Chapter 9Moving Reference Frames and Fictitious Forces — Centrifugal and Coriolis Forces

Derivation: Equation of motion in a translationally accelerating frame

Let $\bm R(t)$ be the position vector of the origin O′ of S′ as seen from the origin O of S, $\bm r$ the position vector of the particle P as seen from S, and $\bm r'$ its position vector as seen from S′ (that is, the vector from O′ to P). Then there is the relation

$$ \bm r=\bm R+\bm r' $$

Since S′ does not rotate, differentiating the components of $\bm r'$ (that is, $x',y',z'$ measured along the coordinate axes fixed in S′) directly with respect to time gives the velocity and the acceleration (the subtle difference in the rotating case is treated in Section 9.2). Differentiating both sides twice with respect to the time $t$ (in nonrelativistic mechanics, the same time is assumed to flow in S and in S′), we obtain

$$ \ddot{\bm r}=\ddot{\bm R}+\ddot{\bm r}' $$

Here $\ddot{\bm R}\equiv\bm A(t)$ is the acceleration, as seen from S, of the origin of S′ (= the train or the elevator itself). Suppose that the real force $\bm F$ (the sum of the forces that can be explained as some kind of interaction, such as gravity and the normal force) acts on the particle P. Since Newton's equation of motion in S (an inertial frame) is $m\ddot{\bm r}=\bm F$, we get

$$ m(\bm A+\ddot{\bm r}')=\bm F \qquad\therefore\quad m\ddot{\bm r}'=\bm F-m\bm A $$

The left-hand side $m\ddot{\bm r}'$ is the acceleration of the particle measured by an observer riding in S′, multiplied by the mass. That is, for an observer in S′, the “apparent net force” is the real force $\bm F$ plus the extra term $-m\bm A$.

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Derivation (Part 1): Time derivative of the rotating basis vectors $\dd\bm e_i'/\dd t=\bm\omega\times\bm e_i'$

First consider the concrete case in which the axis of rotation coincides with the $z$ axis ($\bm\omega=\omega\bm e_z$, where $\omega=\dot\psi$ is the time derivative of the rotation angle $\psi(t)$). Then the basis vectors of S′ can be written as

$$ \bm e_x'(t)=\cos\psi\,\bm e_x+\sin\psi\,\bm e_y,\qquad \bm e_y'(t)=-\sin\psi\,\bm e_x+\cos\psi\,\bm e_y,\qquad \bm e_z'=\bm e_z $$

(these are exactly the same expressions as those in Chapter 3 and Chapter 6 that gave the polar-coordinate basis vectors $\bm e_r,\bm e_\theta$ in terms of the basis of rectangular (Cartesian) coordinates, with the replacement $\psi\leftrightarrow\theta$). Differentiating with respect to time, we obtain

$$ \dot{\bm e}_x'=\dot\psi(-\sin\psi\,\bm e_x+\cos\psi\,\bm e_y)=\dot\psi\,\bm e_y'=\omega\,\bm e_y' $$ $$ \dot{\bm e}_y'=\dot\psi(-\cos\psi\,\bm e_x-\sin\psi\,\bm e_y)=-\dot\psi\,\bm e_x'=-\omega\,\bm e_x' $$

(see Figure 9.3). On the other hand, computing the cross products of $\bm\omega=\omega\bm e_z$ with each basis vector (the relations $\bm e_z\times\bm e_x'=\bm e_y'$, $\bm e_z\times\bm e_y'=-\bm e_x'$, and $\bm e_z\times\bm e_z'=\bm 0$ hold because $\bm e_x',\bm e_y',\bm e_z'$, like $\bm e_x,\bm e_y,\bm e_z$, form a right-handed orthonormal basis), we obtain

$$ \bm\omega\times\bm e_x'=\omega\,\bm e_y',\qquad \bm\omega\times\bm e_y'=-\omega\,\bm e_x',\qquad \bm\omega\times\bm e_z'=\bm 0=\dot{\bm e}_z' $$

Thus it has been verified that $\dot{\bm e}_i'=\bm\omega\times\bm e_i'$ holds for all three basis vectors.

