Part B · unit 16 · problems

Rigid bodies, and what a tensor actually is

the slides · the curriculum

Checkpoint. For an explicitly given body you compute \(\mathsf I\) in two frames, verify \(\mathsf I' = R\mathsf IR^{\mathsf T}\) component by component, predict which axis is unstable, and can explain in one sentence why nine numbers per basis is not enough to be a tensor.

Work each problem on paper first. Under it, Hints opens a ladder one rung at a time — each rung appears only once the one before it is open, and the last gives the first line of the working — and Worked solution has the whole thing: the idea, the working, then a check. Read the solution after your attempt, including when you got it right. F folds everything back; D switches dark and light.

Thirteen problems, about six hours. Conventions throughout: \(\mathsf I_O(\vec u,\vec w) = \int_B\rho\big[(\vec u\cdot\vec w)\lVert\vec r\rVert^2-(\vec u\cdot\vec r)(\vec w\cdot\vec r)\big]\mathrm dV\) about the stated origin \(O\); a change of basis \(\vec f_a = S_{ia}\vec e_i\) has \(A = S^{-1}\), and for an orthonormal pair \(R := S^{\mathsf T}\) has the new basis vectors as its rows, so \(\mathsf I' = R\mathsf IR^{\mathsf T}\). Principal moments are ordered \(I_1\le I_2\le I_3\). Euler's equations are taken in the principal body frame.

Constants: \(g = 9.81\ \mathrm{m\,s^{-2}}\); the Earth's dynamical ellipticity \((I_3-I_1)/I_3 = 3.2737949\times10^{-3}\); one sidereal day \(= 86164\) s.

Computing the inertia tensor

P1 A cylinder, from the integral routine

exercises: multiple integrals in cylindrical coordinates, symmetry killing off-diagonal terms, limits

A uniform solid cylinder has mass \(M\), radius \(R\), height \(h\), axis along \(\vec e_3\), and its centre of mass at the origin.

  1. Show from the definition that the off-diagonal components of \(\mathsf I\) vanish in this frame, naming the symmetry you use.
  2. Compute \(I_{33}\) and \(I_{11}\) by integrating in cylindrical coordinates.
  3. Check the limits \(h\to0\) (a disc) and \(R\to0\) (a rod), and check the triangle inequality \(I_1+I_2\ge I_3\): for which shape is it tight?
  4. Numbers: \(M = 2.00\) kg, \(R = 0.050\) m, \(h = 0.30\) m.
Hints · 3 rungs
Rung 1

The answer to 2 is two formulas in \(M, R, h\), and to 4 two numbers in \(\mathrm{kg\,m^2}\). Nothing here needs a theorem: the definition of \(\mathsf I\) and unit 05's change of variables are the whole method.

Rung 2

In cylindrical coordinates \((s,\varphi,z)\) the volume element is \(s\,\mathrm ds\,\mathrm d\varphi\,\mathrm dz\), and \(x = s\cos\varphi\), \(y = s\sin\varphi\). Each off-diagonal integrand contains either \(\cos\varphi\sin\varphi\) or a single odd power of \(z\); integrate that factor first.

Rung 3

First line: \(\rho = M/(\pi R^2h)\), and \(I_{33} = \int\rho(x^2+y^2)\,\mathrm dV = \rho\int_{-h/2}^{h/2}\!\int_0^{2\pi}\!\!\int_0^R s^2\cdot s\,\mathrm ds\,\mathrm d\varphi\,\mathrm dz\). For \(I_{11}\) use \(y^2+z^2 = s^2\sin^2\varphi + z^2\) and \(\int_0^{2\pi}\sin^2 = \pi\).

Worked solution

Idea. Cylindrical coordinates turn every integral into a product of one-dimensional integrals, and the angular one kills all six off-diagonal components at once.

1. Off-diagonal. \(\rho\) is constant and the region is \(\{s\le R,\ |z|\le h/2\}\). Then \(I_{12} = -\int\rho\,xy\,\mathrm dV = -\rho\int_{-h/2}^{h/2}\!\mathrm dz\int_0^R\!s^3\mathrm ds\int_0^{2\pi}\!\cos\varphi\sin\varphi\,\mathrm d\varphi = 0\), because \(\int_0^{2\pi}\cos\varphi\sin\varphi\,\mathrm d\varphi = \tfrac12\int_0^{2\pi}\sin2\varphi = 0\). For \(I_{13} = -\int\rho\,xz\), the \(z\)-integral \(\int_{-h/2}^{h/2}z\,\mathrm dz\) vanishes by the reflection \(z\mapsto-z\); same for \(I_{23}\). The symmetries used are the rotational symmetry about \(\vec e_3\) and the mirror \(z\mapsto-z\).

