously have definite values ($\hat L_x,\hat L_y,\hat L_z$ themselves do not commute with one another, so they cannot simultaneously have definite values).

Solution

(1) We construct a canonical transformation under which the new Hamiltonian $K$ vanishes identically, using a generating function of the type $F_2(q,P,t)=S(q,P,t)$. Setting $K=0$ in the transformation relations $p=\pdiff{S}{q},\ Q=\pdiff{S}{P},\ K=H+\pdiff{S}{t}$ and substituting $p=\pdiff{S}{q}$ into $H$, we obtain the Hamilton–Jacobi equation

$$ \pdiff{S}{t}+H\!\left(q,\pdiff{S}{q},t\right)=0 $$

$S$ is a function corresponding to the action $\int L\,\dd t$ evaluated along the motion. Since $K=0$, we have $\dot Q=\pdiff{K}{P}=0,\ \dot P=-\pdiff{K}{Q}=0$, so the new coordinate $Q$ and the new momentum $P$ are both constants, independent of time (instead of solving $2n$ canonical equations, solving a single first-order partial differential equation for $S$ determines the entire motion in terms of $2n$ constants). As for its place in the theory, what is important is that, in the limit $\hbar\to0$, the wave function $\psi=\exp(iS/\hbar)$ leads to this equation in the semiclassical approximation (WKB approximation) of quantum mechanics, and that the equation has the same form as the equation of geometrical optics.

(2) Since $H=p^2/(2m)$, the Hamilton–Jacobi equation is $\pdiff{S}{t}+\dfrac{1}{2m}\left(\pdiff{S}{q}\right)^2=0$. For $S=Pq-\dfrac{P^2}{2m}t$ we have $\pdiff{S}{q}=P,\ \pdiff{S}{t}=-\dfrac{P^2}{2m}$, so

$$ \pdiff{S}{t}+\frac{1}{2m}\left(\pdiff{S}{q}\right)^2=-\frac{P^2}{2m}+\frac{P^2}{2m}=0 $$

and it is indeed a solution. In this case $p=\pdiff{S}{q}=P$ (the momentum is constant), and since

$$ Q=\pdiff{S}{P}=q-\frac{P}{m}t $$

is a constant independent of time, we obtain $q=Q+\dfrac{P}{m}t$; that is, uniform rectilinear motion with speed $P/m$.

Solution

Substituting $n=5$ into the formula for the volume of an $n$-dimensional ball, $V_n(R)=\pi^{n/2}R^n/\Gamma(n/2+1)$, we find that we need $\Gamma(5/2+1)=\Gamma(7/2)$. We use the recurrence $\Gamma(\alpha+1)=\alpha\Gamma(\alpha)$ repeatedly until we reach $\Gamma(1/2)=\sqrt\pi$:

$$ \Gamma\!\left(\frac72\right)=\frac52\,\Gamma\!\left(\frac52\right)=\frac52\times\frac32\,\Gamma\!\left(\frac32\right)=\frac52\times\frac32\times\frac12\,\Gamma\!\left(\frac12\right)=\frac{15}{8}\sqrt\pi $$

Therefore, we obtain

$$ V_5(R)=\frac{\pi^{5/2}}{\dfrac{15}8\sqrt\pi}R^5=\frac{\pi^{5/2}}{\pi^{1/2}}\times\frac8{15}R^5=\frac8{15}\pi^2R^5 $$

For $n=1,2,3,4$ the volumes were $2R,\ \pi R^2,\ \frac43\pi R^3,\ \frac{\pi^2}2R^4$; compared with these, we can see that both the denominator and the numerator of the coefficient keep growing larger.

Solution

Starting from $\Gamma(1)=1$ and using the recurrence $\Gamma(\alpha+1)=\alpha\Gamma(\alpha)$ four times, we get

$$ \Gamma(2)=1\cdot\Gamma(1)=1,\quad\Gamma(3)=2\cdot\Gamma(2)=2,\quad\Gamma(4)=3\cdot\Gamma(3)=6,\quad\Gamma(5)=4\cdot\Gamma(4)=24 $$

(this agrees with $4!=24$). In general, for a natural number $m$, using the recurrence repeatedly in the order $m-1,m-2,\ldots,1$ gives

$$ \Gamma(m)=(m-1)\,\Gamma(m-1)=(m-1)(m-2)\,\Gamma(m-2)=\cdots=(m-1)(m-2)\cdots2\cdot1\cdot\Gamma(1) $$

and, using $\Gamma(1)=1$, we get

$$ \Gamma(m)=(m-1)(m-2)\cdots2\cdot1=(m-1)! $$

This is what we set out to show. It is nothing other than the relation “$\Gamma(n+1)=n!$” verified in the main text, written with the index shifted to $n=m-1$.

Solution

The trapezoidal rule is a method that approximates the area over the interval $[k-1,k]$ by the width $1$ multiplied by the average of the heights at the two ends, $\ln(k-1)$ and $\ln k$, that is, by $\frac12\bigl(\ln(k-1)+\ln k\bigr)$. Since the interval of integration is $[1,n]$, we add this up over the $n-1$ intervals $k=2,3,\ldots,n$ (we do not use the interval $[0,1]$ for $k=1$ because it contains $\ln0$):

$$ \int_1^n\ln x\,\dd x\approx\sum_{k=2}^n\frac12\bigl(\ln(k-1)+\ln k\bigr)=\frac12\sum_{j=1}^{n-1}\ln j+\frac12\sum_{k=2}^{n}\ln k $$

On the right-hand side, each of $\ln2,\ln3,\ldots,\ln(n-1)$ appears once in each of two trapezoids and so receives the weight $\frac12+\frac12=1$; the endpoint $\ln1=0$ adds nothing; and only the endpoint $\ln n$ receives the weight $\frac12$. Therefore (writing $\sum_{k=2}^n\ln k=\sum_{k=1}^n\ln k$, which holds because $\ln1=0$), we obtain

$$ \int_1^n\ln x\,\dd x\approx\sum_{k=1}^n\ln k-\frac12\ln n $$

The left-hand side equals $n\ln n-n+1$, as computed in Section 47.3, so

$$ \sum_{k=1}^n\ln k\approx n\ln n-n+\frac12\ln n+1 $$

This is the result. Compared with the right-endpoint approximation in the main text ($\sum_{k=1}^n\ln k\approx n\ln n-n+1$, from which the constant $1$ is later dropped to give $n\ln n-n$), it has the extra correction term $+\frac12\ln n$. This corresponds to the fact that the total area of the right-endpoint rectangles, $\sum_{k=2}^n\ln k$, is larger than the trapezoidal-rule value by $\frac12(\ln n-\ln1)=\frac12\ln n$.

Comparing this with the logarithm of the more precise version $n!\approx\sqrt{2\pi n}\,n^ne^{-n}$ learned in University Mathematics, Section 32.3, namely $\ln n!\approx n\ln n-n+\frac12\ln n+\frac12\ln(2\pi)$, we see that the trapezoidal-rule result differs from it only in the constant term: what remains is the difference between $1$ and $\frac12\ln(2\pi)\approx0.919$, namely $1-\frac12\ln(2\pi)\approx0.081$. In other words, the trapezoidal-rule correction correctly captures the part of the correction term of the precise version that depends on $n$, namely $\frac12\ln n$.

A numerical comparison at $n=10,50,100$ gives the following values for the true value, the simple version (the logarithm of Formula 47.3 in the main text), and the trapezoidal-rule approximation (including the constant $1$), respectively.

$$ \begin{array}{c|ccc} n & \text{True value }\ln n! & \text{Simple version} & \text{Trapezoidal rule}\\\hline 10 & 15.104 & 13.026\ (\text{error}\ 2.079) & 15.177\ (\text{error}\ {-0.073})\\ 50 & 148.478 & 145.601\ (\text{error}\ 2.877) & 148.557\ (\text{error}\ {-0.079})\\ 100 & 363.739 & 360.517\ (\text{error}\ 3.222) & 363.820\ (\text{error}\ {-0.080}) \end{array} $$

(The error is the true value $-$ the approximate value.) The errors of the simple version, $2.08,\ 2.88,\ 3.22$, are nearly equal to $\frac12\ln(2\pi n)$ (for $n=10,50,100$ these are $2.07,\ 2.87,\ 3.22$; see the Supplement in Section 47.3 of the main text), and they increase slowly with $n$. In contrast, the error of the trapezoidal-rule approximation has an absolute value of about $0.08$ or less for every $n$, is nearly constant independent of $n$ (approaching the constant $-0.081$), and is roughly $1/30$ of the error of the simple version. Moreover, if we also drop the constant $1$ and use $n\ln n-n+\frac12\ln n$, the errors are $0.927,\ 0.921,\ 0.920$, and about $1-0.081=0.919$ remains. In either case, the trapezoidal-rule approximation is consistently closer to the true value than the simple version.

Solution

Let $H_n=\displaystyle\sum_{k=1}^n\frac1k$ be the partial sum of the harmonic series. We treat the first term $1$ separately and group the remaining terms so that the number of terms in each group is a power of $2$ (the $j$th group has $2^{j-1}$ terms, $j=1,2,\ldots,k$):

$$ H_{2^k}=1+\underbrace{\frac12}_{1\text{ term}}+\underbrace{\left(\frac13+\frac14\right)}_{2\text{ terms}}+\underbrace{\left(\frac15+\frac16+\frac17+\frac18\right)}_{4\text{ terms}}+\cdots+\underbrace{\left(\frac1{2^{k-1}+1}+\cdots+\frac1{2^k}\right)}_{2^{k-1}\text{ terms}} $$

Every term in a group is at least as large as the last term of that group (its smallest value). For example, $\left(\frac13+\frac14\right)\ge\left(\frac14+\frac14\right)=\frac12$ and $\left(\frac15+\frac16+\frac17+\frac18\right)\ge\left(\frac18+\frac18+\frac18+\frac18\right)=\frac12$; in the same way, in the $j$th group, which consists of $2^{j-1}$ terms ($j=1,2,\ldots,k$), every term is at least the minimum value $1/2^j$, so the sum of that group is at least $2^{j-1}\times\frac1{2^j}=\frac12$ (the group with $j=1$ is exactly $\frac12$). Since there are $k$ groups,

$$ H_{2^k}\ge1+\underbrace{\frac12+\frac12+\cdots+\frac12}_{k\text{ terms}}=1+\frac k2 $$

This inequality holds. As $k\to\infty$ the right-hand side $1+k/2$ becomes arbitrarily large, so $H_{2^k}$ also becomes arbitrarily large, and $H_n$ (a monotonically increasing sequence) is not bounded above. Therefore $\zeta(1)=\displaystyle\lim_{n\to\infty}H_n$ diverges. Checking numerically, for $k=1,2,3,4,5$ ($n=2,4,8,16,32$) we get $H_n=1.500,\,2.083,\,2.718,\,3.381,\,4.058$, which are at least the lower bounds $1+k/2=1.5,\,2.0,\,2.5,\,3.0,\,3.5$ (equality holds only for $k=1$).

Solution

Substituting $p=1$ into the integral formula in the main text (the explicit form of the Bose-type integral) $\displaystyle\int_0^\infty\frac{x^p}{e^x-1}\,\dd x=\Gamma(p+1)\zeta(p+1)$, we obtain

$$ \int_0^\infty\frac{x}{e^x-1}\,\dd x=\Gamma(2)\,\zeta(2) $$

Since $\Gamma(2)=1!=1$ and $\zeta(2)=\pi^2/6$,

$$ \int_0^\infty\frac{x}{e^x-1}\,\dd x=1\times\frac{\pi^2}6=\frac{\pi^2}6 $$

This is the result. Numerical integration also gives $\approx1.6449$, in agreement with $\pi^2/6\approx1.6449$. This integral is the simplest example of a Bose-type integral. Using the same formula with a different $p$, we obtain $2\zeta(3)$ for $p=2$ and $\pi^4/15$ for $p=3$.

Solution

Using the relation $\dfrac{e^x}{(e^x-1)^2}=-\dfrac{\dd}{\dd x}\!\left(\dfrac1{e^x-1}\right)$ and integrating by parts with $u=x^4,\ \dd v=-\dfrac{\dd}{\dd x}\!\left(\dfrac1{e^x-1}\right)\dd x$ (that is, $v=-\dfrac1{e^x-1}$, $\dd u=4x^3\,\dd x$), we get

$$ \int_0^\infty\frac{x^4e^x}{(e^x-1)^2}\,\dd x=\left[-\frac{x^4}{e^x-1}\right]_0^\infty+4\int_0^\infty\frac{x^3}{e^x-1}\,\dd x $$

For $x\to\infty$, the exponential function approaches $0$ faster than a power function does, so the boundary term is $0$; for $x\to0$, since $e^x-1\approx x$ we have $x^4/(e^x-1)\approx x^3\to0$, so it is also $0$ there. Therefore,

$$ \int_0^\infty\frac{x^4e^x}{(e^x-1)^2}\,\dd x=4\int_0^\infty\frac{x^3}{e^x-1}\,\dd x $$

The integral on the right-hand side is the form obtained by setting $p=3$ in the integral formula in the main text (the explicit form of the Bose-type integral), that is, it equals $\Gamma(4)\zeta(4)$. Since $\Gamma(4)=3!=6$,

$$ \int_0^\infty\frac{x^4e^x}{(e^x-1)^2}\,\dd x=4\times6\times\zeta(4)=24\,\zeta(4)=4!\,\zeta(4) $$

Substituting $\zeta(4)=\pi^4/90$, we obtain

$$ \int_0^\infty\frac{x^4e^x}{(e^x-1)^2}\,\dd x=24\times\frac{\pi^4}{90}=\frac{24\pi^4}{90}=\frac{4\pi^4}{15} $$

This is the formula we were asked to show. Numerically the value is $\approx25.976$, which agrees with $4\pi^4/15\approx25.976$.

Solution

Using $\dfrac{e^x}{(e^x+1)^2}=-\dfrac{\dd}{\dd x}\!\left(\dfrac1{e^x+1}\right)$ and integrating by parts with $u=x^3,\ \dd v=-\dfrac{\dd}{\dd x}\!\left(\dfrac1{e^x+1}\right)\dd x$ ($v=-\dfrac1{e^x+1}$, $\dd u=3x^2\dd x$), we get

$$ \int_0^\infty\frac{x^3e^x}{(e^x+1)^2}\,\dd x=\left[-\frac{x^3}{e^x+1}\right]_0^\infty+3\int_0^\infty\frac{x^2}{e^x+1}\,\dd x $$

Both boundary terms are $0$, for the same reason as in the discussion in the main text. The integral on the right-hand side is the form obtained from the general formula for Fermi-type integrals, $\displaystyle\int_0^\infty\frac{x^{s-1}}{e^x+1}\dd x=(1-2^{1-s})\Gamma(s)\zeta(s)$, by setting $s=3$, so

$$ \int_0^\infty\frac{x^2}{e^x+1}\,\dd x=\left(1-2^{1-3}\right)\Gamma(3)\,\zeta(3)=\left(1-\frac14\right)\times2\times\zeta(3)=\frac32\,\zeta(3) $$

(We used $1-2^{1-3}=1-2^{-2}=1-\frac14=\frac34$ and $\Gamma(3)=2!=2$.) Substituting this, we obtain

$$ \int_0^\infty\frac{x^3e^x}{(e^x+1)^2}\,\dd x=3\times\frac32\,\zeta(3)=\frac92\,\zeta(3) $$

This is the result. On the other hand, if we substitute $p=3$ into the formula in the main text (integral formula (iii)), which has the form $\left(1-\dfrac1{2^{p-1}}\right)p!\,\zeta(p)$, we get $\left(1-\dfrac1{2^2}\right)3!\,\zeta(3)=\dfrac34\times6\times\zeta(3)=\dfrac{18}4\zeta(3)=\dfrac92\zeta(3)$, and we can confirm that the two agree. Numerically, since $\zeta(3)\approx1.20206$, we have $\dfrac92\zeta(3)\approx5.4093$, which agrees with the numerical integration result $\approx5.4093$.

Solution

In $V_n(1)=\pi^{n/2}/\Gamma(n/2+1)$, the numerator $\pi^{n/2}=(\sqrt\pi)^n$ grows on the order of a geometric sequence (an exponential function) with common ratio $\sqrt\pi\approx1.772$. The denominator $\Gamma(n/2+1)$, on the other hand, is $\Gamma(n/2+1)=(n/2)!$ when $n$ is even (the factorial of a natural number itself), and applying Stirling's formula (Formula 47.3) $m!\approx m^me^{-m}$ with $m=n/2$ gives

$$ \Gamma\!\left(\frac n2+1\right)\approx\left(\frac n2\right)^{n/2}e^{-n/2} $$

This approximation shows that when $n$ becomes large, the denominator grows on the order of “a power whose exponent is half of $n$ itself,” that is, on the order of a power whose base grows without bound together with $n$, namely $(n/2)$—this is called super-exponential growth. In contrast, the numerator $(\sqrt\pi)^n$ grows only on the order of an exponential function whose base remains the fixed constant $\sqrt\pi$. Therefore the ratio

$$ V_n(1)=\frac{\pi^{n/2}}{\Gamma(n/2+1)}\approx\frac{(\sqrt\pi)^n}{(n/2)^{n/2}e^{-n/2}} $$

Because the denominator grows much faster than the numerator, as $n\to\infty$ this ratio necessarily converges to $0$. Checking numerically, for $n=10,50,100,200$ the values of $V_n(1)$ are $\approx2.55,\ 1.73\times10^{-13},\ 2.37\times10^{-40},\ 5.56\times10^{-109}$ respectively, and we can clearly see the number of digits decreasing at an accelerating rate. The behavior seen in Figure 47.2, that “the maximum is at $n=5$,” is a phenomenon that occurs because, while $n$ is small, the growth of the numerator dominates, but as $n$ becomes larger the super-exponential growth of the denominator catches up and eventually overwhelms it.

Solution

We use Formula 47.2 (with $n=2$). Since $(2n-1)!!=3!!=1\cdot3=3$, $2^{n+1}=2^3=8$, and $\sqrt{\pi/a^{2n+1}}=\sqrt{\pi/a^5}$,

$$ \int_0^\infty x^4e^{-ax^2}\dd x=\frac{3}{8}\sqrt{\frac\pi{a^5}} $$

This is the result (another way to verify it: using $\Gamma(\frac52)=\frac32\cdot\frac12\sqrt\pi=\frac34\sqrt\pi$, we get $\dfrac{\Gamma(5/2)}{2a^{5/2}}=\dfrac{3\sqrt\pi}{8a^{5/2}}$).

Next we find the averages. The denominator is $\displaystyle\int_{-\infty}^{\infty}e^{-ap^2}\dd p=\sqrt{\pi/a}=\sqrt\pi\,a^{-1/2}$. Since $p^2$ and $p^4$ are even functions, each numerator is twice the integral from $0$ to $\infty$, and from the entry for $n=1$ in Table 47.1 and the result above we find that

$$ \int_{-\infty}^\infty p^2e^{-ap^2}\dd p=2\cdot\frac14\sqrt{\frac\pi{a^3}}=\frac12\sqrt\pi\,a^{-3/2},\qquad \int_{-\infty}^\infty p^4e^{-ap^2}\dd p=2\cdot\frac38\sqrt{\frac\pi{a^5}}=\frac34\sqrt\pi\,a^{-5/2} $$

These are the values of the numerators. Therefore,

$$ \langle p^2\rangle=\frac{\frac12\sqrt\pi\,a^{-3/2}}{\sqrt\pi\,a^{-1/2}}=\frac1{2a},\qquad \langle p^4\rangle=\frac{\frac34\sqrt\pi\,a^{-5/2}}{\sqrt\pi\,a^{-1/2}}=\frac3{4a^2} $$

and the ratio is

$$ \frac{\langle p^4\rangle}{\langle p^2\rangle^2}=\frac{3/(4a^2)}{1/(4a^2)}=3 $$

which is a constant, $3$, independent of $a$. This is the general relation between the fourth moment and the second moment of a Gaussian distribution (the kurtosis of a Gaussian distribution is $3$), and it expresses the fact that even when the temperature changes (even when $a=\beta/2m$ changes), the “shape” of the momentum distribution of the gas molecules does not change.

Solution

The phase-space volume of a monatomic ideal gas shown in Chapter 48 has the form $\Omega(E)=C_0E^{3N/2}$ ($C_0$ is a constant determined only by $N,V,m,h$ and independent of the energy, $C_0=\dfrac{V^N}{N!\,h^{3N}}\cdot\dfrac{\pi^{3N/2}\,(2m)^{3N/2}}{\Gamma(3N/2+1)}=\dfrac{V^N}{N!\,h^{3N}}\cdot\dfrac{(2\pi m)^{3N/2}}{\Gamma(3N/2+1)}$). If we increase the energy from $E$ to $2E$,

$$ \frac{\Omega(2E)}{\Omega(E)}=\frac{C_0(2E)^{3N/2}}{C_0E^{3N/2}}=2^{3N/2} $$

results. Even for $N=1$ (a single molecule) this is a factor of $2^{3/2}\approx2.83$, and the larger $N$ is (the larger the exponent $3N/2$ is), the more explosively this factor grows—for example, for $N=10$ it is a factor of $2^{15}=32768$, and if $N$ is about Avogadro's number, the exponent $3N/2$ itself is of the order of magnitude of $10^{23}$, and the factor goes far beyond our everyday sense of numbers.

From another point of view, differentiating $\Omega(E)$ with respect to $E$ gives

$$ \frac{\dd\Omega}{\dd E}=C_0\cdot\frac{3N}{2}E^{\frac{3N}{2}-1}=\frac{3N}{2E}\,\Omega(E) $$

so that $\dd\Omega/\dd E$ (the “density” of the states newly added when the energy increases slightly) also shows an extremely rapid increase, proportional to $\Omega(E)$ itself. This is the quantitative content of the property that was drawn schematically in Figure 48.3: “for the same number of particles, the larger the energy, the more rapidly the number of microstates increases.” Also, the number of states in the energy shell $[E,E+\Delta E]$ can be written as $W=\dfrac{\dd\Omega}{\dd E}\Delta E=\dfrac{3N}{2}\dfrac{\Delta E}{E}\,\Omega(E)$.

Solution

The correct definition of the time average is

$$ \bar A_{\text{time}}=\lim_{T\to\infty}\frac1T\int_0^TA(l(t))\,\dd t $$

(the equation in Chapter 48). Here, consider the simplest case $A(t)=A_0$ (a physical quantity that is constant in time), and suppose we take the expression $\displaystyle\int_0^TA(t)\,\dd t$, with $1/T$ removed, as the definition of the “time average.” Then

$$ \int_0^TA_0\,\dd t=A_0\,\big[t\big]_0^T=A_0T $$

results. As long as $A_0\neq0$, as $T\to\infty$ we get $A_0T\to\pm\infty$ (depending on the sign of $A_0$): it diverges and does not converge to a finite value. This completely contradicts the intuitive meaning of an average, namely that “the average of a quantity that keeps a constant value should be that value itself.” On the other hand, multiplying by $1/T$ as in the correct definition gives

$$ \frac1T\int_0^TA_0\,\dd t=\frac1T\cdot A_0T=A_0 $$

and the reasonable value $A_0$ is always obtained, independent of $T$. That is, the factor $1/T$ is indispensable for “dividing back” the integral $\int_0^TA(t)\dd t$, which keeps growing with time, by the length $T$ of the observation time, and thereby recovering the original meaning of an average: the mean value per unit time.

Solution

(1) If the three particles are distinguishable, the number of ways of assigning “which particle goes into cell A, which into cell B, and which into cell C” equals the number of one-to-one correspondences (bijections) between the three particles and the three cells, that is, the number of permutations arranging $3$ distinct objects in a row, which is

$$ 3!=3\times2\times1=6\ \text{ways} $$

(For example, think of arranging particles $1,2,3$ in the order $(\text{cell A},\text{cell B},\text{cell C})$; there are $6$ ways to do so.) (2) If the three particles are indistinguishable (identical particles), the situation “one particle in each of cells A, B, and C” looks like exactly the same physical state no matter how we exchange which particle is in which cell, so the number of microstates is

$$ 1\ \text{way} $$

and no more. (3) Dividing the answer $6$ of (1) by $N!=3!=6$ gives

$$ \frac{6}{3!}=\frac66=1 $$

which indeed agrees with the answer $1$ of (2). This is a check of exactly the same structure as Example 48.3 of the main text (the case $N=2$), and it confirms, also for $N=3$, the meaning of $N!$: “dividing the number of cases counted on the assumption that the particles are distinguishable by the total number $N!$ of exchanges of identical particles gives the correct way of counting for indistinguishable particles.”

Solution

As an example, consider two one-dimensional harmonic oscillators that exert no force on each other (angular frequencies $\omega_1,\omega_2$, with the ratio $\omega_1/\omega_2$ irrational). The Hamiltonian is $H=H_1+H_2$, $H_i=\dfrac{p_i^2}{2m}+\dfrac12m\omega_i^2x_i^2$ ($i=1,2$). The phase space is four-dimensional, $(x_1,p_1,x_2,p_2)$, and the surface of constant total energy $E=E_1+E_2$ is a three-dimensional hypersurface. If this system were ergodic, the trajectory would visit (uniformly) the whole of this three-dimensional constant-energy surface as time passes.

However, since the two oscillators are independent of each other, the energies $E_1$ and $E_2$ of the individual oscillators are conserved separately. The trajectory is therefore confined to the surface on which $E_1,E_2$ are both held constant, that is, a two-dimensional surface (a torus, a surface like that of a doughnut) that is the direct product of an ellipse in the $(x_1,p_1)$ plane and an ellipse in the $(x_2,p_2)$ plane. The constant-energy surface is a collection of infinitely many tori obtained by varying $E_1$ from $0$ to $E$, and the trajectory keeps moving on just one of them (if $\omega_1/\omega_2$ is irrational, it fills that torus without gaps, but it never goes to a torus with a different value of $E_1$).

That the time average and the phase average do not coincide can be checked with $A=E_1$. Since $E_1$ is conserved, the time average is just the initial value $E_1(0)$. On the other hand, for the phase average (uniform within the energy shell), we can write $\dd x_i\,\dd p_i=\dd E_i\,\dd\phi_i/\omega_i$, so that $\dd x_1\dd p_1\dd x_2\dd p_2\propto\dd E_1\dd E_2\dd\phi_1\dd\phi_2$, and under the condition $E_1+E_2=E$, $E_1$ is distributed uniformly from $0$ to $E$. The average is $E/2$. Therefore, when $E_1(0)\neq E/2$, the time average $\neq$ the phase average, and this system is not ergodic.

In general, an integrable system, which has many conserved quantities besides the energy (quantities that stay constant during the motion, as seen in Chapter 46) and whose motion is regular, is a typical example in which the ergodic hypothesis fails for the same reason (the unperturbed two-body Kepler motion is another such case). Note that for a system with one degree of freedom (such as a simple pendulum), the constant-energy surface is the trajectory itself, so time average = phase average holds trivially. As a matter of time scales, there are also systems, such as glass, that can reach only a very small part of the microstates within the observable time.

Solution

The ergodic hypothesis (Law 48.1) is the claim that “for a system in thermal equilibrium, the time average and the phase average of a physical quantity are equal,” and it states that the values of two kinds of averages coincide: the average obtained by observing a single system for a long time (the time average), and the average calculated over a hypothetical collection of all the microstates corresponding to the same macrostate (the phase average). The hypothesis itself says nothing about what probability $P_i$ should be assigned to each microstate when the phase average is computed.

On the other hand, the principle of equal a priori probability (Law 48.2) is a claim that, for an isolated system ($E,V,N$ constant), specifies concretely the way of assigning probabilities itself: “all allowed microstates are realized with equal probability.”

The relationship between the two can be understood as follows, as stated in Section 48.4 of the main text: if we suppose that the ergodic hypothesis holds (the system follows, one after another, all the allowed microstates over a long time), there is no reason why only particular microstates should be visited more often than others, so the principle of equal a priori probability, “all microstates should be found with equal probability,” is motivated as a natural consequence—but this is not a “proof” of the principle of equal a priori probability, only an intuitive justification. Logically, the ergodic hypothesis asserts the identity of the two averages, and the principle of equal a priori probability asserts the concrete form of the probability distribution in an isolated system, each independently, and statistical mechanics adopts these two as separate (mutually complementary) fundamental assumptions.

Solution

(1) Substituting $N=6$ into Formula 48.1, $W(N,n)=\dbinom Nn=\dfrac{N!}{n!(N-n)!}$, gives

$n$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
$W(6,n)$	$1$	$6$	$15$	$20$	$15$	$6$	$1$

(It can also be confirmed that the sum $1+6+15+20+15+6+1=64=2^6$ agrees with the total number of microstates $2^N=2^6=64$.) (2) There are $W(6,3)=20$ microstates corresponding to $E=3\varepsilon$ ($n=3$), so by the principle of equal a priori probability, the probability that each of them is realized is

$$ P_i=\frac{1}{W(6,3)}=\frac{1}{20}\qquad(i=1,\dots,20) $$

This is the probability of each microstate in this microcanonical distribution.

Solution

(1) Substituting $N=4$, $q=3$ into Formula 48.2, $W_{\text{Einstein}}(N,q)=\dfrac{(q+N-1)!}{q!\,(N-1)!}$, gives

$$ W_{\text{Einstein}}(4,3)=\frac{(3+4-1)!}{3!\,(4-1)!}=\frac{6!}{3!\,3!}=\frac{720}{6\times6}=20=\binom{6}{3} $$

(2) Ignoring order, the ways of distributing the three quanta fall into the following three patterns.

• “$3$” (all three on one oscillator): we choose which oscillator, so there are $4$ ways, such as $(3,0,0,0)$.
• “$2+1$” (two on one oscillator and one on another oscillator): we choose the oscillator holding two from the $4$, and then choose the oscillator holding one from the remaining three, so there are $4\times3=12$ ways (such as $(2,1,0,0)$).
• “$1+1+1$” (one each on three oscillators): we choose the oscillator holding no quantum, so there are $4$ ways (such as $(1,1,1,0)$).

The total is $4+12+4=20$ ways, which agrees with the result of (1).

Solution

In Formula 48.3 of Chapter 48 ($\Omega(E)=\dfrac{V^N}{N!\,h^{3N}}\dfrac{\pi^{3N/2}}{\Gamma(3N/2+1)}(2mE)^{3N/2}$), only $V^N$ depends on $V$. Keeping $N,E$ fixed and letting $V\to2V$ gives

$$ \frac{\Omega(2V)}{\Omega(V)}=\frac{(2V)^N}{V^N}=2^N $$

For $N=1$ this is a factor of $2$. For $N=6.0\times10^{23}$ it is a factor of $2^{6.0\times10^{23}}=10^{6.0\times10^{23}\times\log_{10}2}\approx10^{1.8\times10^{23}}$ ($\log_{10}2\approx0.301$, $6.0\times10^{23}\times0.301\approx1.8\times10^{23}$), an unimaginable factor: “a number in which the digit $1$ is followed by $1.8\times10^{23}$ digits $0$.”

The number of microstates in which all the molecules are gathered in the left half of the container (volume $V/2$) has a position integral of just $(V/2)^N$ while the momentum integral is unchanged, so its ratio to the total number of states $\Omega(V)$ is $(V/2)^N/V^N=2^{-N}$. By the principle of equal a priori probability, this is its probability. Even for $N=10$ it is $2^{-10}=1/1024$, and for $N=6.0\times10^{23}$ it is $10^{-1.8\times10^{23}}$, which in effect never happens, not even once. This is the reason a gas spreads uniformly throughout the whole container (it has the same structure as the “all heads” of the coins in Example 48.3).

