Chapter 44Applications of the Lagrangian Formalism and Variational Principles

Example 44.1: Lagrange's equations of motion contain Newton's equation of motion

A particle (point mass) of mass $m$ moves along the $x$ axis, subject only to a conservative force $F=-\dd U/\dd x$ (Section 4.4) described by the potential energy $U(x)$. Apply Lagrange's equations of motion to $L=\frac12m\dot x^2-U(x)$ and verify that Newton's equation of motion $m\ddot x=F$ is obtained. Then write down the result when this is extended to three-dimensional rectangular coordinates $(x,y,z)$.

Solution. We differentiate $L=\frac12m\dot x^2-U(x)$ partially. The only term containing $\dot x$ is $\frac12m\dot x^2$, so $\partial L/\partial\dot x=m\dot x$. The only term containing $x$ is $-U(x)$, so $\partial L/\partial x=-\dd U/\dd x$ ($q$ and $\dot q$ are treated as independent variables; see Illustration 44.1 below). Substituting into Lagrange's equations of motion, we obtain

(since $m$ is constant, $\diff{}{t}(m\dot x)=m\ddot x$). Hence $m\ddot x=-\dd U/\dd x=F$, which is indeed Newton's equation of motion. For example, for a spring $U=\frac12kx^2$, so $m\ddot x=-kx$, and for gravity $U=mgx$, so $m\ddot x=-mg$.

In three dimensions, if we take $L=\frac12m(\dot x^2+\dot y^2+\dot z^2)-U(x,y,z)$, the same calculation can be carried out for each of $q=x,y,z$, giving $m\ddot x=-\partial U/\partial x$, $m\ddot y=-\partial U/\partial y$, and $m\ddot z=-\partial U/\partial z$. Written together, this is $m\bm a=-\nabla U=\bm F$, which is precisely Newton's equation of motion in vector form (Chapter 2). The unit of the right-hand side is $\mathrm{J/m}=\mathrm N$, and that of the left-hand side is also $\mathrm{kg\,m/s^2}=\mathrm N$, so they agree.

Illustration 44.1: Taking partial derivatives with $q$ and $\dot q$ as independent variables

Find $\partial L/\partial\dot q$ and $\partial L/\partial q$ for $L=\dfrac12 m\dot q^2-mgq$.

Solution. When computing $\partial L/\partial\dot q$, we treat $q$ as a constant and differentiate with respect to $\dot q$ alone. The only term in $L$ that contains $\dot q$ is $\frac12m\dot q^2$, so

On the other hand, when computing $\partial L/\partial q$, we treat $\dot q$ as a constant and differentiate with respect to $q$ alone. The only term in $L$ that contains $q$ is $-mgq$, so

Note that $\partial L/\partial\dot q=m\dot q$ (the dimension of momentum) and $\partial L/\partial q=-mg$ (the dimension of force) are completely different expressions—since the term containing $\dot q$ and the term containing $q$ are different terms, their partial derivatives can be computed independently of each other.

Illustration 44.2: Falling motion (Lagrangian formalism)

Solve the free-fall motion with the $x$ axis taken vertically upward using Lagrange's equations of motion.

Solution. It goes exactly as in the main text, Eqs. \eqref{eq:44-fall-U}–\eqref{eq:44-fall-eom}, and the result is $\diff{}{t}(m\dot x)=-mg$, that is, $\ddot x=-g$ (motion with constant acceleration).

Example 44.2: Falling motion with air resistance — the case including a nonconservative force

A particle of mass $m$ falls freely, with the coordinate $x$ taken vertically upward. When air resistance proportional to the velocity (proportionality constant $k>0$, always opposite in direction to the velocity) acts, find the terminal velocity (the constant velocity after a sufficiently long time has elapsed).

Solution. Air resistance is a nonconservative force, so it cannot be included in the potential energy $U$. We use Lagrange's equations of motion in the form that includes nonconservative forces, learned in Chapter 43,

We keep $L$ as Eq. \eqref{eq:44-fall-L} ($L=\frac12m\dot x^2-mgx$), which includes only gravity, and put the air resistance on the right-hand side as the generalized force $Q'$ (by the definition $Q'=\sum_j F_j'\,\partial x_j/\partial q$ in Chapter 43; when the rectangular coordinate $x$ is used as it is, $Q'=F'_x$). Since the resistive force is proportional to the velocity $\dot x$ and opposite in direction, $Q'=-k\dot x$ (a resistive force of the same form was treated in Section 7.3. Since $k\dot x$ is a force, the unit of $k$ is $\mathrm{N/(m/s)}=\mathrm{kg/s}$). Therefore,

At the terminal velocity $v_{\text{terminal}}$, the acceleration is 0 ($\ddot x=0$), so substituting this gives

(Dimension check: the unit of $mg/k$ is $\mathrm{kg\,(m/s^2)/(kg/s)}=\mathrm{m/s}$, which agrees with the unit of velocity.) The negative sign indicates that, since we took $x$ to be positive upward, the terminal velocity is directed downward (with magnitude $mg/k$)—the larger the air resistance (the larger $k$) or the lighter the particle (the smaller $m$), the smaller the terminal velocity becomes (this is why raindrops and parachutes keep falling at a constant speed).

Illustration 44.3: Simple harmonic motion (Lagrangian formalism)

Solve the spring pendulum of Figure 44.2 using Lagrange's equations of motion.