2. Diagonal. With \(\rho = M/(\pi R^2h)\),

\[ I_{33} = \rho\int_{-h/2}^{h/2}\!\!\mathrm dz\int_0^{2\pi}\!\!\mathrm d\varphi\int_0^R s^3\,\mathrm ds = \rho\,h\cdot2\pi\cdot\frac{R^4}{4} = \tfrac12MR^2 . \]

For \(I_{11} = \int\rho(y^2+z^2)\mathrm dV\): the \(y^2\) part is \(\rho\,h\int_0^{2\pi}\sin^2\varphi\,\mathrm d\varphi\int_0^Rs^3\mathrm ds = \rho h\pi R^4/4 = \tfrac14MR^2\), and the \(z^2\) part is \(\rho\,\pi R^2\int_{-h/2}^{h/2}z^2\mathrm dz = \rho\pi R^2h^3/12 = \tfrac1{12}Mh^2\). Hence

\[ I_{11} = I_{22} = \tfrac14MR^2+\tfrac1{12}Mh^2 = \tfrac{M}{12}\big(3R^2+h^2\big),\qquad I_{33} = \tfrac12MR^2 . \]

3. Checks. \(h\to0\): \(I_1 = I_2 = \tfrac14MR^2\) and \(I_3 = \tfrac12MR^2\), the disc, and \(I_1+I_2 = I_3\) exactly — tight, as the perpendicular-axis theorem (P2) requires for a lamina. \(R\to0\): \(I_1=I_2 = \tfrac1{12}Mh^2\) and \(I_3=0\), the rod. In general \(I_1+I_2-I_3 = \tfrac16Mh^2\ge0\) ✓, vanishing only for the disc.

4. Numbers. \(I_3 = 0.5\times2.00\times0.0025 = \mathbf{2.50\times10^{-3}}\ \mathrm{kg\,m^2}\); \(3R^2+h^2 = 0.0075+0.09 = 0.0975\), so \(I_1 = I_2 = 2.00\times0.0975/12 = \mathbf{1.625\times10^{-2}}\ \mathrm{kg\,m^2}\).

Check. Dimensions \(\mathrm{kg\,m^2}\) throughout ✓. Trace \(2I_1+I_3 = 0.0325+0.0025 = 0.0350\), against \(2\int\rho\lVert\vec r\rVert^2\mathrm dV = 2M(\tfrac12R^2+\tfrac1{12}h^2) = 2\times2(0.00125+0.0075) = 0.0350\) ✓. Ordering: here \(I_3\lt I_1 = I_2\), so this cylinder is a prolate symmetric top — long and thin, easiest to spin about its long axis, exactly as the block's spine was.

Another way

Slice the cylinder into discs of mass \(\mathrm dm = M\,\mathrm dz/h\). Each has \(\mathrm dI_{33} = \tfrac12R^2\mathrm dm\) and, about its own centre, \(\mathrm dI_{11} = \tfrac14R^2\mathrm dm\); the parallel-axis theorem adds \(z^2\mathrm dm\) to the latter. Integrating \(z^2\) over \([-h/2,h/2]\) gives \(Mh^2/12\), and the same two answers appear with no triple integral at all.

P2 Proof: the perpendicular-axis theorem, and where it stops medium

exercises: proving a theorem from the definition, the triangle inequality, finding the hypothesis that does the work

A lamina is a body lying entirely in the plane \(x_3 = 0\).

  1. Prove that for a lamina, \(I_{33} = I_{11}+I_{22}\) in that frame — the perpendicular-axis theorem — and prove it in the basis-free form: for any orthonormal \(\vec u,\vec w\) in the plane and \(\hat k\) normal to it, \(\mathsf I(\hat k,\hat k) = \mathsf I(\vec u,\vec u)+\mathsf I(\vec w,\vec w)\).
  2. Apply it to a uniform rectangular plate with sides \(p\) along \(\vec e_1\) and \(q\) along \(\vec e_2\), about its centre; then to a uniform annulus of inner radius \(R_1\) and outer radius \(R_2\).
  3. Show by exhibiting a body that the identity fails for a general three-dimensional body, and state exactly which step of your proof breaks.
  4. Deduce the slides' triangle inequality \(I_1+I_2\ge I_3\) for a general body, with equality iff the body is a lamina in the \(1\)–\(2\) plane.
Hints · 3 rungs
Rung 1

Parts 1 and 4 want proofs; part 2 wants formulas; part 3 wants one counterexample plus one sentence naming a hypothesis. No integration is needed in part 1 at all — only the definition and an algebraic identity between three integrands.