Solution

Substitute $n!\approx\sqrt{2\pi n}\,n^ne^{-n}$ into each factorial of $\dbinom{2M}{M}=\dfrac{(2M)!}{(M!)^2}$. The numerator is $\sqrt{2\pi\cdot2M}\,(2M)^{2M}e^{-2M}=\sqrt{4\pi M}\,2^{2M}M^{2M}e^{-2M}$, and the denominator is $\bigl(\sqrt{2\pi M}\,M^Me^{-M}\bigr)^2=2\pi M\cdot M^{2M}e^{-2M}$. Canceling $M^{2M}e^{-2M}$, we obtain

$$ \binom{2M}{M}\approx\frac{\sqrt{4\pi M}}{2\pi M}\,2^{2M}=\frac{2\sqrt{\pi M}}{2\pi M}\,2^{2M}=\frac{2^{2M}}{\sqrt{\pi M}} $$

Therefore,

$$ \frac{\binom{2M}{M}}{2^{2M}}\approx\frac{1}{\sqrt{\pi M}} $$

holds. For 100 coins, $M=50$, so $1/\sqrt{50\pi}=1/\sqrt{157.08}\approx1/12.53\approx0.0798$, which agrees with the exact value $0.0796$ to within an error of about $0.25\%$. Note that if we use only the simplified version $n!\approx n^ne^{-n}$, the factor $\sqrt{2\pi n}$ disappears in the above calculation and the ratio becomes $1$. In other words, the factor $\sqrt{2\pi n}$ gives the width over which the distribution spreads around $N/2$ (proportional to $\sqrt{M}$).

Solution

(1) For a monatomic ideal gas at constant volume, $W(E)=C_0E^{3N/2-1}$ ($C_0$ is a constant independent of $E$). Therefore,

$$ \frac{W(2E)}{W(E)}=\frac{C_0(2E)^{3N/2-1}}{C_0E^{3N/2-1}}=2^{\frac{3N}{2}-1} $$

Substituting $N=N_{\mathrm A}=6.022\times10^{23}$, we find that the natural logarithm is $\left(\frac{3N}{2}-1\right)\ln2\approx9.03\times10^{23}\times0.6931\approx6.26\times10^{23}$, and the common logarithm is $\left(\frac{3N}{2}-1\right)\log_{10}2\approx9.03\times10^{23}\times0.3010\approx2.72\times10^{23}$. That is, $W$ increases by a factor of $10^{2.72\times10^{23}}$, an enormous factor whose number of digits is itself of order $10^{23}$.

(2) By Boltzmann's principle $S=k_{\mathrm B}\ln W$, the entropy change is

$$ \Delta S=k_{\mathrm B}\ln W(2E)-k_{\mathrm B}\ln W(E)=k_{\mathrm B}\ln\frac{W(2E)}{W(E)}=k_{\mathrm B}\left(\frac{3N}{2}-1\right)\ln2\fallingdotseq\frac32Nk_{\mathrm B}\ln2=\frac32R\ln2 $$

(since $N$ is large, we neglected the $-1$ and used $Nk_{\mathrm B}=N_{\mathrm A}k_{\mathrm B}=R$). The numerical value is $\frac32\times8.314\times0.6931\approx8.64\ \mathrm{J/K}$. (Even if the $-1$ is kept, the relative difference is only about $10^{-24}$, which does not show up in the significant figures.)

(3) Since $E=\frac32Nk_{\mathrm B}T\propto T$, the change $E\to2E$ means $T\to2T$. In Formula 29.1 of Chapter 29, $S=C_V\ln T+nR\ln V+\alpha_0$, take constant volume, $n=1\ \mathrm{mol}$, and $C_V=\frac32R$. Then

$$ \Delta S=C_V\ln(2T)-C_V\ln T=C_V\ln2=\frac32R\ln2\approx8.64\ \mathrm{J/K} $$

which agrees with the result of (2). The entropy change “heat capacity $\times\ln$ (temperature ratio)” taught by thermodynamics is explained in statistical mechanics as “the logarithm of the factor by which the number of states increases.”

Solution

(1) $m=83.798\times1.660\,539\times10^{-27}\approx1.3915\times10^{-25}\ \mathrm{kg}$. The thermal de Broglie wavelength is

$$ \lambda_{\mathrm T}=\frac{h}{\sqrt{2\pi mk_{\mathrm B}T}}=\frac{6.626\,070\times10^{-34}}{\sqrt{2\pi\times1.3915\times10^{-25}\times1.380\,649\times10^{-23}\times298.15}}\approx1.104\times10^{-11}\ \mathrm m=11.0\ \mathrm{pm} $$

(the quantity under the square root in the denominator is $2\pi\times1.3915\times10^{-25}\times4.116\times10^{-21}=3.599\times10^{-45}$, and its square root is $6.00\times10^{-23}$). The volume per molecule is $V/N=k_{\mathrm B}T/p=4.116\times10^{-26}\ \mathrm{m^3}$ (the same as in Example 49.8). Since $\lambda_{\mathrm T}^3=(1.1045\times10^{-11})^3=1.347\times10^{-33}\ \mathrm{m^3}$, we have $V/(N\lambda_{\mathrm T}^3)=4.116\times10^{-26}/1.347\times10^{-33}=3.06\times10^{7}$ and $\ln(3.06\times10^7)=17.23$. Therefore,

$$ S_{\mathrm m}=R\left[\ln\frac{V}{N\lambda_{\mathrm T}^3}+\frac52\right]=8.314\times(17.23+2.5)\approx164.1\ \mathrm{J/(mol\,K)} $$

which is in good agreement with the experimental value (about $164.1\ \mathrm{J/(mol\,K)}$). The value is larger than that of argon ($154.8$) because krypton is heavier (larger $m$), so $\lambda_{\mathrm T}$ is smaller (and $V/(N\lambda_{\mathrm T}^3)$ is larger).

(2) At constant pressure, $V/N=k_{\mathrm B}T/p\propto T$ and $\lambda_{\mathrm T}^{-3}\propto(2\pi mk_{\mathrm B}T)^{3/2}\propto T^{3/2}$, so $V/(N\lambda_{\mathrm T}^3)\propto T^{5/2}$. Therefore, the change in molar entropy on heating from $T_1=298.15\ \mathrm K$ to $T_2=600\ \mathrm K$ is

$$ \Delta S_{\mathrm m}=R\ln\left(\frac{T_2}{T_1}\right)^{5/2}=\frac52R\ln\frac{600}{298.15}=2.5\times8.314\times0.6994\approx14.5\ \mathrm{J/(mol\,K)} $$

(here $\ln(600/298.15)=\ln2.0124=0.6994$). In Chapter 29, $\Delta S=C_p\ln(T_2/T_1)$ (isobaric process); with $C_p=C_V+R=\frac52R$ for a monatomic ideal gas, this gives $\Delta S=\frac52R\ln(T_2/T_1)$, in complete agreement. The Sackur–Tetrode equation also contains within it the value of the heat capacity $C_p=\frac52R$.

Solution

Following the hint, let $E_i=N_ie$ and $V_i=N_iv$ ($i=1,2$). After the partition is removed, $E=Ne$ and $V=Nv$ ($N=N_1+N_2$).

(1) The Sackur–Tetrode equation, which includes the factor $1/N!$, is $S(E,V,N)=Nk_{\mathrm B}\left[\ln\left\{\dfrac VN\left(\dfrac{4\pi m}{3h^2}\dfrac EN\right)^{3/2}\right\}+\dfrac52\right]$. Substituting $V/N=v$ and $E/N=e$, the contents of $[\ \ ]$ become $\ln\{v(4\pi me/(3h^2))^{3/2}\}+\frac52\equiv s$ (a constant independent of $N$), so $S=Nk_{\mathrm B}s$; that is, $S$ is proportional to $N$. Therefore,

$$ \Delta S=S(E,V,N)-S(E_1,V_1,N_1)-S(E_2,V_2,N_2)=Nk_{\mathrm B}s-N_1k_{\mathrm B}s-N_2k_{\mathrm B}s=0 $$

(2) Substitute $V=Nv$ and $E=Ne$ into the expression without the factor $1/N!$, $S^{(0)}=Nk_{\mathrm B}\left[\ln V+\frac32\ln\frac{4\pi mE}{3Nh^2}+\frac32\right]$. Since $\dfrac{E}{N}=e$,

$$ S^{(0)}=Nk_{\mathrm B}\left[\ln(Nv)+\frac32\ln\frac{4\pi me}{3h^2}+\frac32\right]=Nk_{\mathrm B}\ln N+Nk_{\mathrm B}s'\quad\left(s'\equiv\ln v+\frac32\ln\frac{4\pi me}{3h^2}+\frac32\right) $$

(where $s'$ is independent of $N$). Hence,

$$ \Delta S^{(0)}=(Nk_{\mathrm B}\ln N+Nk_{\mathrm B}s')-(N_1k_{\mathrm B}\ln N_1+N_1k_{\mathrm B}s')-(N_2k_{\mathrm B}\ln N_2+N_2k_{\mathrm B}s')=k_{\mathrm B}\bigl(N\ln N-N_1\ln N_1-N_2\ln N_2\bigr) $$

Rewriting $N\ln N=N_1\ln N+N_2\ln N$, we obtain

$$ \Delta S^{(0)}=k_{\mathrm B}\left(N_1\ln\frac{N}{N_1}+N_2\ln\frac{N}{N_2}\right)=-k_{\mathrm B}\left[N_1\ln\frac{N_1}{N}+N_2\ln\frac{N_2}{N}\right] $$

Letting $x_i=N_i/N$, we get $\Delta S^{(0)}=-Nk_{\mathrm B}(x_1\ln x_1+x_2\ln x_2)$, which has exactly the same form as the entropy of mixing for different gases, $\Delta S_{\mathrm{mix}}=-Nk_{\mathrm B}(x_A\ln x_A+x_B\ln x_B)$. That the entropy increases by the same amount as when different gases are mixed, even though the gas is the same, is the Gibbs paradox.

Numerical value: for $N_1=1\ \mathrm{mol}$, $N_2=2\ \mathrm{mol}$, and $N=3\ \mathrm{mol}$, we have $x_1=\frac13$ and $x_2=\frac23$, so

$$ \Delta S^{(0)}=-R\left[1\times\ln\frac13+2\times\ln\frac23\right]=R\left[\ln3+2\ln\frac32\right]=8.314\times(1.0986+0.8109)\approx15.9\ \mathrm{J/K} $$

(with the correct way of counting, $\Delta S=0$).

Solution

(1) The total energy $E_1+E_2$ is conserved. For a monatomic ideal gas, $E_i=\frac32N_ik_{\mathrm B}T_i$, so the initial energy is $\frac32k_{\mathrm B}(N_1T_1+N_2T_2)$ and the energy after equilibrium is reached is $\frac32k_{\mathrm B}(N_1+N_2)T_\mathrm f$. Setting the two equal gives

$$ T_\mathrm f=\frac{N_1T_1+N_2T_2}{N_1+N_2}=\frac{2.0\times10^{22}\times300+8.0\times10^{22}\times400}{2.0\times10^{22}+8.0\times10^{22}}=\frac{6.0\times10^{24}+3.2\times10^{25}}{1.0\times10^{23}}=380\ \mathrm K $$

(this is the average temperature weighted by the numbers of molecules; it corresponds to the condition “$E_i^*/N_i=$ the same for both systems” in Example 49.6).

(2) The entropy change of each system is

$$ \Delta S_1=\frac32N_1k_{\mathrm B}\ln\frac{380}{300}=1.5\times2.0\times10^{22}\times1.380\,649\times10^{-23}\times0.23639\approx+9.79\times10^{-2}\ \mathrm{J/K} $$ $$ \Delta S_2=\frac32N_2k_{\mathrm B}\ln\frac{380}{400}=1.5\times8.0\times10^{22}\times1.380\,649\times10^{-23}\times(-0.05129)\approx-8.50\times10^{-2}\ \mathrm{J/K} $$

The total is

$$ \Delta S=\Delta S_1+\Delta S_2\approx9.79\times10^{-2}-8.50\times10^{-2}=1.29\times10^{-2}\ \mathrm{J/K}\ \gt0 $$

so the total entropy increases. This is because the entropy increase on the low-temperature side exceeds the entropy decrease on the high-temperature side. We see that “the entropy of an isolated system increases” (Chapter 29) and “the system moves toward the most probable partition” (Section 49.4) are two ways of saying the same thing.

Solution

(1) Differentiate $\ln\Omega_{\mathrm{total}}=N_1\ln V_1+N_2\ln(V-V_1)+(\text{constant})$ with respect to $V_1$ and set the result equal to $0$ (differentiating the composite function, where $\dd(V-V_1)/\dd V_1=-1$):

$$ \frac{N_1}{V_1}-\frac{N_2}{V-V_1}=0\quad\Longrightarrow\quad N_1(V-V_1)=N_2V_1\quad\Longrightarrow\quad V_1^*=\frac{N_1}{N_1+N_2}V $$

For $N_1=1\ \mathrm{mol}$, $N_2=3\ \mathrm{mol}$, and $V=10\ \mathrm L$, we have $V_1^*=\frac14\times10\ \mathrm L=2.5\ \mathrm L$, and the volume on the right is $V-V_1^*=7.5\ \mathrm L$. The pressures on the two sides (with the same $T$ on both sides) are then

$$ p_1=\frac{N_1RT}{V_1^*},\quad p_2=\frac{N_2RT}{V-V_1^*}=\frac{3N_1RT}{3V_1^*}=p_1 $$

(using $V-V_1^*=3V_1^*$), which confirms $p_1=p_2$. Concretely, for $T=300\ \mathrm K$, $p_1=1\times8.314\times300/(2.5\times10^{-3})\approx9.98\times10^5\ \mathrm{Pa}$. That is, the volume is divided in proportion to the number of molecules, and $V_1/N_1=V_2/N_2$ holds.

(2) Differentiating once more,

$$ \frac{\dd^2}{\dd V_1^2}\ln\Omega_{\mathrm{total}}=-\frac{N_1}{V_1^2}-\frac{N_2}{(V-V_1)^2}\lt0 $$

which is negative for every $V_1$, so the value at $V_1^*$ is a local maximum (and therefore the maximum).

Solution

(1) Solve $f=1/[\exp(\varepsilon/k_{\mathrm B}T)+1]$, obtained in Chapter 49, for $T$. From $\dfrac1f=\exp\!\left(\dfrac{\varepsilon}{k_{\mathrm B}T}\right)+1$ we get $\exp\!\left(\dfrac\varepsilon{k_{\mathrm B}T}\right)=\dfrac{1-f}{f}$; taking the logarithm of both sides gives $\dfrac{\varepsilon}{k_{\mathrm B}T}=\ln\dfrac{1-f}{f}$, that is,

$$ k_{\mathrm B}T=\frac{\varepsilon}{\ln\{(1-f)/f\}}=\frac{0.020\ \mathrm{eV}}{\ln9}=\frac{0.020}{2.1972}\ \mathrm{eV}=9.10\times10^{-3}\ \mathrm{eV} $$

For $f=0.1$, $(1-f)/f=9$ and $\ln9=2.1972$. Therefore,

$$ T=\frac{9.10\times10^{-3}\ \mathrm{eV}}{8.617\,333\times10^{-5}\ \mathrm{eV/K}}\approx105.6\ \mathrm K $$

(2) From $S/(Nk_{\mathrm B})=-[f\ln f+(1-f)\ln(1-f)]$, obtained in Chapter 49,

$$ \frac{S}{Nk_{\mathrm B}}=-\bigl[0.1\ln0.1+0.9\ln0.9\bigr]=0.1\times2.3026+0.9\times0.10536=0.23026+0.09482=0.3251 $$

Therefore, $S_{\mathrm m}=R\times0.3251=8.314\times0.3251\approx2.70\ \mathrm{J/(mol\,K)}$. (For reference: the total energy in this case is $E=0.1\times N_{\mathrm A}\times0.020\ \mathrm{eV}=0.1\times6.022\times10^{23}\times0.020\times1.602\times10^{-19}\ \mathrm J\approx193\ \mathrm J/\mathrm{mol}$.)

(3) As $T\to0$, $f\to0$ (all particles are in the ground state), and the number of states is $W(N,0)=\binom N0=1$; that is, there is only one way to realize the ground state, so $S=k_{\mathrm B}\ln1=0$. The entropy approaches $0$ (this is the case $W_0=1$ of Law 49.3, and it is consistent with the third law).

Solution

(1) The total amount of substance is $n=4\ \mathrm{mol}$, and the mole fractions are $x_{\mathrm{He}}=\frac14$ and $x_{\mathrm{Ne}}=\frac34$. By Formula 49.3,

$$ \Delta S_{\mathrm{mix}}=-nR\left(x_{\mathrm{He}}\ln x_{\mathrm{He}}+x_{\mathrm{Ne}}\ln x_{\mathrm{Ne}}\right)=-4\times8.314\times\left(\frac14\ln\frac14+\frac34\ln\frac34\right) $$

Since $\frac14\ln\frac14=-0.34657$ and $\frac34\ln\frac34=-0.21576$, the sum is $-0.56233$, and

$$ \Delta S_{\mathrm{mix}}=4\times8.314\times0.56233\approx18.7\ \mathrm{J/K} $$

(2) Since $\Delta H=0$ on mixing, $\Delta G_{\mathrm{mix}}=\Delta H-T\Delta S_{\mathrm{mix}}=-300\times18.70\approx-5.61\times10^3\ \mathrm J=-5.61\ \mathrm{kJ}$. Since this is negative, the mixing proceeds spontaneously.

(3) Since we merely remove the partition between two samples of the same $\mathrm{He}$, Law 49.2 ($S$ including the factor $1/N!$) gives $\Delta S=0$. With the same ratio of amounts of substance, $1:3$, $\Delta S$ is either $18.7\ \mathrm{J/K}$ or $0$, depending only on whether the components are different or the same.

Solution

(1) Each molecule has two possible orientations and the $N$ molecules are independent, so $W_0=2\times2\times\cdots\times2=2^N$ (the product rule for counting). By Boltzmann's principle,

$$ S_0=k_{\mathrm B}\ln2^N=Nk_{\mathrm B}\ln2,\qquad S_{0,\mathrm m}=R\ln2=8.314\times0.6931\approx5.76\ \mathrm{J/(mol\,K)} $$

(2) $W_0$ is the number of ways to choose, from the $N$ lattice sites, the $N/2$ sites on which to place A (B is placed on the remaining sites):

$$ W_0=\binom{N}{N/2}=\frac{N!}{\{(N/2)!\}^2} $$

Using Stirling's formula $\ln m!\fallingdotseq m\ln m-m$,

$$ \ln W_0\fallingdotseq(N\ln N-N)-2\left\{\frac N2\ln\frac N2-\frac N2\right\}=N\ln N-N-N\ln\frac N2+N=N\ln N-N\ln\frac N2=N\ln2 $$

(the $-N$ and $+N$ cancel, and we used $\ln N-\ln(N/2)=\ln2$). Therefore $S_0=k_{\mathrm B}\ln W_0\fallingdotseq Nk_{\mathrm B}\ln2$, the same value as in (1). This is the same expression as the entropy of mixing in Formula 49.3 ($\Delta S_{\mathrm{mix}}=Nk_{\mathrm B}\ln2$ when $x_A=x_B=\frac12$).

(3) For $N=100$, $\ln\binom{100}{50}=\ln100!-2\ln50!=363.739-2\times148.478=66.784$ (more precisely, $\ln\binom{100}{50}=66.7838$ and $\binom{100}{50}\approx1.009\times10^{29}$). On the other hand, $N\ln2=100\times0.69315=69.315$. The difference is $2.53$, about $3.8\ \%$ in relative terms; the error of Stirling's formula (the simple version) becomes relatively small as $N\to\infty$ (the absolute error is of order $\ln N$, which becomes more negligible compared with $N\ln 2$ the larger $N$ is). For $N=N_{\mathrm A}$, the relative error is of order $10^{-23}$, which is no problem at all.

Solution

Let us examine $\Omega(E)$ of Formula 49.1 factor by factor ($q_i$ are position coordinates with the unit $\mathrm m$, and $p_i$ are momentum coordinates with the unit $\mathrm{kg\,m/s}$).

• $V^N$: the volume of position space, which expresses that the $N$ molecules may be anywhere in the box (in the $3N$-dimensional position space, the integral for each molecule gives $V$, and there is one such factor for each of the $N$ molecules). Its unit is $\mathrm{m^{3N}}$.
• $\dfrac{\pi^{3N/2}}{\left(\frac{3N}{2}\right)\Gamma\left(\frac{3N}{2}\right)}$: the volume of a $3N$-dimensional ball of radius $1$ (Formula 47.1 of Chapter 47 with $n=3N$ and $R=1$). It is the dimensionless coefficient, determined by shape alone, in the volume of momentum space. It has no unit.
• $(2mE)^{3N/2}=\left(\sqrt{2mE}\right)^{3N}$: the $3N$th power of the radius $\sqrt{2mE}$ of the sphere in momentum space. It expresses the energy condition $\sum p_i^2\le2mE$. Since $2mE$ has the unit $\mathrm{kg\cdot J}=\mathrm{kg\cdot kg\,m^2/s^2}=(\mathrm{kg\,m/s})^2$, the radius $\sqrt{2mE}$ has the unit $\mathrm{kg\,m/s}$ (the unit of momentum). Hence the unit of $(2mE)^{3N/2}$ is $(\mathrm{kg\,m/s})^{3N}$. Together with the previous factor, this is the volume of momentum space.
• $\dfrac{1}{N!}$: the factor that prevents states differing only by an exchange of identical particles from being counted repeatedly as different states (dimensionless).
• $\dfrac{1}{h^{3N}}$: the unit of quantization that converts a volume into “a number of states” by dividing by $h^{3N}$, the phase-space volume (cell size) occupied by one microstate. Its unit is $(\mathrm{J\,s})^{-3N}=(\mathrm{kg\,m^2/s})^{-3N}$.

Combining position space $\mathrm{m^{3N}}$, momentum space $(\mathrm{kg\,m/s})^{3N}$, and $1/h^{3N}$, the unit of the whole is

$$ \frac{\mathrm{m^{3N}}\cdot(\mathrm{kg\,m/s})^{3N}}{(\mathrm{kg\,m^2/s})^{3N}}=\left(\frac{\mathrm{m}\cdot\mathrm{kg\,m/s}}{\mathrm{kg\,m^2/s}}\right)^{3N}=1 $$

so $\Omega(E)$ is dimensionless (a pure number, the number of states).

Solution

(a) Since $k_{\mathrm B}T=2\varepsilon$, we have $E_J/(k_{\mathrm B}T)=J(J+1)/2$. The weight $g_Je^{-E_J/(k_{\mathrm B}T)}$ of each level is

$$ J=0:\ 1,\qquad J=1:\ 3e^{-1}=1.1036,\qquad J=2:\ 5e^{-3}=0.2489,\qquad J=3:\ 7e^{-6}=0.01735 $$

Hence $Z=1+1.1036+0.2489+0.01735=2.3699$.

(b) From $P_J=(\text{weight})/Z$,

$$ P_0=\frac{1}{2.3699}=0.4220,\quad P_1=\frac{1.1036}{2.3699}=0.4657,\quad P_2=\frac{0.2489}{2.3699}=0.1050,\quad P_3=\frac{0.01735}{2.3699}=0.00732 $$

(the sum is $1.0000$).

(c) The maximum is at $J=1$ ($P_1=0.466$). The most probable level is $J=1$ rather than the lowest-energy level $J=0$ because $J=1$ has $g_1=3$ states. The probability per microstate, $P_J/g_J$, is $0.422,\ 0.155,\ 0.021,\ 0.0010$, which decreases as the energy increases.

(d) $P_2/P_1=0.1050/0.4657=0.2256$. On the other hand, $(g_2/g_1)e^{-(E_2-E_1)/(k_{\mathrm B}T)}=\frac53e^{-(6\varepsilon-2\varepsilon)/(2\varepsilon)}=\frac53e^{-2}=0.2256$, which agrees (since $Z$ in the denominators cancels, $P_2/P_1$ can be found without knowing $Z$).

∎

Solution

We use the formula $n(z)/n(0)=e^{-mgz/(k_{\mathrm B}T)}$ with $z=1.0\times10^4\ \mathrm m$ and $T=250\ \mathrm K$. Then $k_{\mathrm B}T=1.380649\times10^{-23}\times250=3.4516\times10^{-21}\ \mathrm J$.

Nitrogen: $\dfrac{mgz}{k_{\mathrm B}T}=\dfrac{4.65\times10^{-26}\times9.8\times10^4}{3.4516\times10^{-21}}=1.320$, so $n/n(0)=e^{-1.320}=0.267$.

Helium: $\dfrac{6.65\times10^{-27}\times9.8\times10^4}{3.4516\times10^{-21}}=0.1888$, so $n/n(0)=e^{-0.1888}=0.828$.

The factor by which the abundance ratio (helium/nitrogen) is magnified relative to the ground is $\dfrac{n_{\mathrm{He}}(z)/n_{\mathrm{N_2}}(z)}{n_{\mathrm{He}}(0)/n_{\mathrm{N_2}}(0)}=\dfrac{0.828}{0.267}=3.10$. This is also equal to $\exp\!\left[(m_{\mathrm{N_2}}-m_{\mathrm{He}})gz/(k_{\mathrm B}T)\right]=e^{1.320-0.189}=e^{1.131}=3.10$. In other words, in a static isothermal atmosphere, a light gas is distributed more abundantly at high altitudes than a heavy gas, and at $10\ \mathrm{km}$ the helium abundance ratio becomes about three times its value at the ground. However, the real atmosphere is well mixed by convection and turbulence up to an altitude of about $100\ \mathrm{km}$, so the composition ratio stays nearly constant (the separation in which light gases become more abundant higher up occurs at much greater altitudes, where mixing is no longer effective).

∎

Solution

Let $\beta\varepsilon=1$ and $y=e^{-1}=0.36788$. Then $Z=1+y+y^2=1+0.36788+0.13534=1.50321$. The probabilities are $P_0=1/Z=0.66524$, $P_1=y/Z=0.24473$, $P_2=y^2/Z=0.09003$ (the sum is $1$).

(a) $\langle E\rangle=0\cdot P_0+\varepsilon P_1+2\varepsilon P_2=(0.24473+0.18006)\,\varepsilon=0.42479\,\varepsilon$. $\langle E^2\rangle=\varepsilon^2P_1+4\varepsilon^2P_2=(0.24473+0.36012)\,\varepsilon^2=0.60485\,\varepsilon^2$. Therefore

$$ \langle\Delta E^2\rangle=\langle E^2\rangle-\langle E\rangle^2=0.60485\,\varepsilon^2-0.18045\,\varepsilon^2=0.42440\,\varepsilon^2 $$

(b) From $Z=1+y+y^2$ ($y=e^{-\beta\varepsilon}$, $\dd y/\dd\beta=-\varepsilon y$),

$$ \frac{\partial\ln Z}{\partial\beta}=\frac{-\varepsilon(y+2y^2)}{1+y+y^2} $$

(the numerator comes from $\partial Z/\partial\beta=-\varepsilon y-2\varepsilon y^2$). We differentiate this once more with respect to $\beta$. By the quotient rule, with numerator $N=y+2y^2$ ($\partial N/\partial\beta=-\varepsilon y(1+4y)$) and denominator $D=1+y+y^2$ ($\partial D/\partial\beta=-\varepsilon y(1+2y)$),

$$ \frac{\partial}{\partial\beta}\frac ND=\frac{-\varepsilon y(1+4y)D+\varepsilon y(1+2y)N}{D^2}=\frac{-\varepsilon y\left[(1+4y)(1+y+y^2)-(1+2y)(y+2y^2)\right]}{D^2}=\frac{-\varepsilon y(1+4y+y^2)}{D^2} $$

(the difference of $(1+4y)(1+y+y^2)=1+5y+5y^2+4y^3$ and $(1+2y)(y+2y^2)=y+4y^2+4y^3$ is $1+4y+y^2$). Hence

$$ \frac{\partial^2\ln Z}{\partial\beta^2}=-\varepsilon\cdot\frac{-\varepsilon y(1+4y+y^2)}{D^2}=\varepsilon^2\frac{y(1+4y+y^2)}{(1+y+y^2)^2} $$

Substituting $y=0.36788$ gives $y(1+4y+y^2)=0.36788\times2.60684=0.95903$ and $D^2=1.50321^2=2.25964$, so $\varepsilon^2\times0.95903/2.25964=0.42440\,\varepsilon^2$. This agrees with (a).

(c) By Formula 50.3, $C_V=\langle\Delta E^2\rangle/(k_{\mathrm B}T^2)$. When $k_{\mathrm B}T=\varepsilon$, $\varepsilon^2/(k_{\mathrm B}T^2)=k_{\mathrm B}$, so $C_V=0.4244\,k_{\mathrm B}$.

∎

Solution

(a) The logarithm of $P_j/g_j=e^{-\beta E_j}/Z$ is $\ln(P_j/g_j)=-\beta E_j-\ln Z$. Multiplying this by $-k_{\mathrm B}P_j$ and summing over $j$, we obtain

$$ -k_{\mathrm B}\sum_jP_j\ln\frac{P_j}{g_j}=k_{\mathrm B}\beta\sum_jP_jE_j+k_{\mathrm B}\ln Z\sum_jP_j $$

Using $\sum_jP_jE_j=\langle E\rangle$, $\sum_jP_j=1$, and $k_{\mathrm B}\beta=1/T$, the right-hand side becomes $\dfrac{\langle E\rangle}{T}+k_{\mathrm B}\ln Z$, which equals $S=k_{\mathrm B}\ln Z+\langle E\rangle/T$. Thus we have obtained $S=-k_{\mathrm B}\sum_jP_j\ln(P_j/g_j)$.

(b) For two independent systems, $P_{jk}/g_{jk}=\dfrac{P_j^{(1)}}{g_j^{(1)}}\cdot\dfrac{P_k^{(2)}}{g_k^{(2)}}$, so $\ln\dfrac{P_{jk}}{g_{jk}}=\ln\dfrac{P_j^{(1)}}{g_j^{(1)}}+\ln\dfrac{P_k^{(2)}}{g_k^{(2)}}$ (the logarithm of a product is the sum of the logarithms). Therefore

$$ S=-k_{\mathrm B}\sum_{j,k}P_j^{(1)}P_k^{(2)}\left[\ln\frac{P_j^{(1)}}{g_j^{(1)}}+\ln\frac{P_k^{(2)}}{g_k^{(2)}}\right] =-k_{\mathrm B}\sum_jP_j^{(1)}\ln\frac{P_j^{(1)}}{g_j^{(1)}}\underbrace{\sum_kP_k^{(2)}}_{=1}-k_{\mathrm B}\sum_kP_k^{(2)}\ln\frac{P_k^{(2)}}{g_k^{(2)}}\underbrace{\sum_jP_j^{(1)}}_{=1}=S_1+S_2 $$

so the entropy is additive (extensivity).

(c) For two levels ($E_1=0,E_2=\varepsilon$, $g=1$) with $k_{\mathrm B}T=\varepsilon$, we have $P_1=e/(e+1)=0.73106$, $P_2=1/(e+1)=0.26894$, $Z=1+e^{-1}=1.36788$, and $\langle E\rangle/T=k_{\mathrm B}P_2=0.26894\,k_{\mathrm B}$. With the expression $S=k_{\mathrm B}\ln Z+\langle E\rangle/T$,

$$ \frac{S}{k_{\mathrm B}}=\ln1.36788+0.26894=0.31326+0.26894=0.58220 $$

With the expression in (a), since $g=1$,

$$ \frac{S}{k_{\mathrm B}}=-P_1\ln P_1-P_2\ln P_2=-0.73106\times(-0.31326)-0.26894\times(-1.31326)=0.22901+0.35319=0.58220 $$

so the two agree (and the value is smaller than the high-temperature limit $\ln2=0.693$).