Solution. It goes exactly as in the main text, and the result is $\diff{}{t}(m\dot x)=-k(x-x_0)$ (Eq. \eqref{eq:44-sho-eom}), that is, if $x_0$ is taken as the origin, the equation of motion of simple harmonic motion, $m\ddot x=-kx$ (Theorem 7.1).

Example 44.3: Period and maximum speed of a spring pendulum

A particle of mass $m=0.20\ \mathrm{kg}$ is attached to a spring of spring constant $k=80\ \mathrm{N/m}$ (Figure 44.2). At time $t=0$, it is pulled to the right by $A=0.050\ \mathrm m$ from the natural-length position and released gently. (1) Find the angular frequency $\omega$ and the period. (2) Find the position $x(t)$. (3) Find the maximum speed of the particle.

Solution. (1) Using the displacement from the natural-length position, $X\equiv x-x_0$, Eq. \eqref{eq:44-sho-eom} becomes $m\ddot X=-kX$, that is, $\ddot X=-\omega^2X$ ($\omega\equiv\sqrt{k/m}$) (since $x_0$ is a constant, $\ddot X=\ddot x$). Thus

Here $\mathrm{N/(m\,kg)}=\mathrm{s^{-2}}$, so taking the square root gives $\mathrm{s^{-1}}$. The period is $T_0=2\pi/\omega=2\pi/(20\ \mathrm{s^{-1}})=0.314\ \mathrm s$ (about $0.31\ \mathrm s$) (the same as $T_0=2\pi\sqrt{m/k}$ of high school physics).

(2) The general solution is $X=C_1\cos\omega t+C_2\sin\omega t$ (Theorem 7.1). From the initial condition $X(0)=A$, we get $C_1=A$. Putting the initial condition $\dot X(0)=0$ into the velocity $\dot X=-C_1\omega\sin\omega t+C_2\omega\cos\omega t$ gives $C_2\omega=0$, so $C_2=0$. Therefore

is the position.

(3) The velocity is $\dot x=-A\omega\sin\omega t$, and since $\abs{\sin\omega t}\le1$, the maximum speed is $A\omega=0.050\ \mathrm m\times20\ \mathrm{s^{-1}}=1.0\ \mathrm{m/s}$ (attained when the particle passes the natural-length position). Checking with energy, the initial elastic energy is $\frac12kA^2=\frac12\times80\times0.050^2=0.10\ \mathrm J$, and the kinetic energy at the maximum speed is $\frac12m(A\omega)^2=\frac12\times0.20\times1.0^2=0.10\ \mathrm J$; the two indeed agree.

Illustration 44.4: Simple pendulum (Lagrangian formalism)

Solve the simple pendulum of Figure 44.3 using Lagrange's equations of motion with the polar coordinate $\theta$, which satisfies the constraint $r=\ell$ from the start.

Solution. It goes exactly as in the main text, and the result is $\ddot\theta=-(g/\ell)\sin\theta$ (Eq. \eqref{eq:44-pend-eom-final}). Note that the tension in the string of the pendulum does not appear in the equation of motion at all—because we used only the coordinate $\theta$, which satisfies the constraint $r=\ell$, there was no need to consider the constraint force (tension) from the start, as in advantage (IV) of Section 44.1. If you want to know the magnitude of the tension itself, use the method of Lagrange multipliers in Section 44.7.

Example 44.4: Energy conservation for the simple pendulum and the speed at the lowest point

A simple pendulum of length $\ell=1.0\ \mathrm m$ and mass $m=0.50\ \mathrm{kg}$ is released from rest at a position tilted by $\theta_0=60^\circ$ from the vertically downward direction. (1) From the equation of motion \eqref{eq:44-pend-eom}, show that $\frac12m\ell^2\dot\theta^2+mg\ell(1-\cos\theta)$ is constant in time. (2) Find the speed of the particle at the lowest point ($\theta=0$).

Solution. (1) Eq. \eqref{eq:44-pend-eom} reads $m\ell^2\ddot\theta=-mg\ell\sin\theta$, that is, $m\ell^2\ddot\theta+mg\ell\sin\theta=0$. Multiplying both sides by $\dot\theta$ gives

Each term on the left-hand side can be written, by the rule for differentiating a composite function (University Mathematics, Section 6.4), as $\diff{}{t}\left(\frac12m\ell^2\dot\theta^2\right)=m\ell^2\dot\theta\ddot\theta$ and $\diff{}{t}\big(-mg\ell\cos\theta\big)=mg\ell\sin\theta\,\dot\theta$. Adding the constant $mg\ell$ does not change the time derivative, so

That is, $E=T+U=\frac12m\ell^2\dot\theta^2+mg\ell(1-\cos\theta)$ is constant (conservation of mechanical energy, Section 4.5).

(2) At the instant of release, $\dot\theta=0$ and $\theta=\theta_0$, so $E=mg\ell(1-\cos\theta_0)$. At the lowest point $\theta=0$ we have $U=0$, so

Substituting the numerical values, with $\cos60^\circ=0.5$, we obtain

The mass $m$ cancels and drops out, so the result does not depend on the mass (it is the same result as $\frac12mv^2=mgh$ in high school physics, with $h=\ell(1-\cos\theta_0)=0.50\ \mathrm m$). The kinetic energy is $\frac12\times0.50\times9.8=2.45\ \mathrm J$.

Illustration 44.5: Planetary motion (Lagrangian formalism)

Find the equations of motion of a planet moving around the Sun using Lagrange's equations of motion in polar coordinates.

