Rigid Bodies and Rotating Frames

Derives Proposition 29.2. Fix three non-collinear particles of the body. Locating the first costs three coordinates. The second is then confined by Equation (29.1) to a sphere about the first, costing two more; the third is confined to a circle about the axis through the first two, costing one more — six in all. Every further particle is then determined, because its distances from three non-collinear points fix it up to a reflection, and a reflection cannot be reached continuously from the identity. The six coordinates split as the position of the chosen point, an element of \(\R^{3}\), and the orientation of the body frame relative to the space frame, which by Definition 14.26 is an element of \(\SO(3)\).

Derives Theorem 29.3. The displacement preserves all mutual distances and one point, so in Cartesian axes centred on that point it acts as a matrix \(R\) with \(R\transpose R=\identity\), and it is continuously connected to the identity, so \(\det R=+1\): that is, \(R\in\SO(3)\) (Definition 14.26). By the Rodrigues formula of Proposition 14.30 the exponential map onto \(\SO(3)\) is surjective, so \(R=\exp\left(\theta\,\hat{\vect{n}}\cdot\vect{J}\right)\) for some unit \(\hat{\vect{n}}\) and some angle \(\theta\); this is by construction the rotation by \(\theta\) about \(\hat{\vect{n}}\), and \(R\hat{\vect{n}}=\hat{\vect{n}}\). The general rotation formulae in this form are Euler's [Euler:1776].

Derives Proposition 29.5. Write the position of a material point as \(\vect{R}(t)=\vect{R}_{O}(t)+R(t)\vect{r}_{0}\), with \(\vect{r}_{0}\) its fixed body coordinates — fixed precisely because of Equation (29.1). Differentiating, \(\dot{\vect{R}}=\vect{v}_{O}+\dot{R}\vect{r}_{0} =\vect{v}_{O}+\dot{R}R\transpose\left(R\vect{r}_{0}\right)\), and by Equation (29.2) the second term is \(\vect{\omega}\times\vect{r}\) with \(\vect{r}=R\vect{r}_{0}\).

Derives Proposition 29.6. Let \(\vect{d}\) be the vector from \(O\) to \(O'\). Applying Equation (29.3) about each point and subtracting, \(\left(\vect{\omega}-\vect{\omega}'\right)\times \left(\vect{r}-\vect{d}\right)=\vect{0}\) for every material point. A body with three non-collinear points supplies vectors \(\vect{r}-\vect{d}\) spanning \(\R^{3}\), and only the zero vector has vanishing cross product with a spanning set, so \(\vect{\omega}=\vect{\omega}'\).

Derives Theorem 29.7. Expand \(\vect{Q}=Q_{i}\hat{\vect{e}}_{i}\) in the rotating basis and differentiate in the inertial frame:

The first sum is by definition the derivative computed by an observer who regards the basis as fixed. For the second, \(\hat{\vect{e}}_{i}=R\hat{\vect{e}}_{i}(0)\) gives \(\dd\hat{\vect{e}}_{i}/\dd t=\dot{R}R\transpose\hat{\vect{e}}_{i} =\vect{\omega}\times\hat{\vect{e}}_{i}\) by Equation (29.2), so the second sum is \(\vect{\omega}\times\vect{Q}\).

Derives Proposition 29.8. By Theorem 29.3 the displacement is \(\vect{r}\mapsto R\vect{r}+\vect{a}\) with \(R\) a rotation by \(\theta\) about a unit vector \(\hat{\vect{n}}\) through the origin. Split \(\vect{a}=a_{\parallel}\hat{\vect{n}}+\vect{a}_{\perp}\). Shifting the axis to pass through a point \(\vect{c}\) replaces \(\vect{a}\) by \(\vect{a}+\left(\identity-R\right)\vect{c}\). Now \(\identity-R\) annihilates \(\hat{\vect{n}}\) and, for \(\theta\) not a multiple of \(2\pi\), is invertible on the plane orthogonal to \(\hat{\vect{n}}\); choose \(\vect{c}\) in that plane with \(\left(\identity-R\right)\vect{c}=-\vect{a}_{\perp}\). About the shifted axis the translation is \(a_{\parallel}\hat{\vect{n}}\), parallel to the axis. If \(\theta\) is a multiple of \(2\pi\) the displacement is a pure translation, the degenerate screw of infinite pitch.

Derives Proposition 29.11. The three elementary rotations are performed about the space \(z\) axis, about the line of nodes \(\hat{\vect{e}}_{N}\), and about the body axis \(\hat{\vect{e}}_{3}\), so the angular velocities add:

It remains to resolve each of the three axes along the body frame. The third column of Equation (29.6) gives the body components of \(\hat{\vect{z}}\), namely \(\left(\sin\theta\sin\psi,\ \sin\theta\cos\psi,\ \cos\theta\right)\). The line of nodes is the \(x\) axis of the intermediate frame, so applying only the last rotation \(R_{3}(\psi)\) to \((1,0,0)\) gives \(\left(\cos\psi,\ -\sin\psi,\ 0\right)\). Finally \(\hat{\vect{e}}_{3}=(0,0,1)\) in the body frame. Collecting components yields Equations (29.7), (29.8) and (29.9).

Derives Proposition 29.14. With \(O\) fixed, Equation (29.3) gives \(\vect{v}=\vect{\omega}\times\vect{r}\), so

using the triple-product expansion \(\vect{a}\times(\vect{b}\times\vect{c}) =\vect{b}(\vect{a}\cdot\vect{c})-\vect{c}(\vect{a}\cdot\vect{b})\), which is the contraction \(\epsilon_{ijk}\epsilon_{ilm} =\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}\) of the Levi-Civita symbol of Lemma 14.62. That contraction is quoted here and not proved: Part II states the invariance of the symbol but carries no labelled form of the identity, so it is used on its own authority. In components the result is exactly Equation (29.10) contracted with \(\omega_{k}\), which is Equation (29.11). For the energy,

by the same identity. The separation of the general motion is the centre-of-mass decomposition of Newtonian Dynamics: the cross term carries the factor \(\int_{V}\rho\,\vect{r}\,\dd^{3}r\), which vanishes when \(\vect{r}\) is measured from the centre of mass.

Derives Proposition 29.15. Contract Equation (29.10) with \(\hat{\vect{n}}\): \(n_{i}n_{k}\left(r^{2}\delta_{ik}-x_{i}x_{k}\right) =r^{2}-\left(\vect{r}\cdot\hat{\vect{n}}\right)^{2} =\abs{\vect{r}}^{2}\sin^{2}\alpha=d^{2}\), with \(\alpha\) the angle between \(\vect{r}\) and \(\hat{\vect{n}}\). The integrand is non-negative, and vanishes identically only where \(\rho\) is supported on the line \(d=0\).