∎

Solution

By the equipartition theorem (Theorem 50.4), $\langle\frac12m\omega^2x^2\rangle=\frac12k_{\mathrm B}T$, that is, $\langle x^2\rangle=k_{\mathrm B}T/(m\omega^2)$. The numerical values are as follows.

Mass of an atom: $m=\dfrac{63.55\times10^{-3}\ \mathrm{kg/mol}}{6.022\times10^{23}\ \mathrm{mol^{-1}}}=1.055\times10^{-25}\ \mathrm{kg}$; angular frequency: $\omega=2\pi\times5.0\times10^{12}\ \mathrm{s^{-1}}=3.14\times10^{13}\ \mathrm{s^{-1}}$; $k_{\mathrm B}T=1.380649\times10^{-23}\times300=4.14\times10^{-21}\ \mathrm J$.

$$ m\omega^2=1.055\times10^{-25}\times(3.14\times10^{13})^2=104\ \mathrm{N/m},\qquad \langle x^2\rangle=\frac{4.14\times10^{-21}}{104}\ \mathrm{m^2}=3.98\times10^{-23}\ \mathrm{m^2} $$

($\mathrm{kg\,s^{-2}=N/m}$, $\mathrm{J/(N/m)=m^2}$). Therefore $x_{\mathrm{rms}}=\sqrt{3.98\times10^{-23}}\ \mathrm m=6.3\times10^{-12}\ \mathrm m=0.063\ \text{Å}$. The ratio to the nearest-neighbor interatomic distance $2.56\ \text{Å}$ is $0.063/2.56=0.025$, that is, about $2.5$%. Even at room temperature, atoms vibrate over a range of a few percent of the interatomic distance. As the temperature is raised, the vibration amplitude grows according to $x_{\mathrm{rms}}\propto\sqrt T$, and there is an empirical rule that a crystal melts when the amplitude reaches around $10$% of the interatomic distance (the Lindemann melting criterion).

(Checking the conditions of applicability) $\theta=h\nu/k_{\mathrm B}=\dfrac{6.626\times10^{-34}\times5.0\times10^{12}}{1.3806\times10^{-23}}\ \mathrm K=240\ \mathrm K$, and $T=300\ \mathrm K$ is only $1.25$ times $\theta$. For this reason, the approximation by the equipartition theorem (classical statistical mechanics) has an accuracy of only a few percent. Using the quantum-mechanical harmonic oscillator (levels $\hbar\omega(n+\frac12)$), we get $\langle x^2\rangle=\dfrac{\hbar}{2m\omega}\coth\dfrac{\hbar\omega}{2k_{\mathrm B}T}$, and the ratio to the classical value $k_{\mathrm B}T/(m\omega^2)$ is $\dfrac{\hbar\omega}{2k_{\mathrm B}T}\coth\dfrac{\hbar\omega}{2k_{\mathrm B}T}$. Here $\hbar\omega=h\nu$, so $\dfrac{\hbar\omega}{2k_{\mathrm B}T}=\dfrac{240}{2\times300}=0.400$, and the ratio is $0.400\times\coth0.400=1.053$. That is, the quantum-mechanical $\langle x^2\rangle$ is about $1.05$ times the classical value, $x_{\mathrm{rms}}$ is larger by about $2.6$% (about $6.5\times10^{-12}\ \mathrm m$), and the classical estimate contains an error of a few percent even at room temperature.

∎

Solution

(a) The volume of the spherical shell of momentum magnitude from $p$ to $p+\dd p$ is $4\pi p^2\dd p$, and, letting $\alpha=\beta c$,

$$ \langle E\rangle=\frac{\displaystyle\int_0^\infty cp\,e^{-\alpha p}\,4\pi p^2\dd p}{\displaystyle\int_0^\infty e^{-\alpha p}\,4\pi p^2\dd p}=c\,\frac{\int_0^\infty p^3e^{-\alpha p}\dd p}{\int_0^\infty p^2e^{-\alpha p}\dd p}=c\cdot\frac{3!/\alpha^4}{2!/\alpha^3}=\frac{3c}{\alpha}=\frac{3}{\beta}=3k_{\mathrm B}T $$

(we used the formula $\int_0^\infty p^ne^{-\alpha p}\dd p=n!/\alpha^{n+1}$).

(b) Summing the general formula $\langle p_i\,\partial E/\partial p_i\rangle=k_{\mathrm B}T$ ($i=1,2,3$) over the three components gives $\left\langle\sum_ip_i\,\partial E/\partial p_i\right\rangle=3k_{\mathrm B}T$. Since $E=c|\bm p|$ is a homogeneous function of degree one in $\bm p$, Euler's theorem gives $\sum_ip_i\,\partial E/\partial p_i=E$ (indeed, $\partial E/\partial p_i=cp_i/|\bm p|$, and $\sum_ip_i\cdot cp_i/|\bm p|=c|\bm p|=E$). Hence $\langle E\rangle=3k_{\mathrm B}T$.

Both methods give $3k_{\mathrm B}T$, not $\frac32k_{\mathrm B}T$. An energy that is linear in momentum is not a quadratic form, so each component contributes $k_{\mathrm B}T$ rather than $\frac12k_{\mathrm B}T$ (the case $s=1$ in the Advanced Topic box: $\langle a|\xi|^s\rangle=k_{\mathrm B}T/s$). For $N$ particles, $U=3Nk_{\mathrm B}T$ and $C_V=3Nk_{\mathrm B}$.

∎

Solution

Applying the result $n_j=g_je^{-\alpha}e^{-\beta E_j}$ of Section 50.8 to two levels ($g_1=g_2=1$, $E_1=0$, $E_2=\varepsilon$), the distribution that maximizes $W$ satisfies the ratio $n_2/n_1=e^{-\beta(E_2-E_1)}=e^{-\beta\varepsilon}$. Conversion: $\varepsilon/k_{\mathrm B}=\dfrac{0.050\times1.602\,176\,634\times10^{-19}\ \mathrm J}{1.380\,649\times10^{-23}\ \mathrm{J/K}}=580.2\ \mathrm K$.

(a) Since $n_2=100$ and $n_1=400$, $e^{-\beta\varepsilon}=100/400=\frac14$ and $\beta\varepsilon=\ln4=1.386$. $T=\dfrac{\varepsilon/k_{\mathrm B}}{\beta\varepsilon}=\dfrac{580.2\ \mathrm K}{1.386}=418.5\ \mathrm K$.

(b) Since $n_2=200$ and $n_1=300$, $e^{-\beta\varepsilon}=200/300=\frac23$ and $\beta\varepsilon=\ln\frac32=0.4055$. $T=\dfrac{580.2\ \mathrm K}{0.4055}=1431\ \mathrm K$. When the number of excitations is increased from $100$ to $200$, a factor of $2$, the temperature rises by a factor of about $3.4$, from $418.5\ \mathrm K$ to $1431\ \mathrm K$. The reason a large temperature increase is needed relative to the increase in the number of excitations is that $\ln(n_1/n_2)$ changes only logarithmically with $n_2/n_1$, and $T\propto1/\ln(n_1/n_2)$. As $n_2\to n_1$ (the excited fraction is $\frac12$), $\beta\to0$, that is, $T\to\infty$.

∎

Solution

(a) When $n$ systems are in the excited state, the total energy is $n\varepsilon$, and the number of microstates for it (the number of ways of choosing which of the $N$ systems are in the excited state) is $W(n)=\binom Nn$. The total partition function is $Z=(1+e^{-\beta\varepsilon})^N=\sum_n\binom Nne^{-n\beta\varepsilon}$ (binomial theorem), so

$$ P_n=\frac{W(n)e^{-\beta n\varepsilon}}{Z}=\binom Nn\frac{e^{-n\beta\varepsilon}}{(1+e^{-\beta\varepsilon})^N}=\binom Nn\left(\frac{e^{-\beta\varepsilon}}{1+e^{-\beta\varepsilon}}\right)^n\left(\frac{1}{1+e^{-\beta\varepsilon}}\right)^{N-n}=\binom NnP_2^{\,n}P_1^{\,N-n} $$

which coincides with the binomial distribution for the case in which “each system is in the excited state with probability $P_2$.”

(b) For $k_{\mathrm B}T=\varepsilon$, $P_2=1/(e+1)=0.26894$ and $P_1=0.73106$. The mean of the binomial distribution is $NP_2=26.89$, and its variance is $NP_1P_2=100\times0.19661=19.66$. In terms of energy, $\langle E\rangle=\varepsilon\langle n\rangle$ and $\langle\Delta E^2\rangle=19.66\,\varepsilon^2$. On the other hand, by Formula 50.3 (the result of Example 50.11), for a single system $k_{\mathrm B}T^2C_V=\varepsilon^2P_1P_2=0.19661\,\varepsilon^2$, and for $N=100$ systems this becomes 100 times as large, $19.66\,\varepsilon^2$, in agreement. The standard deviation is $\sigma_n=\sqrt{19.66}=4.43$.

(c) The mode of the binomial distribution is $\lfloor(N+1)P_2\rfloor=\lfloor101\times0.26894\rfloor=\lfloor27.16\rfloor=27$ ($\lfloor x\rfloor$ is the largest integer not exceeding $x$). Computing $P_{27}=\binom{100}{27}P_2^{27}P_1^{73}$ via logarithms, $\ln\binom{100}{27}=55.913$, $27\ln P_2=27\times(-1.31326)=-35.458$, and $73\ln P_1=73\times(-0.31326)=-22.868$, so $\ln P_{27}=55.913-35.458-22.868=-2.413$, that is, $P_{27}=e^{-2.413}=0.0895$ (this is the maximum, compared with $P_{26}=0.0888$ and $P_{28}=0.0859$). In the Gaussian approximation, $1/(\sigma\sqrt{2\pi})=1/(4.434\times2.5066)=0.0900$, which differs from the exact value by only about $0.5$%. We see that even for a system as small as $N=100$, the energy distribution is already nearly Gaussian (Theorem 50.3).

∎

Solution

(1) The phase space is the four-dimensional space $(x,y,p_x,p_y)$, and the volume corresponding to one state is $h^2$. Therefore

$$ Z_1=\frac{1}{h^2}\int_{L^2}\dd x\,\dd y\int_{-\infty}^{\infty}\dd p_x\int_{-\infty}^{\infty}\dd p_y\ e^{-\beta(p_x^2+p_y^2)/(2m)} $$

The position integral gives the area $L^2$. The momentum integral is the square of the Gaussian integral $\int_{-\infty}^{\infty}e^{-ap^2}\dd p=\sqrt{\pi/a}$ ($a=\beta/(2m)$), namely $\left(\sqrt{2\pi m/\beta}\right)^2=2\pi m/\beta=2\pi mk_{\mathrm B}T$. Hence

$$ Z_1=\frac{L^2}{h^2}\,2\pi mk_{\mathrm B}T=\frac{L^2}{\lambda_{\mathrm T}^2},\qquad\lambda_{\mathrm T}=\frac{h}{\sqrt{2\pi mk_{\mathrm B}T}} $$

(Dimensions: multiplying $[mk_{\mathrm B}T/h^2]=\mathrm{m^{-2}}$ by $L^2$ gives a dimensionless quantity.)

(2) From $Z=Z_1^N/N!$ and Stirling's formula $\ln N!\approx N\ln N-N$,

$$ \ln Z=N\ln\frac{L^2}{\lambda_{\mathrm T}^2}-N\ln N+N=N\left[\ln\frac{L^2}{N\lambda_{\mathrm T}^2}+1\right],\qquad A=-k_{\mathrm B}T\ln Z=-Nk_{\mathrm B}T\left[\ln\frac{L^2}{N\lambda_{\mathrm T}^2}+1\right] $$

$\lambda_{\mathrm T}$ does not depend on $L$. When we differentiate with respect to the area $\mathcal A=L^2$ as the variable, the only term in $A$ that contains $\mathcal A$ is the $\ln\mathcal A$ term, so

$$ \Pi=-\left(\pdiff{A}{\mathcal A}\right)_T=Nk_{\mathrm B}T\cdot\frac{1}{\mathcal A}\qquad\therefore\quad\Pi\,L^2=Nk_{\mathrm B}T $$

which is the equation of state of the two-dimensional ideal gas. The unit of $\Pi$ is $\mathrm{J/m^2}=\mathrm{N/m}$ (force per unit length), and $\Pi L^2$ is in $\mathrm J$, consistent with $Nk_{\mathrm B}T$.

(3) From $Z_1=(L^2/h^2)\cdot2\pi m/\beta\propto\beta^{-1}$ we have $\ln Z_1=-\ln\beta+\text{(a quantity independent of }\beta\text{)}$, and since $\ln N!$ does not depend on $\beta$,

$$ U=-\pdiff{\ln Z}{\beta}=-N\pdiff{\ln Z_1}{\beta}=\frac{N}{\beta}=Nk_{\mathrm B}T $$

By the equipartition theorem, the kinetic energy has two quadratic terms, $p_x^2/(2m)$ and $p_y^2/(2m)$, each of which contributes $\tfrac12k_{\mathrm B}T$; this gives $k_{\mathrm B}T$ per particle and $Nk_{\mathrm B}T$ for $N$ particles, in agreement with the result above. The heat capacity is $C=Nk_{\mathrm B}$.

Solution

(1) Since $\lambda_{\mathrm T}\propto T^{-1/2}$, we have $\ln\dfrac{V}{N\lambda_{\mathrm T}^3}=\ln\dfrac VN+\dfrac32\ln T+c$ ($c$ is a constant independent of $T$). Hence, differentiating $A=-Nk_{\mathrm B}T\left[\ln\dfrac VN+\dfrac32\ln T+c+1\right]$ with respect to $T$ by the product rule, we obtain

$$ \pdiff{A}{T}=-Nk_{\mathrm B}\left[\ln\frac{V}{N\lambda_{\mathrm T}^3}+1\right]-Nk_{\mathrm B}T\cdot\frac{3}{2T}=-Nk_{\mathrm B}\left[\ln\frac{V}{N\lambda_{\mathrm T}^3}+\frac52\right] $$

Therefore $S=-(\partial A/\partial T)_V=Nk_{\mathrm B}\left[\ln\dfrac{V}{N\lambda_{\mathrm T}^3}+\dfrac52\right]$.

(2) The mass of a neon atom is $m=20.180\times1.6605\times10^{-27}=3.351\times10^{-26}\ \mathrm{kg}$.

$$ \lambda_{\mathrm T}=\frac{6.626\times10^{-34}}{\sqrt{2\pi\times3.351\times10^{-26}\times1.3806\times10^{-23}\times298.15}}=2.251\times10^{-11}\ \mathrm m=22.5\ \mathrm{pm} $$

The volume per molecule is $V/N=k_{\mathrm B}T/p=1.3806\times10^{-23}\times298.15/(1\times10^5)=4.116\times10^{-26}\ \mathrm{m^3}$ and $\lambda_{\mathrm T}^3=1.140\times10^{-32}\ \mathrm{m^3}$, so $V/(N\lambda_{\mathrm T}^3)=3.61\times10^{6}$ and $\ln(3.61\times10^6)=15.10$. For $1\ \mathrm{mol}$, using $N k_{\mathrm B}=R$,

$$ S_{\mathrm m}=R\left[15.10+2.5\right]=8.314\times17.60\approx146.3\ \mathrm{J/(mol\,K)} $$

which agrees with the value $146.33\ \mathrm{J/(mol\,K)}$ in standard thermodynamic tables. Note also that $V/(N\lambda_{\mathrm T}^3)\approx3.6\times10^6\gg1$, so the condition for classical statistical mechanics to be applicable is satisfied.

Solution

(1) Let $X\equiv\beta\hbar\omega$ and $x\equiv e^{-X}$. The probability of the level $E_n=\hbar\omega(n+\tfrac12)$ is $P_n=e^{-\beta E_n}/Z_{\mathrm{qm}}=e^{-X/2}x^n/\left[e^{-X/2}/(1-x)\right]=(1-x)x^n$ (the zero-point-energy factor $e^{-X/2}$ cancels between the numerator and the denominator). Therefore

$$ \langle n\rangle=\sum_{n=0}^\infty nP_n=(1-x)\sum_{n=0}^\infty nx^n $$

Differentiating both sides of the geometric series $\sum_nx^n=1/(1-x)$ with respect to $x$ gives $\sum_nnx^{n-1}=1/(1-x)^2$, and multiplying both sides by $x$ gives $\sum_nnx^n=x/(1-x)^2$. Hence

$$ \langle n\rangle=(1-x)\frac{x}{(1-x)^2}=\frac{x}{1-x}=\frac{e^{-X}}{1-e^{-X}}=\frac{1}{e^{X}-1} $$

The mean energy is $\langle E\rangle=\sum_nE_nP_n=\hbar\omega\left(\langle n\rangle+\tfrac12\right)=\tfrac12\hbar\omega+\dfrac{\hbar\omega}{e^{\beta\hbar\omega}-1}$, which agrees with the formula in the main text (the result derived from $-\partial\ln Z_{\mathrm{qm}}/\partial\beta$).

(2) We compute $\langle E\rangle/(k_{\mathrm B}T)=X\left(\langle n\rangle+\tfrac12\right)$.

The classical value is $\langle E\rangle/(k_{\mathrm B}T)=1$. The quantum value is always larger than the classical value (by the amount of the zero-point energy $\tfrac12\hbar\omega$). The larger $X$ is (the lower the temperature), the further the value departs from the classical one. At $X=5$, $\langle n\rangle\approx0.007$; that is, the oscillator is almost in the ground state, and $\langle E\rangle\approx\tfrac12\hbar\omega=0.5\hbar\omega$ ($0.507\,\hbar\omega$). The deviation of $8.2\%$ from the classical value at $X=1$ nearly agrees with the expansion $1+X^2/12=1.083$ of Example 51.5.

(3) Planck's mean energy of an oscillator, $\dfrac{h\nu}{e^{h\nu/(k_{\mathrm B}T)}-1}$, coincides with $\hbar\omega\langle n\rangle=\dfrac{\hbar\omega}{e^{\hbar\omega/(k_{\mathrm B}T)}-1}$ if we put $h\nu=\hbar\omega$ ($\omega=2\pi\nu$, $h=2\pi\hbar$). That is, Planck's formula is the part ($\hbar\omega\langle n\rangle$) of the mean energy of the quantum harmonic oscillator that remains after the zero-point energy $\tfrac12\hbar\omega$ is removed. $\langle n\rangle$ can be read as the mean number of quanta (photons or phonons) in that mode.

Solution

(1) Since $e^x/(e^x-1)^2=1/(e^{x/2}-e^{-x/2})^2=1/(4\sinh^2(x/2))$, we have $f_{\mathrm E}(x)=\dfrac{x^2}{4\sinh^2(x/2)}=\left[\dfrac{y}{\sinh y}\right]^2$ ($y=x/2$). From $\sinh y=y\left(1+\tfrac{y^2}{6}+\tfrac{y^4}{120}+\cdots\right)$, putting $t=\tfrac{y^2}{6}+\tfrac{y^4}{120}$ and computing $\dfrac{1}{1+t}=1-t+t^2-\cdots$ up to $y^4$, we get

$$ \frac{y}{\sinh y}=1-\frac{y^2}{6}-\frac{y^4}{120}+\frac{y^4}{36}+\cdots=1-\frac{y^2}{6}+\frac{7y^4}{360}+\cdots $$

($-\tfrac1{120}+\tfrac1{36}=\tfrac{-3+10}{360}=\tfrac7{360}$). Squaring, using $(1-a+b)^2=1-2a+(a^2+2b)+\cdots$ ($a=y^2/6$, $b=7y^4/360$), we obtain

$$ f_{\mathrm E}=1-\frac{y^2}{3}+\left(\frac{1}{36}+\frac{7}{180}\right)y^4+\cdots=1-\frac{y^2}{3}+\frac{y^4}{15}+\cdots $$

Substituting back $y=x/2$, we have $y^2=x^2/4$ and $y^4=x^4/16$, so

$$ f_{\mathrm E}(x)=1-\frac{x^2}{12}+\frac{x^4}{240}-\cdots $$

At $x=0.8$: with two terms, $1-0.64/12=0.94667$; with three terms, $0.94667+0.4096/240=0.94837$; the exact value is $0.94833$. Three terms are accurate to within $10^{-4}$.

(2) For $x\gg1$, rewrite $\dfrac{e^x}{(e^x-1)^2}=\dfrac{e^{-x}}{(1-e^{-x})^2}$ and use $\dfrac{1}{(1-t)^2}=1+2t+3t^2+\cdots$ for $t=e^{-x}\ll1$; this gives $f_{\mathrm E}=x^2e^{-x}\left(1+2e^{-x}+3e^{-2x}+\cdots\right)$. At $x=5$, $e^{-5}=0.006738$ and $x^2e^{-x}=0.16845$; with the correction factor $1+2e^{-5}=1.01348$ this gives $0.17072$, while the exact value is $0.17074$. Even the approximation $x^2e^{-x}$ alone, without the correction, is accurate to within $1.3\%$, and at $x=10$ the relative error is $0.01\%$.

Solution

(1) At $T=10\ \mathrm K$, $x=\theta_{\mathrm E}/T=24$. Since $x\gg1$, $f_{\mathrm E}\approx x^2e^{-x}=576\times e^{-24}=576\times3.775\times10^{-11}=2.17\times10^{-8}$ (for the molar heat capacity, $3R\times2.17\times10^{-8}=5.4\times10^{-7}\ \mathrm{J/(mol\,K)}$).

(2) The $T^3$ law: $\dfrac{4\pi^4}{5}=77.9$ and $(T/\theta_{\mathrm D})^3=(10/343)^3=2.48\times10^{-5}$, so $C_V/(3Nk_{\mathrm B})=77.9\times2.48\times10^{-5}=1.93\times10^{-3}$ (for the molar heat capacity, $3R\times1.93\times10^{-3}=0.048\ \mathrm{J/(mol\,K)}$).

The ratio of the two values is $1.93\times10^{-3}/(2.17\times10^{-8})\approx9\times10^{4}$, a difference of about five orders of magnitude. Experiment agrees with the $T^3$ law (in metals, a heat capacity proportional to the temperature, due to the conduction electrons, is added to this). The $T^3$ law agrees well with measurements at extremely low temperatures, $T\lesssim\theta_{\mathrm D}/50\approx7\ \mathrm K$; $10\ \mathrm K$ lies slightly outside this range, but it is enough to see a difference of five orders of magnitude.

(3) In the Einstein model, all the oscillators have the same $\omega$ ($\hbar\omega=k_{\mathrm B}\theta_{\mathrm E}=20.7\ \mathrm{meV}$), and the minimum energy required for excitation is uniformly this value. At $T=10\ \mathrm K$ the thermal energy is $k_{\mathrm B}T=0.86\ \mathrm{meV}$, and the probability of exciting this quantum is $e^{-24}$, so the heat capacity freezes out exponentially. In a real solid, however, the atoms vibrate in concert with one another, and, like long-wavelength sound waves, there are arbitrarily many vibrational modes whose frequency is close to $0$. Such modes can be excited by a small quantum $\hbar\omega$, and no matter how low the temperature, a certain number of modes with $\hbar\omega\lesssim k_{\mathrm B}T$ are excited. That number increases as $T^3$, in proportion to the volume of the $3$-dimensional wavenumber space. Consequently, the energy is proportional to $T^4$ and the heat capacity to $T^3$ (the Debye model of Chapter 53). The Einstein model disagrees with experiment at extremely low temperatures because it ignores this “existence of low-frequency modes.” On the other hand, the high-temperature value $3Nk_{\mathrm B}$ (Dulong–Petit) does not depend on the distribution of frequencies (because all modes are classically excited), so it is correctly reproduced by both models.

Solution

(1) For $x\ll1$, $e^x\approx1$, so $\dfrac{x^2e^x}{(e^x+1)^2}\approx\dfrac{x^2}{(1+1)^2}=\dfrac{x^2}{4}$. That is, $C_V\approx Nk_{\mathrm B}x^2/4=Nk_{\mathrm B}(\varepsilon/2k_{\mathrm B}T)^2\propto T^{-2}$ (at $x=0.1$, the approximate value is $0.002500$ and the exact value is $0.002494$). For $x\gg1$, $e^x+1\approx e^x$, so $\dfrac{x^2e^x}{e^{2x}}=x^2e^{-x}$ (at $x=8$, the approximate value is $0.02147$ and the exact value is $0.02146$). Both tend to $0$.

(2) The peak is at the solution $x=2.3994$ of $x\tanh(x/2)=2$ (Example 51.10), so $k_{\mathrm B}T=\varepsilon/2.3994$. If $\varepsilon/k_{\mathrm B}=5.0\ \mathrm K$, then $T_{\mathrm{peak}}=5.0/2.3994=2.08\ \mathrm K$. The height is $0.4392\,R=0.4392\times8.314=3.65\ \mathrm{J/(mol\,K)}$.

(3) Integrating the thermodynamic relation (Chapter 29) $\dd S=\dd'Q/T=C_V\dd T/T$ (at constant volume) from $0$ to $\infty$ gives $S(\infty)-S(0)=\int_0^\infty(C_V/T)\dd T$. On the other hand, we examine the limits of the canonical-distribution entropy $S=k_{\mathrm B}\ln Z+U/T$ ($Z=(1+e^{-\beta\varepsilon})^N$, $U=N\varepsilon/(e^{\beta\varepsilon}+1)$). As $T\to0$: $Z\to1$ and $U\to0$, so $S\to0$. As $T\to\infty$: $Z\to2^N$, $\ln Z\to N\ln2$, and $U\to N\varepsilon/2$ so that $U/T\to0$; hence $S\to Nk_{\mathrm B}\ln2$. Therefore

$$ \int_0^\infty\frac{C_V}{T}\dd T=Nk_{\mathrm B}\ln2 $$

and this does not depend on $\varepsilon$ (even if the level spacing changes, the integral of $C_V/T$ over $T$, that is, the area under the curve of $C_V/T$, is the same $Nk_{\mathrm B}\ln2$). For $1\ \mathrm{mol}$, $R\ln2=8.314\times0.6931=5.76\ \mathrm{J/(mol\,K)}$. Numerical integration (with $t=k_{\mathrm B}T/\varepsilon$, $\int_0^\infty f_{\mathrm S}(1/t)\,\dd t/t$) also gives $0.69314$ ($=\ln2$). This $\ln2$ corresponds to the entropy of $1$ bit of freedom per two-level system, namely “which of $2$ states it is in.”

Solution

(1) The two levels are $E_\uparrow=-\mu B$ and $E_\downarrow=+\mu B$. If we shift the origin of energy to the lower level $E_\uparrow$, the levels become $0$ and $\varepsilon=2\mu B$. The mean energy per spin (measured from this origin) is then given by the two-level formula $\varepsilon/(e^{\beta\varepsilon}+1)$. To return the origin to $-\mu B$ we only need to add $-\mu B$, so

$$ \langle E\rangle=-\mu B+\frac{2\mu B}{e^{2\beta\mu B}+1}=\mu B\,\frac{-\left(e^{2\beta\mu B}+1\right)+2}{e^{2\beta\mu B}+1}=-\mu B\,\frac{e^{2\beta\mu B}-1}{e^{2\beta\mu B}+1}=-\mu B\tanh(\beta\mu B) $$

(in the last step we divided the numerator and the denominator by $e^{\beta\mu B}$). Since the energy in a magnetic field is $-\bm m\cdot\bm B$, we have $\langle E\rangle=-\langle m\rangle B$. Hence $\langle m\rangle=\mu\tanh(\beta\mu B)$, and for $N$ spins, $M=N\mu\tanh(\mu B/k_{\mathrm B}T)$.

(2) $Z=z^N$ with $z=e^{\beta\mu B}+e^{-\beta\mu B}=2\cosh x$ ($x=\beta\mu B=\mu B/(k_{\mathrm B}T)$). Since $U=-N\mu B\tanh x$, we have $U/T=-N(\mu B/T)\tanh x=-Nk_{\mathrm B}x\tanh x$ ($\mu B/T=k_{\mathrm B}x$). Hence

$$ S=k_{\mathrm B}\ln Z+\frac UT=Nk_{\mathrm B}\left[\ln(2\cosh x)-x\tanh x\right] $$

For $x\to0$ (weak field or high temperature): $\cosh x\to1$ and $x\tanh x\approx x^2\to0$, so $S\to Nk_{\mathrm B}\ln2$ (the directions are completely random). For $x\to\infty$ (strong field or low temperature): $\cosh x\approx e^x/2$, so $\ln(2\cosh x)\to x$, and $\tanh x\to1$, so $x-x\to0$ and $S\to0$ (all the spins are aligned in parallel).

(3) Initially, $x_{\mathrm i}=\mu_{\mathrm B}B_{\mathrm i}/(k_{\mathrm B}T_{\mathrm i})=9.274\times10^{-24}\times4.0/(1.381\times10^{-23}\times2.0)=1.343$. The magnetization is $M/(N\mu_{\mathrm B})=\tanh1.343=0.872$ (about $87\%$ of saturation). The entropy is $S/(Nk_{\mathrm B})=\ln(2\cosh1.343)-1.343\tanh1.343=\ln4.093-1.172=0.237$ (about $34\%$ of the zero-field value $\ln2=0.693$). In an adiabatic reversible process $S$ is constant, and since $S$ is a function of $x$ alone, $x_{\mathrm f}=x_{\mathrm i}$; that is, $B/T$ is constant:

$$ T_{\mathrm f}=T_{\mathrm i}\frac{B_{\mathrm f}}{B_{\mathrm i}}=2.0\ \mathrm K\times\frac{0.040}{4.0}=0.020\ \mathrm K=20\ \mathrm{mK} $$

(Note, however, that this is the result for the ideal case in which there is no interaction between the spins.)

Solution

(1) Since the upper level holds $60\%$ of the whole, $n/N=0.6$ and $(N-n)/n=0.4/0.6=2/3$. From the temperature formula $1/T=(k_{\mathrm B}/\varepsilon)\ln[(N-n)/n]$,

$$ \frac{1}{T}=\frac{k_{\mathrm B}}{\varepsilon}\ln\frac23=-0.4055\,\frac{k_{\mathrm B}}{\varepsilon}\qquad\therefore\quad T=-\frac{\varepsilon}{0.4055\,k_{\mathrm B}}=-2.47\,\frac{\varepsilon}{k_{\mathrm B}} $$

Bring this system (system 1, $1/T_1\lt0$) into contact with a heat bath at positive temperature (system 2, $T_2\gt0$, $1/T_2\gt0$). Since the total energy is conserved, $\dd E_2=-\dd E_1$, and the change in the total entropy is

$$ \dd S=\frac{\dd E_1}{T_1}+\frac{\dd E_2}{T_2}=\left(\frac1{T_1}-\frac1{T_2}\right)\dd E_1 $$

Since $1/T_1\lt0\lt1/T_2$, the quantity in parentheses is negative, and to satisfy the law of increase of entropy, $\dd S\ge0$, we need $\dd E_1\lt0$. That is, the energy of system 1 (negative temperature) decreases, and heat flows from the negative-temperature system to the positive-temperature heat bath. This holds no matter how high the temperature $T_2$ of the heat bath is (even for $1/T_2\to0^+$). A system at negative temperature is “hotter” than any system at positive temperature.