Solution. As in the main text (Eqs. \eqref{eq:44-planet-abstract}–\eqref{eq:44-planet-final}), the result is $m(\ddot r-r\dot\theta^2)=-GMm/r^2$ and $m(2\dot r\dot\theta+r\ddot\theta)=0$, which agree exactly with the result obtained by Newtonian mechanics in Chapter 8.

Example 44.5: Circular orbits and Kepler's third law

A planet moves around the Sun on a circular orbit of radius $R$ ($r=R$ constant). Find the orbital angular velocity $\Omega\equiv\dot\theta$ and the orbital period $P$. Also confirm that $P^2$ is proportional to $R^3$ (Kepler's third law).

Solution. On a circular orbit $r=R$ is constant, so $\dot r=0,\ \ddot r=0$. Substituting these into the first equation of Eq. \eqref{eq:44-planet-final}, $m(\ddot r-r\dot\theta^2)=-GMm/r^2$, gives

(Note that the mass $m$ cancels from both sides and disappears—regardless of the planet's mass, circular orbits of the same radius have the same angular velocity.) From the definition of angular velocity, the period is $P=2\pi/\Omega$, so

The constant of proportionality $4\pi^2/(GM)$ does not depend on the planet (it is determined by the Sun's mass $M$ alone), so $P^2\propto R^3$; that is, Kepler's third law is confirmed (in Section 8.3 we study it in detail, including the case of an elliptical orbit, where $R$ is replaced by the semi-major axis $a$).

Example 44.6: Action of a free particle — Does the straight path really give a stationary value (minimum value)?

A free particle of mass $m=1\ \mathrm{kg}$ (a particle on which no force acts; since $U=0$, $L=T=\frac12m\dot x^2$) is at $x=0$ at time $t=0$ and at $x=X=1\ \mathrm m$ at time $t=\tau=1\ \mathrm s$. Keeping the endpoints fixed, we displace the path to

($a$ is a constant expressing the size of the displacement of the path, in units of $\mathrm m$; $a=0$ is the uniform rectilinear motion that connects the endpoints by a straight line). Find the action $S(a)=\int_0^\tau\frac12m\dot x^2\,\dd t$ and verify that it takes its minimum value at $a=0$. Also find the action of the polygonal path made of two straight line segments, $(0,0)\to(\tau/2,\ 0.8X)\to(\tau,X)$, and compare the results.

Solution. Differentiating $x(t)$ with respect to time, and noting that the derivative of $\sin(\pi t/\tau)$ is $(\pi/\tau)\cos(\pi t/\tau)$, we obtain

We square this and substitute it into the expression for the action:

We integrate each term. The first term gives $(X/\tau)^2\cdot\tau=X^2/\tau$. The integral in the second term, $\int_0^\tau\cos(\pi t/\tau)\,\dd t=\dfrac{\tau}{\pi}\Big[\sin\dfrac{\pi t}{\tau}\Big]_0^\tau=0$, vanishes. For the third term, the half-angle formula $\cos^2\phi=(1+\cos2\phi)/2$ gives $\int_0^\tau\cos^2(\pi t/\tau)\,\dd t=\tau/2$, so the term is $(a\pi/\tau)^2\cdot\tau/2=a^2\pi^2/(2\tau)$. Therefore

(the unit is $\mathrm{kg\,m^2/s}=\mathrm{J\,s}$). Substituting numerical values gives $S(a)=0.500+2.47\,a^2\ \mathrm{J\,s}$; at $a=0$ we have $S(0)=0.500\ \mathrm{J\,s}$, and at $a=\pm0.5\ \mathrm m$ we have $S=1.117\ \mathrm{J\,s}$. As in the graph on the right of Figure 44.7, it is an upward-opening parabola.

Let us consider the meaning of Eq. \eqref{eq:44-free-action}. Since the first-order term in the size $a$ of the displacement (the integral of the second term) is zero, near $a=0$ the value of $S$ does not increase or decrease in proportion to $a$; it only increases as $a^2$. This is precisely the content of “the first-order change in the action is zero even if the path is displaced slightly,” that is, $\delta S=0$ (stationarity), and since the coefficient of the second-order term is positive, the stationary value is a minimum. In other words, the uniform rectilinear motion that a particle with no external force actually follows (the law of inertia of Chapter 2) minimizes the action within this family of paths.

For the polygonal path, the speed in the first half, $0\le t\le\tau/2$, is $0.8X/(\tau/2)=1.6\ \mathrm{m/s}$ and the speed in the second half is $0.2X/(\tau/2)=0.4\ \mathrm{m/s}$, each constant, so

which is larger than the $0.500\ \mathrm{J\,s}$ of the straight path. Even with the same endpoints, a path whose speed changes along the way has a larger action.

Finally, let us show in general that “the straight path is the minimum among all paths.” We write an arbitrary path with fixed endpoints as $x(t)=x_0(t)+\eta(t)$. Here $x_0(t)=Xt/\tau$ is the straight path, and $\eta(t)$ is an arbitrary displacement that vanishes at the endpoints ($\eta(0)=\eta(\tau)=0$). Since $\dot x=\dot x_0+\dot\eta$, we have

Since $\dot x_0=X/\tau$ is a constant, the integral in the second term vanishes: $\int_0^\tau\dot\eta\,\dd t=\eta(\tau)-\eta(0)=0$. Hence $S=S_0+\int_0^\tau\frac12m\dot\eta^2\,\dd t\ge S_0$, and equality holds only when $\dot\eta\equiv0$ (that is, $\eta\equiv0$). For any displacement whatever, the straight path has the least action.