Next consider the general case, in which the direction of the axis of rotation may change with time. Here too we show that an “instantaneous angular velocity vector” $\bm\omega(t)$ exists at each instant and that the same equation holds. Write the basis as $\bm e_1',\bm e_2',\bm e_3'$ ($=\bm e_x',\bm e_y',\bm e_z'$). Differentiating with respect to time the condition that the basis is orthogonal and of length 1, $\bm e_i'\cdot\bm e_j'=\delta_{ij}$ (equal to 1 if $i=j$ and to 0 if $i\neq j$), we obtain

$$ \dot{\bm e}_i'\cdot\bm e_j'+\bm e_i'\cdot\dot{\bm e}_j'=0 $$

Now let $\Omega_{ij}\equiv\dot{\bm e}_i'\cdot\bm e_j'$. Then the equation above becomes $\Omega_{ij}+\Omega_{ji}=0$, that is, $\Omega_{ji}=-\Omega_{ij}$ (in particular, $\Omega_{ii}=0$ when $i=j$). Since $\Omega_{ij}$ is the coefficient when $\dot{\bm e}_i'$ is expanded in the basis $\bm e_j'$, the only independent components are the three $\Omega_{12},\Omega_{23},\Omega_{31}$; putting $\omega_3\equiv\Omega_{12},\ \omega_1\equiv\Omega_{23},\ \omega_2\equiv\Omega_{31}$, we can write

$$ \dot{\bm e}_1'=\omega_3\bm e_2'-\omega_2\bm e_3',\qquad \dot{\bm e}_2'=-\omega_3\bm e_1'+\omega_1\bm e_3',\qquad \dot{\bm e}_3'=\omega_2\bm e_1'-\omega_1\bm e_2' $$

On the other hand, let us compute the cross products of $\bm\omega\equiv\omega_1\bm e_1'+\omega_2\bm e_2'+\omega_3\bm e_3'$ with each basis vector, using relations such as $\bm e_1'\times\bm e_2'=\bm e_3'$. We find $\bm\omega\times\bm e_1'=\omega_2\bm e_2'\times\bm e_1'+\omega_3\bm e_3'\times\bm e_1'=-\omega_2\bm e_3'+\omega_3\bm e_2'$, and similarly $\bm\omega\times\bm e_2'=\omega_1\bm e_3'-\omega_3\bm e_1'$ and $\bm\omega\times\bm e_3'=\omega_2\bm e_1'-\omega_1\bm e_2'$, which agree exactly with the three equations above. Therefore, in general, the following holds:

\begin{equation} \diff{\bm e_i'}{t}=\bm\omega\times\bm e_i' \qquad(i=x,y,z) \label{eq:9-basis-derivative} \end{equation}

This $\bm\omega$ is the instantaneous angular velocity vector of S′. Indeed, a vector parallel to $\bm\omega$ does not change its direction at that instant, because $\bm\omega\times\bm\omega=\bm 0$. That is, the direction of $\bm\omega$ is the instantaneous axis of rotation, and its magnitude is the rotation angle per unit time. The case treated earlier, in which the axis is fixed along the $z$ axis, is nothing other than $\bm\omega=\omega\bm e_z$.

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Derivation (Part 2): Proof of Theorem 9.1

Express the vector $\bm Q(t)$ in components with respect to the basis of S′:

$$ \bm Q(t)=Q_x'(t)\,\bm e_x'(t)+Q_y'(t)\,\bm e_y'(t)+Q_z'(t)\,\bm e_z'(t) $$

The time derivative seen from the inertial frame S (applying the product rule as it stands) is

$$ \left(\diff{\bm Q}{t}\right)_{\text{inertial}} =\underbrace{\dot Q_x'\bm e_x'+\dot Q_y'\bm e_y'+\dot Q_z'\bm e_z'}_{\text{(i)}} +\underbrace{Q_x'\dot{\bm e}_x'+Q_y'\dot{\bm e}_y'+Q_z'\dot{\bm e}_z'}_{\text{(ii)}} $$

Term (i) is obtained by differentiating only the components $Q_i'(t)$ with respect to time and multiplying them back onto the same basis vectors $\bm e_i'$; this is precisely “the rate of change of $\bm Q$ measured by an observer riding on S′,” that is, the time derivative seen from the rotating frame, $(\dd\bm Q/\dd t)_{\text{rotating}}$. Substituting $\dot{\bm e}_i'=\bm\omega\times\bm e_i'$ from Eq. \eqref{eq:9-basis-derivative} into term (ii), we obtain

$$ \text{(ii)}=Q_x'(\bm\omega\times\bm e_x')+Q_y'(\bm\omega\times\bm e_y')+Q_z'(\bm\omega\times\bm e_z') =\bm\omega\times(Q_x'\bm e_x'+Q_y'\bm e_y'+Q_z'\bm e_z')=\bm\omega\times\bm Q $$

(here we used the distributive law of the cross product—that is, that $\bm\omega\times(\ )$ is a linear operation). Combining the above, Eq. \eqref{eq:9-rotating-derivative} (Theorem 9.1) holds for an arbitrary vector $\bm Q(t)$.