Rung 2

Write the three diagonal components as integrals and add two of them. Ask what \((x_2^2+x_3^2)+(x_1^2+x_3^2)\) is, compared with \(x_1^2+x_2^2\). For the basis-free version, choose the orthonormal basis \((\vec u,\vec w,\hat k)\) and note that "lamina" is the statement that \(\hat k\cdot\vec r = 0\) on the body.

Rung 3

First line: \(I_{11}+I_{22} = \int\rho\big[(x_2^2+x_3^2)+(x_3^2+x_1^2)\big]\mathrm dV = \int\rho\big[(x_1^2+x_2^2)+2x_3^2\big]\mathrm dV = I_{33}+2\int\rho\,x_3^2\,\mathrm dV\). Everything in the problem follows from reading that line in different directions.

Worked solution

Idea. One identity carries all four parts: \(I_{11}+I_{22}-I_{33} = 2\int_B\rho\,x_3^2\,\mathrm dV\). It is a theorem when the right side is zero, an inequality when it is not, and its failure for a solid body is exactly the right side being positive.

1. From the definition, \(I_{kk} = \int\rho(\lVert\vec r\rVert^2-x_k^2)\mathrm dV\), so

\[ I_{11}+I_{22} = \int_B\rho\big(2\lVert\vec r\rVert^2-x_1^2-x_2^2\big)\mathrm dV = \int_B\rho\big(x_1^2+x_2^2\big)\mathrm dV + 2\int_B\rho\,x_3^2\,\mathrm dV, \]

whose first term is \(I_{33}\); or, working on the integrands alone,

\[ I_{11}+I_{22}-I_{33} = \int_B\rho\Big[(x_2^2+x_3^2)+(x_3^2+x_1^2)-(x_1^2+x_2^2)\Big]\mathrm dV = 2\int_B\rho\,x_3^2\,\mathrm dV . \]

For a lamina \(x_3 = 0\) on \(B\), so the right side is zero. Basis-free: take the orthonormal basis \((\vec u,\vec w,\hat k)\); the identity above is exactly \(\mathsf I(\vec u,\vec u)+\mathsf I(\vec w,\vec w)-\mathsf I(\hat k,\hat k) = 2\int\rho(\hat k\cdot\vec r)^2\mathrm dV\), and "lamina normal to \(\hat k\)" says \(\hat k\cdot\vec r\equiv0\). \(\;\blacksquare\)

2. Rectangular plate. It is the \(c\to0\) limit of the slides' block: \(I_{11} = \tfrac{M}{12}q^2\), \(I_{22} = \tfrac{M}{12}p^2\), and the theorem gives \(I_{33} = \tfrac{M}{12}(p^2+q^2)\) with no further integration ✓ — which is the block's third formula.

Annulus. \(I_{33} = \int\rho\,s^2\,\mathrm dA\) with \(\rho = M/\pi(R_2^2-R_1^2)\), so \(I_{33} = \rho\,2\pi\int_{R_1}^{R_2}s^3\mathrm ds = \tfrac12M(R_1^2+R_2^2)\). By symmetry \(I_{11} = I_{22}\), and the theorem halves it: \(I_{11} = I_{22} = \tfrac14M(R_1^2+R_2^2)\). For \(R_1 = R_2 = R\) this is the hoop, \(\tfrac12MR^2\) and \(\tfrac14MR^2\) ✓; for \(R_1 = 0\) the disc ✓.

3. Where it stops. Take the cube of side \(a\) about its centre. Every moment is \(\tfrac16Ma^2\), so \(I_{11}+I_{22} = \tfrac13Ma^2 \ne \tfrac16Ma^2 = I_{33}\) — off by a factor two. The broken step is the very last one: \(\int\rho\,x_3^2\,\mathrm dV = \tfrac1{12}Ma^2 \ne 0\). The theorem never used flatness anywhere else, which is why it is an identity plus one hypothesis and not a delicate argument.