Example 44.7: The Euler–Lagrange equation for simple harmonic motion (check by the calculus of variations)

Apply the Euler–Lagrange equation, Eq. \eqref{eq:44-el-var}, to $L=\frac12m\dot x^2-\frac12kx^2$ and derive the equation of motion of simple harmonic motion, $m\ddot x=-kx$.

Solution. We have $\partial L/\partial\dot x=m\dot x$ (from the term $\frac12m\dot x^2$ alone) and $\partial L/\partial x=-kx$ (from the term $-\frac12kx^2$ alone). Substituting these into Eq. \eqref{eq:44-el-var} gives

which agrees exactly with Theorem 7.1 and with the result of Section 44.2.2. In Section 44.2 we obtained exactly the same result by the procedure “form $L=T-U$ and substitute it into Lagrange's equations derived from Newton's laws”; here we have confirmed that the same equation is obtained even when we start from the variational principle “make the action $S=\int L\,\dd t$ stationary.”

Example 44.8: Why is the shortest path between two points a straight line?

Apply the Euler–Lagrange equation (Eq. \eqref{eq:44-el-var} of Section 44.5, read with $q\to y,\ t\to x$) to the functional $s[y]=\int_{x_1}^{x_2}\sqrt{1+y'^2}\,\dd x$ (Eq. \eqref{eq:44-shortest-func}), and show that the shortest path is a straight line.

Solution. Since $F=\sqrt{1+y'^2}$ does not contain $y$ explicitly (it is a function of $y'$ alone), $\partial F/\partial y=0$. On the other hand,

(the derivative of a composite function: differentiating $F=(1+y'^2)^{1/2}$ with respect to $y'$ gives $\frac12(1+y'^2)^{-1/2}\cdot2y'=y'/\sqrt{1+y'^2}$). Substituting into the Euler–Lagrange equation $\diff{}{x}(\partial F/\partial y')-\partial F/\partial y=0$, we obtain

This says that “the quantity in parentheses is constant, independent of $x$,” so $y'/\sqrt{1+y'^2}=C$ (constant). When we solve this equation for $y'$, then, since the right-hand side $C$ is a constant, the left-hand side $y'/\sqrt{1+y'^2}$ (a monotonically increasing function of $y'$) must also be constant, and therefore $y'$ itself is constant (let $y'\equiv a$). Integrating $y'=a$ (constant), the general solution is

(where $a,b$ are constants determined by the endpoint conditions $y(x_1)=y_1,\ y(x_2)=y_2$), which is the straight line joining the two points. This is the derivation, by the calculus of variations, of the fact that the shortest path between two points is a straight line. (What the Euler–Lagrange equation gives is the stationary path; that it really is the shortest can also be confirmed from the triangle inequality, “the sum of two sides is longer than the remaining side.”)

Example 44.9: The angles of a light ray passing through three layers

Air (refractive index $n_1=1.00$), water ($n_2=1.33$), and glass ($n_3=1.50$) are stacked, in this order, as flat layers. Light is incident from the air at an angle $\theta_1=30^\circ$ to the normal to the layers. (1) Find the angles $\theta_2,\theta_3$ that the ray makes with the normal in the water and in the glass. (2) What is the angle $\theta_3'$ if there is no water layer and the light is incident directly from the air onto the glass? (3) Light can pass from the glass into a layer of air only when the angle of incidence in the glass is smaller than what value?

Solution. (1) By Eq. \eqref{eq:44-snell-layer}, $n\sin\theta$ has a common value in all the layers. On the incident side the value is $n_1\sin\theta_1=1.00\times\sin30^\circ=0.500$. In the water, $1.33\sin\theta_2=0.500$, so

In the glass, $1.50\sin\theta_3=0.500$, so

(2) Even without the water layer, the conserved quantity $n\sin\theta=0.500$ does not change, so $\theta_3'=19.5^\circ$, which is equal to $\theta_3$. The refractive indices of the intermediate layers do not affect the final angle (it is determined by the conserved quantity $n\sin\theta$ alone). (3) On exiting into a layer of air, we must have $\sin\theta_{\text{air}}=n_3\sin\theta_3/1.00\le1$ (since $\sin$ cannot exceed 1). Hence $\sin\theta_3\le1.00/1.50=0.667$, and $\theta_3\le\arcsin0.667=41.8^\circ$. For angles of incidence larger than this, the light cannot exit into the air layer and is totally internally reflected (the critical angle for total internal reflection in Chapter 14).

Example 44.10: The brachistochrone curve (the brachistochrone problem)

Suppose a particle slides down a smooth, frictionless wire under gravity alone. Take the starting point as the origin $(0,0)$, and take $y$ positive vertically downward. The particle starts from the origin at rest (with zero initial speed) and slides down to a point $(x_1,y_1)$ ($x_1,y_1\gt0$). Find the curve $y(x)$ that minimizes the time taken (the brachistochrone). (This is one of the most famous problems in the history of the calculus of variations; it was posed by Johann Bernoulli in 1696.)