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Derivation: Acceleration and the equation of motion in a rotating frame

As in Section 9.1, suppose that the origin O′ of S′ undergoes translational motion with respect to the origin O of the inertial frame S, with position vector $\bm R(t)$ (velocity $\bm V=\dot{\bm R}$, acceleration $\bm A=\ddot{\bm R}$), and that S′ in addition rotates relative to S with angular velocity vector $\bm\omega(t)$. Let $\bm r'$ be the position vector of the particle P as seen from O′; then the position vector of P as seen from S is $\bm r=\bm R+\bm r'$. We define the relative velocity and relative acceleration of P, as measured by an observer riding on S′, to be

$$ \bm v_{\text{rel}}\equiv\left(\diff{\bm r'}{t}\right)_{\text{rotating}},\qquad \bm a_{\text{rel}}\equiv\left(\diff{\bm v_{\text{rel}}}{t}\right)_{\text{rotating}} $$

(these are the velocity and acceleration in the ordinary sense, measured with coordinate axes fixed to S′). First, applying Theorem 9.1 (Eq. \eqref{eq:9-rotating-derivative}) to $\bm r'$, the time derivative of $\bm r'$ as seen from S is

$$ \left(\diff{\bm r'}{t}\right)_{\text{inertial}}=\bm v_{\text{rel}}+\bm\omega\times\bm r' $$

Therefore, the velocity of P as seen from S is

\begin{equation} \bm v=\dot{\bm r}=\bm V+\bm v_{\text{rel}}+\bm\omega\times\bm r' \label{eq:9-velocity-composition} \end{equation}

Next, we differentiate this whole equation with respect to time once more, as seen from S. Using the ordinary product rule for differentiation (which holds regardless of the frame) for each term on the right-hand side, we obtain

$$ \dot{\bm v}=\dot{\bm V}+\left(\diff{\bm v_{\text{rel}}}{t}\right)_{\text{inertial}}+\left(\diff{\bm\omega}{t}\right)_{\text{inertial}}\times\bm r'+\bm\omega\times\left(\diff{\bm r'}{t}\right)_{\text{inertial}} $$

Here, $\dot{\bm V}=\bm A$; $\left(\dd\bm\omega/\dd t\right)_{\text{inertial}}=\left(\dd\bm\omega/\dd t\right)_{\text{rotating}}+\bm\omega\times\bm\omega=\dot{\bm\omega}$ (since $\bm\omega\times\bm\omega=\bm 0$, the time derivative of $\bm\omega$ itself is the same whichever frame we view it from; we write it as $\dot{\bm\omega}$); applying Theorem 9.1 to $\bm v_{\text{rel}}$ gives $\left(\dd\bm v_{\text{rel}}/\dd t\right)_{\text{inertial}}=\bm a_{\text{rel}}+\bm\omega\times\bm v_{\text{rel}}$; and further, from the intermediate equation above, $\left(\dd\bm r'/\dd t\right)_{\text{inertial}}=\bm v_{\text{rel}}+\bm\omega\times\bm r'$. Substituting all of these, we obtain

$$ \bm a=\bm A+\left(\bm a_{\text{rel}}+\bm\omega\times\bm v_{\text{rel}}\right)+\dot{\bm\omega}\times\bm r'+\bm\omega\times\left(\bm v_{\text{rel}}+\bm\omega\times\bm r'\right) $$

and, noting that the term $\bm\omega\times\bm v_{\text{rel}}$ appears twice, rearranging gives

\begin{equation} \bm a=\bm A+\bm a_{\text{rel}}+2\,\bm\omega\times\bm v_{\text{rel}}+\bm\omega\times(\bm\omega\times\bm r')+\dot{\bm\omega}\times\bm r' \label{eq:9-general-accel} \end{equation}