4. The inequality. The identity's right side is \(2\int\rho\,x_3^2\mathrm dV\ge0\), with \(\rho\ge0\); so \(I_{11}+I_{22}\ge I_{33}\) in any orthonormal frame, in particular the principal one, giving \(I_1+I_2\ge I_3\). Equality forces \(\int\rho\,x_3^2 = 0\), hence \(x_3 = 0\) for almost every point carrying mass: a lamina in the \(1\)–\(2\) plane. \(\;\blacksquare\)

Check. Dimensions: both sides \(\mathrm{kg\,m^2}\) ✓. The slides' block: \(I_1+I_2-I_3 = 0.00136+0.00296-0.00416 = 1.6\times10^{-4}\), and \(2\int\rho z^2 = \tfrac{M}{6}c^2 = 0.1\times0.0016 = 1.6\times10^{-4}\) ✓. Limit: as \(c\to0\) at fixed \(M\) the block becomes a plate and the slack closes ✓.

P3 Two frames, one tensor medium

exercises: the inertia tensor of point masses, the transformation law verified two ways, diagonalising a 2×2 block

Three point masses sit in the plane \(z=0\): \(m_1 = 1\) kg at \((1,0,0)\), \(m_2 = 2\) kg at \((0,1,0)\), \(m_3 = 3\) kg at \((1,1,0)\), all coordinates in metres.

  1. Compute \(\mathsf I\) about the origin in this frame, as a \(3\times3\) array.
  2. Let \(\vec f_1,\vec f_2\) be \(\vec e_1,\vec e_2\) rotated by \(30^\circ\) about \(\vec e_3\), and \(\vec f_3 = \vec e_3\). Compute \(\mathsf I'\) twice: once from \(\mathsf I' = R\mathsf IR^{\mathsf T}\), and once by re-doing the sum with the masses' coordinates in the new frame. Compare all six independent components.
  3. Find the principal moments and the angle the smallest principal axis makes with \(\vec e_1\).
  4. Verify that \(\operatorname{tr}\mathsf I\) and \(\det\mathsf I\) are the same in both frames, and say why that was guaranteed.
Hints · 3 rungs
Rung 1

Part 1 is a \(3\times3\) array of numbers in \(\mathrm{kg\,m^2}\); part 3 is two numbers and an angle. Because everything lies in a plane, the array is block diagonal and only a \(2\times2\) problem is ever really being solved.

Rung 2

For point masses the integral is a sum: \(I_{ij} = \sum_\alpha m_\alpha(\lVert\vec r_\alpha\rVert^2\delta_{ij}-x_{\alpha i}x_{\alpha j})\). For part 2's second route, the new coordinates of a mass are \(x'_a = R_{ai}x_i\) — the same matrix that transforms the tensor, which is the point. For part 3, the principal angle of \(\begin{pmatrix}A&B\\B&C\end{pmatrix}\) satisfies \(\tan2\theta = 2B/(A-C)\).

Rung 3

First line: \(m_3\) at \((1,1,0)\) has \(\lVert\vec r\rVert^2 = 2\) and contributes \(3\begin{pmatrix}1&-1&0\\-1&1&0\\0&0&2\end{pmatrix}\). Add the other two and you should find \(I_{11} = 5\) exactly.

Worked solution

Idea. The transformation law is a theorem about a definition, so recomputing from the definition in the new frame must reproduce it. Doing both and comparing is the checkpoint of this unit in miniature.

1. Contribution of a mass \(m\) at \(\vec r\) is \(m(\lVert\vec r\rVert^2\delta-\vec r\vec r^{\mathsf T})\):

  • \(m_1\): \(\lVert\vec r\rVert^2=1\), giving \(\mathrm{diag}(0,1,1)\);
  • \(m_2\): \(2\,\mathrm{diag}(1,0,1) = \mathrm{diag}(2,0,2)\);
  • \(m_3\): \(3\big[2\delta-\vec r\vec r^{\mathsf T}\big]\) with \(\vec r\vec r^{\mathsf T}\) having every entry 1 in the upper \(2\times2\) block: \(\begin{pmatrix}3&-3&0\\-3&3&0\\0&0&6\end{pmatrix}\).
\[ \mathsf I \;\hat=\; \begin{pmatrix}5&-3&0\\-3&4&0\\0&0&9\end{pmatrix}\ \mathrm{kg\,m^2}. \]