Derives Theorem 29.16. \(I_{ik}\) is a real symmetric matrix, hence self-adjoint with respect to the Euclidean inner product. The spectral theorem (Theorem 5.75) supplies an orthonormal basis of eigenvectors with real eigenvalues, and Corollary 5.27 makes the decomposition orthogonal, so the change of basis is itself a rotation once the triad is ordered to be right-handed. The eigenvalues are the values of the form \(\hat{\vect{n}}\cdot I\hat{\vect{n}}\) on the eigenvectors, and these are non-negative by Proposition 29.15. The remaining formulae are Equations (29.11) and (29.12) evaluated in the diagonalizing basis. Historically the existence of three orthogonal principal axes is Segner's [Segner:1755]; the statement here is the modern one, and the mathematics belongs to Linear Algebra and Representation Theory rather than to mechanics.

Derives Proposition 29.17. In principal axes \(I_{1}=\int\rho\left(y^{2}+z^{2}\right)\), \(I_{2}=\int\rho\left(z^{2}+x^{2}\right)\) and \(I_{3}=\int\rho\left(x^{2}+y^{2}\right)\), all integrals over \(V\) with measure \(\dd^{3}r\). Hence \(I_{1}+I_{2}-I_{3}=2\int_{V}\rho\,z^{2}\,\dd^{3}r\geq0\), which vanishes precisely when \(\rho\) is supported on \(z=0\). The other two follow by relabelling.

Derives Theorem 29.20. Write \(\vect{r}=\vect{a}+\vect{r}'\), with \(\vect{r}'\) measured from the centre of mass, so that \(\int_{V}\rho\,\vect{r}'\,\dd^{3}r=\vect{0}\) by Definition 19.47. Substituting into Equation (29.10),

Every term linear in \(\vect{r}'\) integrates to zero. What remains is the \(\vect{r}'\)-quadratic part, which is \(I^{\text{cm}}_{ik}\), plus the \(\vect{a}\)-quadratic part times the total mass, which is \(M\left(a^{2}\delta_{ik}-a_{i}a_{k}\right)\). Contracting Equation (29.15) with \(\hat{\vect{n}}\) twice gives \(Ma^{2}-\left(\vect{a}\cdot\hat{\vect{n}}\right)^{2}M=Md^{2}\) for the added term.

Derives Corollary 29.21. The added term \(Md^{2}\) of Theorem 29.20 is non-negative and vanishes only for \(d=0\).

Derives Proposition 29.22. With \(z=0\) throughout the body, Equation (29.10) gives \(I_{xx}=\int\rho y^{2}\), \(I_{yy}=\int\rho x^{2}\) and \(I_{zz}=\int\rho\left(x^{2}+y^{2}\right)\), all over \(V\) with measure \(\dd^{3}r\); adding the first two gives the third. It is the equality case of Proposition 29.17.

Derives Proposition 29.24. Write \(U=-GM\int_{V}\rho(\vect{r})\,\abs{\vect{R}-\vect{r}}^{-1} \dd^{3}r\) and expand the kernel in Legendre polynomials by the generating function of Theorem 9.113, with \(\cos\gamma=\hat{\vect{R}}\cdot\hat{\vect{r}}\):

The \(n=0\) term integrates to \(m/R\). The \(n=1\) term vanishes because \(\vect{r}\) is measured from the centre of mass. For \(n=2\), note from Equation (29.10) that \(I_{1}+I_{2}+I_{3}=\tr I=2\int_{V}\rho\,r^{2}\,\dd^{3}r\) and that \(I_{R}=\int_{V}\rho\left(r^{2} -\left(\vect{r}\cdot\hat{\vect{R}}\right)^{2}\right)\dd^{3}r\), whence

Multiplying by \(-GM/(2R^{3})\) gives Equation (29.17).

Derives Theorem 29.25. The rotational law of Newtonian Dynamics holds in an inertial frame: \(\left(\dd\vect{L}/\dd t\right)_{\text{space}} =\vect{\tau}\) (Equation (19.14)). The inertia tensor is constant only in the body frame, so transport the derivative there with Equation (29.4):

From Equations (19.14) and (29.4) (transporting the inertial rotational law into the body frame, where the inertia tensor is constant). In principal axes \(L_{i}=I_{i}\omega_{i}\) with constant \(I_{i}\) (Theorem 29.16), so the first term has components \(I_{i}\dot{\omega}_{i}\); and \(\left(\vect{\omega}\times\vect{L}\right)_{1} =\omega_{2}L_{3}-\omega_{3}L_{2} =\left(I_{3}-I_{2}\right)\omega_{2}\omega_{3}\), with the other two components obtained by cyclic permutation. Substituting gives Equations (29.18), (29.19) and (29.20).

Derives Proposition 29.26. Multiply Equations (29.18), (29.19) and (29.20) by \(\omega_{1},\omega_{2},\omega_{3}\) and add: the right-hand side vanishes and the products \(\left(I_{2}-I_{3}\right)\omega_{1}\omega_{2}\omega_{3}\) cancel in pairs, leaving \(\dd\left(\sum_{i}I_{i}\omega_{i}^{2}\right)/\dd t=0\). Multiplying instead by \(I_{1}\omega_{1},I_{2}\omega_{2},I_{3}\omega_{3}\) and adding cancels the same products and leaves \(\dd\left(\sum_{i}I_{i}^{2}\omega_{i}^{2}\right)/\dd t=0\). The second sum is \(\abs{\vect{L}}^{2}\) by Theorem 29.16. Both statements also follow directly in the inertial frame, where \(\vect{L}\) is constant and \(\dd T/\dd t=\vect{\omega}\cdot\vect{\tau}\).

Derives Proposition 29.28. With \(I_{1}=I_{2}\) the third of Equation (29.20) reads \(C\dot{\omega}_{3}=0\), so \(\omega_{3}\) is constant. The first two become \(A\dot{\omega}_{1}=\left(A-C\right)\omega_{2}\omega_{3}\) and \(A\dot{\omega}_{2}=\left(C-A\right)\omega_{3}\omega_{1}\). Setting \(\zeta:=\omega_{1}+\ii\omega_{2}\) combines them into the single equation \(\dot{\zeta}=\ii\Omega_{\text{body}}\zeta\) with \(\Omega_{\text{body}}\) as in Equation (29.23), whose solution \(\zeta=\zeta_{0}\ee^{\ii\Omega_{\text{body}}t}\) has constant modulus: the equatorial part of \(\vect{\omega}\) has fixed length and turns uniformly, which is statement (i).