Substituting Eq. \eqref{eq:9-general-accel} into Newton's equation of motion $m\bm a=\bm F$ ($\bm F$ is the sum of the real forces) and solving for $\bm a_{\text{rel}}$ (that is, the acceleration that an observer riding on S′ actually measures), we obtain

\begin{equation} m\bm a_{\text{rel}}=\bm F\ \underbrace{-\,m\bm A}_{\text{translational fictitious force}}\ \underbrace{-\,2m\,\bm\omega\times\bm v_{\text{rel}}}_{\text{Coriolis force}}\ \underbrace{-\,m\bm\omega\times(\bm\omega\times\bm r')}_{\text{centrifugal force}}\ \underbrace{-\,m\dot{\bm\omega}\times\bm r'}_{\text{Euler force}} \label{eq:9-rotating-eom} \end{equation}

In addition to the translational fictitious force $-m\bm A$ of Section 9.1, three apparent forces peculiar to rotation have appeared.

The “2” in $2\,\bm\omega\times\bm v_{\text{rel}}$ in Eq. \eqref{eq:9-general-accel} corresponds to (a) and (b) in the “Intuition” above. The first is the effect of $\bm v_{\text{rel}}$ itself changing direction because of the rotation of the frame (the term that arises when Theorem 9.1 is applied to $\bm v_{\text{rel}}$; it corresponds to (b)). The second is the effect that appears inside $\bm\omega\times(\dd\bm r'/\dd t)$: the position $\bm r'$ is carried along by the rotation and so has the velocity $\bm\omega\times\bm r'$, and this $\bm r'$ changes by the amount $\bm v_{\text{rel}}$ (it corresponds to (a)). Since these two have exactly the same magnitude, together they give a factor of 2. In the following Section 9.5, this factor appears once more, as the $2\dot r\dot\theta$ term in the acceleration in polar coordinates.

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Derivation: Direction and magnitude of the centrifugal force

Let us find concretely the direction and magnitude of Eq. \eqref{eq:9-centrifugal-def}. To make the calculation as simple as possible, we choose the coordinate axes anew, just for this instant, so that $\bm\omega=\omega\bm e_z$ (the $z$ axis coincides with the axis of rotation), and let the position of the particle be $\bm r=(x,y,z)$. First,

$$ \bm\omega\times\bm r=\omega\bm e_z\times(x\bm e_x+y\bm e_y+z\bm e_z)=\omega(x\,\bm e_z\times\bm e_x+y\,\bm e_z\times\bm e_y)=\omega(x\bm e_y-y\bm e_x) $$

(we used $\bm e_z\times\bm e_x=\bm e_y$, $\bm e_z\times\bm e_y=-\bm e_x$, and $\bm e_z\times\bm e_z=\bm 0$). Taking the cross product with $\bm\omega=\omega\bm e_z$ once more, we obtain

$$ \bm\omega\times(\bm\omega\times\bm r)=\omega\bm e_z\times\omega(x\bm e_y-y\bm e_x)=\omega^2(x\,\bm e_z\times\bm e_y-y\,\bm e_z\times\bm e_x)=\omega^2(-x\bm e_x-y\bm e_y)=-\omega^2\bm\rho $$

Here $\bm\rho\equiv x\bm e_x+y\bm e_y$ is the vector obtained by projecting $\bm r$ orthogonally onto the plane perpendicular to the axis of rotation (the $z$ axis), that is, “the vector from the axis of rotation to the particle, perpendicular to the axis” (its magnitude $\rho=\sqrt{x^2+y^2}$ is the distance from the axis of rotation. Note that this $\rho$ is different from the radius of curvature $\rho$ in Chapter 3; in this chapter it denotes only the distance from the axis of rotation). Therefore,

\begin{equation} \bm F_{\text{cf}}=-m\bm\omega\times(\bm\omega\times\bm r)=m\omega^2\bm\rho \label{eq:9-centrifugal-result} \end{equation}

and we see that the centrifugal force has magnitude $m\omega^2\rho$ ($\rho$ is the distance from the axis of rotation) and is directed from the axis of rotation toward the particle (outward from the axis). Recall that in Chapter 5 the magnitude of the centripetal force in uniform circular motion was $m\omega^2r$ ($r$ is the radius of rotation); the centrifugal force has exactly the same magnitude as that centripetal force and the opposite direction. For an observer riding in the rotating frame and performing uniform circular motion together with the particle, the real centripetal force (such as the tension of a string) and the centrifugal force balance, and the particle appears to be at rest; we see that the centrifugal force is precisely a device that makes the description consistent in this way.