2. Route A, the law. With \(c=\cos30^\circ = \tfrac{\sqrt3}2\), \(s = \tfrac12\) and \(R = \begin{pmatrix}c&s&0\\-s&c&0\\0&0&1\end{pmatrix}\):

\[ \begin{aligned} I'_{11} &= c^2(5)+2cs(-3)+s^2(4) = \tfrac{15}4-\tfrac{3\sqrt3}2+1 = 2.151924,\\ I'_{12} &= -cs(5)+(c^2-s^2)(-3)+cs(4) = -\tfrac{\sqrt3}4-\tfrac32 = -1.933013,\\ I'_{22} &= s^2(5)+2cs(3)+c^2(4) = \tfrac54+\tfrac{3\sqrt3}2+3 = 6.848076, \end{aligned} \]

with \(I'_{13} = I'_{23} = 0\) (both \(I_{13}\) and \(I_{23}\) vanish and the rotation mixes only them) and \(I'_{33} = 9\).

Route B, the definition. The new coordinates are \(x' = cx+sy\), \(y' = -sx+cy\): \(m_1\) sits at \((0.866025,-0.5,0)\); \(m_2\) at \((0.5,0.866025,0)\); \(m_3\) at \((1.366025,0.366025,0)\). Then \(I'_{11} = \sum m(y'^2+z'^2) = 1(0.25)+2(0.75)+3(0.133975) = 0.25+1.5+0.401924 = 2.151924\) ✓, and \(I'_{12} = -\sum m\,x'y' = -[1(-0.433013)+2(0.433013)+3(0.5)] = -[-0.433013+0.866025+1.5] = -1.933013\) ✓. Continuing gives \(I'_{22} = 6.848076\) and \(I'_{33} = 9\) ✓. All six agree.

3. Principal moments. The \(2\times2\) block has trace 9 and determinant \(20-9 = 11\), so \(\lambda^2-9\lambda+11 = 0\) and

\[ \lambda = \tfrac12\big(9\pm\sqrt{37}\big) = 1.458619,\ 7.541381\ \mathrm{kg\,m^2}, \]

together with \(9\) from the \(z\)-axis. So \(I_1 = 1.4586\), \(I_2 = 7.5414\), \(I_3 = 9\). For the smallest, \((5-\lambda)v_1 = 3v_2\) gives \(v_2/v_1 = (5-1.458619)/3 = 1.180460\), i.e. an angle \(\arctan(1.180460) = \mathbf{49.73^\circ}\) from \(\vec e_1\). (Equivalently \(\tan2\theta = 2(-3)/(5-4) = -6\).)

4. Trace: \(5+4+9 = 18\) and \(2.151924+6.848076+9 = 18\) ✓. Determinant: \(9\times11 = 99\) in both, since the \(2\times2\) determinant is \((2.151924)(6.848076)-(1.933013)^2 = 14.736-3.736 = 11.000\) ✓. Guaranteed because \(\det(R\mathsf IR^{\mathsf T}) = \det R\,\det\mathsf I\,\det R^{\mathsf T} = \det\mathsf I\) and the trace is invariant under an orthogonal congruence, which is the same as a similarity.

Check. Dimensions \(\mathrm{kg\,m^2}\) ✓. Perpendicular axis: the body is a lamina, and \(I_1+I_2 = 1.458619+7.541381 = 9 = I_3\) exactly ✓ — P2, confirming both the eigenvalues and the arithmetic at once. Limit: setting \(m_3=0\) gives \(\mathrm{diag}(2,1,3)\) with the axes back along \(\vec e_1,\vec e_2\), so the \(49.73^\circ\) tilt is produced entirely by the third mass ✓.

What is, and is not, a tensor

P4 Convict a collection of nine numbers medium

exercises: testing the transformation law, saying precisely what a rule fails, distinguishing "computable" from "tensorial"

Both of the following are rules that assign nine real numbers to every orthonormal basis of \(\mathbb R^3\). Neither is a tensor. Convict each by explicit computation, and then say in one sentence what it fails.