For (ii), decompose \(\vect{\omega}=\vect{\omega}_{\perp}+\omega_{3}\hat{\vect{e}}_{3}\), so that \(\vect{L}=A\vect{\omega}_{\perp}+C\omega_{3}\hat{\vect{e}}_{3}\) by Theorem 29.16. Eliminating \(\vect{\omega}_{\perp}\),

which exhibits \(\vect{\omega}\) as a combination of \(\vect{L}\) and \(\hat{\vect{e}}_{3}\) — hence the coplanarity. Since \(\hat{\vect{e}}_{3}\) is carried by the body, \(\dd\hat{\vect{e}}_{3}/\dd t =\vect{\omega}\times\hat{\vect{e}}_{3}\), and substituting Equation (29.25) kills the second term:

which is precisely a rotation of \(\hat{\vect{e}}_{3}\) about the fixed vector \(\vect{L}\) at the rate \(\abs{\vect{L}}/A\). The angle between \(\hat{\vect{e}}_{3}\) and \(\vect{L}\) is constant because \(L_{3}\) and \(\abs{\vect{L}}\) are, so the motion is a cone.

Derives Theorem 29.29. Put \(\vect{\rho}:=\vect{\omega}/\sqrt{2T}\). Then \(\vect{\rho}\cdot I\vect{\rho} =\left(\vect{\omega}\cdot I\vect{\omega}\right)/(2T)=1\) by Equation (29.12), so \(\vect{\rho}\) lies on the inertia ellipsoid Equation (29.14) at every instant.

The outward normal to the ellipsoid at \(\vect{\rho}\) is the gradient of the quadratic form, \(\tfrac{1}{2}\nabla \left(\vect{x}\cdot I\vect{x}\right)=I\vect{\rho} =\vect{L}/\sqrt{2T}\), which is parallel to \(\vect{L}\) — a direction fixed in space, because the motion is torque-free. The tangent plane at the contact point is therefore perpendicular to \(\vect{L}\) for all time. Its distance from the centre is

constant by Proposition 29.26. A plane of fixed orientation and fixed distance from a fixed point is a fixed plane.

Finally, the contact point \(\vect{\rho}\) is parallel to \(\vect{\omega}\), so it lies on the instantaneous axis of rotation; by Equation (29.3) its material velocity vanishes. A surface touching a plane at a point of zero velocity rolls without slipping.

Derivation. Derives Phenomenon 29.31. Take Euler's torque-free equations in the principal frame (Equation (29.18)),

and perturb the steady rotation \(\vect{\omega}=(\Omega,0,0)\) by writing \(\omega_{1}=\Omega+\varepsilon_{1}\), \(\omega_{2}=\varepsilon_{2}\) and \(\omega_{3}=\varepsilon_{3}\) with the \(\varepsilon_{i}\) small. To first order the first equation gives \(\dot{\varepsilon}_{1}=0\), and the other two give

Differentiating the first and substituting the second,

The coefficient on the right is negative — so that \(\varepsilon_{2}\) merely oscillates, and the axis is stable — when \(I_{1}\) is larger than both of \(I_{2},I_{3}\) or smaller than both; and it is positive — so that \(\varepsilon_{2}\) grows exponentially, and the axis is unstable — exactly when \(I_{1}\) lies strictly between them. The unstable axis is therefore the intermediate one, and only that one. A body with two equal moments has no intermediate axis and shows no such instability, which is why a symmetric top misbehaves in other ways but never in this one.

The growth rate follows from the same coefficient: writing \(\varepsilon_{2}\propto\ee^{\lambda t}\) about the intermediate axis \(I_{2}<I_{1}<I_{3}\),

which is of the order of the spin rate itself unless two of the moments nearly coincide. That is why the flip is fast, and why it recurs at a regular interval rather than settling: the linearization governs only the departure, and the exact motion returns periodically to the neighbourhood of the intermediate axis, as Remark 29.30 shows geometrically.

Derives Proposition 29.33. Equation (29.28) is Equation (29.12) written in terms of \(L_{i}=I_{i}\omega_{i}\). At fixed \(L^{2}=\sum_{i}L_{i}^{2}\) the right-hand side is a weighted average of the \(1/I_{i}\) with weights \(L_{i}^{2}/L^{2}\) summing to one, so it lies between \(L^{2}/(2I_{\max})\) and \(L^{2}/(2I_{\min})\) and attains the lower bound exactly when all the weight sits on the largest moment. Internal forces exert no external torque (Postulate 19.48), so \(\vect{L}\) is conserved while \(T\) decreases; a decreasing quantity bounded below by \(L^{2}/(2I_{\max})\) converges, and the only stationary configurations compatible with the bound are rotations about the axis of largest moment.

Derivation. Derives Phenomenon 29.34. For a rigid symmetric body with equatorial moment \(A\) and polar moment \(C\), Euler's torque-free equations (Equation (29.18)) with \(\omega_{3}\) constant read

Setting \(\zeta:=\omega_{1}+\ii\omega_{2}\) combines the two into \(\dot{\zeta}=\ii\Omega_{\text{free}}\zeta\), so the equatorial part of \(\vect{\omega}\) rotates uniformly within the body at the rate

that is, one turn in \(A/(C-A)\) rotation periods of the body — which is Equation (29.23) specialized to the Earth. The observed dynamical ellipticity is \(H:=\left(C-A\right)/C=3.274\times 10^{-3}\) (Table 29.2), so

and the rigid prediction is a free wobble of about 305 sidereal days — Euler's number [Euler:1765], and one that no observation matches. Chandler's analysis of the latitude records gave a free period near 427 days [Chandler:1891], and modern determinations place it near 433 days; the amplitude corresponds to a displacement of the pole of a few metres at the surface.