Solution. By the law of conservation of energy (Section 4.5), the speed after descending a distance $y$ from the starting point follows from $\frac12mv^2=mgy$ as $v=\sqrt{2gy}$. The arc-length element is $\dd s=\sqrt{1+y'^2}\,\dd x$, so the time taken is the functional obtained by summing $\dd t=\dd s/v$:

The integrand $F=\sqrt{1+y'^2}/\sqrt{2gy}$ does not contain $x$ explicitly, so Formula 44.1 (the Beltrami identity) can be used. Since $\partial F/\partial y'=y'/\big(\sqrt{2gy}\sqrt{1+y'^2}\big)$,

This is the result (note that after combining over a common denominator, the $y'^2$ terms cancel each other). From the condition of Formula 44.1 that this equals a constant $C$,

we obtain an ordinary differential equation ($R$ is a positive constant with the dimension of length). Let us solve it. Since the left side contains $1+y'^2$, we would like to put it in the form of the reciprocal of the square of something. We therefore recall the trigonometric identity $1+\cot^2\phi=1/\sin^2\phi$ and, using a new variable $\theta$ (theta), set $y'=\dd y/\dd x=\cot(\theta/2)$ (we consider the part with $y'\ge0$, in the range $0\lt\theta\le\pi$). Then $1+y'^2=1+\cot^2(\theta/2)=1/\sin^2(\theta/2)$, so from Eq. \eqref{eq:44-brach-ode} we obtain

(by the half-angle formula $\sin^2(\theta/2)=(1-\cos\theta)/2$). Next we find $x$. We have $\dd y=\dfrac R2\sin\theta\,\dd\theta=R\sin\dfrac\theta2\cos\dfrac\theta2\,\dd\theta$, and $\dd x=\dd y/y'$, so

Integrating, we obtain $x=\frac R2(\theta-\sin\theta)+(\text{constant})$; since the curve passes through the origin $(x,y)=(0,0)$ ($\theta=0$), the constant is 0. Setting $a\equiv R/2$, we have

Let us verify that this is the cycloid (the locus of a point on the circumference of a circle as the circle rolls). Suppose a circle of radius $a$ rolls without slipping along the line $y=0$ (since $y$ is positive downward, the circle lies below this line). When the circle has rotated through an angle $\theta$, its center has moved to $(a\theta,\ a)$ (the distance rolled equals the length $a\theta$ of the circular arc). The point on the circumference that was at the origin at the start is then displaced from the center by $(-a\sin\theta,\ -a\cos\theta)$, so its coordinates are $x=a\theta-a\sin\theta,\ y=a-a\cos\theta$, which agree with Eq. \eqref{eq:44-cycloid}. Therefore the brachistochrone is a cycloid. Here we have considered only the part with $y'\ge0$ (up to the lowest point $\theta=\pi$). When the endpoint lies at a shallow, distant position with $y_1/x_1\lt2/\pi$, the same Eq. \eqref{eq:44-cycloid} also holds for the part in which the particle passes the lowest point and then rises again ($\pi\lt\theta\lt2\pi$, $y'\lt0$). In either case, the constant $a$ and the value $\theta_1$ at the endpoint are determined by solving $x_1=a(\theta_1-\sin\theta_1),\ y_1=a(1-\cos\theta_1)$ simultaneously.

The descent time can also be found exactly. On the cycloid, $\dd x/\dd\theta=a(1-\cos\theta)$ and $\dd y/\dd\theta=a\sin\theta$, so $\dd s/\dd\theta=a\sqrt{(1-\cos\theta)^2+\sin^2\theta}=a\sqrt{2-2\cos\theta}=2a\sin(\theta/2)$, and the speed is $v=\sqrt{2gy}=\sqrt{2ga(1-\cos\theta)}=2\sqrt{ga}\,\sin(\theta/2)$. Hence $\dd t=\dd s/v=\sqrt{a/g}\,\dd\theta$, so the integrand is a constant independent of $\theta$, and the descent time from the starting point to $\theta_1$ is

For example, take $a=1\ \mathrm m$ and the endpoint $\theta_1=\pi$ (the lowest point of the cycloid). Then the endpoint is $(x_1,y_1)=(\pi,\,2)\ \mathrm m$, and $t_{\text{cycloid}}=\sqrt{\dfrac{1\ \mathrm m}{9.8\ \mathrm{m/s^2}}}\times\pi=1.004\ \mathrm s$. For descent along the straight line $y=cx$ ($c=y_1/x_1=2/\pi$) joining the same two points, using $v=\sqrt{2gcx}$ we get $t_{\text{line}}=\int_0^{x_1}\sqrt{1+c^2}\,\dd x/\sqrt{2gcx}=\sqrt{2(1+c^2)x_1/(gc)}=1.190\ \mathrm s$, so the particle reaches the endpoint sooner along the cycloid (Figure 44.9). Intuitively, this is because gaining speed first on a steep slope and then easing off gives a larger average speed than descending at a constant slope throughout. It is also known that the cycloid has the property that the time to reach the lowest point is the same wherever the particle is released from rest (it is an isochronous curve, or tautochrone); this property was discovered in the 17th century by Christiaan Huygens.

Example 44.11: Finding the tension of a simple pendulum by the method of Lagrange multipliers

Treat the simple pendulum (Figure 44.3) without using the constraint $r=\ell$, in rectangular coordinates $(x,y)$ with the origin at the fixed point $\mathrm O$ ($y$ positive vertically downward). The constraint is that the length of the rod is constant, $g(x,y)=x^2+y^2-\ell^2=0$. Set up the equations of motion using the method of Lagrange multipliers, and find the tension $N$ in the string (the magnitude of the force with which the rod pulls the particle) as a function of the angle $\theta$ (the displacement angle measured from the vertically downward direction) and the angular velocity $\dot\theta$. Furthermore, when a pendulum with $m=0.50\ \mathrm{kg}$ and $\ell=1.0\ \mathrm m$ is released from rest at $\theta_0=60^\circ$, find the tension at the lowest point and the tension at the instant of release.