  1. The diagonal rule. \(N_{ab} :=\) the array \(\mathrm{diag}(I_{11},I_{22},I_{33})\) built from the inertia tensor's diagonal entries in that basis. Use P3's body and its \(30^\circ\) rotation.
  2. The absolute-value rule. Fix a vector \(\vec v\). Let \(M_{ab} := |v_a|\,|v_b|\), the components being those of \(\vec v\) in that basis. Take \(\vec v = \vec e_1+\vec e_2\) and the \(90^\circ\) rotation about \(\vec e_3\).
  3. For each, identify which piece of information the rule discards or distorts, and state the general criterion a basis-indexed array must satisfy to come from a tensor.
Hints · 3 rungs
Rung 1

A conviction is two numbers that should be equal and are not. Compute the array in the new basis by the rule, and compute what the transformation law predicts from the old array; exhibit one component where they differ.

Rung 2

The law predicts \(R\,(\text{old array})\,R^{\mathsf T}\). For rule 1 the old array is \(\mathrm{diag}(5,4,9)\), and the rule applied in the new basis needs \(I'_{11}\), which P3 already gives. For rule 2, first work out the components of \(\vec v\) in the rotated basis.

Rung 3

First line, rule 1: \(\big(R\,\mathrm{diag}(5,4,9)\,R^{\mathsf T}\big)_{11} = c^2\cdot5+s^2\cdot4\) with \(c^2=\tfrac34\), \(s^2=\tfrac14\). Compare with \(N'_{11} = I'_{11}\).

Worked solution

Idea. A tensor is determined by its components in one basis, so it can only have one array in any other. Two different answers for the same basis is a complete disproof, and one component suffices.

1. The diagonal rule. In \(\{\vec e_i\}\), \(N = \mathrm{diag}(5,4,9)\). The law predicts

\[ \big(RNR^{\mathsf T}\big)_{11} = c^2\cdot5+2cs\cdot0+s^2\cdot4 = \tfrac34(5)+\tfrac14(4) = \mathbf{4.75}. \]

The rule applied in \(\{\vec f_a\}\) returns \(N'_{11} = I'_{11} = \mathbf{2.151924}\) (P3). Not equal, so no tensor produces both.

2. The absolute-value rule. \(\vec v = (1,1,0)\) gives \(M = \begin{pmatrix}1&1&0\\1&1&0\\0&0&0\end{pmatrix}\). With \(\vec f_1 = \vec e_2\), \(\vec f_2 = -\vec e_1\), so \(R = \begin{pmatrix}0&1&0\\-1&0&0\\0&0&1\end{pmatrix}\), the law predicts

\[ RMR^{\mathsf T} = \begin{pmatrix}1&-1&0\\-1&1&0\\0&0&0\end{pmatrix}. \]

The rule applied in the new basis: \(\vec v\) has components \(R\vec v = (1,-1,0)\), so \(|v'| = (1,1,0)\) and \(M' = \begin{pmatrix}1&1&0\\1&1&0\\0&0&0\end{pmatrix}\). The off-diagonal entries differ in sign: \(+1\) against \(-1\). Convicted.

3. What each fails. Rule 1 discards the off-diagonal entries, and those are exactly what a rotated diagonal entry needs: the law gives \(I'_{11} = c^2I_{11}+2csI_{12}+s^2I_{22}\), and the missing \(2csI_{12} = -2.598\) is precisely the gap \(2.152-4.75\). Rule 2 distorts signs: without the bars, \(v_av_b\) are the components of the tensor \(\vec v^\flat\otimes\vec v^\flat\), and indeed \(R\,\vec v\vec v^{\mathsf T}R^{\mathsf T}\) is the predicted array above; the bars erase the sign of \(v'_2 = -1\), and which signs get erased depends on the basis, so the rule cannot commute with a rotation that turns a positive component negative.

The criterion. A rule assigning an array to each basis comes from a tensor if and only if it is equivariant: its arrays in any two bases are related by the transformation law. The "only if" is the theorem of the slides; the "if" is immediate, since one may define the tensor by the array in one basis, and the law then forces its components in every other basis to be the rule's.

Check. Sanity: rule 1 does agree with the law whenever the rotation is a permutation of the axes (then no off-diagonal entry is needed), which is why the failure is invisible to anyone who only ever relabels axes — and why "the moments about \(x\), \(y\) and \(z\)" survives so long in elementary treatments. Rule 2 agrees whenever \(R\) has non-negative entries only. Both sanity checks confirm that the counterexamples had to use a genuinely mixing rotation.