(2) $x=\theta_{\mathrm{vib}}/T=470/300=1.567$. The vibrational contribution is

$$ f_{\mathrm E}(1.567)=1.567^2\times\frac{e^{1.567}}{(e^{1.567}-1)^2}=2.455\times\frac{4.791}{(3.791)^2}=2.455\times\frac{4.791}{14.37}\approx0.818 $$

Hence $C_{V,\mathrm m}/R=\tfrac32+1+0.818=3.318$, that is, $C_{V,\mathrm m}=3.318\times8.314\approx27.6\ \mathrm{J/(mol\,K)}$. From Mayer's relation $C_p=C_V+R$, the heat capacity ratio is $\gamma=(3.318+1)/3.318=1.30$. This agrees well with the values for bromine gas in standard thermodynamic tables, $C_p\approx36.0\ \mathrm{J/(mol\,K)}$ ($C_V\approx27.7$, $\gamma\approx1.30$). It is smaller than the value $\gamma=1.40$ for nitrogen and oxygen because the vibration is considerably excited.

Solution

(1) By the grand canonical distribution, the weights of the two states of an adsorption site are $e^{0}=1$ for empty and $e^{-\beta(-\varepsilon_0-\mu)}=e^{\beta(\varepsilon_0+\mu)}$ for adsorbed. Therefore the probability of being adsorbed (the coverage) is

$$ \theta=\frac{e^{\beta(\varepsilon_0+\mu)}}{1+e^{\beta(\varepsilon_0+\mu)}}=\frac1{1+e^{-\beta(\varepsilon_0+\mu)}} $$

The chemical potential of a monatomic ideal gas is $\mu=k_{\mathrm B}T\ln(n\lambda_{\mathrm T}^3)$, and $n=p/k_{\mathrm B}T$, so $e^{\beta\mu}=n\lambda_{\mathrm T}^3=p\lambda_{\mathrm T}^3/k_{\mathrm B}T$ and hence $e^{-\beta\mu}=k_{\mathrm B}T/(p\lambda_{\mathrm T}^3)$. Substituting this into $\theta$ gives

$$ \theta=\frac1{1+\dfrac{k_{\mathrm B}T}{p\lambda_{\mathrm T}^3}e^{-\varepsilon_0/k_{\mathrm B}T}}=\frac1{1+p_0/p}=\frac p{p+p_0},\qquad p_0=\frac{k_{\mathrm B}T}{\lambda_{\mathrm T}^3}e^{-\varepsilon_0/k_{\mathrm B}T} $$

Here we multiplied the numerator and denominator by $p$. For $p\ll p_0$, $\theta\approx p/p_0$ (proportional to the pressure), and for $p\gg p_0$, $\theta\to1$ (saturation). $p_0$ is the pressure at which $\theta=\tfrac12$.

(2) $m=39.948\times1.6605\times10^{-27}=6.634\times10^{-26}\ \mathrm{kg}$ and $k_{\mathrm B}T=1.3806\times10^{-23}\times300=4.142\times10^{-21}\ \mathrm J=0.02585\ \mathrm{eV}$. The thermal de Broglie wavelength is

$$ \lambda_{\mathrm T}=\frac h{\sqrt{2\pi mk_{\mathrm B}T}}=\frac{6.626\times10^{-34}}{\sqrt{2\pi\times6.634\times10^{-26}\times4.142\times10^{-21}}}\ \mathrm m=1.595\times10^{-11}\ \mathrm m $$

Hence $\lambda_{\mathrm T}^3=4.056\times10^{-33}\ \mathrm{m^3}$ and $k_{\mathrm B}T/\lambda_{\mathrm T}^3=4.142\times10^{-21}/4.056\times10^{-33}=1.021\times10^{12}\ \mathrm{Pa}$. The exponential factor is $e^{-0.40/0.02585}=e^{-15.47}=1.907\times10^{-7}$, so

$$ p_0=1.021\times10^{12}\ \mathrm{Pa}\times1.907\times10^{-7}=1.95\times10^{5}\ \mathrm{Pa}\approx1.9\ \mathrm{atm} $$

At $1\ \mathrm{atm}=1.013\times10^5\ \mathrm{Pa}$ the coverage is $\theta=1.013\times10^5/(1.013\times10^5+1.947\times10^5)=0.342$. The pressure at which $\theta=0.9$ follows from $p/(p+p_0)=0.9$: $p=9p_0=1.75\times10^6\ \mathrm{Pa}\approx17\ \mathrm{atm}$.

Solution

(1) The probability that the number of particles is exactly $N$ is the sum of the probabilities of all states $j$ of the $N$-particle system, $P(N)=\sum_je^{-\beta(E_{j,N}-\mu N)}/\Xi=z^NZ_N/\Xi$. Using $Z_N=Z_1^N/N!$, $\Xi=e^{zZ_1}$, and $\bar N=zZ_1$, we obtain

$$ P(N)=\frac{(zZ_1)^N}{N!}e^{-zZ_1}=\frac{\bar N^N}{N!}e^{-\bar N}=\frac{4^N}{N!}e^{-4} $$

(This is the Poisson distribution.)

(2) $e^{-4}=0.018316$. $P(0)=e^{-4}=0.0183$ and $P(4)=\dfrac{4^4}{4!}e^{-4}=\dfrac{256}{24}\times0.018316=10.667\times0.018316=0.1954$. $P(N\ge8)=1-\sum_{N=0}^{7}P(N)$. The sum of $P(0),\ldots,P(7)=0.0183,\ 0.0733,\ 0.1465,\ 0.1954,\ 0.1954,\ 0.1563,\ 0.1042,\ 0.0595$ is $0.9489$, so $P(N\ge8)=1-0.9489=0.0511$. For a mean of $4$, the probability that $8$ or more particles are present is about $5\,\%$.

(3) $\langle N(N-1)\rangle=e^{-\bar N}\sum_{N\ge2}N(N-1)\dfrac{\bar N^N}{N!}=e^{-\bar N}\sum_{N\ge2}\dfrac{\bar N^N}{(N-2)!}=e^{-\bar N}\bar N^2\sum_{M\ge0}\dfrac{\bar N^M}{M!}=\bar N^2$ (we put $N-2=M$; the terms with $N=0,1$ vanish because $N(N-1)=0$). Therefore $\langle N^2\rangle=\langle N(N-1)\rangle+\langle N\rangle=\bar N^2+\bar N$, and the variance is $\langle N^2\rangle-\bar N^2=\bar N$.

(4) The number density of air is $n=p/k_{\mathrm B}T=2.446\times10^{25}\ \mathrm{m^{-3}}$. From $\bar N=nV=4$, $V=4/n=1.635\times10^{-25}\ \mathrm{m^3}$. The edge length of the cube is $V^{1/3}=5.47\times10^{-9}\ \mathrm m\approx5.5\ \mathrm{nm}$. In such a tiny region, the relative fluctuation of the number of particles is as large as $1/\sqrt{\bar N}=50\,\%$.

Solution

(1)–(3) The energy of the two particles is the sum of the energies of the states they occupy. (1) For distinguishable particles, each particle independently occupies one of the three states (with weights $1,x,x^2$), so $Z=(1+x+x^2)^2=1+2x+3x^2+2x^3+x^4$. (2) For bosons there are the six cases of Example 52.7, $(0,0),(0,\varepsilon),(0,2\varepsilon),(\varepsilon,\varepsilon),(\varepsilon,2\varepsilon),(2\varepsilon,2\varepsilon)$, with energies $0,\varepsilon,2\varepsilon,2\varepsilon,3\varepsilon,4\varepsilon$, so $Z_{\mathrm B}=1+x+2x^2+x^3+x^4$. (3) For fermions there are three cases, $(0,\varepsilon),(0,2\varepsilon),(\varepsilon,2\varepsilon)$, with energies $\varepsilon,2\varepsilon,3\varepsilon$, so $Z_{\mathrm F}=x+x^2+x^3$.

(4) The mean energy is $\langle E\rangle=\sum E\,e^{-\beta E}/Z$. For $k_{\mathrm B}T=\varepsilon$, $x=e^{-1}=0.36788$, $x^2=0.13534$, $x^3=0.04979$, and $x^4=0.01832$. (1): The numerator is $0\cdot1+1\cdot2x+2\cdot3x^2+3\cdot2x^3+4x^4=0.7358+0.8120+0.2987+0.0733=1.9198$ and $Z=1+0.7358+0.4060+0.0996+0.0183=2.2597$, so $\langle E\rangle=0.850\,\varepsilon$. (2): The numerator is $x+2\cdot2x^2+3x^3+4x^4=0.3679+0.5413+0.1494+0.0733=1.1318$ and $Z_{\mathrm B}=1.7067$, so the mean energy is $0.663\,\varepsilon$. (3): The numerator is $x+2x^2+3x^3=0.3679+0.2707+0.1494=0.7879$ and $Z_{\mathrm F}=0.5530$, so the mean energy is $1.425\,\varepsilon$. As $T\to0$ ($x\to0$), the state of lowest energy dominates: in (1) and (2), $\langle E\rangle\to0$ (both particles go into the ground state $0$); in (3), the exclusion principle forbids both particles from entering the ground state, so the system settles in the lowest allowed state $(0,\varepsilon)$ and $\langle E\rangle\to\varepsilon$.

(5) The states in which the two particles are in the same state are, in (1), the three states $(0,0),(\varepsilon,\varepsilon),(2\varepsilon,2\varepsilon)$, whose weights sum to $1+x^2+x^4=1.1537$, so the probability is $1.1537/2.2597=0.511$. In (2), the same three states also have total weight $1.1537$, so the probability is $1.1537/1.7067=0.676$. For bosons, the probability that the two particles occupy the same state is larger than for distinguishable particles ($0.511$). This is a numerical confirmation of the statement “indistinguishable bosons tend to gather in the same state.”

Solution

At $T=300\ \mathrm K$, $k_{\mathrm B}T=0.02585\ \mathrm{eV}$. The difference between the bottom of the conduction band and $\mu$ is $E_{\mathrm g}/2$, so $x=(E_{\mathrm g}/2)/k_{\mathrm B}T$, and the occupation probability is $f=1/(e^x+1)$.

(1) Silicon $E_{\mathrm g}/2=0.56\ \mathrm{eV}$ and $x=0.56/0.02585=21.66$. Since $e^{21.66}=2.556\times10^{9}$, $f=1/(2.556\times10^{9}+1)=3.91\times10^{-10}$. The Boltzmann factor is $e^{-21.66}=3.91\times10^{-10}$; because $e^x\gg1$, the $1$ in the denominator contributes only a relative error of $10^{-9}$, so the two are practically identical.

(2) Germanium $E_{\mathrm g}/2=0.33\ \mathrm{eV}$, $x=12.77$, and $f=2.86\times10^{-6}$ (the Boltzmann factor is $e^{-12.77}=2.86\times10^{-6}$). The narrower the band gap, the more easily electrons are excited; the value is about $7000$ times larger than for silicon.

Probability of a hole The top of the valence band lies $E_{\mathrm g}/2$ below $\mu$. The probability that it is empty is $1-f(\mu-E_{\mathrm g}/2)$, and by the symmetry $f(\mu+\delta)+f(\mu-\delta)=1$ (Section 52.5), $1-f(\mu-\delta)=f(\mu+\delta)$. That is, for $\delta=E_{\mathrm g}/2$ it has the same value as the occupation probability of the bottom of the conduction band ($3.9\times10^{-10}$ for Si and $2.9\times10^{-6}$ for Ge). Each time one electron is excited into the conduction band, one hole is created in the valence band, so this is a natural result.

Solution

(1) Let $x=(E_{\mathrm d}-\mu)/k_{\mathrm B}T$. There are three states: empty (electron number $0$, energy $0$), and one electron with spin up or spin down (electron number $1$, energy $E_{\mathrm d}$, two ways). The grand partition function is

$$ \Xi_{\mathrm d}=1+2e^{-\beta(E_{\mathrm d}-\mu)}=1+2e^{-x} $$

The probability that one electron is present is $2e^{-x}/\Xi_{\mathrm d}$, and the electron number in that case is $1$, so

$$ \langle n_{\mathrm d}\rangle=\frac{2e^{-x}}{1+2e^{-x}}=\frac{1}{1+\tfrac12e^{x}} $$

(here we divided the numerator and denominator by $2e^{-x}$). The same result is obtained from $\langle n_{\mathrm d}\rangle=k_{\mathrm B}T\,\partial\ln\Xi_{\mathrm d}/\partial\mu$.

(2) $x=0.030/0.02585=1.160$ and $e^x=3.191$, so

$$ \langle n_{\mathrm d}\rangle=\frac1{1+\tfrac12\times3.191}=\frac1{1+1.596}=0.385 $$

The ordinary Fermi distribution, which does not take the spin degeneracy into account, gives $1/(e^x+1)=1/(3.191+1)=0.239$, so the value that counts the degeneracy is about $1.6$ times larger.

(3) Because there are two ways for an electron to occupy the donor level, spin up and spin down, the weight of the “electron present” state is doubled, and the level is more easily occupied. Since this can be written as $\tfrac12e^{x}=e^{x-\ln2}$, it has the same effect as decreasing $E_{\mathrm d}-\mu$ by $k_{\mathrm B}T\ln2=0.018\ \mathrm{eV}$. The ordinary Fermi distribution is the probability that a single state with a specified spin is occupied; for a level like a donor, which “can accept an electron in only one of the two spin states (not in both at once),” this factor of $\tfrac12$ appears.

Solution

(1) For a square of side $L$ (area $L^2$), the allowed wavenumbers are $k_x=n_x\pi/L$ and $k_y=n_y\pi/L$ ($n_x,n_y=1,2,\ldots$), and one lattice point in wavenumber space occupies an area $(\pi/L)^2$. The states whose wavenumber magnitude is at most $k$ are the lattice points in the $1/4$ disk of the first quadrant (area $\pi k^2/4$), so the number of orbitals is $\dfrac{\pi k^2/4}{(\pi/L)^2}=\dfrac{L^2k^2}{4\pi}$. Multiplying by the $2$ spin states gives $N(\varepsilon)=\dfrac{L^2k^2}{2\pi}$. Substituting $k^2=2m\varepsilon/\hbar^2$ gives $N(\varepsilon)=\dfrac{mL^2}{\pi\hbar^2}\varepsilon$, a linear function of $\varepsilon$. Differentiating gives $g(\varepsilon)=\dfrac{\dd N}{\dd\varepsilon}=\dfrac{mL^2}{\pi\hbar^2}$ (constant).

(2) At $T=0$, $N=\int_0^{E_{\mathrm F}}g\dd\varepsilon=gE_{\mathrm F}$. From $N=n_2L^2$, $n_2L^2=\dfrac{mL^2}{\pi\hbar^2}E_{\mathrm F}$, that is, $E_{\mathrm F}=\dfrac{\pi\hbar^2n_2}m$ (here $k_{\mathrm F}^2=2\pi n_2$).

(3) Let $z=e^{\mu/k_{\mathrm B}T}$ and $y=e^{-\varepsilon/k_{\mathrm B}T}$. Then $f=\dfrac1{e^{(\varepsilon-\mu)/k_{\mathrm B}T}+1}=\dfrac{zy}{1+zy}$ (multiplying the numerator and denominator by $y$), and $\dd\varepsilon=-k_{\mathrm B}T\,\dd y/y$. Since $\varepsilon:0\to\infty$ corresponds to $y:1\to0$,

$$ \int_0^\infty f\dd\varepsilon=k_{\mathrm B}T\int_0^1\frac{z}{1+zy}\dd y=k_{\mathrm B}T\bigl[\ln(1+zy)\bigr]_0^1=k_{\mathrm B}T\ln\bigl(1+e^{\mu/k_{\mathrm B}T}\bigr) $$

From $N=g\int f\dd\varepsilon=gE_{\mathrm F}$ we get $k_{\mathrm B}T\ln(1+e^{\mu/k_{\mathrm B}T})=E_{\mathrm F}$. Solving $e^{\mu/k_{\mathrm B}T}=e^{E_{\mathrm F}/k_{\mathrm B}T}-1$ gives $\mu=k_{\mathrm B}T\ln\bigl(e^{E_{\mathrm F}/k_{\mathrm B}T}-1\bigr)$. At low temperatures, $\mu=E_{\mathrm F}+k_{\mathrm B}T\ln(1-e^{-E_{\mathrm F}/k_{\mathrm B}T})\approx E_{\mathrm F}-k_{\mathrm B}Te^{-E_{\mathrm F}/k_{\mathrm B}T}$, and unlike the three-dimensional case (a shift proportional to $T^2$, Section 52.9), the shift is exponentially small (because the density of states is constant, the $T^2$ correction to $\mu$ vanishes since $g'(E_{\mathrm F})=0$). For example, for $k_{\mathrm B}T=0.5E_{\mathrm F}$, $\mu=0.927E_{\mathrm F}$, and for $k_{\mathrm B}T=E_{\mathrm F}$, $\mu=0.541E_{\mathrm F}$.

(4) $n_2=3\times10^{11}\ \mathrm{cm^{-2}}=3\times10^{15}\ \mathrm{m^{-2}}$ and $m^*=0.067\times9.109\times10^{-31}=6.10\times10^{-32}\ \mathrm{kg}$. $E_{\mathrm F}=\pi\hbar^2n_2/m^*=\pi\times(1.0546\times10^{-34})^2\times3\times10^{15}/6.10\times10^{-32}\ \mathrm J=1.72\times10^{-21}\ \mathrm J=10.7\ \mathrm{meV}$. $T_{\mathrm F}=E_{\mathrm F}/k_{\mathrm B}=1.72\times10^{-21}/1.381\times10^{-23}=124\ \mathrm K$. Therefore the gas is strongly degenerate at liquid-helium temperature ($4.2\ \mathrm K$), whereas at room temperature ($300\ \mathrm K>T_{\mathrm F}$) it is no longer degenerate.

Solution

(1) The number density of atoms is $\dfrac{2.70\times10^{3}\ \mathrm{kg/m^3}}{26.98\times10^{-3}\ \mathrm{kg/mol}}\times6.022\times10^{23}\ \mathrm{mol^{-1}}=6.03\times10^{28}\ \mathrm{m^{-3}}$. Multiplying by the number of valence electrons, $3$, gives the electron number density $n=1.81\times10^{29}\ \mathrm{m^{-3}}$.

(2) $k_{\mathrm F}=(3\pi^2n)^{1/3}=(3\times9.870\times1.808\times10^{29})^{1/3}=(5.353\times10^{30})^{1/3}=1.749\times10^{10}\ \mathrm{m^{-1}}$. $E_{\mathrm F}=\dfrac{\hbar^2k_{\mathrm F}^2}{2m}=\dfrac{(1.0546\times10^{-34})^2\times(1.749\times10^{10})^2}{2\times9.109\times10^{-31}}=1.867\times10^{-18}\ \mathrm J=11.7\ \mathrm{eV}$. $T_{\mathrm F}=E_{\mathrm F}/k_{\mathrm B}=1.867\times10^{-18}/1.381\times10^{-23}=1.35\times10^5\ \mathrm K$.

(3) For $1\ \mathrm{mol}$ of atoms, $N=3N_{\mathrm A}$, so $Nk_{\mathrm B}=3R$. $\gamma=\dfrac{\pi^2}2\dfrac{3R}{T_{\mathrm F}}=4.935\times\dfrac{3\times8.314}{1.353\times10^5}=9.10\times10^{-4}\ \mathrm{J/(mol\,K^2)}=0.91\ \mathrm{mJ/(mol\,K^2)}$. The electronic specific heat at $T=300\ \mathrm K$ is $\gamma T=0.273\ \mathrm{J/(mol\,K)}$. This is $0.273/37.4=0.73\,\%$ of the classical value $\tfrac32\times3R=37.4\ \mathrm{J/(mol\,K)}$. It is smaller than for copper ($1.2\,\%$, Example 52.16) because aluminum has a higher $T_{\mathrm F}$ (a higher electron density), so $T/T_{\mathrm F}$ is smaller and the degeneracy is stronger.

Solution

Taking the ratio of the two formulas in the hint, $m,n,\hbar,k_{\mathrm B}$ cancel and we obtain

$$ \frac{T_{\mathrm c}}{T_{\mathrm F}}=\frac{2\pi\,\zeta(\tfrac32)^{-2/3}}{(3\pi^2)^{2/3}/2}=\frac{4\pi}{\bigl(3\pi^2\zeta(\tfrac32)\bigr)^{2/3}} $$

Numerically, $3\pi^2\zeta(\tfrac32)=3\times9.8696\times2.6124=77.35$, $77.35^{2/3}=18.15$ ($77.35^{1/3}=4.26$, squared), and $4\pi=12.57$, so $T_{\mathrm c}/T_{\mathrm F}=12.57/18.15=0.692$.

For helium-3 ($m=3.016\times1.6605\times10^{-27}=5.008\times10^{-27}\ \mathrm{kg}$) with $n=1.6\times10^{28}\ \mathrm{m^{-3}}$, $(3\pi^2n)^{2/3}=(4.738\times10^{29})^{2/3}=6.08\times10^{19}\ \mathrm{m^{-2}}$ and $\dfrac{\hbar^2}{2mk_{\mathrm B}}=\dfrac{(1.0546\times10^{-34})^2}{2\times5.008\times10^{-27}\times1.381\times10^{-23}}=8.04\times10^{-20}\ \mathrm{m^2\,K}$, so $T_{\mathrm F}=8.04\times10^{-20}\times6.08\times10^{19}\ \mathrm K=4.9\ \mathrm K$ (an estimate that treats it as a free Fermi gas).

Conclusion: for the same mass and density, the degeneracy temperature $T_{\mathrm F}$ of a Fermi gas and the condensation temperature $T_{\mathrm c}$ of a Bose gas are of the same order of magnitude ($T_{\mathrm c}\approx0.7\,T_{\mathrm F}$), because in both cases degeneracy begins at the temperature at which “the thermal de Broglie wavelength $\lambda_{\mathrm T}$ becomes comparable to the interparticle distance $n^{-1/3}$” ($n\lambda_{\mathrm T}^3\sim1$). The temperature at which degeneracy begins is determined independently of the type of particle, and the difference between the types shows up in the behavior after degeneracy sets in (fermions: a Fermi sphere with a step-function distribution; bosons: condensation).

Solution

(1) Abbreviate $N_R,V_R,E_R$ as $N,V,E$. Then $\ln\Omega_R=S_R/k_{\mathrm B}=N\bigl[\ln V+\tfrac32\ln E-\tfrac52\ln N+\tfrac32\ln\tfrac{4\pi m}{3h^2}+\tfrac52\bigr]$ (we used $\ln(V/N)=\ln V-\ln N$ and $\tfrac32\ln\tfrac{4\pi mE}{3Nh^2}=\tfrac32\ln E-\tfrac32\ln N+\tfrac32\ln\tfrac{4\pi m}{3h^2}$). Differentiating with respect to $E$ gives

$$ \left(\pdiff{\ln\Omega_R}{E}\right)_{N}=\frac{3N}{2E} $$

Differentiating with respect to $N$, by the product rule we get the contents of $[\ ]$ itself plus $N\times\bigl(-\tfrac5{2N}\bigr)=-\tfrac52$, and the latter cancels the $+\tfrac52$, so that

$$ \left(\pdiff{\ln\Omega_R}{N}\right)_{E}=\ln V+\frac32\ln E-\frac52\ln N+\frac32\ln\frac{4\pi m}{3h^2}=\ln\left\{\frac VN\left(\frac{4\pi mE}{3Nh^2}\right)^{3/2}\right\} $$

(2) Substituting $E=\tfrac32Nk_{\mathrm B}T$, the first equation becomes $\dfrac{3N}{2\cdot\frac32Nk_{\mathrm B}T}=\dfrac1{k_{\mathrm B}T}=\beta$. In the second equation, $\dfrac{4\pi mE}{3Nh^2}=\dfrac{4\pi m\cdot\frac32Nk_{\mathrm B}T}{3Nh^2}=\dfrac{2\pi mk_{\mathrm B}T}{h^2}=\dfrac1{\lambda_{\mathrm T}^2}$, so the argument of the logarithm is $\dfrac VN\dfrac1{\lambda_{\mathrm T}^3}=\dfrac1{n\lambda_{\mathrm T}^3}$, and the second equation is therefore $-\ln(n\lambda_{\mathrm T}^3)=-\beta\mu$ ($\mu=k_{\mathrm B}T\ln(n\lambda_{\mathrm T}^3)$). The first-order term of the Taylor expansion in Section 52.1 is $-E_{j,N}\times\beta-N\times(-\beta\mu)=-\beta(E_{j,N}-\mu N)$, so we obtain $P_{j,N}\propto\Omega_R\propto\exp\bigl[-\beta(E_{j,N}-\mu N)\bigr]$.

(3) If $E_{j,N}=0$ and $N=1$, the change in $\ln\Omega_R$ is $-1\times(-\beta\mu)=\beta\mu$, so the number of states of the reservoir is multiplied by $e^{\beta\mu}=n\lambda_{\mathrm T}^3$. For helium-4 gas ($300\ \mathrm K$, $1\ \mathrm{atm}$), the value in Table 52.3 is $n\lambda_{\mathrm T}^3=3.1\times10^{-6}$ ($\mu=-12.7\,k_{\mathrm B}T$). The number of states of the reservoir is multiplied by about $3\times10^{-6}$; that is, it is reduced to about one part in 320,000 ($32$ × 10,000). When the reservoir gives up one particle, the states (places) available to it decrease by that much, so a state in which system $s$ contains a particle is harder to realize than the empty state.

Solution

(1) There are four pairs of occupation numbers $(n_1,n_2)$ ($n_i=0,1$). $(0,0)$: $N=0$, $E=0$, weight $1$. $(1,0)$: $N=1$, $E=0$, weight $z$. $(0,1)$: $N=1$, $E=\varepsilon$, weight $zx$. $(1,1)$: $N=2$, $E=\varepsilon$, weight $z^2x$ ($e^{-\beta(E-\mu N)}=e^{-\beta\varepsilon}e^{2\beta\mu}$). Therefore the grand partition function factorizes as

$$ \Xi=1+z+zx+z^2x=(1+z)+zx(1+z)=(1+z)(1+zx) $$

This is the product $\Xi=\Xi_1\Xi_2$ of the grand partition functions of the individual single-particle states, $\Xi_1=1+z$ (energy $0$) and $\Xi_2=1+zx=1+e^{-\beta(\varepsilon-\mu)}$; this corresponds to $\Xi=\prod_i\Xi_i$ in Section 52.4.

(2) $\ln\Xi=\ln(1+z)+\ln(1+zx)$. Since $\partial/\partial\mu=\beta z\,\partial/\partial z$, $\langle N\rangle=k_{\mathrm B}T\,\partial\ln\Xi/\partial\mu=z\,\partial\ln\Xi/\partial z=\dfrac z{1+z}+\dfrac{zx}{1+zx}$. Dividing the numerator and denominator by $z$ (by $zx$ for the second term) gives $\dfrac1{z^{-1}+1}+\dfrac1{(zx)^{-1}+1}$. Since $z^{-1}=e^{-\beta\mu}=e^{\beta(0-\mu)}$ and $(zx)^{-1}=e^{\beta(\varepsilon-\mu)}$, this is exactly $f(0)+f(\varepsilon)$.

(3) For $k_{\mathrm B}T=\varepsilon$, $\beta\varepsilon=1$, and for $\mu=\varepsilon/2$, $z=e^{1/2}=1.6487$, $x=e^{-1}=0.36788$, and $zx=0.60653$. $f(0)=z/(1+z)=1.6487/2.6487=0.6225$ and $f(\varepsilon)=zx/(1+zx)=0.60653/1.60653=0.3775$. The total is $\langle N\rangle=0.6225+0.3775=1.0000$. It comes out to exactly $1$ because $\mu=\varepsilon/2$ lies midway between the two levels, so the symmetry $f(\mu+\delta)+f(\mu-\delta)=1$ (Section 52.5, with $\delta=\varepsilon/2$) can be used. The probability that both levels are occupied is obtained by dividing the weight of the state $(1,1)$, $z^2x=e\times e^{-1}=1.0000$, by $\Xi=(1+z)(1+zx)=2.6487\times1.6065=4.2552$, which gives $0.2350$. This agrees with $f(0)f(\varepsilon)=0.6225\times0.3775=0.2350$, which shows that the two levels are independent.

Solution

(1) By the same procedure as in Example 52.2, $\Xi=\sum_Nz^NZ_N=\sum_N\dfrac{(zZ_1)^N}{N!}=e^{zZ_1}=\exp\bigl(zS/\lambda_{\mathrm T}^2\bigr)$ ($z=e^{\beta\mu}$).

(2) $\ln\Xi=zS/\lambda_{\mathrm T}^2$. Since $\partial z/\partial\mu=\beta z$, $N=k_{\mathrm B}T\,\partial\ln\Xi/\partial\mu=k_{\mathrm B}T\cdot\beta zS/\lambda_{\mathrm T}^2=zS/\lambda_{\mathrm T}^2$. Therefore $z=e^{\beta\mu}=N\lambda_{\mathrm T}^2/S=n_2\lambda_{\mathrm T}^2$ and $\mu=k_{\mathrm B}T\ln(n_2\lambda_{\mathrm T}^2)$.

(3) $J=-k_{\mathrm B}T\ln\Xi=-k_{\mathrm B}T\,zS/\lambda_{\mathrm T}^2=-k_{\mathrm B}TN$. On the other hand, $J=-\Pi S$, so $\Pi S=Nk_{\mathrm B}T$, that is, $\Pi=n_2k_{\mathrm B}T$ (the equation of state in two dimensions). Since $J$ is proportional to $S$ when $T,\mu$ are fixed, the same result follows from $\Pi=-(\partial J/\partial S)_{T,\mu}=k_{\mathrm B}Tz/\lambda_{\mathrm T}^2=n_2k_{\mathrm B}T$. As for the units, $n_2k_{\mathrm B}T$ has units $\mathrm{m^{-2}\cdot J}=\mathrm{N/m}$, which agrees with a force per unit length.

(4) $m=39.948\times1.6605\times10^{-27}=6.634\times10^{-26}\ \mathrm{kg}$ and $\lambda_{\mathrm T}=1.595\times10^{-11}\ \mathrm m$ (the same as in Exercise 52.1). $n_2\lambda_{\mathrm T}^2=10^{16}\times(1.595\times10^{-11})^2=2.543\times10^{-6}$. Since $\ln(2.543\times10^{-6})=-12.88$, $\mu=-12.88\,k_{\mathrm B}T=-12.88\times0.02585\ \mathrm{eV}=-0.333\ \mathrm{eV}$. The surface pressure is $\Pi=n_2k_{\mathrm B}T=10^{16}\times4.142\times10^{-21}\ \mathrm{N/m}=4.14\times10^{-5}\ \mathrm{N/m}$ ($0.041\ \mathrm{mN/m}$).

Solution

(1) With $\beta=1/k_{\mathrm B}T$, the exponent is $\beta(\varepsilon-\mu)$. If $\varepsilon\lt\mu$, then $\varepsilon-\mu\lt0$, so as $\beta\to\infty$, $\beta(\varepsilon-\mu)\to-\infty$ and $e^{\beta(\varepsilon-\mu)}\to0$, and therefore $f\to1/(0+1)=1$. If $\varepsilon\gt\mu$, then $\beta(\varepsilon-\mu)\to+\infty$, $e^{\beta(\varepsilon-\mu)}\to\infty$, and $f\to0$. At $\varepsilon=\mu$, the exponent is $0$ at any $T$, so $e^0=1$ and $f=\tfrac12$ (the midpoint of the jump of the step).

(2) Compute $x=(\varepsilon/\mu-1)/(k_{\mathrm B}T/\mu)$ and substitute it into $f=1/(e^x+1)$.