The elastic correction follows from the same equations once the body is allowed to yield. A wobble tilts the rotation axis to \(\vect{\omega}=\Omega\left(m_{1},m_{2},1\right)\), with \(\abs{m}\) of order \(5\times 10^{-7}\) for a pole displacement of a few metres, and the centrifugal potential energy per unit mass, \(\Phi=-\tfrac{1}{2}\abs{\vect{\omega}\times\vect{r}}^{2}\), acquires the perturbation

in \(\mathrm{J}/\mathrm{kg}\): a second-degree harmonic of the same shape as a tide, raised by the wobble itself. The Earth responds to it elastically, and Love's number \(k_{2}\) is defined as the ratio of the extra gravitational potential the deformation produces at the surface to the potential that provoked it [Love:1911]. By Equation (29.17) the orientation-dependent part of the external potential of a body is carried entirely by its inertia tensor, and the part of it belonging to the products of inertia \(c_{13},c_{23}\) is \(3G\left(c_{13}xz+c_{23}yz\right)/r^{5}\). Equating that at \(r=a_{\oplus}\) to \(k_{2}\,\delta\Phi\) gives

with \(\kappa\) carrying \(\mathrm{kg}\,\mathrm{m}^{2}\), as an increment of inertia must.

Now redo the free-wobble calculation with \(I=I_{0}+c\). Angular momentum is conserved in space, so by Equation (29.4) its body components obey \(\dot{\vect{L}}+\vect{\omega}\times\vect{L}=\vect{0}\), and to first order in \(m\) and in \(c\) one has \(L_{1}=\Omega\left(Am_{1}+c_{13}\right)\), \(L_{2}=\Omega\left(Am_{2}+c_{23}\right)\) and \(L_{3}=C\Omega\). Writing \(m:=m_{1}+\ii m_{2}\) and combining the first two components as before,

so the free wobble is still circular, but at the rate

From Equations (29.31) and (29.32) (the elastic increment enters only through the ratio of Love numbers). the approximation using \(\kappa/A\sim10^{-3}\). The dimensionless \(k_{\text{s}}\) is the Love number a body would have if it yielded until its whole bulge were centrifugal; from Table 29.2, with \(C-A=HC=2.629\times 10^{35}\,\mathrm{kg}\,\mathrm{m}^{2}\) and \(a_{\oplus}^{5}=1.0556\times 10^{34}\,\mathrm{m}^{5}\), it is \(k_{\text{s}}=0.938\).

Equation (29.33) therefore lengthens the period by the factor \(\left(1-k_{2}/k_{\text{s}}\right)^{-1}\), and the observed lengthening from \(304.5\) to \(433\) days inverts to

which is the measurement the phenomenon promised: one number describing how far the whole planet yields, read off a period. It is of the size of, and a little below, the degree-two Love number near \(0.30\) that Earth-tide observations give—no source is cited here for that comparison value—and the difference is where the fluid outer core, which does not follow the mantle's wobble, and the ocean pole tide enter. Neither is in this calculation. Neither is the damping: the observed wobble decays in a few decades and must therefore be re-excited continually, by some redistribution of mass among the atmosphere, the oceans and the solid Earth. Which of those dominates is a question this treatise does not settle.

Derives Proposition 29.36. Neither \(\psi\) nor \(\varphi\) appears in Equation (29.35): both are cyclic (Definition 21.35), so their conjugate momenta are conserved by Theorem 21.36. Computing them from Equation (29.35) gives Equations (29.36) and (29.37). The Lagrangian has no explicit time dependence, so the energy function of Theorem 21.43 is conserved, and since the kinetic term is quadratic in the velocities it equals \(T+V\), which is Equation (29.38). That \(p_{\psi}\) and \(p_{\varphi}\) are the stated components of \(\vect{L}\) follows because \(\psi\) generates rotation about \(\hat{\vect{e}}_{3}\) and \(\varphi\) rotation about \(\hat{\vect{z}}\).

Derives Theorem 29.37. Solve Equation (29.37) for \(\dot{\varphi}\), obtaining the first of Equation (29.40), and Equation (29.36) for \(\dot{\psi}\), obtaining the second. Substituting the first into Equation (29.38) and using \(\dot{u}=-\sin\theta\,\dot{\theta}\), so that \(\dot{u}^{2}=\left(1-u^{2}\right)\dot{\theta}^{2}\), gives

which rearranges to Equation (29.39). Evaluating \(f\) at \(u=\pm1\) kills the first product and leaves \(-\left(b\mp a\right)^{2}\), which is negative unless the axis reaches the vertical. Since \(f\) is a cubic with positive leading coefficient \(\beta>0\), it tends to \(+\infty\) at large \(u\); and physical motion requires \(f\geq0\), which the initial condition supplies at some \(u_{0}\in(-1,1)\). The intermediate value theorem (Theorem 7.23) then places a root in \((u_{0},1)\) and another in \((-1,u_{0})\), and a cubic has no more than three roots, the third lying above \(1\) and being unphysical. Between the two roots \(\dot{u}^{2}>0\), and the motion runs from one to the other and back, periodically, because Equation (29.39) is autonomous and \(\dot{u}\) changes sign only at the roots.

Derivation. Derives Phenomenon 29.38. Release from rest at \(\theta_{0}\) means \(\dot{\theta}=\dot{\varphi}=0\) there, so Equation (29.37) gives \(b=au_{0}\) with \(u_{0}=\cos\theta_{0}\), and \(f(u_{0})=0\) in Equation (29.39) forces \(\alpha=\beta u_{0}\). Writing \(u=u_{0}+x\) with \(x\) small, \(b-au=-ax\) and \(\alpha-\beta u=-\beta x\), so

Differentiating once, \(\ddot{x}=-a^{2}x-\tfrac{1}{2}\beta\sin^{2}\theta_{0}\): the variable \(x\) performs simple harmonic motion at the angular frequency

about the displaced centre \(x=-\beta\sin^{2}\theta_{0}/(2a^{2})\), with amplitude equal to that displacement. Since \(x\) starts at zero and must be negative for Equation (29.41) to be positive, \(u\) decreases and \(\theta\) increases: the top falls first. Converting the full swing in \(u\) to an angle through \(\Delta u=\sin\theta_{0} \Delta\theta\),

which falls off as the inverse square of the spin.

The precession follows from the first of Equation (29.40): \(\dot{\varphi}=-ax/(1-u^{2})\), which vanishes exactly where \(x=0\) — at the top of each nutation swing. There the axis is momentarily at rest in both angles, which is the cusp. Averaging over one nutation period, \(\avg{-x}\) is the centre offset \(\beta\sin^{2}\theta_{0}/(2a^{2})\), so

which is exactly the fast-top rate Equation (29.55) derived independently below. The two calculations agreeing is the check that the smooth precession of Phenomenon 29.46 is the nutation-averaged motion and not a different phenomenon.