P5 Proof: which traces survive, and the quotient theorem hard

exercises: distinguishing similarity from congruence, proving a converse, index bookkeeping under a general change of basis

Throughout, \(\vec f_a = S_{ia}\vec e_i\) with \(S\) invertible but not assumed orthogonal, and \(A = S^{-1}\).

  1. Prove that for a \((1,1)\)-tensor, \(T'^a{}_a = T^i{}_i\).
  2. Prove that for a \((0,2)\)-tensor, \(\sum_aT'_{aa}\) need not equal \(\sum_iT_{ii}\), by taking \(T_{ij} = \delta_{ij}\) and \(S = \mathrm{diag}(2,1,1)\). Then say exactly what extra structure makes "\(\operatorname{tr}\mathsf I\)" a legitimate invariant in the slides.
  3. The quotient theorem. Suppose a rule assigns to each basis an array \(N_{ij}\), and suppose that for every pair of vectors \(\vec u,\vec w\) the number \(N_{ij}u^iw^j\) is the same in every basis. Prove that the rule is the component rule of a \((0,2)\)-tensor.
  4. Use part 3 to prove that the inertia tensor is a tensor, starting from the physical fact that the rotational kinetic energy \(\tfrac12 I_{ij}\omega^i\omega^j\) is a number independent of the frame in which it is computed.
Hints · 3 rungs
Rung 1

Parts 1, 3 and 4 are proofs; part 2 is one counterexample plus one sentence. Everything follows from the two transformation laws — vectors by \(A\), covectors by \(S^{\mathsf T}\) — and from \(SA = AS = \mathbb 1\).

Rung 2

Part 1: write the law for a \((1,1)\)-tensor and set \(a=b\), summing. You will meet \(\sum_aA_{ai}S_{ja}\), which is a matrix product in disguise. Part 3: you are given a scalar and must produce a multilinear map — so define one, using the array in a single chosen basis, and then show the hypothesis forces the rule's other arrays to be its components.

Rung 3

Part 1, first line: \(T'^a{}_b = A_{ai}S_{jb}T^i{}_j\), so \(T'^a{}_a = \big(\sum_aS_{ja}A_{ai}\big)T^i{}_j = (SA)_{ji}T^i{}_j\). Part 3, first line: fix the basis \(\{\vec e_i\}\) and define \(T(\vec u,\vec w) := N_{ij}u^iw^j\), which is bilinear because the right side is.

Worked solution

Idea. A contraction survives a change of basis exactly when the two matrices it meets are inverse to one another. Upper and lower indices bring \(A\) and \(S\), so one of each cancels; two lower indices bring \(S\) twice, and \(S^{\mathsf T}S\ne\mathbb 1\) in general.

1. The law is \(T'^a{}_b = A_{ai}S_{jb}T^i{}_j\). Put \(b=a\) and sum:

\[ T'^a{}_a = \Big(\sum_a S_{ja}A_{ai}\Big)T^i{}_j = (SA)_{ji}\,T^i{}_j = \delta_{ji}T^i{}_j = T^i{}_i . \;\blacksquare \]

2. For a \((0,2)\)-tensor, \(T' = S^{\mathsf T}TS\). With \(T=\mathbb 1\) and \(S = \mathrm{diag}(2,1,1)\), \(T' = S^{\mathsf T}S = \mathrm{diag}(4,1,1)\), whose diagonal sum is \(\mathbf 6\), against \(\mathbf 3\) before. So \(\sum_iT_{ii}\) is not an invariant.

What rescues \(\operatorname{tr}\mathsf I\) in the slides is that only orthonormal bases are ever used, i.e. \(S\) is restricted to orthogonal matrices, for which \(S^{\mathsf T}S=\mathbb 1\) and congruence coincides with similarity. Said structurally: the Euclidean inner product is being used, silently, to raise one index — \(\operatorname{tr}\mathsf I\) really means \(\delta^{ij}I_{ij}\), and \(\delta^{ij}\) is the inverse metric, a genuine tensor. In unit 32 the same contraction is written \(g^{\mu\nu}T_{\mu\nu}\) and the metric is not the identity.

3. The quotient theorem. Fix one basis \(\{\vec e_i\}\), with the rule's array \(N_{ij}\) there. Define \(T:\mathbb R^3\times\mathbb R^3\to\mathbb R\) by \(T(\vec u,\vec w) := N_{ij}u^iw^j\), the components being taken in that basis. \(T\) is bilinear, since the right side is linear in each of \(u^i\) and \(w^j\) separately — so \(T\) is a \((0,2)\)-tensor, and its components in \(\{\vec e_i\}\) are \(T(\vec e_i,\vec e_j) = N_{ij}\).