Solution. The kinetic energy is $T=\frac12m(\dot x^2+\dot y^2)$ and the potential energy is (since $y$ is positive downward) $U=-mgy$, so $L=T-U=\frac12m(\dot x^2+\dot y^2)+mgy$. From the constraint $g=x^2+y^2-\ell^2$ we have $\partial g/\partial x=2x,\ \partial g/\partial y=2y$, and also $\partial L/\partial x=0,\ \partial L/\partial y=mg$, so by Theorem 44.2 (Eq. \eqref{eq:44-multiplier-eq}),

We substitute the coordinates that satisfy the constraint, $x=\ell\sin\theta,\ y=\ell\cos\theta$ ($\theta=0$ is the lowest point). The time derivatives are $\dot x=\ell\dot\theta\cos\theta,\ \dot y=-\ell\dot\theta\sin\theta$, and furthermore

so the two equations become

The unknowns are the two quantities $\ddot\theta$ and $\lambda$. Although (i) and (ii) are equations in the $x$ direction and the $y$ direction, we want to arrange them into components along the direction of motion (the tangential direction $(\cos\theta,-\sin\theta)$) and along the string (the radial direction $(\sin\theta,\cos\theta)$); this corresponds to rotating the coordinate axes through the angle $\theta$. The tangential component is (i)$\times\cos\theta-$(ii)$\times\sin\theta$, in which $\lambda$ drops out because the constraint force does not point in the tangential direction. The radial component is (i)$\times\sin\theta+$(ii)$\times\cos\theta$, in which $\ddot\theta$ drops out. First, forming (i)$\times\cos\theta-$(ii)$\times\sin\theta$, the left side becomes $m\ell\ddot\theta(\cos^2\theta+\sin^2\theta)+mg\sin\theta=m\ell\ddot\theta+mg\sin\theta$ (the $\dot\theta^2$ terms vanish, since $-m\ell\dot\theta^2\sin\theta\cos\theta+m\ell\dot\theta^2\sin\theta\cos\theta=0$), and the right side is $2\lambda\ell\sin\theta\cos\theta-2\lambda\ell\cos\theta\sin\theta=0$, so

This agrees exactly with the equation of motion obtained in Section 44.2.3 (Eq. \eqref{eq:44-pend-eom-final}) by using the constraint from the beginning. Next, forming (i)$\times\sin\theta+$(ii)$\times\cos\theta$, the left side becomes $-m\ell\dot\theta^2(\sin^2\theta+\cos^2\theta)-mg\cos\theta$ (the $\ddot\theta$ terms vanish, since $m\ell\ddot\theta\sin\theta\cos\theta-m\ell\ddot\theta\sin\theta\cos\theta=0$), and the right side is $2\lambda\ell(\sin^2\theta+\cos^2\theta)=2\lambda\ell$, so

is obtained (the unit of $\lambda$ is $\mathrm{N/m}$, since $mg/\ell$ has the unit $\mathrm{kg\,(m/s^2)/m}=\mathrm{N/m}$; the unit of the component $2\lambda x$ of the constraint force is $\mathrm{N/m}\times\mathrm m=\mathrm N$, which agrees with the unit of force). The constraint force (in rectangular components) is $(2\lambda x,\ 2\lambda y)=2\lambda\ell(\sin\theta,\ \cos\theta)$; its direction is $(\sin\theta,\cos\theta)$, pointing from the origin toward the particle, and its magnitude is the absolute value of $2\lambda\ell$. The tension $N$ is the force pulling the particle toward the origin, that is, in the direction $-(\sin\theta,\cos\theta)$, so $2\lambda\ell=-N$, that is, $N=-2\lambda\ell$. Substituting the $\lambda$ found above, we obtain

This result can also be confirmed by considering Newton's equation of motion in the radial direction (along the rod) directly. The forces acting on the particle are the tension $N$ (toward the origin) and the radial component of gravity $mg\cos\theta$ (directed away from the origin), and since the particle moves in a circle of radius $\ell$, the net force equals the centripetal force $m\ell\dot\theta^2$ (toward the origin): $N-mg\cos\theta=m\ell\dot\theta^2$. This agrees exactly with Eq. \eqref{eq:44-pendulum-tension}.

Let us find the numerical values. From the law of conservation of energy in Example 44.4, $\frac12m\ell^2\dot\theta^2+mg\ell(1-\cos\theta)=mg\ell(1-\cos\theta_0)$, we have $\ell\dot\theta^2=2g(\cos\theta-\cos\theta_0)$, so Eq. \eqref{eq:44-pendulum-tension} can be written as

At the lowest point ($\theta=0$), $N=mg(3-2\cos60^\circ)=2mg=2\times0.50\times9.8=9.8\ \mathrm N$, and at the instant of release ($\theta=\theta_0$, $\dot\theta=0$), $N=mg\cos\theta_0=0.50\times9.8\times0.50=2.45\ \mathrm N$. The tension at the lowest point is twice the tension $mg=4.9\ \mathrm N$ when the particle is at rest (this is why your body feels heavy at the lowest point of a swing).