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Proof: Theorem 9.2

Place the observer on the $+z$ side (above) and let $\bm\omega=\omega\bm e_z$ ($\omega\gt0$). Writing the velocity in the plane perpendicular to $\bm\omega$ as $\bm v_{\text{rel}}=(v_x,v_y,0)$, we have, from $\bm e_z\times\bm e_x=\bm e_y$ and $\bm e_z\times\bm e_y=-\bm e_x$,

$$ \bm\omega\times\bm v_{\text{rel}}=\omega(v_x\,\bm e_z\times\bm e_x+v_y\,\bm e_z\times\bm e_y)=\omega(-v_y,\ v_x,\ 0) $$

Therefore, the Coriolis force of Eq. \eqref{eq:9-coriolis-def} is given by

\begin{equation} \bm F_{\text{Co}}=-2m\,\bm\omega\times\bm v_{\text{rel}}=2m\omega\,(v_y,\ -v_x,\ 0) \label{eq:9-coriolis-plane} \end{equation}

Its magnitude is $2m\omega\sqrt{v_x^2+v_y^2}=2m\omega v_{\text{rel}}$. Also, since $(v_x,v_y)\cdot(v_y,-v_x)=0$, we have $\bm F_{\text{Co}}\perp\bm v_{\text{rel}}$. As for the direction, it suffices to check that $(v_x,v_y)\mapsto(v_y,-v_x)$ is the operation of “rotating a vector clockwise by $90^\circ$ in a coordinate plane in which counterclockwise is positive.” For example, east (the $+x$ direction) $(1,0)$ goes to $(0,-1)$ (south), north $(0,1)$ goes to $(1,0)$ (east), west $(-1,0)$ goes to $(0,1)$ (north), and south $(0,-1)$ goes to $(-1,0)$ (west) (Figure 9.5). In every case the force is on the right-hand side as one faces the direction of motion. If $\omega\lt0$ (clockwise), the sign of Eq. \eqref{eq:9-coriolis-plane} is reversed, so the force points to the left.

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Derivation: The apparent gravitational acceleration $g_{\text{eff}}(\varphi)$

If the Earth is approximated as a perfect sphere (radius $R_{\text{Earth}}$) with a spherically symmetric density distribution, the true gravitational acceleration $g_0$ has the same magnitude everywhere on the surface and points toward the center of the Earth. The centrifugal acceleration $\omega_{\text{Earth}}^2\rho=\omega_{\text{Earth}}^2R_{\text{Earth}}\cos\varphi$ points perpendicular to the axis of rotation (a direction tilted slightly toward the equator relative to the vertical at the surface). The apparent gravitational acceleration $\bm g_{\text{eff}}$ is the vector sum of the two, and in general it is tilted slightly from the direction toward the center of the Earth (this tilt is used in the very definition of “vertically downward”; the vertical in the geodetic sense refers to this resultant direction).

First we find the magnitude approximately. The centrifugal acceleration is small, at most about 0.3% of the true gravity, so to first-order approximation we need only subtract from $g_0$ the component of the centrifugal acceleration along the vertical (the direction toward the center of the Earth). Since the angle between the vertical and the direction of the centrifugal acceleration is exactly $\varphi$, the vertical component is $\omega_{\text{Earth}}^2R_{\text{Earth}}\cos\varphi\times\cos\varphi=\omega_{\text{Earth}}^2R_{\text{Earth}}\cos^2\varphi$, so that we obtain

\begin{equation} g_{\text{eff}}(\varphi)\approx g_0-\omega_{\text{Earth}}^2R_{\text{Earth}}\cos^2\varphi \label{eq:9-latitude-g} \end{equation}

(the unit of the second term on the right-hand side is $\mathrm{(rad/s)^2\cdot m}=\mathrm{m/s^2}$, which agrees with that of an acceleration; the latitude dependence is shown in Figure 9.8). At $\varphi=0^\circ$ (the equator) the decrease is largest, $\omega_{\text{Earth}}^2R_{\text{Earth}}$, and at $\varphi=90^\circ$ (the poles) the decrease is zero (the poles lie on the axis of rotation, so $\rho=0$ and no centrifugal force acts at all).