Derives Proposition 29.39. The Euler–Lagrange equation (Theorem 16.22) for \(\theta\) applied to Equation (29.35) is

using \(\omega_{3}=\dot{\psi}+\dot{\varphi}\cos\theta\) from Equation (29.9). Setting \(\ddot{\theta}=0\), \(\dot{\varphi}=\Omega\) and \(\sin\theta\neq0\) gives the quadratic in Equation (29.45), whose roots are as stated. Reality of the roots is the discriminant condition. For \(C\omega_{3}\gg\sqrt{AMg\ell}\), expanding the square root to first order gives \(\Omega_{-}\to Mg\ell/(C\omega_{3})\) and \(\Omega_{+}\to C\omega_{3}/(A\cos\theta)\).

Derives Proposition 29.40. At \(\theta=0\) the two conserved momenta coincide, \(L_{z}=L_{3}\), since Equation (29.37) at \(\sin\theta=0\) gives \(L_{z}=L_{3}\cos 0\). Eliminating \(\dot{\varphi}\) from Equation (29.38) with Equation (29.40) puts the energy in the one-dimensional form \(E'=\tfrac{1}{2}A\dot{\theta}^{2}+V_{\text{eff}}(\theta)\) (Definition 27.25) with

With \(L_{z}=L_{3}\) the first term is \(\left(L_{3}^{2}/2A\right)\left(1-\cos\theta\right)/ \left(1+\cos\theta\right)\), which for small \(\theta\) is \(L_{3}^{2}\theta^{2}/(8A)\); and \(Mg\ell\cos\theta =Mg\ell-Mg\ell\theta^{2}/2+O(\theta^{4})\). Hence

so \(\theta=0\) is a minimum — and the motion in \(\theta\) is bounded near it — exactly when \(L_{3}^{2}>4AMg\ell\), which with \(L_{3}=C\omega_{3}\) is Equation (29.46). When the coefficient is negative the effective potential falls away from \(\theta=0\) and no bounded motion near the vertical exists.

Derives Proposition 29.43. Cases (i) and (ii) are read off the results already proved: Proposition 29.26 supplies \(\abs{\vect{L}}^{2}\) when the torque vanishes, and Proposition 29.36 supplies \(L_{3}\) when the body is symmetric about the axis carrying the centre of mass. The existence of the fourth integral in case (iii), and the completeness of the list, are Kovalevskaya's and are not reproved here; the fourth integral itself is quoted in Remark 29.44.

Derives Equation (29.48). Write \(I_{3}=:I\), so that \(I_{1}=I_{2}=2I\). Keep \(\vect{\gamma}\) the upward vertical, and orient the body \(1\) axis from the centre of mass towards the fixed point, so that the centre of mass sits at \(-\ell\hat{\vect{e}}_{1}\) and the gravitational torque about the fixed point is \(\vect{\tau}=\left(-\ell\hat{\vect{e}}_{1}\right)\times \left(-Mg\vect{\gamma}\right) =Mg\ell\,\hat{\vect{e}}_{1}\times\vect{\gamma} =Mg\ell\left(0,-\gamma_{3},\gamma_{2}\right)\). That is the convention of the remark; reversing \(\hat{\vect{e}}_{1}\) reverses the torque and with it the sign of \(c_{0}\), which is the freedom recorded there. Euler's equations Equations (29.18), (29.19) and (29.20) then read

with \(c_{0}=Mg\ell/I\). The vertical is fixed in space, so Equation (29.4) applied to \(\vect{\gamma}\) gives its body components as

Set \(\zeta:=\omega_{1}+\ii\omega_{2}\) and \(\eta:=\gamma_{1}+\ii\gamma_{2}\). Combining the first two components of Equation (29.49), and then of Equation (29.50),

Now differentiate \(\xi:=\zeta^{2}+c_{0}\eta\):

From Equations (29.51) and (29.52) (the two terms carrying \(\gamma_{3}\) cancel identically). The coefficient is purely imaginary, so \(\dd\abs{\xi}^{2}/\dd t=\bar{\xi}\dot{\xi}+\xi\overline{\dot{\xi}} =-\ii\omega_{3}\abs{\xi}^{2}+\ii\omega_{3}\abs{\xi}^{2}=0\), and \(K=\abs{\xi}^{2}\) of Equation (29.48) is conserved.

The ratio of the moments is exactly what makes the cancellation happen. Repeat the calculation with \(I_{1}=I_{2}=A\) and \(I_{3}=C\) arbitrary, the centre of mass still on the body \(1\) axis. Euler's equations now give \(\dot{\zeta}=\ii k\,\omega_{3}\zeta -\ii\left(Mg\ell/A\right)\gamma_{3}\) with \(k:=\left(C-A\right)/A\), while Equation (29.50) is unchanged. For a trial quantity \(\xi=\zeta^{n}+\lambda\eta\),

The two terms in \(\gamma_{3}\) cancel for no choice of \(\lambda\) unless \(n=2\), and then only for \(\lambda=2Mg\ell/A\). What survives, \(\ii\omega_{3}\left(2k\zeta^{2}-\lambda\eta\right)\), is proportional to \(\xi=\zeta^{2}+\lambda\eta\) only if \(2k=-1\), that is only if \(A=2C\); and then \(\lambda=2Mg\ell/(2C)=Mg\ell/I_{3}=c_{0}\). So the quantity displayed in Equation (29.48) is conserved precisely when \(I_{1}=I_{2}=2I_{3}\), which is what singles the Kovalevskaya case out. That no other fourth integral exists for other ratios is Kovalevskaya's own theorem [Kovalevskaya:1889] and is not proved here.

Derivation. Derives Phenomenon 29.46. Let the top be symmetric, spinning at \(\omega_{3}\) about its own axis \(\hat{\vect{e}}_{3}\), and supported at a point a distance \(\ell\) from its centre of mass along that axis. In the fast-top limit the angular momentum is dominated by the spin, \(\vect{L}\simeq I_{3}\omega_{3}\hat{\vect{e}}_{3}\). About the support the weight exerts the torque

with \(\theta\) the tilt from the vertical. This torque is perpendicular to \(\hat{\vect{e}}_{3}\) and hence to \(\vect{L}\), so by \(\dv{\vect{L}}{t}=\vect{\tau}\) it can turn \(\vect{L}\) but cannot lengthen or shorten it — which is the whole reason the top turns instead of falling. Writing the motion of the axis as a rotation about the vertical at rate \(\Omega\), the tip of \(\vect{L}\) travels on a circle of radius \(I_{3}\omega_{3}\sin\theta\) at speed \(\Omega I_{3}\omega_{3}\sin\theta\); equating this to \(\abs{\vect{\tau}}\), the factor \(\sin\theta\) cancels and

independent of the tilt and inversely proportional to the spin, as observed. The one approximation made was to neglect the angular momentum of the precession itself beside that of the spin, which is what “fast” means; releasing it produces the superposed nutation of Phenomenon 29.38, and Equation (29.44) shows that averaging that nutation returns Equation (29.55) exactly.