$k_{\mathrm B}T/\mu$	$\varepsilon/\mu=0.9$	$0.99$	$1.01$	$1.1$
$0.1$ ($x=-1,\,-0.1,\,+0.1,\,+1$)	$0.7311$	$0.5250$	$0.4750$	$0.2689$
$0.01$ ($x=-10,\,-1,\,+1,\,+10$)	$0.99995$	$0.7311$	$0.2689$	$4.5\times10^{-5}$

If the temperature is reduced to $1/10$, the values at the same distance from $\mu$ ($\varepsilon/\mu=0.9$ or $1.1$) stick to $1$ and $0$: $0.73\to0.99995$ and $0.27\to4.5\times10^{-5}$. The width of the transition (from $f=0.9$ to $0.1$; $4.4\,k_{\mathrm B}T$) also becomes $1/10$, and as $T\to0$ the distribution becomes a step function. Moreover, $f(\mu+\delta)+f(\mu-\delta)=1$ (for example, $0.7311+0.2689=1$) holds for every pair of columns in the table.

Solution

(1) The mean number of particles is $N=\int_0^\infty g(\varepsilon)\langle n_\varepsilon\rangle\dd\varepsilon$ ($\langle n_\varepsilon\rangle=1/(e^{(\varepsilon-\mu)/k_{\mathrm B}T}-1)$). Changing the variable to $x=\varepsilon/k_{\mathrm B}T$ and using the integral given in the Hint, we obtain

$$ N=\frac{Am}{2\pi\hbar^2}\,k_{\mathrm B}T\int_0^\infty\frac{\dd x}{z^{-1}e^{x}-1}=\frac{Amk_{\mathrm B}T}{2\pi\hbar^2}\bigl[-\ln(1-z)\bigr] $$

Since $\lambda_{\mathrm T}^2=2\pi\hbar^2/(mk_{\mathrm B}T)$ (using $h=2\pi\hbar$, this is the same as $\lambda_{\mathrm T}^2=h^2/(2\pi mk_{\mathrm B}T)$), it follows that

$$ n=\frac NA=\frac{-\ln(1-z)}{\lambda_{\mathrm T}^2} $$

(The integral given in the Hint is evaluated by noting that $\dfrac{1}{z^{-1}e^x-1}=\dfrac{ze^{-x}}{1-ze^{-x}}$ is equal to $\dfrac{\dd}{\dd x}\ln(1-ze^{-x})$, so that $\bigl[\ln(1-ze^{-x})\bigr]_0^\infty=0-\ln(1-z)$.)

(2) From $-\ln(1-z)=n\lambda_{\mathrm T}^2$ we obtain

$$ z=1-\exp\left(-n\lambda_{\mathrm T}^2\right) $$

For $n\lambda_{\mathrm T}^2\ll1$ we have $z\approx n\lambda_{\mathrm T}^2$ (the same form as the result for a classical ideal gas).

(3) For any finite values of $n$ and $T$, $n\lambda_{\mathrm T}^2$ is finite, so $z=1-e^{-n\lambda_{\mathrm T}^2}\lt1$, that is, $\mu\lt0$. The condition $z\to1$ ($\mu\to0$) is reached only when $n\lambda_{\mathrm T}^2\to\infty$, that is, in the limit $T\to0$. In three dimensions, even when $\mu\to0$, the upper limit $\int g(\varepsilon)/(e^{\varepsilon/k_{\mathrm B}T}-1)\dd\varepsilon$ on the number of particles that can enter the excited states is finite ($\propto\zeta(3/2)$), so the particles in excess of it fall into the ground state (condensation). In two dimensions, the density of states is constant, and $\int\dd\varepsilon/\varepsilon$ diverges logarithmically as $\varepsilon\to0$, so as $\mu\to0$ the excited states alone can accommodate any number of particles. Therefore, at any density the excited states can take in all the particles, and a macroscopic number of particles collecting in the ground state at a finite temperature (condensation) does not occur.

(4) For $n\lambda_{\mathrm T}^2=1$, $z=1-e^{-1}=1-0.3679=0.632$ and $\mu/k_{\mathrm B}T=\ln z=-0.459$. Substituting the same $n\lambda_{\mathrm T}^2=1$ into the classical ideal-gas formula $z=n\lambda_{\mathrm T}^2$ would give $z=1$ ($\mu=0$), so for the Bose distribution $z$ is smaller than in the classical approximation ($\mu$ is lower).

Solution

(a) The chemical potential is $\mu=(\partial A/\partial N)_{T,V}$, that is, the change in the free energy $A$ when one particle is added at constant temperature and volume. For an ordinary gas the number of particles $N$ is given from outside, so $\mu$ is in general not $0$. However, the walls of the cavity freely absorb and emit photons, so the number of photons $N$ is not conserved, and in thermal equilibrium $N$ settles by itself so that the free energy is minimized. At the point where $A(T,V,N)$ is minimized as a function of $N$, $\partial A/\partial N=0$. This directly means $\mu=0$. Looking at it from another viewpoint, the partition function of a single mode, $\sum_{n=0}^\infty e^{-\beta\hbar\omega n}$, is a sum taken with no restriction whatsoever on the photon number $n$, and it has the form of the factor $e^{\beta\mu n}$ of the grand canonical distribution with $\mu=0$.

(b) The mean number of photons is $\langle n\rangle=1/(e^x-1)$, and the mean energy is $h\nu\langle n\rangle=k_{\mathrm B}T\cdot x/(e^x-1)$ ($x=h\nu/k_{\mathrm B}T$).

For $x=1$, with $e^1-1=1.71828$, we have $\langle n\rangle=0.5820$, and the mean energy is $1\times0.5820\,k_{\mathrm B}T=0.582\,k_{\mathrm B}T$.

For $x=5$, with $e^5-1=147.413$, we have $\langle n\rangle=6.78\times10^{-3}$, and the mean energy is $5\times6.78\times10^{-3}\,k_{\mathrm B}T=0.0339\,k_{\mathrm B}T$.

Compared with the equipartition value $k_{\mathrm B}T$, this is only about $58\%$ at $x=1$ and about $3.4\%$ at $x=5$. As $x\to0$, $x/(e^x-1)\to1$, which agrees with the equipartition value, so it has been confirmed that only the modes with $h\nu\ll k_{\mathrm B}T$ behave classically, while the modes with large $h\nu$ are frozen out.

Solution

(a) Substitute $\nu=c/\lambda$ into Planck's formula $u(\nu,T)\,\dd\nu=\dfrac{8\pi h\nu^3}{c^3}\dfrac{1}{e^{h\nu/k_{\mathrm B}T}-1}\dd\nu$. Since $\nu^3=c^3/\lambda^3$ and $\abs{\dd\nu}=(c/\lambda^2)\,\dd\lambda$ (differentiating $\nu=c/\lambda$ with respect to $\lambda$ gives $-c/\lambda^2$), we have

$$ \frac{8\pi h}{c^3}\cdot\frac{c^3}{\lambda^3}\cdot\frac{c}{\lambda^2}\,\dd\lambda=\frac{8\pi hc}{\lambda^5}\,\dd\lambda,\qquad\frac{h\nu}{k_{\mathrm B}T}=\frac{hc}{\lambda k_{\mathrm B}T} $$

Therefore $u(\lambda,T)\,\dd\lambda=\dfrac{8\pi hc}{\lambda^5}\dfrac{1}{e^{hc/(\lambda k_{\mathrm B}T)}-1}\dd\lambda$.

(b) The condition for $u(\nu,T)$ to be at its maximum is $x=3(1-e^{-x})$ ($x=h\nu/k_{\mathrm B}T$), and the solution is $x_\nu=2.8214$. Hence

$$ \nu_{\max}=\frac{x_\nu k_{\mathrm B}T}{h}=\frac{2.8214\times1.380649\times10^{-23}\times5772}{6.62607\times10^{-34}}=3.393\times10^{14}\ \mathrm{Hz} $$

The corresponding wavelength is $c/\nu_{\max}=2.998\times10^{8}/3.393\times10^{14}=8.83\times10^{-7}\ \mathrm m=883\ \mathrm{nm}$ (near infrared). On the other hand, the peak of $u(\lambda,T)$, from the displacement law, is at $\lambda_{\max}=b/T=2.898\times10^{-3}/5772=5.02\times10^{-7}\ \mathrm m=502\ \mathrm{nm}$ (visible light). The values $883\ \mathrm{nm}$ and $502\ \mathrm{nm}$ do not coincide. Their ratio is $883/502=1.76$, which agrees with $x_\lambda/x_\nu=4.965/2.821=1.760$. Even for the same spectrum, the position of the peak changes depending on whether it is viewed per unit frequency or per unit wavelength, because of the Jacobian $\dd\nu=(c/\lambda^2)\,\dd\lambda$.

Solution

(a) Let $x=h\nu/(k_{\mathrm B}T)$. Then $\nu=(k_{\mathrm B}T/h)x$, $\dd\nu=(k_{\mathrm B}T/h)\dd x$, and the range of integration remains $0\to\infty$.

$$ u(T)=\frac{8\pi h}{c^3}\int_0^\infty\frac{\nu^3\,\dd\nu}{e^{h\nu/k_{\mathrm B}T}-1}=\frac{8\pi h}{c^3}\left(\frac{k_{\mathrm B}T}{h}\right)^4\int_0^\infty\frac{x^3}{e^x-1}\dd x=\frac{8\pi(k_{\mathrm B}T)^4}{h^3c^3}\cdot\Gamma(4)\zeta(4) $$

By Formula 47.4 of Chapter 47 ($p=3$), $\Gamma(4)\zeta(4)=3!\times\pi^4/90=\pi^4/15$. Hence $u(T)=\dfrac{8\pi^5k_{\mathrm B}^4}{15h^3c^3}T^4=aT^4$. The reason $T^4$ comes out of the integral is that $(k_{\mathrm B}T/h)^3\cdot(k_{\mathrm B}T/h)$ comes out of $\nu^3\,\dd\nu$, and the remaining integral over $x$ is a number independent of the temperature.

(b) $\sigma=\dfrac c4a=\dfrac c4\cdot\dfrac{8\pi^5k_{\mathrm B}^4}{15h^3c^3}=\dfrac{2\pi^5k_{\mathrm B}^4}{15h^3c^2}$. Numerically, with $a=7.566\times10^{-16}\ \mathrm{J\,m^{-3}K^{-4}}$ and $c/4=7.4948\times10^{7}\ \mathrm{m/s}$, we get $\sigma=7.566\times10^{-16}\times7.4948\times10^{7}=5.670\times10^{-8}\ \mathrm{W\,m^{-2}K^{-4}}$.

(c) $j=\sigma T^4=5.670\times10^{-8}\times(300)^4=5.670\times10^{-8}\times8.1\times10^{9}=459\ \mathrm{W/m^2}$. That is, about $459\ \mathrm W$ per $1\ \mathrm{m^2}$. If the temperature rises by $10\%$, then $T\to1.1\,T$ and $j\to1.1^4j=1.4641\,j$, that is, about $1.46$ times as large (a $46\%$ increase), or about $672\ \mathrm{W/m^2}$. A slight rise in temperature greatly increases the radiated power.

Solution

The number of photons per unit volume is obtained by summing “the number of modes” × “the mean number of photons per mode” and dividing by the volume:

$$ n_\gamma=\frac1V\int_0^\infty g(\omega)\langle n\rangle\dd\omega=\frac{1}{\pi^2c^3}\int_0^\infty\frac{\omega^2\,\dd\omega}{e^{\hbar\omega/k_{\mathrm B}T}-1} $$

Here we used $g(\omega)=V\omega^2/(\pi^2c^3)$ and $\langle n\rangle=1/(e^{\hbar\omega/k_{\mathrm B}T}-1)$. Setting $x=\hbar\omega/k_{\mathrm B}T$, we have $\omega^2\dd\omega=(k_{\mathrm B}T/\hbar)^3x^2\dd x$, and by Formula 47.4 ($p=2$), $\int_0^\infty x^2/(e^x-1)\,\dd x=\Gamma(3)\zeta(3)=2\zeta(3)$, so

$$ n_\gamma=\frac{1}{\pi^2c^3}\left(\frac{k_{\mathrm B}T}{\hbar}\right)^3\cdot2\zeta(3)=\frac{2\zeta(3)}{\pi^2}\left(\frac{k_{\mathrm B}T}{\hbar c}\right)^3 $$

The coefficient is $2\zeta(3)/\pi^2=2\times1.2021/9.8696=0.2436$.

(a) $T=300\ \mathrm K$: $k_{\mathrm B}T=4.142\times10^{-21}\ \mathrm J$, $\hbar c=1.0546\times10^{-34}\times2.998\times10^{8}=3.1615\times10^{-26}\ \mathrm{J\,m}$, $k_{\mathrm B}T/(\hbar c)=1.310\times10^{5}\ \mathrm{m^{-1}}$, and its cube is $2.249\times10^{15}$, so $n_\gamma=0.2436\times2.249\times10^{15}=5.5\times10^{14}\ \mathrm{m^{-3}}$.

(b) $T=5772\ \mathrm K$: $k_{\mathrm B}T=7.969\times10^{-20}\ \mathrm J$, $k_{\mathrm B}T/(\hbar c)=2.521\times10^{6}\ \mathrm{m^{-1}}$, and the cube is $1.602\times10^{19}$, so $n_\gamma=0.2436\times1.602\times10^{19}=3.9\times10^{18}\ \mathrm{m^{-3}}$.

(c) Since $u=aT^4=\dfrac{\pi^2(k_{\mathrm B}T)^4}{15\hbar^3c^3}$ (we rewrote $a=8\pi^5k_{\mathrm B}^4/(15h^3c^3)$ using $h=2\pi\hbar$),

$$ \frac{u}{n_\gamma}=\frac{\pi^2(k_{\mathrm B}T)^4}{15\hbar^3c^3}\cdot\frac{\pi^2}{2\zeta(3)}\frac{(\hbar c)^3}{(k_{\mathrm B}T)^3}=\frac{\pi^4}{30\,\zeta(3)}k_{\mathrm B}T=\frac{97.409}{30\times1.2021}k_{\mathrm B}T=2.70\,k_{\mathrm B}T $$

The mean energy per photon is about $2.7$ times $k_{\mathrm B}T$, independent of the temperature.

Solution

(a) From $\displaystyle\int_0^{\omega_{\mathrm D}}C\omega^2\,\dd\omega=\dfrac{C\omega_{\mathrm D}^3}{3}=3N$ we get $C=\dfrac{9N}{\omega_{\mathrm D}^3}$. Hence $g(\omega)=9N\omega^2/\omega_{\mathrm D}^3$ ($0\le\omega\le\omega_{\mathrm D}$).

(b) Using $c_{\mathrm l}=6.0\ \mathrm{km/s}$ and $c_{\mathrm t}=3.0\ \mathrm{km/s}$ in units of $\mathrm{km/s}$, we have

$$ \frac{3}{c_s^3}=\frac1{6.0^3}+\frac{2}{3.0^3}=4.630\times10^{-3}+7.407\times10^{-2}=7.870\times10^{-2}\ (\mathrm{km/s})^{-3} $$

so $c_s^3=38.12\ (\mathrm{km/s})^3$ and $c_s=3.365\ \mathrm{km/s}=3.365\times10^{3}\ \mathrm{m/s}$. The upper limit of the wavenumber is $k_{\mathrm D}=(6\pi^2n)^{1/3}=(59.22\times6.0\times10^{28})^{1/3}=(3.553\times10^{30})^{1/3}=1.526\times10^{10}\ \mathrm{m^{-1}}$. Therefore

$$ \omega_{\mathrm D}=c_sk_{\mathrm D}=3.365\times10^{3}\times1.526\times10^{10}=5.14\times10^{13}\ \mathrm{rad/s} $$ $$ \theta_{\mathrm D}=\frac{\hbar\omega_{\mathrm D}}{k_{\mathrm B}}=\frac{1.0546\times10^{-34}\times5.14\times10^{13}}{1.3806\times10^{-23}}=3.9\times10^{2}\ \mathrm K $$

(about $392\ \mathrm K$). The shortest wavelength is $2\pi/k_{\mathrm D}=2\pi/(1.526\times10^{10})=4.1\times10^{-10}\ \mathrm m=0.41\ \mathrm{nm}$, which is about 1.6 times the interatomic spacing $n^{-1/3}=0.26\ \mathrm{nm}$.

Solution

(a) We have $U(T)=9Nk_{\mathrm B}T\,(T/\theta_{\mathrm D})^3\int_0^{\theta_{\mathrm D}/T}x^3/(e^x-1)\,\dd x$, and since $T\ll\theta_{\mathrm D}$, we replace the upper limit by $\infty$. The integrand becomes small for large $x$, like $x^3e^{-x}$, so the error is negligible (about $1\%$ at $\theta_{\mathrm D}/T=10$). From Formula 47.4 ($p=3$) the integral is $\pi^4/15$, and

$$ U(T)=9Nk_{\mathrm B}T\left(\frac{T}{\theta_{\mathrm D}}\right)^3\frac{\pi^4}{15}=\frac{3\pi^4}{5}\,\frac{Nk_{\mathrm B}}{\theta_{\mathrm D}^3}\,T^4 $$

(since $9/15=3/5$). Differentiating with respect to $T$, we obtain $C_V=\dfrac{\dd U}{\dd T}=\dfrac{12\pi^4}{5}\dfrac{Nk_{\mathrm B}}{\theta_{\mathrm D}^3}T^3=\dfrac{12\pi^4}{5}Nk_{\mathrm B}\left(\dfrac T{\theta_{\mathrm D}}\right)^3$.

(b) For one mole, $Nk_{\mathrm B}=R$. Since $12\pi^4/5=233.78$, we have $\dfrac{12\pi^4}{5}R=233.78\times8.314=1944\ \mathrm{J/(mol\,K)}$. With $T/\theta_{\mathrm D}=4/400=0.01$,

$$ C_V=1944\times(0.01)^3=1.94\times10^{-3}\ \mathrm{J/(mol\,K)} $$

The quantity of heat required for heating is the increase in internal energy from $0\ \mathrm K$ to $4\ \mathrm K$, $\Delta U=U(4\ \mathrm K)$; substituting into $U=\frac{3\pi^4}{5}RT^4/\theta_{\mathrm D}^3$ gives

$$ \Delta U=58.44\times8.314\times\frac{(4)^4}{(400)^3}=485.9\times\frac{256}{6.4\times10^{7}}=1.94\times10^{-3}\ \mathrm{J/mol} $$

(with $\frac{3\pi^4}{5}=58.44$). From the relation between $U\propto T^4$ and $C_V\propto T^3$, we have $U=C_VT/4=1.94\times10^{-3}\times4/4$, which agrees. Only about $2\ \mathrm{mJ}$ of heat is enough to warm $1\ \mathrm{mol}$ to $4\ \mathrm K$.

Solution

We have $C_V=9Nk_{\mathrm B}y^{-3}\int_0^y x^4e^x/(e^x-1)^2\,\dd x$ ($y=\theta_{\mathrm D}/T$). First we expand $x^2e^x/(e^x-1)^2$. Since $e^x/(e^x-1)^2=1/(4\sinh^2(x/2))$ and $\sinh(x/2)=\frac x2\left(1+\frac{x^2}{24}+\cdots\right)$, we have $\sinh^2(x/2)=\frac{x^2}{4}\left(1+\frac{x^2}{12}+\cdots\right)$. Therefore

$$ \frac{x^2e^x}{(e^x-1)^2}=\frac{x^2}{4\sinh^2(x/2)}=\frac{1}{1+x^2/12+\cdots}=1-\frac{x^2}{12}+\cdots $$

The integrand is $x^4e^x/(e^x-1)^2=x^2\cdot\dfrac{x^2e^x}{(e^x-1)^2}=x^2-\dfrac{x^4}{12}+\cdots$, and integrating gives $\displaystyle\int_0^y\left(x^2-\frac{x^4}{12}\right)\dd x=\frac{y^3}{3}-\frac{y^5}{60}$. Hence

$$ C_V=9Nk_{\mathrm B}y^{-3}\left(\frac{y^3}{3}-\frac{y^5}{60}\right)=3Nk_{\mathrm B}\left(1-\frac{y^2}{20}\right)=3Nk_{\mathrm B}\left[1-\frac{1}{20}\left(\frac{\theta_{\mathrm D}}{T}\right)^2\right] $$

Numerical comparison: at $T=\theta_{\mathrm D}$ ($y=1$), the approximate value is $1-1/20=0.9500$, and the value by numerical integration is $0.9517$ (error $0.2\%$). At $T=2\theta_{\mathrm D}$ ($y=0.5$), the approximate value is $1-0.25/20=0.9875$, and the value by numerical integration is $0.9876$ (error $0.01\%$). Adding the next term $+y^4/560$ gives $0.9518$ at $y=1$, which nearly agrees with the value by numerical integration. It has been confirmed that the approximation is better at higher temperatures.

Solution

(a) If we impose periodic boundary conditions on a chain of length $L$, the wavenumbers are $k=2\pi j/L$ ($j$ is an integer), and the lattice points in the (one-dimensional) wavenumber space lie on a line at intervals of $2\pi/L$. The number of lattice points whose wavenumber magnitude is at most $K$ is the length $2K$ of the range $-K\le k\le K$ divided by the interval $2\pi/L$, namely $KL/\pi$; since there is only one polarization, the number of modes up to $\omega$ is $N(\omega)=L\omega/(\pi c_s)$. Therefore $g(\omega)=\dd N/\dd\omega=L/(\pi c_s)$ (constant). From the normalization condition $\int_0^{\omega_{\mathrm D}}g\,\dd\omega=N$ we get $L\omega_{\mathrm D}/(\pi c_s)=N$, that is, $\omega_{\mathrm D}=\pi c_sN/L=\pi c_s/a$ ($a=L/N$ is the interatomic spacing), so that

$$ g(\omega)=\frac{N}{\omega_{\mathrm D}}\quad(0\le\omega\le\omega_{\mathrm D}) $$

holds.

(b) $U=\displaystyle\int_0^{\omega_{\mathrm D}}\frac{N}{\omega_{\mathrm D}}\,\frac{\hbar\omega}{e^{\hbar\omega/k_{\mathrm B}T}-1}\,\dd\omega$. Setting $x=\hbar\omega/k_{\mathrm B}T$, $\theta_{\mathrm D}=\hbar\omega_{\mathrm D}/k_{\mathrm B}$, and $y=\theta_{\mathrm D}/T$, we have $\omega\,\dd\omega=(k_{\mathrm B}T/\hbar)^2x\,\dd x$, so

$$ U=\frac{N\hbar}{\omega_{\mathrm D}}\left(\frac{k_{\mathrm B}T}{\hbar}\right)^2\int_0^y\frac{x}{e^x-1}\dd x=Nk_{\mathrm B}T\cdot\frac{T}{\theta_{\mathrm D}}\,F_1(y),\qquad F_1(y)\equiv\int_0^y\frac{x}{e^x-1}\dd x $$

Differentiate with respect to $T$. Using $\dd y/\dd T=-y/T$ and $F_1'(y)=y/(e^y-1)$, we obtain

$$ C_V=\diff{U}{T}=Nk_{\mathrm B}\left[\frac{2T}{\theta_{\mathrm D}}F_1(y)+\frac{T^2}{\theta_{\mathrm D}}\cdot\frac{y}{e^y-1}\cdot\left(-\frac{y}{T}\right)\right]=Nk_{\mathrm B}\frac{T}{\theta_{\mathrm D}}\left[2F_1(y)-\frac{y^2}{e^y-1}\right] $$

By integration by parts, $\displaystyle\int_0^y\frac{x^2e^x}{(e^x-1)^2}\dd x=\left[-\frac{x^2}{e^x-1}\right]_0^y+2\int_0^y\frac{x}{e^x-1}\dd x=-\frac{y^2}{e^y-1}+2F_1(y)$, so the expression in the brackets is equal to this integral, giving

$$ C_V=Nk_{\mathrm B}\frac{T}{\theta_{\mathrm D}}\int_0^{\theta_{\mathrm D}/T}\frac{x^2e^x}{(e^x-1)^2}\dd x $$

as the result.

(c) At low temperatures ($T\ll\theta_{\mathrm D}$), we set the upper limit to $\infty$ and use Formula 47.5 ($p=2$), $\int_0^\infty x^2e^x/(e^x-1)^2\,\dd x=2!\,\zeta(2)=2\cdot\pi^2/6=\pi^2/3$, which gives $C_V=\dfrac{\pi^2}{3}Nk_{\mathrm B}\dfrac T{\theta_{\mathrm D}}$, proportional to the temperature (the $T^3$ of three dimensions becomes $T^1$ in one dimension). At high temperatures ($T\gg\theta_{\mathrm D}$), $y\ll1$ and the integrand $x^2e^x/(e^x-1)^2\approx1$, so the integral becomes $\int_0^y\dd x=y=\theta_{\mathrm D}/T$, and we obtain $C_V=Nk_{\mathrm B}\dfrac T{\theta_{\mathrm D}}\cdot\dfrac{\theta_{\mathrm D}}T=Nk_{\mathrm B}$. This is the Dulong–Petit value for a one-dimensional chain ($k_{\mathrm B}$ per atom, since there is one vibrational degree of freedom). Checking by numerical integration, the slope on the low-temperature side is $C_V/(Nk_{\mathrm B}T/\theta_{\mathrm D})\to3.2899=\pi^2/3$, and on the high-temperature side $C_V/Nk_{\mathrm B}\to0.99993$.

Solution

(a) With $y=\theta_{\mathrm D}/T$, we can write $\int_0^yx^3/(e^x-1)\,\dd x=\pi^4/15-\int_y^\infty x^3/(e^x-1)\,\dd x$. The error caused by setting the upper limit to $\infty$ is the second term on the right-hand side (the remaining integral). Since the integrand becomes exponentially small for large $x$, like $x^3e^{-x}$, the remaining integral is very small. By numerical integration, it is $0.062$ at $y=10$ (about $0.96\%$ of the total value $\pi^4/15=6.494$) and $1.9\times10^{-5}$ at $y=20$ (about $3\times10^{-6}$ times the total). $T\ll\theta_{\mathrm D}$ means that $y$ is large, so the error from setting the upper limit to $\infty$ becomes small rapidly as the temperature is lowered. Physically, this means that the modes of high frequency with $\hbar\omega\gg k_{\mathrm B}T$ (near $\omega_{\mathrm D}$) are not excited by thermal energy and are frozen out, so whether or not the upper limit $\omega_{\mathrm D}$ exists does not affect the low-temperature specific heat.

(b) In the Einstein model, all $3N$ modes have a single angular frequency $\omega_{\mathrm E}$ (the density of states is concentrated at the single point $\omega_{\mathrm E}$). Since the lowest excitation energy is the finite value $\hbar\omega_{\mathrm E}$ (a gap), for $k_{\mathrm B}T\ll\hbar\omega_{\mathrm E}$ the probability of excitation becomes exponentially small, in proportion to $e^{-\hbar\omega_{\mathrm E}/k_{\mathrm B}T}$, and the specific heat also goes to $0$ exponentially. In contrast, in the density of states of the Debye model, $g\propto\omega^2$, the modes continue all the way down to $\omega\to0$. However low the temperature, there exist long-wavelength modes satisfying $\hbar\omega\lesssim k_{\mathrm B}T$; their number is $\omega^3\propto T^3$, and the energy of each mode is of the order of $k_{\mathrm B}T$, so $U\propto T^4$ and $C_V\propto T^3$. In other words, the difference comes from the difference in the assumption about the density of states: it is concentrated at a single point (Einstein), or it is continuous down to $\omega\to0$ (Debye).

(c) The result is as in the following table.

Item	Photons	Phonons (Debye model)
Similarities	Each mode is a harmonic oscillator with energy $\hbar\omega(n+\tfrac12)$. They are bosons with $\mu=0$ whose number is not conserved. The mean occupation number is $1/(e^{\hbar\omega/k_{\mathrm B}T}-1)$. The calculation proceeds in three steps: “count the modes → Bose distribution → the integral of Chapter 47.”
Differences: type of wave	electromagnetic wave (transverse wave)	elastic wave (one longitudinal, two transverse)
Dispersion relation	$\omega=ck$	$\omega=c_sk$ (speed of sound $c_s$)
Number of polarizations	2	3
Total number of modes	infinite (no upper limit)	$3N$ (cut off at $\omega_{\mathrm D}$)
When the temperature is raised	keeps increasing, with $u\propto T^4$ and $C_V\propto T^3$	saturates at $C_V\to3Nk_{\mathrm B}$ (Dulong–Petit)
Where they exist	exist even in vacuum	exist only inside a crystal, and $\hbar k$ is the crystal momentum

(d) Chapter 49 of University Mathematics (AD 02) showed the road from the failure of the equipartition theorem to explain black-body radiation (the ultraviolet catastrophe) to Planck's radiation law via the quantum hypothesis, as material for mathematics such as the Boltzmann distribution, the counting of standing waves, and Fourier series. This chapter re-derives the same formula from the general theory of statistical mechanics, namely that “photons are bosons with $\mu=0$ whose number is not conserved,” and limits the counting of the density of states to a restatement of the key points. On that basis it treats the topics that are not in University Mathematics, namely the $T^4$ law for the total energy and the Stefan–Boltzmann constant (the integral of Chapter 47), the pressure and entropy of the photon gas, the cosmic microwave background, and phonons and the Debye model.

Solution

(a) The peak in terms of frequency is $\nu_{\max}=2.821\,k_{\mathrm B}T_0/h=2.821\times1.380649\times10^{-23}\times2.725/(6.62607\times10^{-34})=1.60\times10^{11}\ \mathrm{Hz}=160\ \mathrm{GHz}$. The peak in terms of wavelength follows from Wien's displacement law: $\lambda_{\max}=b/T_0=2.898\times10^{-3}\ \mathrm{m\,K}/2.725\ \mathrm K=1.06\times10^{-3}\ \mathrm m=1.06\ \mathrm{mm}$. Note that this does not coincide with $c/\nu_{\max}=1.87\ \mathrm{mm}$ (Section 53.4 of the main text).

(b) Substituting $k_{\mathrm B}T_0/(\hbar c)=1.190\times10^{3}\ \mathrm{m^{-1}}$ into $n_\gamma=\dfrac{2\zeta(3)}{\pi^2}\left(\dfrac{k_{\mathrm B}T_0}{\hbar c}\right)^3$, we obtain $n_\gamma=0.2436\times(1.190\times10^{3})^3=4.1\times10^{8}\ \mathrm{m^{-3}}=4.1\times10^{2}\ \mathrm{cm^{-3}}$ (about $410$ photons per $\mathrm{cm^3}$).

(c) When the length doubles, the wavelength of light is also stretched by a factor of two (redshift), and the temperature is halved because $T\propto1/s$: $T=T_0/2=1.3625\ \mathrm K$. The volume becomes $2^3=8$ times larger, and the total number of photons does not change (photons are neither created nor annihilated), so the photon number density becomes $1/8$: $n_\gamma\to51\ \mathrm{cm^{-3}}$. This also agrees with $n_\gamma\propto T^3$ ($(1/2)^3=1/8$). From $u=aT^4$, the energy density becomes $(1/2)^4=1/16$: $u=4.17\times10^{-14}/16=2.6\times10^{-15}\ \mathrm{J/m^3}$. This is also consistent with the fact that the number of photons becomes $1/8$ and the energy $h\nu$ of each photon becomes $1/2$, so that $u$ becomes $1/8\times1/2=1/16$.

Solution

$\sqrt{\kappa/m}=\sqrt{40/(1.0\times10^{-25})}\ \mathrm{s^{-1}}=2.0\times10^{13}\ \mathrm{s^{-1}}$, and the speed of sound is $c_s=a\sqrt{\kappa/m}=5.0\times10^{3}\ \mathrm{m/s}$ (Example 53.10).