Next we estimate the tilt. The component of the centrifugal acceleration along the ground surface (the horizontal component) is the component perpendicular to the vertical, namely $\omega_{\text{Earth}}^2R_{\text{Earth}}\cos\varphi\sin\varphi$, and this tilts the direction of the apparent gravity toward the equator. It is largest at $\varphi=45^\circ$, where its value is $\omega_{\text{Earth}}^2R_{\text{Earth}}/2\approx0.017\ \mathrm{m/s^2}$. The tilt angle is only $\omega_{\text{Earth}}^2R_{\text{Earth}}\sin\varphi\cos\varphi/g_0\approx1.7\times10^{-3}\ \mathrm{rad}\approx0.1^\circ$.

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Derivation: The period of Foucault's pendulum (outline)

At a point of latitude $\varphi$, take local rectangular coordinates fixed to the ground: an $x$ axis (east) and a $y$ axis (north) in the horizontal plane, and a $z$ axis pointing vertically upward. The motion of the pendulum takes place mainly in the horizontal plane, and its vertical velocity $v_z$ is very small, so from Eq. \eqref{eq:9-coriolis-local} we need to consider only the horizontal Coriolis acceleration, $2\Omega_z(v_y,-v_x)$ with $\Omega_z=\omega_{\text{Earth}}\sin\varphi$ (the term $\omega_{\text{Earth}}\cos\varphi\,v_z$ is neglected). At the North Pole, $\varphi=90^\circ$, we have $\Omega_z=\omega_{\text{Earth}}$, and at the equator, $\varphi=0^\circ$, we have $\Omega_z=0$. Let $x,y$ be the horizontal displacements of the pendulum, and let $\omega_0=\sqrt{g/l}$ be the angular frequency of an ordinary simple pendulum (ignoring the rotation of the Earth; $l$ is the length of the pendulum, and this is the result for the simple pendulum in Chapter 7). Then the equations of motion, obtained by adding the horizontal components of the Coriolis force to the restoring force, are

$$ \ddot x=-\omega_0^2x+2\Omega_z\dot y,\qquad \ddot y=-\omega_0^2y-2\Omega_z\dot x $$

Introducing the complex number $\zeta\equiv x+iy$ (the same complex-number technique as when we solved the forced oscillation in Chapter 7) and combining the two equations, we obtain

$$ \ddot\zeta+2i\Omega_z\dot\zeta+\omega_0^2\zeta=0 $$

This is a single differential equation for a complex variable (verify that $\ddot x+i\ddot y=-\omega_0^2(x+iy)+2\Omega_z(\dot y-i\dot x)=-\omega_0^2\zeta-2i\Omega_z(\dot x+i\dot y)=-\omega_0^2\zeta-2i\Omega_z\dot\zeta$). It can be solved by the same characteristic-equation method as in Chapter 7 (substitute $\zeta=e^{\lambda t}$), and the roots of the characteristic equation $\lambda^2+2i\Omega_z\lambda+\omega_0^2=0$ are $\lambda=-i\Omega_z\pm i\sqrt{\Omega_z^2+\omega_0^2}$. The period of the pendulum is a few seconds to over ten seconds (for the Panthéon, with $l=67\ \mathrm m$, $2\pi/\omega_0\approx16\ \mathrm s$ and $\omega_0\approx0.38\ \mathrm{rad/s}$), whereas the rotation of the Earth is far slower, $\omega_{\text{Earth}}\approx7\times10^{-5}\ \mathrm{rad/s}$, so that $\Omega_z\ll\omega_0$ ($\Omega_z/\omega_0\lt10^{-3}$). Approximating $\sqrt{\Omega_z^2+\omega_0^2}\approx\omega_0$, the general solution takes the form

$$ \zeta(t)\approx e^{-i\Omega_zt}\left(C_1e^{i\omega_0t}+C_2e^{-i\omega_0t}\right) $$