Exercise 44.1: Simple harmonic motion with two springs (in series)

Two springs, with spring constants $k_1$ and $k_2$, are connected in series; one end is fixed to a wall, and a particle of mass $m$ is attached to the other end. Assume that the connection point of the two springs is massless and can move smoothly. Let $x$ be the displacement of the particle from the position at which the springs are at their natural length, and let $s$ be the elongation of the first spring (on the wall side); then the elongation of the second spring is $x-s$. Taking $x,s$ as the two coordinates, set up the Lagrangian $L$, derive the equations of motion, show that this system undergoes simple harmonic motion with angular frequency $\omega=\sqrt{k_{\text{eq}}/m}$, and find the equivalent spring constant $k_{\text{eq}}$. Also find $\omega$ and the period for $k_1=30\ \mathrm{N/m}$, $k_2=60\ \mathrm{N/m}$, $m=0.20\ \mathrm{kg}$.

Hint: $L=\frac12m\dot x^2-\frac12k_1s^2-\frac12k_2(x-s)^2$. There is no kinetic energy corresponding to the coordinate $s$ (the connection point is massless), so Lagrange's equation of motion for $s$ becomes $\partial L/\partial s=0$, which is an equation of force balance.

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Exercise 44.2: Simple pendulum: comparing rectangular and polar coordinates, and finding the tension

Treat a simple pendulum (length $\ell$, mass $m$, with $\theta$ measured from the vertically downward direction) in the following two ways. (1) In rectangular coordinates $(x,y)$ ($y$ positive downward), write down $L$ without taking the constraint $x^2+y^2=\ell^2$ into account. (2) Keep both polar coordinates $(r,\theta)$ as generalized coordinates and set up $L$; treat the tension $N$ of the string (acting in the radial direction and pulling the particle toward the origin) as a generalized force of nonconservative type, $Q_r'=-N$, write down Lagrange's equations of motion in the $r$ direction and the $\theta$ direction, and substitute $r=\ell$ to find the equation of motion for $\theta$ and the tension $N$. Does the result agree with Section 44.2.3 and Example 44.11? (3) Into the $L$ written in (1), substitute $x=\ell\sin\theta,\ y=\ell\cos\theta$, which satisfy the constraint, and compare the result with the $L$ of Section 44.2.3.

Hint: In (2), $T=\frac12m(\dot r^2+r^2\dot\theta^2)$ and $U=-mgr\cos\theta$. Since the tension acts only in the radial direction, $Q_\theta'=0$. Set up the $r$-direction equation first, and then substitute $r=\ell$ ($\dot r=\ddot r=0$). In (3), compute $\dot x^2+\dot y^2$ and use $\sin^2\theta+\cos^2\theta=1$.

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Exercise 44.3: Energy conservation in planetary motion

From the Lagrangian of planetary motion, $L=\frac12m(\dot r^2+(r\dot\theta)^2)+GMm/r$, show, using the equation of motion \eqref{eq:44-planet-final}, that the mechanical energy $E=T+U=\frac12m(\dot r^2+(r\dot\theta)^2)-GMm/r$ is constant in time. Also, for the circular orbit ($r=R$ constant) of Example 44.5, express $E$ in terms of $G,M,m,R$.

Hint: Compute $\dd E/\dd t$, substitute the equation of motion \eqref{eq:44-planet-final}, and show that it vanishes. For the circular orbit, use the fact that $\dot r=0$ and the result $\Omega^2=GM/R^3$ ($\dot\theta=\Omega$) obtained in Example 44.5.

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Exercise 44.4: Equations of motion and conservation of angular momentum for a general central-force potential

In planetary motion we used the gravitational potential $U=-GMm/r$; now set up Lagrange's equations of motion in polar coordinates $(r,\theta)$ for a general potential $U(r)$ that is a function of $r$ only (a central force; Definition 8.1) as well. Write down the four partial derivatives $\partial L/\partial\dot r,\ \partial L/\partial r,\ \partial L/\partial\dot\theta,\ \partial L/\partial\theta$ one by one, and from the $\theta$-direction equation confirm that $p_\theta=mr^2\dot\theta$ is conserved for any central force (conservation of angular momentum).

Hint: Let $L=\frac12m(\dot r^2+(r\dot\theta)^2)-U(r)$; in the calculation of Section 44.3 it suffices to replace $-GMm/r$ by $-U(r)$. Note that, in the $\theta$-direction equation, $\partial L/\partial\theta=0$ regardless of the specific form of $U(r)$. This result is a preview of Noether's theorem (the relation between symmetry and conservation laws), which you will study in Chapter 46.

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Exercise 44.5: Why is the fixed-endpoint condition $\delta q(t_1)=\delta q(t_2)=0$ needed?

In the derivation of Section 44.5, we used the fact that the boundary term $\left[\partial L/\partial\dot q\cdot\delta q\right]_{t_1}^{t_2}$ vanishes because of the fixed-endpoint condition $\delta q(t_1)=\delta q(t_2)=0$. Without this condition, what expression would the variation $\delta S$ of the action take? Taking the free particle ($L=\frac12m\dot x^2$) as an example, explain, by calculating the boundary term, what happens if we demand “$\delta S=0$” without fixing the position at the final time $t_2$.

Hint: Write down $\delta S$ with the boundary term left in. Note that $\partial L/\partial\dot x=m\dot x$ is the momentum. Physically, fixing the endpoints corresponds to the situation in which we know where the particle was at times $t_1,t_2$ (observational data).

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Exercise 44.6: Why can't the frictional force be derived from a potential energy?