(a) Since $ka=\pi/2$, $\sin(ka/2)=\sin(\pi/4)=1/\sqrt2$. The actual angular frequency is $\omega=2\sqrt{\kappa/m}\times\dfrac{1}{\sqrt2}=\sqrt2\times2.0\times10^{13}=2.83\times10^{13}\ \mathrm{rad/s}$.

(b) In the Debye approximation, $\omega=c_sk=a\sqrt{\kappa/m}\cdot\dfrac{\pi}{2a}=\dfrac\pi2\sqrt{\kappa/m}=1.5708\times2.0\times10^{13}=3.14\times10^{13}\ \mathrm{rad/s}$. The difference from the actual value is $3.14\times10^{13}-2.83\times10^{13}=3.1\times10^{12}\ \mathrm{rad/s}$, and as a fraction it is $\dfrac{\pi/2}{\sqrt2}-1=0.111$; that is, the Debye approximation overestimates by about $11\%$.

(c) The wavelength is $\lambda=2\pi/k=2\pi\cdot2a/\pi=4a=1.0\times10^{-9}\ \mathrm m=1.0\ \mathrm{nm}$ (four times the atomic spacing).

(d) At $k=\pi/a$ (wavelength $2a$) in Example 53.10(3), the Debye value $\pi\sqrt{\kappa/m}$ was about $57\%$ larger than the actual maximum $2\sqrt{\kappa/m}$. At $k=\pi/(2a)$ the difference is about $11\%$, and the smaller $k$ is (the longer the wavelength), the smaller the difference becomes. Indeed, for $ka\ll1$, $2\sin(ka/2)\approx ka$ and the difference approaches $0$ (about $1\%$ at $ka=0.5$). The Debye approximation is accurate for long-wavelength modes, which corresponds to the fact that it is the long-wavelength modes that determine the thermal properties at low temperatures.

Solution

In frame S (the coordinate system in which the air is at rest), the equation of motion is $m\ddot{\bm{r}}=-k\dot{\bm{r}}$. Since S′ moves with a constant velocity $\bm{v}$ relative to S, we use the Galilean transformation $\bm{r}'=\bm{r}-\bm{v}t$, that is, $\bm{r}=\bm{r}'+\bm{v}t$. Since $\bm{v}$ is a constant vector, differentiating both sides with respect to $t$ gives

$$ \dot{\bm{r}}=\dot{\bm{r}}'+\bm{v},\qquad\ddot{\bm{r}}=\ddot{\bm{r}}' $$

Substituting this into the equation of motion, the equation in frame S′ is

$$ m\ddot{\bm{r}}'=-k\left(\dot{\bm{r}}'+\bm{v}\right)=-k\dot{\bm{r}}'-k\bm{v} $$

The right-hand side has acquired a constant term $-k\bm{v}$ that was absent in frame S, so the equation does not take exactly the same form as in frame S, namely $m\ddot{\bm{r}}'=-k\dot{\bm{r}}'$. The only exception is $\bm{v}=\bm{0}$ (when S′ is at rest relative to the air).

However, this does not mean that “Galilean relativity is violated.” Seen from frame S′, the air moves with velocity $\bm{u}'_{\text{air}}=\bm{0}-\bm{v}=-\bm{v}$. If we therefore use the particle's “velocity relative to the air,” $\dot{\bm{r}}'-\bm{u}'_{\text{air}}=\dot{\bm{r}}'+\bm{v}$, the equation in frame S′ becomes

$$ m\ddot{\bm{r}}'=-k\left(\dot{\bm{r}}'-\bm{u}'_{\text{air}}\right) $$

This has the same form as the equation in frame S, $m\ddot{\bm{r}}=-k(\dot{\bm{r}}-\bm{u}_{\text{air}})$ ($\bm{u}_{\text{air}}=\bm{0}$). In other words, if the law of the resistive force is written not in terms of “the velocity of the particle” but in terms of “the velocity of the particle relative to the medium (air),” and the air is transformed along with it as part of the physical system, then the form of the equation of motion is unchanged by the Galilean transformation. Even when the force depends on velocity, Galilean relativity holds as long as what matters is the relative velocity. (“$\bm{F}=-k\dot{\bm{r}}$” was an equation of a form that holds only in the coordinate system in which the air is at rest.)

Solution

(a) Since $\beta=0.8$, $1-\beta^2=1-0.64=0.36$ and $\sqrt{0.36}=0.6$, so

$$ \gamma=\frac{1}{0.6}=\frac53\approx1.667 $$

(b) Since $\beta=0.99$, $1-\beta^2=1-0.9801=0.0199$ and $\sqrt{0.0199}=0.14107$, so

$$ \gamma=\frac{1}{0.14107}\approx7.089 $$

(c) Squaring both sides of $\gamma=1/\sqrt{1-\beta^2}$ and taking reciprocals gives $1-\beta^2=1/\gamma^2$, that is, $\beta^2=1-1/\gamma^2$. For $\gamma=10$, $\beta^2=1-1/100=0.99$ and $\beta=\sqrt{0.99}=0.99499$. That is, $v=0.995c\approx2.98\times10^8\ \mathrm{m/s}$. (d) For $\gamma=2$, $\beta^2=1-1/4=3/4$ and $\beta=\sqrt3/2\approx0.866$ ($v\approx0.866c$).

Solution

$\beta=v/c=3.0\times10^4/(3.00\times10^8)=1.0\times10^{-4}$ and $\beta^2=1.0\times10^{-8}$.

(a) The difference in round-trip times is $\Delta t\approx(L/c)\beta^2$. Here

$$ \frac{L}{c}=\frac{2.0\ \mathrm m}{3.00\times10^8\ \mathrm{m/s}}=6.67\times10^{-9}\ \mathrm s $$

and multiplying by $\beta^2$ (dimensionless) gives

$$ \Delta t=\frac{L}{c}\,\beta^2=6.67\times10^{-9}\ \mathrm s\times1.0\times10^{-8}=6.7\times10^{-17}\ \mathrm s $$

(b) The optical path difference is $c\Delta t=3.00\times10^8\times6.67\times10^{-17}=2.0\times10^{-8}\ \mathrm m$. When the arms are interchanged (rotated by $90^\circ$), the sign of $\Delta t$ reverses, so the number of fringes shifted is $2\times c\Delta t/\lambda$:

$$ 2\times\frac{2.0\times10^{-8}\ \mathrm m}{6.0\times10^{-7}\ \mathrm m}=0.067 $$

That is, about $7\ \%$ of the fringe spacing, a small shift that is not clearly visible to the naked eye. (c) The value of $L$ for which the fringe shift satisfies $2L\beta^2/\lambda=0.4$ is

$$ L=\frac{0.4\,\lambda}{2\beta^2}=\frac{0.4\times6.0\times10^{-7}\ \mathrm m}{2\times1.0\times10^{-8}}=12\ \mathrm m $$

Since making the arm itself $12\ \mathrm m$ long is unrealistic, Michelson and Morley made the light travel back and forth many times with mirrors, bringing the effective arm length to about $11\ \mathrm m$.

Solution

For the linear transformation $x'=Ax+B\,ct$, $ct'=Cx+D\,ct$, we use the four conditions in turn.

(i) For light with $x=ct$, $x'=(A+B)\,ct$ and $ct'=(C+D)\,ct$, so the condition for $x'=ct'$ is

$$ A+B=C+D \qquad\cdots\ \text{①} $$

(ii) For light with $x=-ct$, $x'=(-A+B)\,ct$ and $ct'=(-C+D)\,ct$, and the condition for $x'=-ct'$ is $-A+B=-(-C+D)=C-D$, that is,

$$ -A+B=C-D \qquad\cdots\ \text{②} $$

From ①$+$② we get $2B=2C$, that is, $B=C$. From ①$-$② we get $2A=2D$, that is, $A=D$.

(iii) The origin $x'=0$ of S′ corresponds to $x=vt$ in frame S, so $0=Avt+B\,ct$, that is, $B=-\beta A$ ($\beta=v/c$). Therefore $C=B=-\beta A$ and $D=A$, and the transformation has been narrowed down to the form

$$ x'=A\,(x-vt),\qquad ct'=A\,(ct-\beta x) $$

Only the value of $A$ remains.

(iv) By the principle of relativity, the inverse transformation is obtained by the replacement $\beta\to-\beta$ ($v\to-v$), so $x=A\,(x'+\beta\,ct')$. Substituting the above $x',ct'$ here gives

$$ x=A\left[A\,(x-\beta ct)+\beta A\,(ct-\beta x)\right]=A^2\left[x-\beta ct+\beta ct-\beta^2x\right]=A^2(1-\beta^2)\,x $$

Since this holds for arbitrary $x$, $A^2(1-\beta^2)=1$, that is, $A^2=1/(1-\beta^2)=\gamma^2$. We take $A=+\gamma$ so that $x'\to x$ as $v\to0$ ($A\to1$). In the end,

$$ A=D=\gamma,\qquad B=C=-\beta\gamma $$

which agrees with the result in the main text of Chapter 54 (the one obtained by comparing the coefficients of $x^2-c^2t^2$). In this method, without using the invariance of $x^2-c^2t^2$ from the start, we have derived the Lorentz transformation from the invariance of the speed of light (the light-ray conditions (i) and (ii)) and the principle of relativity ((iv)) alone.

Solution

Placing the Galilean transformation and the Lorentz transformation ($\beta=v/c$, $\gamma=1/\sqrt{1-\beta^2}$) side by side gives the following.

• (a) $x'$: Galilean transformation $x'=x-vt$. Lorentz transformation $x'=\gamma\,(x-vt)$.
• (b) $y',z'$: in both, $y'=y$ and $z'=z$.
• (c) $t'$: Galilean transformation $t'=t$. Lorentz transformation $t'=\gamma\left(t-\dfrac{v}{c^2}x\right)$.
• (d) Inverse transformation for $x$: Galilean transformation $x=x'+vt'$. Lorentz transformation $x=\gamma\,(x'+vt')$.
• (e) Inverse transformation for $t$: Galilean transformation $t=t'$. Lorentz transformation $t=\gamma\left(t'+\dfrac{v}{c^2}x'\right)$.
• (f) The speed, in frame S′, of light traveling in the $+x$ direction with speed $c$ in frame S ($x=ct$): with the Galilean transformation, $x'=x-vt=(c-v)t$ and $t'=t$, so $x'/t'=c-v$. With the Lorentz transformation, as in Example 54.9 of Chapter 54, $x'=ct'$, so the speed is $c$.

We examine the limit $\beta\to0$ ($v\ll c$, and the speed of the object $u\ll c$). Since $\gamma=1/\sqrt{1-\beta^2}\to1$, the Lorentz transformations in (a) and (d) become $x'\to x-vt$ and $x\to x'+vt'$, which coincide with the Galilean transformation. The correction terms in (c) and (e) are $vx/c^2$ and $vx'/c^2$, and their ratio to $t$ is $\beta\,u/c\ll1$, so they can be neglected (Section 54.6), and $t'\to t$ and $t\to t'$, which coincide with the Galilean transformation. In (f) the speeds are $c-v$ and $c$, and their relative difference is $v/c\to0$. Therefore, for every quantity, the Lorentz transformation coincides with the Galilean transformation as $v/c\to0$.

Solution

Let $r_n=\bm{r}\cdot\bm{n}$ (the component of $\bm{r}$ in the direction of $\bm{n}$). If we regard the direction of $\bm{n}$ as a new “$x$ axis,” the coordinate $r_n$ parallel to $\bm{n}$ undergoes the same transformation as $x$ in Theorem 54.1 of Chapter 54, and the perpendicular component $\bm{r}_\perp=\bm{r}-r_n\bm{n}$ is unchanged, just like $y,z$, as follows:

$$ r'_n=\gamma\,(r_n-vt),\qquad\bm{r}'_\perp=\bm{r}_\perp,\qquad t'=\gamma\left(t-\frac{v}{c^2}r_n\right) $$

Substituting the above into $\bm{r}'=r'_n\bm{n}+\bm{r}'_\perp$ gives

$$ \bm{r}'=\gamma\,(r_n-vt)\,\bm{n}+\bm{r}-r_n\bm{n}=\bm{r}+(\gamma-1)(\bm{r}\cdot\bm{n})\,\bm{n}-\gamma vt\,\bm{n} $$

For the time, using $v\,r_n=\bm{v}\cdot\bm{r}$ ($\bm{v}=v\bm{n}$), we have $t'=\gamma\left(t-\bm{v}\cdot\bm{r}/c^2\right)$.

Invariant: Since $\bm{r}'^2=(r'_n)^2+\bm{r}'^2_\perp=\gamma^2(r_n-vt)^2+(\bm{r}^2-r_n^2)$ and $c^2t'^2=\gamma^2(ct-\beta r_n)^2$,

$$ c^2t'^2-\bm{r}'^2=\gamma^2\left[(ct-\beta r_n)^2-(r_n-\beta ct)^2\right]-\bm{r}^2+r_n^2 $$

The expression in the brackets is $c^2t^2-2\beta ct\,r_n+\beta^2r_n^2-r_n^2+2\beta ct\,r_n-\beta^2c^2t^2=(1-\beta^2)(c^2t^2-r_n^2)$, so using $\gamma^2(1-\beta^2)=1$ we obtain

$$ c^2t'^2-\bm{r}'^2=(c^2t^2-r_n^2)-\bm{r}^2+r_n^2=c^2t^2-\bm{r}^2 $$

Hence the invariant is preserved.

The case $\bm{n}=\bm{e}_x$: Here $r_n=x$ and $\bm{r}_\perp=(0,y,z)$, so $x'=\gamma(x-vt)$, $y'=y$, $z'=z$, and $t'=\gamma(t-vx/c^2)$, which agrees with Theorem 54.1 of Chapter 54. Note that the assumption $y=y'$, $z=z'$ concerns the components perpendicular to the direction of motion ($\bm{n}$); for motion in a general direction, the same argument holds if it is read as “the components perpendicular to $\bm{n}$ are unchanged.”

Solution

(a) For $\beta=1.2$, $1-\beta^2=1-1.44=-0.44\lt0$. We have $\sqrt{-0.44}=i\sqrt{0.44}=0.663\,i$ ($i$ is the imaginary unit), and $\gamma=1/(0.663\,i)=-1.51\,i$ is not real. Therefore the $x',t'$ of the Lorentz transformation are not real either, and have no meaning as coordinates and times that can actually be measured. That is, the relative velocity of two inertial frames connected by a Lorentz transformation must satisfy $\abs{v}\lt c$.

(b) As in Table 54.1 of Chapter 54, $\gamma$ is $2.3$ at $\beta=0.9$, $7.1$ at $\beta=0.99$, and $22.4$ at $\beta=0.999$; it grows without bound as $\beta\to1$. As will be shown in Chapter 57, the energy of an object of mass $m$ is $E=\gamma mc^2$, that is, it is $\gamma$ times the rest energy $mc^2$ (the momentum also grows, in the form $\gamma mv$). To bring an object close to the speed of light, the energy and momentum of the object grow rapidly for even a slight increase in speed, so the energy that must be supplied keeps increasing, and an infinite amount of energy is required to reach $v=c$. In other words, the mathematical fact that $\gamma\to\infty$ corresponds to the fact that the speed of an object with mass cannot exceed $c$ (cannot even reach it).

Solution

(a) For $\beta=0.6$, $\gamma=5/4$ and $\beta\gamma=3/4$, so $\Lambda(0.6)=\begin{pmatrix}5/4&-3/4\\-3/4&5/4\end{pmatrix}$. The product is

$$ \Lambda(0.6)\Lambda(0.6)=\begin{pmatrix}\frac{25}{16}+\frac{9}{16}&-\frac{15}{16}-\frac{15}{16}\\-\frac{15}{16}-\frac{15}{16}&\frac{9}{16}+\frac{25}{16}\end{pmatrix}=\begin{pmatrix}17/8&-15/8\\-15/8&17/8\end{pmatrix} $$

(b) Comparing this with $\begin{pmatrix}\gamma&-\beta\gamma\\-\beta\gamma&\gamma\end{pmatrix}$, we find $\gamma=17/8=2.125$ and $\beta\gamma=15/8$. Therefore $\beta=(15/8)/(17/8)=15/17\approx0.882$. Verification: $\gamma^2-(\beta\gamma)^2=(289-225)/64=1$, and $1/\sqrt{1-(15/17)^2}=1/\sqrt{64/289}=17/8$, which agrees with $\gamma$.

(c) For general $\beta_1,\beta_2$ ($\gamma_1,\gamma_2$), we have

$$ \Lambda(\beta_1)\Lambda(\beta_2)=\gamma_1\gamma_2\begin{pmatrix}1&-\beta_1\\-\beta_1&1\end{pmatrix}\begin{pmatrix}1&-\beta_2\\-\beta_2&1\end{pmatrix}=\gamma_1\gamma_2\begin{pmatrix}1+\beta_1\beta_2&-(\beta_1+\beta_2)\\-(\beta_1+\beta_2)&1+\beta_1\beta_2\end{pmatrix} $$

Setting this equal to $\Lambda(\beta)$, the $(1,1)$ entry gives $\gamma=\gamma_1\gamma_2(1+\beta_1\beta_2)$ and the $(1,2)$ entry gives $\beta\gamma=\gamma_1\gamma_2(\beta_1+\beta_2)$, so taking the ratio yields

$$ \beta=\frac{\beta_1+\beta_2}{1+\beta_1\beta_2} $$

(The factor $\gamma_1\gamma_2>0$ cancels. Also, since $1-\beta^2=\dfrac{(1+\beta_1\beta_2)^2-(\beta_1+\beta_2)^2}{(1+\beta_1\beta_2)^2}=\dfrac{(1-\beta_1^2)(1-\beta_2^2)}{(1+\beta_1\beta_2)^2}$, we get $\gamma=1/\sqrt{1-\beta^2}=\gamma_1\gamma_2(1+\beta_1\beta_2)$, which is also consistent with the $(1,1)$ entry.)

(d) In the Galilean composition of velocities, $0.6c+0.6c=1.2c$ exceeds the speed of light, but in the composition of Lorentz transformations, $\beta=(0.6+0.6)/(1+0.36)=1.2/1.36=15/17\approx0.882$, which does not exceed the speed of light. In general, if $\abs{\beta_1},\abs{\beta_2}\lt1$, then $1-\beta^2=(1-\beta_1^2)(1-\beta_2^2)/(1+\beta_1\beta_2)^2\gt0$, so $\abs{\beta}\lt1$. Also, when $\beta_1,\beta_2\ll1$, the denominator is close to $1$ and $\beta\approx\beta_1+\beta_2$, which agrees with the composition of velocities in high school physics. This formula appears again in Chapter 55 as the velocity addition law (the Lorentz velocity transformation).

Solution

(a) When $\beta=0.5$, $1-\beta^2=1-0.25=0.75$, so

$$ \frac{\ell'}{\ell}=\sqrt{0.75}=\frac{\sqrt3}{2}=0.866 $$

Thus the length of the rod shrinks to about $86.6\%$. (b) When $\beta=0.9$, $1-\beta^2=1-0.81=0.19$, so

$$ \frac{\ell'}{\ell}=\sqrt{0.19}=0.4359 $$

and the rod shrinks to about $43.6\%$. (c) Setting $\ell'/\ell=1/2$ gives $\sqrt{1-\beta^2}=1/2$. Squaring both sides gives $1-\beta^2=1/4$, that is, $\beta^2=3/4$, so the required speed is

$$ \beta=\frac{\sqrt3}{2}=0.866,\qquad v=0.866\,c $$

(d) Setting $\ell'/\ell=1\ \mathrm{mm}/1\ \mathrm m=10^{-3}$ gives $\sqrt{1-\beta^2}=10^{-3}$, so $1-\beta^2=10^{-6}$. When $\beta$ is very close to 1, we can approximate $1-\beta^2=(1-\beta)(1+\beta)\approx2\,(1-\beta)$ (since $1+\beta\approx2$), so

$$ 1-\beta\approx\frac{10^{-6}}{2}=5\times10^{-7} $$

that is, this happens when $\beta=0.9999995$ ($v=0.9999995\,c$). In this case $\gamma=1/\sqrt{1-\beta^2}=10^{3}$ (and $\ell'/\ell=1/\gamma$).

∎

Solution

From $\beta=0.8$ we have $\gamma=1/\sqrt{1-0.64}=1/0.6=5/3$ and $1/\gamma=0.6$. (a) In Earth's frame, the spacecraft travels a distance of $4$ light-years at speed $0.8c$, so the time taken is

$$ \Delta t=\frac{4\ \text{light-years}}{0.8\,c}=\frac{4}{0.8}\ \text{years}=5\ \text{years} $$

(b) The clock on the spacecraft measures the interval between departure and arrival, which occur at the same place inside the spacecraft, so it reads the proper time $\Delta\tau$. By time dilation ($\Delta t=\gamma\Delta\tau$),

$$ \Delta\tau=\frac{\Delta t}{\gamma}=5\ \text{years}\times0.6=3\ \text{years} $$

(c) In the spacecraft's frame, the Earth (the point of departure) and the destination star move toward the spacecraft (backward) at speed $0.8c$. The distance between the two bodies is the proper length of $4$ light-years in the frame in which the two bodies are at rest (Earth's frame); in the spacecraft's frame it is Lorentz-contracted to

$$ \ell'=\frac{4\ \text{light-years}}{\gamma}=4\times0.6=2.4\ \text{light-years} $$

Since the star approaches at speed $0.8c$ and covers this distance, the time taken is

$$ \Delta\tau=\frac{2.4\ \text{light-years}}{0.8\,c}=3\ \text{years} $$

which agrees with the answer to (b). Time dilation (the explanation in Earth's frame) and length contraction (the explanation in the spacecraft's frame) explain the same result in two ways. (A round trip at the same speed takes $6$ years on the spacecraft's clock and $10$ years on Earth. The treatment of a round trip that includes accelerated motion is examined in Chapter 56.)

∎

Solution

(a) We take the Taylor expansion of $f(\beta)=(1-\beta)^{1/2}(1+\beta)^{-1/2}$ about $\beta=0$, factor by factor. For $(1-\beta)^{1/2}$: since $\dfrac{\dd}{\dd\beta}(1-\beta)^{1/2}=-\dfrac12(1-\beta)^{-1/2}$ equals $-\dfrac12$ at $\beta=0$, we obtain $1-\dfrac{\beta}{2}+\cdots$. For $(1+\beta)^{-1/2}$: since $\dfrac{\dd}{\dd\beta}(1+\beta)^{-1/2}=-\dfrac12(1+\beta)^{-3/2}$ equals $-\dfrac12$ at $\beta=0$, we obtain $1-\dfrac{\beta}{2}+\cdots$. Therefore, to first order in $\beta$,

$$ \sqrt{\frac{1-\beta}{1+\beta}}\approx\left(1-\frac\beta2\right)\left(1-\frac\beta2\right)\approx1-\beta $$

which shows that $\nu'\approx\nu(1-\beta)$ (terms of order $\beta^2$ and higher were neglected). This agrees with the classical Doppler effect (for an observer receding at speed $v$, the frequency is $\nu(1-v/c)$).

(b) For $\beta=0.01$, $(1-\beta)/(1+\beta)=0.99/1.01=0.980198$, and its square root is $0.990050$ (to five significant figures). The approximate value is $1-\beta=0.99$, so the difference is

$$ 0.990050-0.99=5.0\times10^{-5} $$

(more precisely, $4.95\times10^{-5}$). (c) We expand to second order. Since $(1-\beta)^{1/2}=1-\dfrac\beta2-\dfrac{\beta^2}{8}+\cdots$ and $(1+\beta)^{-1/2}=1-\dfrac\beta2+\dfrac{3\beta^2}{8}+\cdots$ (by the binomial expansion $(1+x)^{\alpha}=1+\alpha x+\dfrac{\alpha(\alpha-1)}{2}x^2+\cdots$), the coefficient of $\beta^2$ in the product is $1\cdot\dfrac38+\left(-\dfrac12\right)\left(-\dfrac12\right)+\left(-\dfrac18\right)\cdot1=\dfrac38+\dfrac14-\dfrac18=\dfrac12$. Hence

$$ \sqrt{\frac{1-\beta}{1+\beta}}=1-\beta+\frac{\beta^2}{2}-\cdots $$

and the difference between the exact value and the approximation $1-\beta$ is approximately $\beta^2/2=(0.01)^2/2=5\times10^{-5}$, which agrees with the value in (b) (including the next term, $-\beta^3/2=-5\times10^{-7}$, gives $4.95\times10^{-5}$). The relativistic correction is a small quantity of order $\beta^2/2$.

∎

Solution

Since $\beta=0.8$, $\gamma=1/\sqrt{1-0.64}=5/3$. Let the reading of the rocket's clock be $\tau=1$ hour $=60$ minutes. (a) In Earth's frame, the moving rocket's clock runs at the rate $1/\gamma$, so the rocket's clock reads $\tau$ when the Earth clock reads

$$ t_{\mathrm E}=\gamma\tau=\frac53\times60\ \text{minutes}=100\ \text{minutes}=1\ \text{hour}\,40\ \text{minutes} $$

(b) What we see through the telescope is the moment at which the light that left the rocket at this time (Earth time $t_{\mathrm E}$) reaches Earth. The rocket is at the position $x_{\mathrm E}=vt_{\mathrm E}$, and the light takes a time $x_{\mathrm E}/c=\beta t_{\mathrm E}$ to reach Earth, so the time at which it is seen is

$$ t=t_{\mathrm E}+\beta t_{\mathrm E}=(1+\beta)\,\gamma\tau=\sqrt{\frac{1+\beta}{1-\beta}}\ \tau=\sqrt{\frac{1.8}{0.2}}\times60\ \text{minutes}=3\times60\ \text{minutes}=180\ \text{minutes} $$

that is, $3$ hours after departure (3 hours 00 minutes) (this can also be checked from $(1+\beta)\gamma=1.8\times5/3=3$). (c) The difference between (b) and (a) is $180-100=80$ minutes. On the other hand, the position of the rocket is $x_{\mathrm E}=vt_{\mathrm E}=0.8c\times100\ \text{minutes}=80\ \text{light-minutes}$ (the distance light travels in 80 minutes), so the time for the light to travel from the rocket to Earth is $x_{\mathrm E}/c=80\ \text{minutes}$. These two agree.

∎

Solution

(a) The velocity of S′ relative to S is $v=\beta_1c$, and the velocity of the object in the S′ frame is $u'_x=\beta_2c$. By the first equation of the inverse velocity transformation, the velocity of the object in the S frame is

$$ u_x=\frac{u'_x+v}{1+\dfrac{vu'_x}{c^2}}=\frac{\beta_2c+\beta_1c}{1+\beta_1\beta_2} $$

Dividing both sides by $c$ and writing $\beta=u_x/c$, we obtain

$$ \beta=\frac{\beta_1+\beta_2}{1+\beta_1\beta_2} $$

(b) We compute $1-\beta^2$:

$$ 1-\beta^2=\frac{(1+\beta_1\beta_2)^2-(\beta_1+\beta_2)^2}{(1+\beta_1\beta_2)^2} $$

Expanding the numerator,

$$ \begin{aligned} (1+\beta_1\beta_2)^2-(\beta_1+\beta_2)^2&=1+2\beta_1\beta_2+\beta_1^2\beta_2^2-\beta_1^2-2\beta_1\beta_2-\beta_2^2\\ &=1-\beta_1^2-\beta_2^2+\beta_1^2\beta_2^2=(1-\beta_1^2)(1-\beta_2^2) \end{aligned} $$

(the last equality is the expansion of $(1-\beta_1^2)(1-\beta_2^2)$). Hence

$$ 1-\beta^2=\frac{(1-\beta_1^2)(1-\beta_2^2)}{(1+\beta_1\beta_2)^2} $$

If $0\le\beta_1,\beta_2<1$, then $1-\beta_1^2>0$ and $1-\beta_2^2>0$, so the right-hand side is positive; hence $1-\beta^2>0$, that is, $\beta<1$ (that $\beta\ge0$ is clear from the formula in (a)). No matter how close $\beta_1,\beta_2$ are brought to 1, the composite speed never exceeds the speed of light. (c) For $\beta_1=\beta_2=0.9$,

$$ \beta=\frac{0.9+0.9}{1+0.9\times0.9}=\frac{1.8}{1.81}=\frac{180}{181}=0.9945 $$

The simple sum $1.8$ would exceed the speed of light, but the correct value is $0.9945$, which is 99.45 percent of $c$.

∎

Solution

(a) Since $\beta=0.98$, $1-\beta^2=1-0.9604=0.0396$ and $\sqrt{0.0396}=0.19900$, so

$$ \gamma=\frac{1}{0.19900}=5.025 $$

The mean lifetime as seen from the ground is $\gamma T_\mu=5.025\times2.197\ \mathrm{\mu s}=11.04\ \mathrm{\mu s}$. (b) The mean flight distance as seen from the ground, with $c=2.998\times10^{8}\ \mathrm{m/s}$, is

$$ L=v\,(\gamma T_\mu)=0.98\times2.998\times10^{8}\ \mathrm{m/s}\times11.04\times10^{-6}\ \mathrm s=3.24\times10^{3}\ \mathrm m $$

that is, about $3.2\ \mathrm{km}$. (c) The fraction of muons descending from an altitude $h=12\ \mathrm{km}$ that arrive without decaying is, because the proper time $h/(\gamma v)$ elapses during the descent time $h/v$ in the ground frame,

$$ P=\exp\left(-\frac{h/(\gamma v)}{T_\mu}\right)=\exp\left(-\frac{h}{L}\right)=\exp\left(-\frac{12\,000}{3\,244}\right)=\exp(-3.70)=0.0247 $$

which is about $2.5\%$. (d) In the muon's frame, the thickness of the atmosphere is Lorentz-contracted to $12\,000\ \mathrm m/\gamma=12\,000/5.025=2\,388\ \mathrm m$. Since the ground approaches at speed $v=0.98c$, the time taken to descend (the proper time) is

$$ \tau=\frac{2\,388\ \mathrm m}{0.98\times2.998\times10^{8}\ \mathrm{m/s}}=8.13\times10^{-6}\ \mathrm s=8.13\ \mathrm{\mu s} $$

The surviving fraction is $\exp(-\tau/T_\mu)=\exp(-8.13/2.197)=\exp(-3.70)=0.0247$, which agrees with (c).