Derives Proposition 29.47. Use local axes at latitude \(\lambda\) with \(x\) east, \(y\) north and \(z\) up, in which the Earth's angular velocity is \(\vect{\Omega}=\Omega\left(0,\cos\lambda,\sin\lambda\right)\). Let \(\alpha\) be the azimuth of the spin axis measured from north towards east, so that \(\hat{\vect{e}}=\left(\sin\alpha,\cos\alpha,0\right)\). A rotation carrying north into east is a rotation by \(-\alpha\) about \(\hat{\vect{z}}\), so the slew contributes \(-A\dot{\alpha}\hat{\vect{z}}\) to the angular momentum and

Apply Equation (29.4) in the frame rotating with the Earth: \(\left(\dd\vect{L}/\dd t\right)_{\text{Earth}} +\vect{\Omega}\times\vect{L}=\vect{\tau}\). The rotor is balanced, so gravity exerts no torque about the support; the constraint that keeps the axis horizontal acts about a horizontal axis and so has no vertical component. Taking the vertical component of the equation,

whence \(A\ddot{\alpha}=-C\omega_{3}\Omega\cos\lambda\sin\alpha\). This is the pendulum equation with \(\alpha=0\) — the spin axis pointing north — as the stable equilibrium, and linearizing about it gives Equation (29.56). At \(\cos\lambda=0\) the restoring term vanishes identically and no direction is preferred.

Derivation. Derives Phenomenon 29.49. Treat the Earth as a symmetric top with moments \(A=B\) and \(C\) and symmetry axis \(\hat{\vect{e}}_{3}\), and a perturber of mass \(M\) on a circular orbit of radius \(r\) in a plane with unit normal \(\hat{\vect{n}}\). By Proposition 29.24 the orientation dependence of the potential energy is

since for a symmetric top \(\tr I-3I_{r}=\left(C-A\right) \left(1-3\cos^{2}\Theta\right)\) with \(\cos\Theta=\hat{\vect{e}}_{3}\cdot\hat{\vect{r}}\). The torque on the Earth is minus the rotational gradient of Equation (29.57),

Now average over one orbit. Let \(\hat{\vect{e}}_{3}=\cos\varepsilon\,\hat{\vect{n}} +\sin\varepsilon\,\hat{\vect{u}}\) with \(\hat{\vect{u}}\) in the orbit plane, and \(\hat{\vect{v}}=\hat{\vect{n}}\times\hat{\vect{u}}\), and write \(\hat{\vect{r}}=\cos\phi\,\hat{\vect{u}} +\sin\phi\,\hat{\vect{v}}\). Then \(\hat{\vect{e}}_{3}\cdot\hat{\vect{r}}=\sin\varepsilon\cos\phi\) and

Using \(\avg{\cos^{2}\phi}=\tfrac{1}{2}\) and \(\avg{\cos\phi\sin\phi}=0\), only the \(\hat{\vect{v}}\) term survives:

Insert this into \(\dd\vect{L}/\dd t=\avg{\vect{\tau}}\) with \(\vect{L}=C\omega_{3}\hat{\vect{e}}_{3}\), exactly as in Phenomenon 29.46. A precession of \(\hat{\vect{e}}_{3}\) about \(\hat{\vect{n}}\) at rate \(\Omega_{p}\) means \(\dd\hat{\vect{e}}_{3}/\dd t =\Omega_{p}\hat{\vect{n}}\times\hat{\vect{e}}_{3} =\Omega_{p}\sin\varepsilon\,\hat{\vect{v}}\), so the common factor \(\sin\varepsilon\) cancels and

the sign showing the drift to be retrograde — westward along the ecliptic — as observed.

Now put in the numbers of Table 29.2. For the Sun, \(GM_{\odot}/a_{\odot}^{3}=3.964\times 10^{-14}\,/\mathrm{s}^{2}\), and with \(H=3.2738\times 10^{-3}\), \(\cos\varepsilon=0.91748\) and \(\omega_{3}=\Omega\),

For the Moon, \(GM_{\text{M}}/a_{\text{M}}^{3} =8.632\times 10^{-14}\,/\mathrm{s}^{2}\), larger than the Sun's by the factor \(2.177\), so the coplanar estimate is \(34.71\) seconds of arc a year. The lunar orbit is not in the ecliptic but inclined by \(i=5.145^\circ\), and its node regresses in \(18.6\,\mathrm{yr}\), which is short compared with the precession; averaging Equation (29.57) a second time over that cone multiplies the orientation-dependent part by \(1-\tfrac{3}{2}\sin^{2}i=0.98794\), giving \(34.29\). The two contributions add to

against the observed \(50.29\): agreement to one part in a thousand, from the mass and orbit of the Moon, the mass and orbit of the Sun, and one number describing the shape of the Earth. The residual is of the size of the terms deliberately dropped — the orbital eccentricities, the higher multipoles of the Earth's figure, and the periodic terms discarded by the averaging, which are the subject of the nutation.

Derives Proposition 29.50. Return to Equation (29.60) before the second averaging. What that equation describes is a precession of \(\hat{\vect{e}}_{3}\) about the normal \(\hat{\vect{n}}_{\text{M}}\) of the lunar orbit, at a rate proportional to the cosine of the angle \(\varepsilon_{\text{M}}\) between the two:

The lunar orbit normal is not the ecliptic pole \(\hat{\vect{n}}\): it lies on a cone of half-angle \(i\) about it and travels once round that cone in \(18.6\,\mathrm{yr}\) as the nodes regress. In axes with \(\hat{\vect{n}}=\hat{\vect{z}}\) and \(\hat{\vect{e}}_{3}=\left(\sin\varepsilon,0,\cos\varepsilon\right)\),

so that \(\cos\varepsilon_{\text{M}}=\hat{\vect{e}}_{3}\cdot \hat{\vect{n}}_{\text{M}} =\cos i\cos\varepsilon+\sin i\sin\varepsilon\cos\phi\). To first order in \(i\), Equation (29.62) splits into a secular part and a periodic one,

with \(\Omega_{\text{M}}:=\Lambda\cos\varepsilon\) the mean rate already computed. Both periodic terms matter, and they are different things: the second is the wobble of the axis that the torque precesses the Earth about, the first the modulation of the rate at which it does so.