In disadvantage (I) of Section 44.1, we stated that a nonconservative force such as friction cannot be expressed in Lagrange's equations of motion (the form with $L=T-U$). Based on the definition of a conservative force (Section 4.4), explain why a potential energy cannot be defined for the frictional force. For example, suppose that an object of mass $m=1.0\ \mathrm{kg}$ is moved on a horizontal floor with coefficient of kinetic friction $\mu=0.30$ from a point A to a point B, $2.0\ \mathrm m$ away, (a) along a straight line, and (b) along a semicircle of diameter $2.0\ \mathrm m$. How much work does the kinetic friction do in each case?

Hint: A conservative force was defined as a force for which the work is the same regardless of the path taken between two points (otherwise $U$ could not be defined as a function of position only). The magnitude of the kinetic friction is $\mu mg$, and its direction is always opposite to the direction of motion.

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Exercise 44.7: An example where “stationary,” not “minimum,” is the right name (Advanced)

Hamilton's principle has historically been called the “principle of least action,” but, as stated in Section 44.4, $\delta S=0$ in general only guarantees a stationary value (a local minimum, a local maximum, or a saddle point). Consider a harmonic oscillator of mass $m$ and spring constant $k$ ($\omega=\sqrt{k/m}$), with $L=\frac12m\dot x^2-\frac12m\omega^2x^2$, subject to the endpoint conditions that it is at $x=0$ at time $0$ and again at $x=0$ at time $\tau$. (1) The only solution of the equation of motion satisfying these conditions is $x=0$ (when $\omega\tau\ne n\pi$). Find the action $S(\varepsilon)$ when the displacement $x_\varepsilon(t)=\varepsilon\sin(\pi t/\tau)$, which keeps the endpoints fixed, is added to it. (2) For the case $\omega\tau\lt\pi$ and the case $\omega\tau\gt\pi$, determine whether $x=0$ gives the minimum value of the action.

Hint: Substitute $\dot x_\varepsilon=\varepsilon(\pi/\tau)\cos(\pi t/\tau)$ into the expression for the action and use $\int_0^\tau\cos^2(\pi t/\tau)\,\dd t=\int_0^\tau\sin^2(\pi t/\tau)\,\dd t=\tau/2$. The action at $x=0$ is $S=0$. The case $\omega\tau=\pi$ corresponds to a special time interval, equal to half a period of the spring's natural oscillation.

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Exercise 44.8: The shape of a soap film (catenoid) — an application of the Beltrami identity

A soap film stretched between two circular rings with a common axis forms the surface of revolution with the minimum surface area (= surface energy). Take the axis as the $x$ axis and let the curve of the film's cross section be $y(x)$ (the distance from the axis); then the lateral area is $A[y]=2\pi\int_{x_1}^{x_2}y\sqrt{1+y'^2}\,\dd x$. Using the Beltrami identity (Formula 44.1), derive the first-order differential equation satisfied by $y(x)$, and show that its solution is $y=C\cosh\dfrac{x-x_0}{C}$ ($C,x_0$ are constants).

Hint: $F=y\sqrt{1+y'^2}$ does not contain $x$ explicitly. From $F-y'\partial F/\partial y'=y/\sqrt{1+y'^2}=C$ we obtain $y'=\sqrt{y^2/C^2-1}$. Separate the variables and integrate, and use the definition of $\cosh$ and the inverse hyperbolic function $\operatorname{arcosh}$ (University Mathematics, Sections 2.5–2.6).

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Exercise 44.9: Planetary motion: correspondence between the abstract form and the substituted form

For the Lagrangian of planetary motion, $L=\frac12m(\dot r^2+(r\dot\theta)^2)+GMm/r$, (1) compute explicitly each of the four quantities $\diff{}{t}(\partial L/\partial\dot r)$, $\partial L/\partial r$, $\diff{}{t}(\partial L/\partial\dot\theta)$, $\partial L/\partial\theta$ that appear in the abstract form \eqref{eq:44-planet-abstract}. (2) Match the three terms of the $r$-direction equation $m\ddot r-mr\dot\theta^2+GMm/r^2=0$ with “the rate of change of momentum,” “the term corresponding to the centrifugal force ($\partial T/\partial r$),” and “the gravitational attraction (the generalized force $Q_r=-\partial U/\partial r$),” and check the signs. (3) Regard the Earth's revolution as motion on a circular orbit of radius $r=1.496\times10^{11}\ \mathrm m$ centered on the Sun ($M=1.989\times10^{30}\ \mathrm{kg}$), and find the angular velocity $\Omega$ and the orbital period ($G=6.674\times10^{-11}\ \mathrm{N\,m^2/kg^2}$).

Hint: (3) For a circular orbit, $\dot r=\ddot r=0$. Substituting into the $r$-direction equation gives $\Omega^2=GM/r^3$. One year is about $3.16\times10^7\ \mathrm s$.

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Exercise 44.10: Does the principle of least action mean that “nature has a purpose”?

(1) The path of light from A to B can be determined by Fermat's principle (stationarity of the time), or it can be determined step by step by applying Snell's law at each boundary. How do these two ways of putting it differ, and how are they related? Explain. (2) Hamilton's principle sounds as if it says, “the particle knows its final point and chooses the path that minimizes the action.” Explain why this way of putting it easily leads to misunderstanding, on the basis of the fact that “the actual motion is determined, from the initial position and velocity, by the equation of motion at each instant alone.”

Hint: Specifying the initial position and velocity and specifying the positions of the initial point and the final point are just two different sets of conditions for specifying the same motion. The remark in Section 44.4 that “it is stationary, not necessarily a minimum” is also helpful.

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