∎

Solution

(a) The speed in a circular orbit is $v=\sqrt{GM/r}$ (universal gravitation provides the centripetal force), so

$$ v=\sqrt{\frac{3.986\times10^{14}}{6.778\times10^{6}}}\ \mathrm{m/s}=\sqrt{5.881\times10^{7}}\ \mathrm{m/s}=7.67\times10^{3}\ \mathrm{m/s} $$

(b) The fractional delay due to special relativity is $v^2/(2c^2)$, and with $c^2=8.988\times10^{16}\ \mathrm{m^2/s^2}$,

$$ \frac{v^2}{2c^2}=\frac{5.881\times10^{7}}{2\times8.988\times10^{16}}=3.27\times10^{-10} $$

The delay over one day, $86\,400\ \mathrm s$, is $3.27\times10^{-10}\times86\,400\ \mathrm s=2.83\times10^{-5}\ \mathrm s=28.3\ \mathrm{\mu s}$. (c) The fractional gain due to general relativity is $\dfrac{GM}{c^2}\left(\dfrac1R-\dfrac1r\right)$, and with $GM/c^2=4.435\times10^{-3}\ \mathrm m$ and $1/R-1/r=1.5679\times10^{-7}-1.4754\times10^{-7}=9.25\times10^{-9}\ \mathrm{m^{-1}}$,

$$ \frac{GM}{c^2}\left(\frac1R-\frac1r\right)=4.435\times10^{-3}\times9.25\times10^{-9}=4.10\times10^{-11} $$

The gain over one day is $4.10\times10^{-11}\times86\,400\ \mathrm s=3.5\times10^{-6}\ \mathrm s=3.5\ \mathrm{\mu s}$. (d) The net result is

$$ -28.3\ \mathrm{\mu s}+3.5\ \mathrm{\mu s}=-24.7\ \mathrm{\mu s} $$

that is, the clock on the space station falls behind the clock on the ground by about $25\ \mathrm{\mu s}$ per day. In contrast to a GPS satellite (whose altitude is high, so that the gain of $45.6\ \mathrm{\mu s}$ is larger than the delay of $7.2\ \mathrm{\mu s}$), for the low-altitude space station the delay due to motion (special relativity) is the stronger effect. The two effects cancel when $\dfrac{GM}{2c^2r}=\dfrac{GM}{c^2}\left(\dfrac1R-\dfrac1r\right)$, that is, at $r=\dfrac32R\approx9.6\times10^{6}\ \mathrm m$ (an altitude of about $3\,200\ \mathrm{km}$); in orbits lower than this the clock falls behind, and in higher orbits it gains.

∎

Solution

(a) Let $D$ be the distance from the point where the first signal was emitted to the observer. The second signal is emitted, after a time $\Delta t$, at the point to which the blob has advanced by $\beta c\,\Delta t$. Since the observer is sufficiently far away, the light rays can be regarded as parallel to the line of sight, and that point is closer to the observer by $\beta c\,\Delta t\cos\theta$. Therefore the times at which the signals reach the observer are

$$ t_1=\frac{D}{c},\qquad t_2=\Delta t+\frac{D-\beta c\,\Delta t\cos\theta}{c} $$

and their difference is

$$ \Delta t_{\mathrm{obs}}=t_2-t_1=\Delta t-\beta\,\Delta t\cos\theta=\Delta t\,(1-\beta\cos\theta) $$

If $\theta=0$, then $\Delta t_{\mathrm{obs}}=(1-\beta)\Delta t$, and the interval between the signals reaching the observer appears shortened. (b) The displacement perpendicular to the line of sight is $\beta c\,\Delta t\sin\theta$, so the apparent speed is

$$ \beta_{\mathrm{app}}c=\frac{\beta c\,\Delta t\sin\theta}{\Delta t\,(1-\beta\cos\theta)}\quad\Longrightarrow\quad\beta_{\mathrm{app}}=\frac{\beta\sin\theta}{1-\beta\cos\theta} $$

(c) We substitute $\beta=0.98$ and $\theta=10^\circ$. Since $\sin10^\circ=0.173648$ and $\cos10^\circ=0.984808$,

$$ \beta_{\mathrm{app}}=\frac{0.98\times0.173648}{1-0.98\times0.984808}=\frac{0.170175}{0.034888}=4.88 $$

Although the actual speed is $0.98c$, the apparent speed is about $4.9c$. (d) Differentiating $\beta_{\mathrm{app}}(\theta)$ with the quotient rule, we obtain

$$ \frac{\dd\beta_{\mathrm{app}}}{\dd\theta}=\frac{\beta\cos\theta\,(1-\beta\cos\theta)-\beta\sin\theta\cdot\beta\sin\theta}{(1-\beta\cos\theta)^2}=\frac{\beta\cos\theta-\beta^2}{(1-\beta\cos\theta)^2}=\frac{\beta\,(\cos\theta-\beta)}{(1-\beta\cos\theta)^2} $$

(we used $\cos^2\theta+\sin^2\theta=1$ in the numerator). This vanishes at $\cos\theta=\beta$, is positive for $\cos\theta\gt\beta$ (the side of smaller $\theta$), and is negative for $\cos\theta\lt\beta$, so the apparent speed is maximal when $\cos\theta=\beta$. In this case $\sin\theta=\sqrt{1-\beta^2}=1/\gamma$ and $1-\beta\cos\theta=1-\beta^2=1/\gamma^2$, so

$$ \beta_{\mathrm{app,max}}=\frac{\beta\,(1/\gamma)}{1/\gamma^2}=\gamma\beta $$

(for $\beta=0.98$, $\gamma\beta=4.92$, and the corresponding angle is $\theta=\arccos0.98=11.5^\circ$). We have $\gamma\beta\gt1$ when $\beta^2\gt1-\beta^2$, that is, when $\beta\gt1/\sqrt2\approx0.707$ (for $\beta=0.9$, $\gamma\beta=2.06$).

This does not contradict relativity. The speed of the blob itself remains $\beta c\lt c$; the apparent speed exceeds $c$ because it is an apparent effect of the difference in light travel times: “since the blob emits the signals while approaching the observer, the difference $\Delta t_{\mathrm{obs}}=\Delta t\,(1-\beta\cos\theta)$ between the arrival times of the two signals is much shorter than the actual time difference $\Delta t$.” Nothing has actually been carried faster than light.

∎

Solution

(a) The period of a light clock at rest is the time for light to make one round trip between mirrors separated by a distance $d$, so

$$ \Delta\tau=\frac{2d}{c}=\frac{2\times0.30\ \mathrm m}{3.00\times10^{8}\ \mathrm{m/s}}=2.00\times10^{-9}\ \mathrm s=2.00\ \mathrm{ns} $$

(b) For $\beta=0.8$, $\gamma=1/\sqrt{1-0.64}=1/0.6=5/3$, so the period of the clock that appears to be moving is

$$ T=\gamma\,\Delta\tau=\frac53\times2.00\ \mathrm{ns}=3.33\ \mathrm{ns} $$

(c) The distance traveled by the light during one round trip is $cT=3.00\times10^{8}\ \mathrm{m/s}\times3.33\times10^{-9}\ \mathrm s=1.00\ \mathrm m$, which is $5/3$ times ($=\gamma$ times) the round-trip distance $2d=0.60\ \mathrm m$ when the clock is at rest. Let us check the Pythagorean theorem. The hypotenuse of a one-way trip is $cT/2=0.50\ \mathrm m$, the horizontal displacement of the clock is $vT/2=0.8\times0.50\ \mathrm m=0.40\ \mathrm m$, and the vertical side is $d=0.30\ \mathrm m$, so

$$ d^2+\left(\frac{vT}{2}\right)^2=0.30^2+0.40^2=0.09+0.16=0.25=0.50^2=\left(\frac{cT}{2}\right)^2 $$

which holds (it is a $3:4:5$ right triangle). (d) Q's clock is at rest relative to Q, so its period in Q's frame is the same $\Delta\tau=2.00\ \mathrm{ns}$. By the principle of relativity, P and Q are on an equal footing. From P's point of view, Q's clock is moving at speed $0.8c$, so by the same calculation as in (b), the period of Q's clock as measured by P is $\gamma\,\Delta\tau=3.33\ \mathrm{ns}$. That is, from Q's point of view P's clock runs slow, and from P's point of view Q's clock runs slow. This is not a contradiction, as was confirmed in Example 55.7.

∎

Solution

(a) For an approaching source we replace $v\to-v$, so that $\nu'/\nu=\sqrt{(1+\beta)/(1-\beta)}$. Substituting $\beta=0.5$,

$$ \frac{\nu'}{\nu}=\sqrt{\frac{1+0.5}{1-0.5}}=\sqrt{\frac{1.5}{0.5}}=\sqrt3=1.732 $$

Since the wavelength is inversely proportional to the frequency ($c=\nu\lambda$),

$$ \lambda'=\frac{\lambda}{\nu'/\nu}=\frac{500\ \mathrm{nm}}{1.732}=288.7\ \mathrm{nm}\approx289\ \mathrm{nm} $$

so visible green light is observed as ultraviolet light. (In terms of frequency, $\nu=c/\lambda=2.998\times10^{8}/(500\times10^{-9})=5.996\times10^{14}\ \mathrm{Hz}$ becomes $\nu'=1.732\nu=1.039\times10^{15}\ \mathrm{Hz}$.) (b) With the classical formula, $\nu'/\nu=1/(1-\beta)=1/0.5=2$, so

$$ \lambda''=\frac{500\ \mathrm{nm}}{2}=250\ \mathrm{nm} $$

The difference from (a) is $288.7-250=38.7\ \mathrm{nm}$, and the relativistic formula gives the longer wavelength. This is due to the time dilation factor $1/\gamma=\sqrt{1-\beta^2}=0.866$, which the classical formula does not contain; indeed $2\times0.866=1.732$, which agrees with the value in (a). (c) For a receding source, $\nu'/\nu=\sqrt{(1-\beta)/(1+\beta)}=\sqrt{0.5/1.5}=1/\sqrt3=0.577$, so

$$ \lambda'=\sqrt3\times500\ \mathrm{nm}=866\ \mathrm{nm} $$

which is near-infrared light. Comparing approach and recession at the same speed, the wavelengths are $0.577$ times and $1.732$ times $500\ \mathrm{nm}$, and these two factors multiply to $1$ ($\sqrt{(1+\beta)/(1-\beta)}\times\sqrt{(1-\beta)/(1+\beta)}=1$).

∎

Solution

The velocity transformation law (Theorem 55.4 of Chapter 55; here we write $v$ for the speed of the frame and $u$ for the velocity of the particle) is, with $\beta=v/c=0.3$ and $\sqrt{1-\beta^2}=\sqrt{1-0.09}=\sqrt{0.91}=0.95394$,

$$ u'_x=\frac{u_x-v}{1-vu_x/c^2},\qquad u'_y=\frac{u_y\sqrt{1-\beta^2}}{1-vu_x/c^2} $$

Since $u_x=0$, the denominator is $1-0=1$, and so

$$ u'_x=-v=-0.3c,\qquad u'_y=0.4c\times0.95394=0.38158c\approx0.382c,\qquad u'_z=0 $$

The speed is $u'=c\sqrt{0.3^2+0.38158^2}=c\sqrt{0.09+0.1456}=c\sqrt{0.2356}=0.48539c\approx0.485c$ (we used $u'_y{}^2=0.4^2\times0.91=0.1456$). Verification: the left-hand side of the identity (Theorem 55.5 of Chapter 55) is $1-u'^2/c^2=1-0.2356=0.7644$, and the right-hand side is $(1-\beta^2)(1-u^2/c^2)/(1-\beta u_x/c)^2=0.91\times(1-0.16)/1=0.91\times0.84=0.7644$, so the two agree.

The direction of travel seen in frame S′ has an $x'$ component of $-0.3c$ and a $y'$ component of $0.38158c$, so it is tilted from the $y'$ axis toward the $-x'$ side, with $\tan\alpha=0.3/0.38158=0.7862$, that is, $\alpha=38.2^\circ$. Under the Galilean transformation we would have $u'_y=0.4c$ and $\tan\alpha=0.3/0.4=0.75$ ($\alpha=36.9^\circ$); in relativity, $u'_y$ is shrunk by the factor $\sqrt{1-\beta^2}$, and the tilt becomes slightly larger.

Solution

(a) For a constant speed $v$, $\gamma=1/\sqrt{1-v^2/c^2}$ is a constant independent of time, so

$$ \tau=\int_0^T\frac{\dd t}{\gamma}=\frac{1}{\gamma}\int_0^T\dd t=\frac{T}{\gamma}=T\sqrt{1-\frac{v^2}{c^2}} $$

(the integral of a constant: $\int_0^T\dd t=T$). (b) For $v=0.99c$, $\sqrt{1-0.99^2}=\sqrt{0.0199}=0.14107$. Taking $T=1$ year $=365.25$ days, $\tau=0.14107\times365.25\ \text{days}=51.5$ days. That is, while one year passes on Earth, the spacecraft's clock advances by only about $52$ days. (c) Use the result of (a) on each interval and add. For $\gamma$, in the first half $1/\gamma=\sqrt{1-0.6^2}=0.8$ and in the second half $1/\gamma=\sqrt{1-0.8^2}=0.6$, so

$$ \tau=\int_0^{T/2}0.8\,\dd t+\int_{T/2}^{T}0.6\,\dd t=0.8\times\frac T2+0.6\times\frac T2=0.4T+0.3T=0.7\,T $$

This lies between the values obtained if the whole time $T$ were spent traveling at one speed or the other ($0.8T$ and $0.6T$).

Solution

(a) The transformation formulas for infinitesimal quantities (Theorem 56.1 of Chapter 56) are $c\,\dd t'=\gamma(c\,\dd t-\beta\,\dd x)$, $\dd x'=\gamma(\dd x-\beta c\,\dd t)$, $\dd y'=\dd y$, $\dd z'=\dd z$. For the $t,x$ part,

$$ c^2\dd t'^2-\dd x'^2=\gamma^2\left\{(c\,\dd t-\beta\,\dd x)^2-(\dd x-\beta c\,\dd t)^2\right\}=\gamma^2\left\{(1-\beta^2)c^2\dd t^2-(1-\beta^2)\dd x^2\right\}=c^2\dd t^2-\dd x^2 $$

(on expanding, the cross terms $-2\beta c\,\dd t\,\dd x$ and $+2\beta c\,\dd t\,\dd x$ cancel each other, and $\gamma^2(1-\beta^2)=1$). Since $\dd y'^2+\dd z'^2=\dd y^2+\dd z^2$ clearly holds, we obtain $c^2\dd t'^2-\dd x'^2-\dd y'^2-\dd z'^2=c^2\dd t^2-\dd x^2-\dd y^2-\dd z^2$ also when $\dd y,\dd z$ are not zero, that is, when the particle (point mass) also moves in the $y,z$ directions.

(b) When the velocity of the particle in frame S is $\bm{u}$, $\dd\tau=\dd t\sqrt{1-u^2/c^2}$; when its velocity in frame S′ is $\bm{u}'$, $\dd\tau'=\dd t'\sqrt{1-u'^2/c^2}$. Taking the square root of the identity (Theorem 55.5 of Chapter 55) and setting $D=1-\beta u_x/c\gt0$, we get

$$ \sqrt{1-\frac{u'^2}{c^2}}=\frac{\sqrt{1-\beta^2}\,\sqrt{1-u^2/c^2}}{D}=\frac{1}{\gamma}\,\frac{\sqrt{1-u^2/c^2}}{D} $$

On the other hand, $\dd t'=\gamma(1-vu_x/c^2)\dd t=\gamma D\,\dd t$ (since $vu_x/c^2=\beta u_x/c$). Multiplying the two,

$$ \dd\tau'=\gamma D\,\dd t\times\frac{1}{\gamma}\frac{\sqrt{1-u^2/c^2}}{D}=\dd t\sqrt{1-u^2/c^2}=\dd\tau $$

Because $\gamma$ and $D$ cancel, $\dd\tau'=\dd\tau$; that is, the proper time does not depend on the frame.

Solution

(a) Let $\phi_1=\tanh^{-1}\beta_1$ and $\phi_2=\tanh^{-1}\beta_2$; then $\beta_1=\tanh\phi_1$ and $\beta_2=\tanh\phi_2$. Substituting these into the right-hand side of the addition theorem for the hyperbolic tangent, $\tanh(\phi_1+\phi_2)=\dfrac{\tanh\phi_1+\tanh\phi_2}{1+\tanh\phi_1\tanh\phi_2}$, we obtain

$$ \tanh(\phi_1+\phi_2)=\frac{\beta_1+\beta_2}{1+\beta_1\beta_2}=\beta $$

Applying $\tanh^{-1}$ to both sides, $\tanh^{-1}\beta=\phi_1+\phi_2=\tanh^{-1}\beta_1+\tanh^{-1}\beta_2$. That is, the rapidity of the composite velocity is the sum of the two rapidities.

(b) $\beta_1=0.6$: $\phi_1=\frac12\ln\frac{1.6}{0.4}=\frac12\ln4=\ln2$. $\beta_2=0.8$: $\phi_2=\frac12\ln\frac{1.8}{0.2}=\frac12\ln9=\ln3$. The sum is $\phi=\ln2+\ln3=\ln6\approx1.792$. The composite velocity is $\beta=\tanh(\ln6)=\dfrac{6-1/6}{6+1/6}=\dfrac{35/6}{37/6}=\dfrac{35}{37}\approx0.9459$ ($\tanh x=(e^x-e^{-x})/(e^x+e^{-x})$). A direct computation gives $\beta=\dfrac{0.6+0.8}{1+0.6\times0.8}=\dfrac{1.4}{1.48}=\dfrac{35}{37}\approx0.9459$, which agrees.

Solution

(a) Compute $s^2=(ct)^2-x^2$. $\mathrm{P}_1$ $(4,3)$: $s^2=16-9=7\gt0$ (timelike). $\mathrm{P}_2$ $(3,5)$: $s^2=9-25=-16\lt0$ (spacelike). $\mathrm{P}_3$ $(5,5)$: $s^2=25-25=0$ (lightlike). The events that O can cause are $\mathrm{P}_1$, which is timelike and later than O, and $\mathrm{P}_3$, which is lightlike and later than O (light emitted from O reaches it). $\mathrm{P}_2$ is spacelike and cannot be reached from O even by light, so O cannot be its cause.

(b) $\mathrm{P}_1$: $\beta_0=x/(ct)=3/4$, $\gamma=1/\sqrt{1-9/16}=4/\sqrt7$. In this frame, $x'=\gamma(x-\beta_0ct)=\gamma(3-\tfrac34\times4)=0$ (the same place), $ct'=\gamma(ct-\beta_0x)=\frac{4}{\sqrt7}(4-\tfrac94)=\frac{4}{\sqrt7}\times\frac74=\sqrt7\approx2.646\ \mathrm m$. That is, in the frame of speed $0.75c$ the two events occur at the same place, and the time difference is $\sqrt7\ \mathrm m/c\approx8.8\ \mathrm{ns}$ ($\sqrt{s^2}/c$). $\mathrm{P}_2$: $\beta_0=ct/x=3/5=0.6$, $\gamma=1.25$. $ct'=\gamma(ct-\beta_0x)=1.25\times(3-0.6\times5)=0$ (simultaneous), $x'=\gamma(x-\beta_0ct)=1.25\times(5-0.6\times3)=1.25\times3.2=4\ \mathrm m$. That is, the two events are simultaneous in the frame of speed $0.6c$, and the distance between them is $\sqrt{-s^2}=\sqrt{16}=4\ \mathrm m$.

(c) In a spacetime diagram (horizontal axis $x$, vertical axis $ct$) with O as the origin, draw the two straight lines $ct=\pm x$ at $45^\circ$ (the light cone). $\mathrm{P}_1$ $(x,ct)=(3,4)$ is drawn inside the light cone (on the future side), $\mathrm{P}_3$ $(5,5)$ on the light cone (on the upper-right $45^\circ$ line), and $\mathrm{P}_2$ $(5,3)$ outside the light cone (on the right side). The physical differences are as follows. O and $\mathrm{P}_1$ can be joined by a particle of speed $3/4\,c=0.75c$ (the speed needed from O to $\mathrm{P}_1$ is $x/(ct)=0.75$), so O can be a cause of $\mathrm{P}_1$. As for the temporal order, O comes first in every frame. On the other hand, joining O and $\mathrm{P}_2$ requires a speed $x/(ct)=5/3\,c\gt c$, and no such signal or object exists. The temporal order of $\mathrm{P}_2$ and O is reversed depending on the frame (if $\beta$ is made larger than $0.6$, $\mathrm{P}_2$ comes before O), but since there is no causal relation, there is no contradiction.

Solution

(a) The round trip of $6$ light-years is traveled at $v=0.6c$, so it takes $T=6/0.6=10$ years in Earth time. The elder brother ages $10$ years. For the younger brother, $\gamma=1/\sqrt{1-0.36}=1.25$ gives $\tau_{\mathrm B}=T/\gamma=10\times0.8=8$ years. Let us also check this with the invariant interval. For one leg, the difference between the two events is $(c\Delta t,\Delta x)=(5,3)$ (in units of years and light-years), so $\Delta s=\sqrt{25-9}=4$ years, and $8$ years for the round trip. The younger brother is $2$ years younger than the elder brother.

(b) The turnaround (event T) is, in frame S, at $x_{\mathrm T}=3$ light-years and $t_{\mathrm T}=3/0.6=5$ years. In the younger brother's inertial frame of reference on the outbound leg, the Earth event Q that is simultaneous with T satisfies, from $\Delta t=v\Delta x/c^2$ (with $c=1$ in units of years and light-years), $t_{\mathrm T}-t_{\mathrm Q}=0.6\times3=1.8$ years, so $t_{\mathrm Q}=5-1.8=3.2$ years. In the younger brother's inertial frame on the return leg (velocity $-0.6c$), $t_{\mathrm Q}-5=0.6\times3=1.8$ years, so $t_{\mathrm Q}=6.8$ years. Across the turnaround, the line of simultaneity jumps from Earth time $3.2$ years to $6.8$ years, a jump of $3.6$ years. Verification: during the younger brother's $4$ years on the outbound leg, the elder brother's clock advances $0.8\times4=3.2$ years, and it advances $3.2$ years on the return leg as well. Adding the jump of $3.6$ years, $3.2+3.6+3.2=10$ years, which agrees with the elder brother's age.

(c) The argument that “from the younger brother's point of view, the elder brother is the one moving” is correct as an argument about the period during which the younger brother stays in a single inertial frame (the outbound leg only, or the return leg only), and from the younger brother's point of view the elder brother's clock runs at $1/\gamma=0.8$ times the rate. However, at the turnaround the younger brother switches from the inertial frame of the outbound leg to that of the return leg. Across this switch, the Earth time that corresponds to the younger brother's “now” jumps from $3.2$ years to $6.8$ years. This jump is due to the relativity of simultaneity. The elder brother stays in a single inertial frame from beginning to end, and there is no such jump. Therefore the elder and younger brothers are not on an equal footing, and the proper time of the younger brother, who switched inertial frames (whose world line bends), is the shorter one. In the spacetime diagram, the elder brother's world line is the straight line joining the two events, and the proper time along a straight line is the maximum (Theorem 56.6). The younger brother can also tell that he is not in an inertial frame because he feels acceleration at the turnaround.

Solution

Take units of years and light-years ($c=1$); $\beta=0.6$, $\gamma=1.25$, the distance to the star is $3$ light-years, and the younger brother turns around at $5$ years of Earth time ($4$ years on his clock).

(a) Signals from the younger brother's clock at the end of his $n$th year. Outbound leg ($n=1,2,3,4$; we include $n=4$, the moment of turnaround): the younger brother's event occurs at Earth time $t=\gamma n=1.25n$ and position $x=0.6t=0.75n$. This signal reaches Earth at time $t+x=2n$ years. That is, at $2,4,6,8$ years, an interval of $2$ years (the younger brother's $1$-year interval is stretched to $2$ years for the elder brother). Return leg ($n=5,6,7,8$): with $m=n-4$, $t=5+1.25m$ and $x=3-0.6\times1.25m=3-0.75m$. The signal arrives at $t+x=8+0.5m$ years: $8.5,9,9.5,10$ years, an interval of $0.5$ years. The elder brother receives the first four signals at $2$-year intervals (years $2\to8$) and the remaining four at $0.5$-year intervals (years $8\to10$), for a total of $8$ signals, the last of which arrives at the reunion at year $10$.

(b) The signal that the elder brother sends at Earth time $n$ years ($n=1,\dots,10$) is light along $x=t-n$. A signal reaches the younger brother while he is on the outbound leg ($x=0.6t$, $0\le t\le5$) when $t-n=0.6t$, that is, at $t=n/0.4=2.5n$. The condition $t\le5$ gives $n=1,2$: at $2.5,\,5.0$ years of Earth time, and at $2.0,\,4.0$ years on the younger brother's clock ($\tau=t/\gamma=0.8t$). The interval is $2$ years. On the return leg ($x=3-0.6(t-5)=6-0.6t$), from $t-n=6-0.6t$ we get $t=(6+n)/1.6$. For $n=3,\dots,10$, $t=5.625,\,6.25,\,\dots,\,10$, and on the younger brother's clock $\tau=4+0.8(t-5)=4.5,\,5.0,\,\dots,\,8.0$. The interval is $0.5$ years. The younger brother receives $2$ signals on the outbound leg ($2$-year intervals) and $8$ signals on the return leg ($0.5$-year intervals) (the signal at the moment of turnaround, $\tau=4$, was counted on the outbound leg).

(c) The elder brother received $8$ signals from the younger brother, who sent them at intervals of $1$ year, so the younger brother's trip took $8$ years on his own clock. The younger brother received $2+8=10$ signals from the elder brother, who sent them at intervals of $1$ year, so the elder brother has aged $10$ years. The result that the elder brother has aged $10$ years and the younger brother $8$ years was thus confirmed by a method that merely counts light signals and does not use simultaneity. The reception intervals are $2$ years on the outbound leg and $0.5$ years on the return leg, a ratio of $4$ (the square of the Doppler factor $\sqrt{(1+\beta)/(1-\beta)}=2$), and this is the same whether seen from the elder brother's side or from the younger brother's.

Solution

(a) The velocity addition law: as seen from the rest frame of one particle (velocity $+0.9c$), the speed of the other (velocity $-0.9c$) would be $1.8c$ in the Galilean treatment, but by the addition law,

$$ \beta=\frac{0.9+0.9}{1+0.9\times0.9}=\frac{1.8}{1.81}=0.99448 $$

which is about $99.4\ \%$ of the speed of light. Now we find it using rapidity. $\phi=\tanh^{-1}0.9=\frac12\ln\frac{1.9}{0.1}=\frac12\ln19\approx1.4722$. Since the rapidity is $+\phi$ for one particle and $-\phi$ for the other, the rapidity of the relative velocity is the difference, $2\phi=2.9444$ (Theorem 56.7). Then $\beta=\tanh(2\phi)=\dfrac{e^{2\phi}-e^{-2\phi}}{e^{2\phi}+e^{-2\phi}}=\dfrac{19-1/19}{19+1/19}=\dfrac{360}{362}=\dfrac{180}{181}=0.99448$ (we used $e^{2\phi}=19$). This agrees with the result of the addition law, $1.8/1.81=180/181$.

(b) $\gamma_{\mathrm{rel}}=1/\sqrt{1-\beta^2}$, and since $1-\beta^2=1-\dfrac{180^2}{181^2}=\dfrac{181^2-180^2}{181^2}=\dfrac{361}{181^2}$, we have $\sqrt{1-\beta^2}=19/181$ and $\gamma_{\mathrm{rel}}=181/19\approx9.526$.

(c) In the laboratory, $\gamma=\cosh\phi=1/\sqrt{1-0.81}=1/\sqrt{0.19}$, so $\gamma^2=1/0.19=5.263$. Then $\cosh2\phi=2\cosh^2\phi-1=2\times5.263-1=9.526$. This agrees with $\gamma_{\mathrm{rel}}=\cosh2\phi=9.526$ from (b) (we applied $\gamma=\cosh\phi$ of Formula 56.2 to the rapidity of the relative velocity, $2\phi$). Colliding two $0.9c$ beams head-on is equivalent, in the rest frame of one particle, to a particle with $\gamma\approx9.5$ striking a particle at rest.

Solution

(a) Substituting $x^\mu=(ct,x,0,0)$ into the definition of the inner product $x\cdot x=(x^0)^2-(x^1)^2-(x^2)^2-(x^3)^2$ gives

$$ x\cdot x=(ct)^2-x^2-0^2-0^2=c^2t^2-x^2 $$

(b) Since all the components are in $\mathrm{m}$, the unit of $x\cdot x$ is $\mathrm{m^2}$.

$$ \begin{aligned} (ct,x)=(4,1):&\quad x\cdot x=4^2-1^2=15\ \mathrm{m^2}\gt0\quad\text{(timelike)}\\ (ct,x)=(1,4):&\quad x\cdot x=1^2-4^2=-15\ \mathrm{m^2}\lt0\quad\text{(spacelike)}\\ (ct,x)=(3,3):&\quad x\cdot x=3^2-3^2=0\quad\text{(lightlike)} \end{aligned} $$

(c) In the inertial frame S′ in which the event $(4,1)$ and the event at the origin occur at the same place, $x'^1=0$, so $x'\cdot x'=(ct')^2$. Since the inner product is invariant (Theorem 57.1), $x'\cdot x'=x\cdot x=15\ \mathrm{m^2}$, and $ct'=\sqrt{15}\ \mathrm{m}=3.873\ \mathrm{m}$. Therefore the proper time is

$$ \Delta\tau=\frac{\sqrt{15}\ \mathrm{m}}{3.00\times10^8\ \mathrm{m/s}}=1.29\times10^{-8}\ \mathrm{s}\quad(12.9\ \mathrm{ns}) $$

Check: this frame S′ is the frame in which the origin and the event are regarded as being at the same place, so it moves relative to frame S with velocity $v=x/t=c\times1/4$, that is, $\beta=0.25$, and $\gamma=1/\sqrt{1-1/16}=4/\sqrt{15}$. By the Lorentz transformation, $ct'=\gamma(ct-\beta x)=\dfrac{4}{\sqrt{15}}(4-0.25\times1)=\dfrac{4\times3.75}{\sqrt{15}}=\dfrac{15}{\sqrt{15}}=\sqrt{15}\ \mathrm{m}$, which agrees. Also, $x'=\gamma(x-\beta ct)=\dfrac{4}{\sqrt{15}}(1-0.25\times4)=0$.

Solution

(a) Since $\gamma=1/\sqrt{1-0.8^2}=1/0.6=5/3$ and $\beta\gamma=0.8\times5/3=4/3$,

$$ \Lambda=\begin{pmatrix}5/3&-4/3\\-4/3&5/3\end{pmatrix},\qquad g=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\qquad g\Lambda=\begin{pmatrix}5/3&-4/3\\4/3&-5/3\end{pmatrix} $$

Since $\Lambda$ is a symmetric matrix, $\Lambda^{\mathsf T}=\Lambda$, and therefore

$$ \Lambda^{\mathsf T}g\Lambda=\begin{pmatrix}5/3&-4/3\\-4/3&5/3\end{pmatrix}\begin{pmatrix}5/3&-4/3\\4/3&-5/3\end{pmatrix}=\begin{pmatrix}\frac{25}{9}-\frac{16}{9}&-\frac{20}{9}+\frac{20}{9}\\-\frac{20}{9}+\frac{20}{9}&\frac{16}{9}-\frac{25}{9}\end{pmatrix}=\begin{pmatrix}1&0\\0&-1\end{pmatrix}=g $$

(b) From $a'^0=\gamma(a^0-\beta a^1)=\dfrac53(3-0.8\times1)=\dfrac53\times2.2=\dfrac{11}{3}$, $a'^1=\gamma(a^1-\beta a^0)=\dfrac53(1-0.8\times3)=\dfrac53\times(-1.4)=-\dfrac73$, $a'^2=2$, and $a'^3=0$, we obtain $a'^\mu=(11/3,\,-7/3,\,2,\,0)$. Similarly, from $b'^0=\dfrac53(2-0.8\times1)=2$, $b'^1=\dfrac53(1-0.8\times2)=\dfrac53\times(-0.6)=-1$, $b'^2=0$, and $b'^3=1$, we obtain $b'^\mu=(2,\,-1,\,0,\,1)$. The inner product is

$$ a'\cdot b'=\frac{11}{3}\times2-\left(-\frac73\right)(-1)-2\times0-0\times1=\frac{22}{3}-\frac73=\frac{15}{3}=5 $$

which agrees with the value in frame S, $a\cdot b=3\times2-(1\times1+2\times0+0\times1)=5$ (Theorem 57.1).