The response is obtained by integrating the periodic terms directly, because \(\Omega_{\text{M}}/\Omega_{\text{N}}\approx5\times 10^{-4}\): the displacement \(\vect{\delta}\) they build up varies at the node rate, so the action of the secular term upon \(\vect{\delta}\) is smaller than \(\dot{\vect{\delta}}\) by that factor and may be dropped. With \(\hat{\vect{n}}\times\hat{\vect{e}}_{3} =\left(0,\sin\varepsilon,0\right)\) and \(\hat{\vect{w}}\times\hat{\vect{e}}_{3} =\left(\sin\phi\cos\varepsilon,-\cos\phi\cos\varepsilon, -\sin\phi\sin\varepsilon\right)\), the bracket of Equation (29.64) is

and integrating in time, with \(\int\sin\phi\,\dd t =\cos\phi/\Omega_{\text{N}}\) and \(\int\cos\phi\,\dd t =-\sin\phi/\Omega_{\text{N}}\) from \(\phi=-\Omega_{\text{N}}t\),

From Equations (29.63) and (29.64) (integrating the periodic forcing at the node frequency). The obliquity is read from the third component: \(\cos\left(\varepsilon+\Delta\varepsilon\right) =\hat{\vect{z}}\cdot\left(\hat{\vect{e}}_{3}+\vect{\delta}\right) =\cos\varepsilon+\delta_{3}\) gives \(\Delta\varepsilon=-\delta_{3}/\sin\varepsilon =\left(\Omega_{\text{M}}\sin i/\Omega_{\text{N}}\right)\cos\phi\), on using \(\Lambda\cos\varepsilon=\Omega_{\text{M}}\). The longitude is read from the second, since displacing \(\psi\) moves \(\hat{\vect{e}}_{3}\) by \(\sin\varepsilon\,\Delta\psi\) in the \(y\) direction: \(\Delta\psi=\delta_{2}/\sin\varepsilon =\left(\Omega_{\text{M}}\sin i/\Omega_{\text{N}}\right) \left(2\cos2\varepsilon/\sin2\varepsilon\right)\sin\phi\). The amplitudes are Equation (29.61), quoted as magnitudes: \(\Lambda\) is negative, since the precession it drives is retrograde, and its sign fixes only the phase of each term relative to the node. Note where each amplitude came from: the obliquity term needs only the wobble of the torque axis, while the longitude term needs the rate modulation as well, and dropping that would leave \(\cot\varepsilon\) in place of \(2\cot2\varepsilon\) and an answer a quarter too large.

Derives Theorem 29.51. Apply Equation (29.4) to the position vector: \(\left(\dd\vect{r}/\dd t\right)_{\text{space}} =\vect{v}+\vect{\omega}\times\vect{r}\). Apply it again, to that whole vector:

Expanding, the first bracket gives \(\vect{a}+\dot{\vect{\omega}}\times\vect{r} +\vect{\omega}\times\vect{v}\), so

Newton's second law holds in the inertial frame (Postulate 19.8), so \(m\left(\dd^{2}\vect{r}/\dd t^{2}\right)_{\text{space}}=\vect{F}\); solving for \(m\vect{a}\) gives Equation (29.66).

Derivation. Derives Phenomenon 29.53. Work in the frame rotating with the bucket at angular velocity \(\omega\) about the vertical, in which the water is at rest. The centrifugal term of Equation (29.66) is a gradient and contributes to the effective potential energy per unit mass the amount \(-\tfrac{1}{2}\omega^{2}r^{2}\), so a fluid element at height \(z\) and distance \(r\) from the axis carries

A free surface in equilibrium must be a surface of constant \(\Phi\) — otherwise there would be an unbalanced force tangent to it and the fluid would not be at rest — so the surface obeys

a paraboloid of revolution whose curvature reports \(\omega^{2}\) directly, from measurements made entirely inside the bucket. Since \(\omega\) enters quadratically, the reading is blind to the sense of the rotation but never vanishes for \(\omega\neq0\): there is no rotation rate that mimics rest. Contrast the derivation of Kinematics, where the constancy of the relative velocity made the accelerations — and with them every mechanical observable inside the cabin — identical in the two frames. The difference is that the map to a rotating frame is not constant in time, and its second time derivative does not vanish.

Derivation. Derives Phenomenon 29.55. In the frame rotating with the Earth, a body at rest on the surface experiences, besides the gravitational attraction, a centrifugal acceleration \(\omega^{2}\rho\) directed away from the rotation axis, where \(\rho=R\cos\lambda\) is the distance from that axis at geographic latitude \(\lambda\). Its component along the local outward vertical is \(\omega^{2}R\cos^{2}\lambda\), and the measured free-fall acceleration is reduced by exactly that amount,

largest in effect at the equator and zero at the poles. With the sidereal rotation rate \(\omega=7.292\times 10^{-5}\,/\mathrm{s}\) and the equatorial radius \(R=6.378\times 10^{6}\,\mathrm{m}\),

about two thirds of the observed equator-to-pole difference of \(5.2\times 10^{-2}\,\mathrm{m}/\mathrm{s}^{2}\). The remaining third is not a defect of the calculation but a second consequence of the same rotation: the Earth is not a sphere, its equatorial radius exceeding the polar by some \(21\,\mathrm{km}\), so equatorial ground stands farther from the mass that attracts it. There is also a component of the centrifugal acceleration tangent to the surface, \(\omega^{2}R\sin\lambda\cos\lambda\), which is what tilts the plumb line; it is largest near \(45^\circ\), where it amounts to about \(1.7\times 10^{-2}\,\mathrm{m}/\mathrm{s}^{2}\) and deflects the vertical by roughly a tenth of a degree.

Derivation. Derives Phenomenon 29.56. Use local axes at latitude \(\lambda\) with \(x\) east, \(y\) north and \(z\) up, so that \(\vect{\Omega}=\Omega\left(0,\cos\lambda,\sin\lambda\right)\). The centrifugal term is absorbed into the effective gravity by Equation (29.68), and the Euler term is negligible (Example 29.52), so Equation (29.66) leaves

Solve perturbatively in \(\Omega\). At zeroth order \(\dot{\vect{r}}=\left(0,0,-gt\right)\); substituting this into the Coriolis term,

which points due east. Integrating twice from rest,

using \(h=\tfrac{1}{2}gt^{2}\) for the drop. There is no first-order deflection in the north–south direction: the Coriolis term acting on a purely vertical velocity has no northward component, and the southward term generated at second order is smaller by a further factor \(\Omega t\).