(where $C_1,C_2$ are complex constants). If the pendulum is displaced along the $x$ axis and released gently ($\zeta=x_0$ and $\dot\zeta=0$ at $t=0$), then $C_1=\tfrac{x_0}2(1+\Omega_z/\omega)$ and $C_2=\tfrac{x_0}2(1-\Omega_z/\omega)$ (with $\omega=\sqrt{\Omega_z^2+\omega_0^2}$), so that $C_1\approx C_2$ (the difference is of order $\Omega_z/\omega_0$). When $C_2=C_1^*$, the expression in the parentheses becomes the real number $2\mathrm{Re}(C_1e^{i\omega_0t})$, which represents the oscillation of an ordinary simple pendulum in which the Coriolis force is ignored (a back-and-forth linear motion in a fixed direction). The factor $e^{-i\Omega_zt}$ multiplying it is nothing other than the operation of rotating everything by an angle $-\Omega_zt$ in the complex plane. That is, the plane of oscillation of the pendulum (the direction of the linear back-and-forth motion) rotates slowly with angular velocity $\Omega_z=\omega_{\text{Earth}}\sin\varphi$ (clockwise in the Northern Hemisphere). The time required for the plane of oscillation to make one revolution is

\begin{equation} T_{\text{F}}=\frac{2\pi}{\Omega_z}=\frac{T_{\text{day}}}{\sin\varphi} \label{eq:9-foucault-period} \end{equation}

In other words, this time is the period of Foucault's pendulum.

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Derivation: Recovering the polar-coordinate formula of Chapter 6 from the general formula of this chapter

Let the position of the particle be $\bm r'=r(t)\bm e_r$, and use Eq. \eqref{eq:9-general-accel} in the rotating frame with angular velocity $\bm\omega=\dot\theta\bm e_z$ (the origin of the rotating frame is held fixed, that is, $\bm A=\bm 0$; in this case it suffices to compute the motion of the particle itself relative to this rotating frame, as $\bm v_{\text{rel}},\bm a_{\text{rel}}$). Since the particle always moves only in the radial direction within the rotating frame,

$$ \bm v_{\text{rel}}=\dot r\,\bm e_r,\qquad \bm a_{\text{rel}}=\ddot r\,\bm e_r $$

(to an observer riding in the rotating frame, the particle appears simply to move one-dimensionally along the $x'$ axis). Also, $\dot{\bm\omega}=\ddot\theta\bm e_z$. We now compute each term of Eq. \eqref{eq:9-general-accel} (with $\bm A=\bm 0$). First, by the same calculation as in the derivation of Eq. \eqref{eq:9-centrifugal-result} (with $\bm\omega=\dot\theta\bm e_z$ and $\bm r'=r\bm e_r$, and since $\bm e_r$ is a unit vector perpendicular to the axis of rotation $\bm e_z$, we have $\bm\rho=r\bm e_r$ itself), we obtain

$$ \bm\omega\times(\bm\omega\times\bm r')=-\dot\theta^2\,r\bm e_r $$

Next, using $\bm e_z\times\bm e_r=\bm e_\theta$ (which holds because the polar basis is also an orthogonal right-handed system), we obtain

$$ 2\,\bm\omega\times\bm v_{\text{rel}}=2\dot\theta\bm e_z\times\dot r\bm e_r=2\dot r\dot\theta\,\bm e_\theta, \qquad \dot{\bm\omega}\times\bm r'=\ddot\theta\bm e_z\times r\bm e_r=r\ddot\theta\,\bm e_\theta $$

Substituting all of these into Eq. \eqref{eq:9-general-accel}, we get

$$ \bm a=\bm a_{\text{rel}}+2\,\bm\omega\times\bm v_{\text{rel}}+\bm\omega\times(\bm\omega\times\bm r')+\dot{\bm\omega}\times\bm r' =\ddot r\bm e_r-\dot\theta^2r\bm e_r+2\dot r\dot\theta\bm e_\theta+r\ddot\theta\bm e_\theta $$

That is,

\begin{equation} \bm a=(\ddot r-r\dot\theta^2)\,\bm e_r+(2\dot r\dot\theta+r\ddot\theta)\,\bm e_\theta \label{eq:9-polar-correspondence} \end{equation}

and we have obtained exactly the same expression as the polar-coordinate acceleration formula of Chapter 6. The terms correspond as follows: $\ddot r\bm e_r=\bm a_{\text{rel}}$, $-r\dot\theta^2\bm e_r=\bm\omega\times(\bm\omega\times\bm r')$ (the centrifugal-force term), $2\dot r\dot\theta\bm e_\theta=2\bm\omega\times\bm v_{\text{rel}}$ (the Coriolis-force term), and $r\ddot\theta\bm e_\theta=\dot{\bm\omega}\times\bm r'$ (the Euler-force term).

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