Solution

(a) Since $u^\mu=\gamma(c,\bm{v})$, we have $u\cdot u=(\gamma c)^2-\gamma^2\bm{v}\cdot\bm{v}=\gamma^2(c^2-v^2)$. Substituting $\gamma^2=1/(1-v^2/c^2)$ gives

$$ u\cdot u=\frac{c^2-v^2}{1-v^2/c^2}=\frac{c^2\left(1-v^2/c^2\right)}{1-v^2/c^2}=c^2 $$

which does not depend on the speed $v$. (b) If $v=0.8c$, then $\gamma=1/\sqrt{1-0.64}=5/3$, and $u^0=\gamma c=\tfrac53c$, $u^1=\gamma v=\tfrac53\times0.8c=\tfrac43c$, $u^2=u^3=0$; that is, $u^\mu=(5/3,\ 4/3,\ 0,\ 0)\,c$. Check: $u\cdot u=(25/9-16/9)c^2=c^2$. (c) Since $\dd t/\dd\tau=u^0/c=\gamma=5/3$, $\Delta\tau=\Delta t/\gamma=30\ \mathrm{s}\times3/5=18\ \mathrm{s}$. A clock moving with the particle advances by only $18\ \mathrm{s}$ (time dilation).

Solution

(a) Since the particle is at rest, $\bm{v}=\bm{0}$ and $\gamma=1$, so $p^\mu=mu^\mu=m(c,\bm{0})=(mc,\bm{0})$. Comparing with $p^0=E/c$ gives $E=mc^2$, and the spatial components give $\bm{p}=\bm{0}$.

(b) Writing $E=\sqrt{(pc)^2+(mc^2)^2}=mc^2\sqrt{1+x}$ with $x=(p/mc)^2\ll1$ and using $\sqrt{1+x}=1+\tfrac12x-\tfrac18x^2+\cdots$, we obtain

$$ E\approx mc^2\left[1+\frac12\frac{p^2}{m^2c^2}-\frac18\frac{p^4}{m^4c^4}\right]=mc^2+\frac{p^2}{2m}-\frac{p^4}{8m^3c^2} $$

Keeping only the first two terms gives $E\approx mc^2+p^2/2m$. For an electron with $pc=0.05\ \mathrm{MeV}$, the exact value is $E=\sqrt{0.05^2+0.511^2}=\sqrt{0.263621}=0.513440\ \mathrm{MeV}$, so $T=E-mc^2=0.002440\ \mathrm{MeV}=2.44\ \mathrm{keV}$. On the other hand, $p^2/2m=(pc)^2/(2mc^2)=0.0025/1.022=0.002446\ \mathrm{MeV}$. The relative deviation is $0.002446/0.002440-1\approx0.24\%$, which is nearly equal to the correction $p^2/(4m^2c^2)=(0.05/0.511)^2/4=0.0024$ caused by the third term. Including the third term, we subtract $p^4/(8m^3c^2)=(pc)^4/[8(mc^2)^3]=6.25\times10^{-6}/(8\times0.1334)=5.86\times10^{-6}\ \mathrm{MeV}$ to get $0.002446-0.000006=0.002440\ \mathrm{MeV}$, which agrees with the exact value.

(c) If $m=0$, then $E^2=(pc)^2$, and since $E\gt0$, $E=pc$. In this case $\beta=v/c=pc/E=1$, that is, $v=c$ (a massless particle moves at the speed of light).

Solution

(a) $E=mc^2=(1\ \mathrm{kg})(2.998\times10^8\ \mathrm{m/s})^2=8.99\times10^{16}\ \mathrm{J}$. In TNT equivalent, this is $8.99\times10^{16}/(4.184\times10^9)\approx2.15\times10^7\ \mathrm{t}$, about $21.5$ megatons ($1\ \mathrm{Mt}=10^6\ \mathrm{t}$).

(b) The required quantity of heat is $Q=cm\Delta T=4186\ \mathrm{J/(kg\,K)}\times1\ \mathrm{kg}\times100\ \mathrm{K}=4.19\times10^5\ \mathrm{J}$, and since this is the increase $\Delta E$ in the energy of the water, the increase in mass is

$$ \Delta m=\frac{\Delta E}{c^2}=\frac{4.186\times10^5\ \mathrm{J}}{\left(2.998\times10^8\ \mathrm{m/s}\right)^2}=4.66\times10^{-12}\ \mathrm{kg} $$

(about $4.7\ \mathrm{ng}$; the fractional increase is $4.7\times10^{-12}$). In everyday exchanges of energy, the change in mass is too small to be measured.

(c) Expanding $\gamma=(1-\beta^2)^{-1/2}$ with $(1-x)^{-1/2}=1+\tfrac12x+\tfrac38x^2+\cdots$ ($x=\beta^2$) gives $\gamma-1=\tfrac12\beta^2+\tfrac38\beta^4+\cdots$. Since $\beta^2mc^2=mv^2$ and $\beta^4mc^2=mv^4/c^2$,

$$ T=(\gamma-1)mc^2=\frac12mv^2+\frac38\,\frac{mv^4}{c^2}+\cdots $$

The first term is the kinetic energy $\tfrac12mv^2$ of high school physics. When $v=0.05c$, $\beta^2=0.0025$, $\gamma-1=1.25235\times10^{-3}$, and $\tfrac12\beta^2=1.25\times10^{-3}$, so the relative deviation is $1.25235/1.25-1\approx1.9\times10^{-3}$, about $0.19\%$ (nearly equal to $\tfrac34\beta^2=0.1875\%$).

Solution

(a) In the rest frame, the four-momentum before the decay is $P^\mu=(Mc,\bm{0})$, and after the decay it is $p_1^\mu+p_2^\mu$. Taking the inner product of each side of $p_2^\mu=P^\mu-p_1^\mu$ with itself gives $p_2\cdot p_2=P\cdot P-2P\cdot p_1+p_1\cdot p_1$, and since $p_2\cdot p_2=p_1\cdot p_1=m^2c^2$, $P\cdot P=M^2c^2$, and $P\cdot p_1=Mc\cdot E_1/c=ME_1$,

$$ m^2c^2=M^2c^2-2ME_1+m^2c^2\quad\Longrightarrow\quad E_1=\frac{Mc^2}{2} $$

From conservation of energy, $E_1+E_2=Mc^2$, we get $E_2=Mc^2/2$, which shows that $E_1=E_2=Mc^2/2$. From $p^2c^2=E_1^2-m^2c^4=M^2c^4/4-m^2c^4$, the magnitude of the momentum is

$$ p=\frac{c}{2}\sqrt{M^2-4m^2} $$

(this requires $M\gt2m$).

(b) Taking $M=91.19\ \mathrm{GeV}/c^2$ and $m=m_\mu=105.658\ \mathrm{MeV}/c^2$, the energy is $E=Mc^2/2=45.595\ \mathrm{GeV}$. The momentum is $pc=\sqrt{E^2-(m_\mu c^2)^2}=\sqrt{45595^2-105.658^2}\ \mathrm{MeV}=45594.9\ \mathrm{MeV}\approx45.595\ \mathrm{GeV}$ (it differs from $E$ by only $0.12\ \mathrm{MeV}$), so $p\approx45.595\ \mathrm{GeV}/c$. Also, $\gamma=E/(m_\mu c^2)=45595/105.658\approx431.5$, so the muon travels at nearly the speed of light ($1-\beta\approx1/(2\gamma^2)=2.7\times10^{-6}$).

Solution

(a) The sum of the masses after the reaction is $2m_{\mathrm p}+m_{\pi^0}$, and the threshold condition is $M_{\mathrm{inv}}c^2=(2m_{\mathrm p}+m_{\pi^0})c^2$. The invariant mass for a fixed target is $M_{\mathrm{inv}}^2c^4=2m_{\mathrm p}^2c^4+2Em_{\mathrm p}c^2$ (Example 57.22), so

$$ 2m_{\mathrm p}^2c^4+2Em_{\mathrm p}c^2=\left(2m_{\mathrm p}+m_{\pi^0}\right)^2c^4=\left(4m_{\mathrm p}^2+4m_{\mathrm p}m_{\pi^0}+m_{\pi^0}^2\right)c^4 $$

Solving this for $E$ gives $E=\left(m_{\mathrm p}+2m_{\pi^0}+\dfrac{m_{\pi^0}^2}{2m_{\mathrm p}}\right)c^2$, so the kinetic energy is $T=E-m_{\mathrm p}c^2=\left(2m_{\pi^0}+\dfrac{m_{\pi^0}^2}{2m_{\mathrm p}}\right)c^2$. Substituting the numerical values,

$$ T=2\times134.977+\frac{134.977^2}{2\times938.272}=269.954+\frac{18218.8}{1876.544}=269.954+9.709=279.66\ \mathrm{MeV}\approx280\ \mathrm{MeV} $$

(b) In the colliding-beam case, the total momentum is $0$ and $M_{\mathrm{inv}}c^2=2E$, so $2E=(2m_{\mathrm p}+m_{\pi^0})c^2$ and $E=m_{\mathrm p}c^2+\tfrac12m_{\pi^0}c^2$. Therefore the kinetic energy per beam is $T=\tfrac12m_{\pi^0}c^2=67.49\ \mathrm{MeV}$ ($135\ \mathrm{MeV}$ for the two beams).

(c) In the fixed-target case, the momentum carried by the beam proton does not disappear after the reaction, so the particles produced continue to move forward as a whole (the same situation as in Figure 57.7 of Section 57.8). Part of the kinetic energy is therefore used for the motion of the center of mass and cannot be used to create mass. To compensate, $T$ must be $280\ \mathrm{MeV}$, about twice the rest energy $135\ \mathrm{MeV}$ of the $\pi^0$. In the colliding-beam case, the total momentum is zero, and the total kinetic energy $2\times67.49=135\ \mathrm{MeV}$ is exactly the rest energy of the $\pi^0$.

Solution

At $\theta=180^\circ$, $1-\cos\theta=2$, so from the energy formula for Compton scattering in Section 57.9, $E'=E/[1+(E/m_{\mathrm e}c^2)(1-\cos\theta)]$, we obtain

$$ E'=\frac{662\ \mathrm{keV}}{1+2\times662/511}=\frac{662}{1+2.591}\ \mathrm{keV}=\frac{662}{3.591}\ \mathrm{keV}=184.4\ \mathrm{keV}\approx184\ \mathrm{keV} $$

From conservation of energy, the kinetic energy received by the electron is $T_{\mathrm e}=E-E'=662-184.4=477.6\ \mathrm{keV}\approx478\ \mathrm{keV}$, and this is the maximum energy the electron can receive in the scattering (the Compton edge). The direction of the electron is determined by conservation of momentum, $\bm{p}_{\mathrm e}'=\bm{q}-\bm{q}'$. In backscattering the scattered photon travels opposite to the incident direction ($\bm{q}'=-(E'/c)\bm{n}$), so $\bm{p}_{\mathrm e}'=[(E+E')/c]\,\bm{n}$; that is, the electron flies in the same direction as the incident photon ($\phi=0^\circ$). The magnitude of the momentum is $p_{\mathrm e}c=E+E'=846.4\ \mathrm{keV}$, and the energy is $\sqrt{846.4^2+511^2}=989\ \mathrm{keV}=511+477.6\ \mathrm{keV}$, which is consistent with conservation of energy. As another check, we compute with wavelengths: $\lambda=hc/E=1239.84\ \mathrm{keV\,pm}/662\ \mathrm{keV}=1.873\ \mathrm{pm}$ and $\lambda'=\lambda+2\lambda_{\mathrm C}=1.873+4.853=6.726\ \mathrm{pm}$, so $E'=1239.84/6.726=184.3\ \mathrm{keV}$, which agrees.

Solution

(a) The energy corresponding to a length $\ell$ is $E=\hbar c/\ell$. With $\ell=0.84\ \mathrm{fm}$,

$$ E=\frac{197.327\ \mathrm{MeV\,fm}}{0.84\ \mathrm{fm}}=2.35\times10^{2}\ \mathrm{MeV} $$

(here $197.327/0.84=234.9$). The world at the size of the proton lies at an energy scale of a few hundred $\mathrm{MeV}$.

(b) In natural units, $1\ \mathrm{GeV}^{-1}$ becomes the length $\hbar c/(1\ \mathrm{GeV})$ when multiplied by $\hbar c$, and the time $\hbar/(1\ \mathrm{GeV})$ when multiplied by $\hbar$ (Formula 58.1). Using $1\ \mathrm{GeV}=10^{3}\ \mathrm{MeV}$, we obtain

$$ \frac{\hbar c}{1\ \mathrm{GeV}}=\frac{197.327\ \mathrm{MeV\,fm}}{10^{3}\ \mathrm{MeV}}=0.197\,327\ \mathrm{fm}=1.973\times10^{-16}\ \mathrm{m},\qquad \frac{\hbar}{1\ \mathrm{GeV}}=\frac{6.582\,12\times10^{-22}\ \mathrm{MeV\,s}}{10^{3}\ \mathrm{MeV}}=6.582\times10^{-25}\ \mathrm{s} $$

(c) Since this is a quantity with the dimension of an area, we square the length $1\ \mathrm{GeV}^{-1}$: $(0.197\,327\ \mathrm{fm})^2=0.038\,94\ \mathrm{fm^2}$. Because $1\ \mathrm{fm^2}=10^{-30}\ \mathrm{m^2}=10\ \mathrm{mb}$ (since $1\ \mathrm{mb}=10^{-31}\ \mathrm{m^2}$, we have $10^{-30}/10^{-31}=10$),

$$ 1\ \mathrm{GeV}^{-2}=0.038\,94\times10\ \mathrm{mb}=0.389\ \mathrm{mb} $$

Conversely, $1\ \mathrm{mb}$ is about $2.57\ \mathrm{GeV}^{-2}$.

(d) The SI unit of $\hbar c$ is $[\mathrm{J\,s}]\cdot[\mathrm{m/s}]=\mathrm{J\,m}$ (energy × length). In natural units $\hbar c=1$, so “energy × length $=1$”; that is, length $=1/$energy, and the dimension of length is “energy$^{-1}$”. Similarly, from the unit of $\hbar$, $\mathrm{J\,s}$, we get time $=\hbar/$energy, so the dimension of time is also “energy$^{-1}$”.

∎

Solution

(a) We have $G_{\mathrm F}^2=(1.166\,379\times10^{-5})^2\ \mathrm{GeV}^{-4}=1.3604\times10^{-10}\ \mathrm{GeV}^{-4}$, $m_\mu^5=(0.105\,658)^5\ \mathrm{GeV}^5=1.3168\times10^{-5}\ \mathrm{GeV}^5$, and $192\pi^3=192\times31.006=5953.2$. Therefore

$$ \Gamma=\frac{1.3604\times10^{-10}\times1.3168\times10^{-5}}{5953.2}\ \mathrm{GeV}=\frac{1.7914\times10^{-15}}{5953.2}\ \mathrm{GeV}=3.009\times10^{-19}\ \mathrm{GeV} $$

(since $\mathrm{GeV}^{-4}\times\mathrm{GeV}^{5}=\mathrm{GeV}$, the result has the dimension of energy; it corresponds to $3.009\times10^{-10}\ \mathrm{eV}$).

(b) In $\tau=\hbar/\Gamma$, substituting $\hbar=6.582\,12\times10^{-25}\ \mathrm{GeV\,s}$ gives

$$ \tau=\frac{6.582\,12\times10^{-25}\ \mathrm{GeV\,s}}{3.009\times10^{-19}\ \mathrm{GeV}}=2.187\times10^{-6}\ \mathrm{s} $$

The difference from the measured value $2.197\times10^{-6}\ \mathrm{s}$ is about $0.45\%$, which nearly matches the difference caused by this simple formula's neglect of the effect of the electron mass (about $0.02\%$) and of the radiative correction of quantum electrodynamics (which reduces the decay width by about $0.4\%$). Merely by calculating with the formula in natural units, we obtain a lifetime of order $10^{-6}\ \mathrm{s}$ with three significant figures.

(c) Noting that $G_{\mathrm F}/(\hbar c)^3$ is a quantity measured in $\mathrm{GeV}^{-2}$, and making the replacement $m_\mu\to m_\mu c^2$ for the mass, we obtain

$$ \Gamma=\frac{1}{192\pi^3}\left[\frac{G_{\mathrm F}}{(\hbar c)^3}\right]^2(m_\mu c^2)^5 $$

(dimensions: $[\mathrm{GeV}^{-2}]^2[\mathrm{GeV}]^5=\mathrm{GeV}$). The lifetime is $\tau=\hbar/\Gamma$, and if $G_{\mathrm F}$ is given in SI units ($\mathrm{J\,m^3}$), then $\Gamma=G_{\mathrm F}^2(m_\mu c^2)^5/[192\pi^3(\hbar c)^6]$ (dimensions: $(\mathrm{J\,m^3})^2\mathrm{J}^5/(\mathrm{J\,m})^6=\mathrm{J}$).

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Solution

(a) In frame S, $A^0=\varphi/c=-E_0x/c$ and $A^1=A_x=0$. To write the result in the coordinates of frame S′, we use the inverse transformation $x=\gamma(x'+\beta ct')$. From $\beta=0.5$, we get $\gamma=1/\sqrt{1-0.25}=2/\sqrt3$, $\gamma^2=4/3$, $\gamma^2\beta=2/3$, and $\gamma^2\beta^2=1/3$. Using the same transformation as for $x^\mu$, namely $A'^0=\gamma(A^0-\beta A^1)$ and $A'^1=\gamma(A^1-\beta A^0)$, we obtain

$$ A'^0=\gamma A^0=-\frac{\gamma E_0}{c}\,x=-\frac{\gamma^2E_0}{c}\,(x'+\beta ct'),\qquad A'^1=-\gamma\beta A^0=\frac{\gamma\beta E_0}{c}\,x=\frac{\gamma^2\beta E_0}{c}\,(x'+\beta ct') $$

Therefore $\varphi'=cA'^0$, $A'_x=A'^1$, and $A'_y=A'_z=0$, and substituting the numerical values gives

$$ \varphi'=-\frac43E_0\left(x'+\frac{ct'}2\right),\qquad A'_x=\frac23\,\frac{E_0}{c}\left(x'+\frac{ct'}2\right) $$

(dimensions: $E_0x'\ [\mathrm V]$ has the dimension of electric potential, and $E_0x'/c\ [\mathrm{V\,s/m}]$ has the dimension of $\bm A$).

(b) The gradient of the electric potential is $\partial_{x'}\varphi'=-\dfrac43E_0$, and the time derivative of the vector potential is $\partial_{t'}A'_x=\dfrac23\dfrac{E_0}{c}\cdot\dfrac c2=\dfrac13E_0$, so

$$ E'_x=-\pdiff{\varphi'}{x'}-\pdiff{A'_x}{t'}=\frac43E_0-\frac13E_0=E_0 $$

Also, $E'_y=-\partial_{y'}\varphi'-\partial_{t'}A'_y=0$ and $E'_z=0$. Furthermore, for the curl of $\bm A'=(A'_x(t',x'),0,0)$ we have $(\rot'\bm A')_y=\partial_{z'}A'_x-\partial_{x'}A'_z=0$, $(\rot'\bm A')_z=\partial_{x'}A'_y-\partial_{y'}A'_x=0$, and $(\rot'\bm A')_x=0$, so $\bm B'=\bm0$. This agrees with the result obtained by setting $E_y=E_z=0$, $\bm B=\bm0$ in Theorem 58.5, namely $E'_x=E_x=E_0$, $\bm E'_\perp=\bm0$, $\bm B'=\bm0$. An electric field parallel to the direction of motion is unchanged even for a moving observer. Note also that $\gamma^2=4/3$ appears in the electric potential and $\gamma^2\beta^2=1/3$ in the time variation of $A'_x$, and that the difference $\gamma^2(1-\beta^2)=1$ remains; this illustrates well a property of the Lorentz transformation.

(c) Since $\dfrac1{c^2}\partial_{t'}\varphi'=\dfrac1{c^2}\left(-\dfrac43E_0\cdot\dfrac c2\right)=-\dfrac23\dfrac{E_0}c$ and $\partial_{x'}A'_x=\dfrac23\dfrac{E_0}c$,

$$ \partial'_\mu A'^\mu=\frac1{c^2}\pdiff{\varphi'}{t'}+\pdiff{A'_x}{x'}=-\frac23\frac{E_0}c+\frac23\frac{E_0}c=0 $$

so the Lorenz condition also holds in frame S′ (in frame S we had $\partial_\mu A^\mu=\dfrac1{c^2}\partial_t\varphi+\nabla\cdot\bm A=0+0=0$; this is an example of the covariance of the Lorenz condition).

∎

Solution

(a) We have $A^0=\varphi/c=0$, $A^1=-B_0y/2$, $A^2=B_0x/2$, $A^3=0$, $\partial^0=\dfrac1c\partial_t$, and $\partial^i=-\partial_i$. The time derivative is zero (the field does not depend on time), and $A^0=0$, so $F^{0i}=\partial^0A^i-\partial^iA^0=0$. The spatial components are

$$ F^{12}=\partial^1A^2-\partial^2A^1=-\pdiff{A^2}{x}+\pdiff{A^1}{y}=-\frac{B_0}{2}+\left(-\frac{B_0}{2}\right)=-B_0 $$

Moreover, $F^{13}=-\partial_xA^3+\partial_zA^1=0$ and $F^{23}=-\partial_yA^3+\partial_zA^2=0$. From antisymmetry, $F^{21}=+B_0$. Therefore

$$ F^{\mu\nu}=\begin{pmatrix}0&0&0&0\\0&0&-B_0&0\\0&B_0&0&0\\0&0&0&0\end{pmatrix} $$

This agrees with the relation $F^{12}=-B_z$, with $B_z=B_0$ and all the other components being $0$. Also, $F_{\mu\nu}F^{\mu\nu}=2B_0^2$.

(b) We have $\beta=0.6$, $\gamma=1.25$, and $v=0.6c$. Substituting $E_y=0$, $B_z=B_0$ into Theorem 58.5 gives

$$ E'_y=\gamma(E_y-vB_z)=-1.25\times0.6\times2.998\times10^{8}\ \mathrm{m/s}\times1.0\ \mathrm{T}=-2.25\times10^{8}\ \mathrm{V/m},\qquad B'_z=\gamma\left(B_z-\frac{v}{c^2}E_y\right)=1.25\ \mathrm{T} $$

Also, $E'_x=E'_z=0$ and $B'_x=B'_y=0$. The invariant is

$$ E'^2-c^2B'^2=(2.248\times10^{8})^2-(2.998\times10^{8}\times1.25)^2=5.06\times10^{16}-1.404\times10^{17}=-8.99\times10^{16}\ \mathrm{V^2/m^2} $$

which agrees with $-c^2B_0^2=-(2.998\times10^{8}\times1.0)^2=-8.99\times10^{16}\ \mathrm{V^2/m^2}$. An observer moving through a uniform magnetic field observes a strong electric field (in this example, $2.2\times10^{8}\ \mathrm{V/m}$) and a slightly strengthened magnetic field.

∎

Solution

(a) The equation for $\nu=2$ is $\partial_0F^{02}+\partial_1F^{12}+\partial_2F^{22}+\partial_3F^{32}=\mu_0j^2$. From the matrix of the field-strength tensor, $F^{02}=-E_y/c$, $F^{12}=-B_z$, $F^{22}=0$, and $F^{32}=B_x$, so

$$ -\frac1{c^2}\pdiff{E_y}{t}-\pdiff{B_z}{x}+\pdiff{B_x}{z}=\mu_0j_y $$

Since $\partial_zB_x-\partial_xB_z=(\rot\bm B)_y$, we obtain $(\rot\bm B)_y=\mu_0j_y+\dfrac1{c^2}\pdiff{E_y}{t}$.

(b) In the equation $\partial_\mu F^{\mu\nu}=\mu_0j^\nu$, apply $\partial_\nu$ to both sides and sum over $\nu$ from $0$ to $3$:

$$ \partial_\nu\partial_\mu F^{\mu\nu}=\mu_0\,\partial_\nu j^\nu $$

The value of the left-hand side does not change when the names of the summed indices $\mu,\nu$ are interchanged, so $\partial_\nu\partial_\mu F^{\mu\nu}=\partial_\mu\partial_\nu F^{\nu\mu}$. By the antisymmetry of $F$, $F^{\nu\mu}=-F^{\mu\nu}$, the chain continues as $=-\partial_\mu\partial_\nu F^{\mu\nu}$, and by interchanging the order of partial differentiation, $\partial_\mu\partial_\nu=\partial_\nu\partial_\mu$, it continues as $=-\partial_\nu\partial_\mu F^{\mu\nu}$. This is equal to the original left-hand side with its sign reversed, so it is $0$. Therefore $\partial_\nu j^\nu=0$, that is, $\dfrac1c\partial_t(c\rho)+\nabla\cdot\bm j=\partial_t\rho+\nabla\cdot\bm j=0$ (conservation of charge).

∎

Solution

(a) In frame S, $E_y=E_0$, $B_z=E_0/c$, and the other components are $0$. With $\beta=0.6$ and $\gamma=1.25$, and letting $v=\beta c$, Theorem 58.5 gives

$$ E'_y=\gamma(E_y-vB_z)=\gamma\left(E_0-\beta c\cdot\frac{E_0}{c}\right)=\gamma(1-\beta)E_0=1.25\times0.4\times1.0\times10^{3}\ \mathrm{V/m}=5.0\times10^{2}\ \mathrm{V/m} $$ $$ B'_z=\gamma\left(B_z-\frac{v}{c^2}E_y\right)=\gamma\left(\frac{E_0}{c}-\frac{\beta E_0}{c}\right)=\gamma(1-\beta)\frac{E_0}{c}=\frac{5.0\times10^{2}\ \mathrm{V/m}}{2.998\times10^{8}\ \mathrm{m/s}}=1.67\times10^{-6}\ \mathrm{T} $$

and the other components remain $0$. The relation $E'_y/B'_z=c$, that is, $E'=cB'$, is preserved (a consequence of the invariant $E^2-c^2B^2=0$).

(b) We have $E'/E_0=\gamma(1-\beta)=\dfrac{1-\beta}{\sqrt{(1-\beta)(1+\beta)}}=\sqrt{\dfrac{1-\beta}{1+\beta}}=\sqrt{\dfrac{0.4}{1.6}}=0.5$, which is the same factor as $\omega'/\omega$ for the wave four-vector in Example 58.7. To an observer moving away together with the wave, the amplitude and the frequency both appear smaller by the same factor (this corresponds to the energy of a photon decreasing while the number of photons is preserved).

(c) If the velocity of S′ is in the $-x$ direction (the direction approaching the light source), then replacing $\beta\to-\beta$ gives $E'/E_0=\gamma(1+\beta)=\sqrt{\dfrac{1+\beta}{1-\beta}}=2$ and $E'=2.0\times10^{3}\ \mathrm{V/m}$ ($B'=E'/c$). The amplitude doubles, and the frequency also doubles (blueshift).

∎

Solution

(a) From $p\ [\mathrm{GeV}/c]=0.2998\,B\ [\mathrm{T}]\,r\ [\mathrm{m}]$ in Formula 58.2,

$$ r=\frac{p}{0.2998\,B}=\frac{10}{0.2998\times5\times10^{-5}}\ \mathrm{m}=6.67\times10^{5}\ \mathrm{m}=6.7\times10^{2}\ \mathrm{km} $$

Compared with the Earth's radius, $6.4\times10^{3}\ \mathrm{km}$, this is about $1/10$ ($667/6400=0.104$), so the geomagnetic field bends cosmic-ray protons of around $10\ \mathrm{GeV}/c$ on a radius far smaller than the size of the Earth. This is the reason why the geomagnetic field keeps low-energy cosmic rays away from the Earth's surface.

(b) The trajectory radius $r=p/(qB)$ is determined only by the momentum and the magnitude $\abs{q}$ of the electric charge, and does not depend on the mass. An electron has the same magnitude of electric charge as a proton, so for the same momentum the radius is the same, $6.7\times10^{2}\ \mathrm{km}$ (since the sign of the charge is opposite, the direction of bending is opposite).
If instead the kinetic energy is the same, $T=10\ \mathrm{GeV}$, the momentum is determined by $pc=\sqrt{T^2+2Tmc^2}$. For the proton ($m_{\mathrm p}c^2=0.938\ \mathrm{GeV}$), $pc=\sqrt{100+18.77}\ \mathrm{GeV}=10.90\ \mathrm{GeV}$ and $r=10.90/(0.2998\times5\times10^{-5})\ \mathrm{m}=7.27\times10^{5}\ \mathrm{m}$. For the electron ($m_{\mathrm e}c^2=0.511\times10^{-3}\ \mathrm{GeV}$), $pc=\sqrt{100+0.0102}\ \mathrm{GeV}=10.0005\ \mathrm{GeV}$ and $r=6.67\times10^{5}\ \mathrm{m}$, so the difference is only about $9\%$. When the energy is sufficiently high, the difference in mass has almost no effect (the ultrarelativistic limit, where $pc\approx T$).

∎

Solution

(a) We have $p=\partial L/\partial v=\gamma mv$ and $H=pv-L=\gamma mv^2+mc^2/\gamma=\gamma mc^2$ (main text, Section 58.9.2). We now write this in terms of $p$. Since $p^2c^2=\gamma^2m^2v^2c^2=\gamma^2m^2c^4\beta^2$ and $\gamma^2\beta^2=\gamma^2-1$ (from $\gamma^2(1-\beta^2)=1$), we have $p^2c^2=(\gamma^2-1)m^2c^4$, that is, $\gamma^2m^2c^4=p^2c^2+m^2c^4$. Taking the square root, with $\gamma mc^2\gt0$, we obtain

$$ H=\gamma mc^2=\sqrt{p^2c^2+m^2c^4} $$

This expresses the Hamiltonian in terms of the momentum.

(b) Writing $H=mc^2\sqrt{1+x}$ with $x=p^2/(mc)^2$, for $p\ll mc$ we have $x\ll1$. From $\sqrt{1+x}=1+\dfrac x2-\dfrac{x^2}8+\cdots$ (Taylor expansion),

$$ H=mc^2\left(1+\frac{p^2}{2m^2c^2}-\frac{p^4}{8m^4c^4}+\cdots\right)=mc^2+\frac{p^2}{2m}-\frac{p^4}{8m^3c^2}+\cdots $$

The first term is the rest energy, the second term is the nonrelativistic kinetic energy, and the third term is the first relativistic correction (it acts to decrease the kinetic energy).

(c) Let $pc=0.10\ \mathrm{MeV}$ and $mc^2=0.511\ \mathrm{MeV}$. The exact kinetic energy is

$$ T=\sqrt{(pc)^2+(mc^2)^2}-mc^2=\sqrt{0.0100+0.2611}\ \mathrm{MeV}-0.511\ \mathrm{MeV}=9.693\times10^{-3}\ \mathrm{MeV} $$

The nonrelativistic value is $p^2/(2m)=(pc)^2/(2mc^2)=0.0100/1.022\ \mathrm{MeV}=9.785\times10^{-3}\ \mathrm{MeV}$. The difference is $9.785\times10^{-3}-9.693\times10^{-3}=9.2\times10^{-5}\ \mathrm{MeV}=92\ \mathrm{eV}$ (about $1\%$ of $T$). The third term in (b) is $-\dfrac{(pc)^4}{8(mc^2)^3}=-\dfrac{1.0\times10^{-4}}{8\times0.1334}\ \mathrm{MeV}=-9.4\times10^{-5}\ \mathrm{MeV}=-94\ \mathrm{eV}$, which agrees well with the magnitude of the difference (the remaining few $\mathrm{eV}$ come from still higher-order terms).

∎