For Reich's shaft, \(h=158.5\,\mathrm{m}\) gives \(t=5.68\,\mathrm{s}\), and with \(\lambda=50.9^\circ\) and \(\Omega=7.292\times 10^{-5}\,/\mathrm{s}\), Equation (29.69) predicts \(27.6\,\mathrm{mm}\), against the measured mean of \(28.3\,\mathrm{mm}\) [Reich:1832] — agreement to \(3\,\mathrm{\%}\), which for a measurement of a \(28\,\mathrm{mm}\) displacement at the bottom of a \(160\,\mathrm{m}\) shaft in 1831 is remarkable. The prediction contains no adjustable parameter: \(h\) and \(\lambda\) are properties of the shaft and \(\Omega\) is the rotation rate of the Earth.

Derives Proposition 29.57. In the local axes of Phenomenon 29.56, with \(\dot{\vect{r}}=\left(v_{E},v_{N},0\right)\),

The vertical component is \(2\Omega v_{E}\cos\lambda\), upward for \(v_{E}>0\); a gravimeter, which weighs against the local vertical, therefore reads low by that amount, which is Equation (29.70). The component \(v_{N}\) contributes nothing vertical.

Derivation. Derives Phenomenon 29.58. For horizontal motion only the vertical component of \(\vect{\Omega}\) matters, since the horizontal component produces a vertical Coriolis force that the pressure field balances (this is the same separation used in Proposition 29.57). Define the Coriolis parameter

which at \(45^\circ\) is \(1.03\times 10^{-4}\,/\mathrm{s}\) and vanishes on the equator. In the local horizontal plane the equation of motion of a fluid element of density \(\rho\) is, from Equation (29.66) with the pressure-gradient force of Fluid Dynamics,

When the flow is steady and slowly varying in space the left-hand side is negligible against either term on the right, and the two balance:

This is geostrophic balance, and it says three things at once. The velocity is perpendicular to the pressure gradient, hence along the isobars. Its sense is fixed by the sign of \(f\), so it reverses across the equator. And its magnitude is \(\abs{\vect{u}}=\abs{\nabla_{h}p}/(\rho f)\): for a typical gradient of one hectopascal per \(100\,\mathrm{km}\), that is \(10^{-3}\,\mathrm{N}/\mathrm{m}^{3}\), with \(\rho=1.2\,\mathrm{kg}/\mathrm{m}^{3}\) and \(f\) at \(45^\circ\), the wind is \(8.1\,\mathrm{m}/\mathrm{s}\).

Whether the balance holds is decided by the ratio of the neglected term to the retained one, the Rossby number \(\mathrm{Ro}=U/(fL)\) for a flow of speed \(U\) over a length \(L\). A mid-latitude weather system has \(U\sim10\,\mathrm{m}/\mathrm{s}\) and \(L\sim10^{6}\,\mathrm{m}\), so \(\mathrm{Ro}\approx0.10\) and the balance is good. Near the equator \(f\to0\) and \(\mathrm{Ro}\) diverges, which is why organized cyclonic circulation does not form there.

Derives Theorem 29.60. Equation (29.73) is Equation (29.66) written in the plane: for \(\vect{\Omega}=\Omega\hat{\vect{z}}\) the Coriolis term \(-2\vect{\Omega}\times\vect{v}\) becomes \(-2\ii\Omega\dot{\zeta}\) and the centrifugal term \(-\vect{\Omega}\times \left(\vect{\Omega}\times\vect{r}\right)\) becomes \(+\Omega^{2}\zeta\). Substituting \(\zeta=\ee^{-\ii\Omega t}\eta\),

The right-hand side of Equation (29.73) becomes \(\ee^{-\ii\Omega t}\left[\left(F/m\right)\eta -2\ii\Omega\dot{\eta}-2\Omega^{2}\eta+\Omega^{2}\eta\right]\), and the factor \(F\) is unchanged because \(\abs{\zeta}=\abs{\eta}\). Cancelling \(\ee^{-\ii\Omega t}\) and subtracting, the terms in \(\dot{\eta}\) and the terms in \(\Omega^{2}\eta\) cancel identically on both sides, and \(\ddot{\eta}=\left(F/m\right)\eta\) remains. No approximation in \(\Omega\) was made.

Derivation. Derives Phenomenon 29.61. Use the local axes of Phenomenon 29.56, in which \(\vect{\Omega}=\Omega\left(0,\cos\lambda,\sin\lambda\right)\), and let the bob swing with small amplitude about the lowest point of its suspension. Three reductions bring the problem into the form of Theorem 29.60.

First, the centrifugal term of Equation (29.66) is independent of time and is absorbed once and for all into the effective gravity and the local vertical, exactly as in Equation (29.68); the swing is then about that vertical.

Second, split \(\vect{\Omega}\) into its vertical and horizontal parts, \(\Omega_{z}=\Omega\sin\lambda\) and \(\Omega_{y}=\Omega\cos\lambda\). For small oscillations the bob's velocity is horizontal to first order, since the vertical excursion is second order in the amplitude. The Coriolis force built from \(\Omega_{y}\) acting on a horizontal velocity is then vertical, and is taken up by the tension in the wire, which is free to adjust; it does not move the bob horizontally. Only \(\Omega_{z}\) survives in the horizontal equations, and this is why the experiment reports the component of \(\vect{\Omega}\) along the local vertical rather than the whole of it.

Third, for small amplitude the horizontal restoring force is central and linear, \(\vect{F}=-m\omega_{0}^{2}\vect{r}\) with \(\omega_{0}^{2}=g/\ell\) (Equation (28.7)). Theorem 29.60 now applies verbatim with \(\Omega\to\Omega_{z}\): the substitution \(\zeta=\ee^{-\ii\Omega_{z}t}\eta\) reduces the motion to \(\ddot{\eta}=-\omega_{0}^{2}\eta\), a plane pendulum whose plane is fixed. Seen from the room, that plane therefore turns at

clockwise from above for \(\lambda>0\), one turn per sidereal day at the pole and none on the equator. Note what does not appear: the length, the mass and the amplitude of the pendulum are all absent, so the effect cannot be confused with a property of the particular apparatus. The numerical value at the latitude of Paris, and the data that confirm it, are Equation (34.39) and Section 34.5.