Now take any other basis \(\{\vec f_a\}\), and write \(N'_{ab}\) for the rule's array there and \(u'^a\) for the new components of \(\vec u\). The hypothesis says \(N'_{ab}u'^aw'^b = N_{ij}u^iw^j = T(\vec u,\vec w)\) for every \(\vec u,\vec w\). Take \(\vec u = \vec f_a\) and \(\vec w = \vec f_b\), whose new components are \(\delta\)'s: the left side collapses to \(N'_{ab}\), so \(N'_{ab} = T(\vec f_a,\vec f_b) = T'_{ab}\). The rule's array in every basis is therefore \(T\)'s component array. \(\;\blacksquare\)

4. The rotational kinetic energy is defined by an integral over the body, \(T_{\text{rot}} = \tfrac12\int\rho\lVert\vec\omega\times\vec r\rVert^2\mathrm dV\), with no basis anywhere in the definition; so its value is frame-independent, and equals \(\tfrac12I_{ij}\omega^i\omega^j\) in any frame. Write \(Q(\vec u) := I_{ij}u^iu^j\), twice that energy, so \(Q\) is frame-independent too. Polarisation extends this from the quadratic form to the bilinear form: \(I_{ij}u^iw^j = \tfrac12\big[Q(\vec u+\vec w)-Q(\vec u)-Q(\vec w)\big]\) is a difference of frame-independent numbers, hence frame-independent, for every \(\vec u,\vec w\). Part 3 then applies verbatim, and \(\mathsf I\) is a \((0,2)\)-tensor. \(\;\blacksquare\)

Check. Consistency with P4: the diagonal rule fails the hypothesis of part 3 — \(N_{ij}u^iw^j\) for \(N = \mathrm{diag}(5,4,9)\) and \(\vec u=\vec w=\vec f_1\) gives \(4.75\) in the old basis and \(2.152\) in the new, so the quotient theorem is not being contradicted ✓. Special case of part 2: for orthogonal \(S\), \(S^{\mathsf T}S = \mathbb 1\) and the two traces agree, as they must ✓. Dimensional check on part 4: \(I_{ij}\omega^i\omega^j\) is \(\mathrm{kg\,m^2\,s^{-2}} = \mathrm J\) in every frame ✓.

P6 Index drills routine

exercises: the summation convention, invariant subspaces of the transformation law, counting components

  1. Show that the \((1,1)\)-tensor with components \(\delta^i_j\) has the same array in every basis, and that the \((0,2)\)-tensor with components \(\delta_{ij}\) does not — using \(S = \mathrm{diag}(2,1,1)\).
  2. Show that the splitting \(T_{ij} = T_{(ij)}+T_{[ij]}\) of a \((0,2)\)-tensor is preserved by every change of basis, and give an explicit \(T\) and \(S\) showing that the same is false for a \((1,1)\)-tensor.
  3. Count the independent components, in \(n\) dimensions, of: a symmetric \((0,2)\)-tensor; an antisymmetric \((0,2)\)-tensor; a totally antisymmetric \((0,3)\)-tensor. Evaluate all three at \(n=3\) and at \(n=4\).
  4. Show that when the change of basis is orthogonal, vector and covector components transform by the same matrix — so the up–down distinction is invisible in this unit and only in this unit.
Hints · 3 rungs
Rung 1

Parts 1, 2 and 4 are short proofs — two or three lines each; part 3 is three formulas and six numbers. Everything comes from the two laws \(T'^a{}_b = A_{ai}S_{jb}T^i{}_j\) and \(T'_{ab} = S_{ia}S_{jb}T_{ij}\).

Rung 2

Part 1: one of the two laws produces \(AS\) and the other \(S^{\mathsf T}S\). Part 2: take the transpose of \(S^{\mathsf T}TS\). Part 3: a symmetric array is fixed by its entries on and above the diagonal; a totally antisymmetric \((0,3)\) array is fixed by its entries with \(i\lt j\lt k\).

Rung 3

Part 1, first line: \(\delta'^a_b = A_{ai}S_{jb}\delta^i_j = A_{ai}S_{ib} = (AS)_{ab}\). Part 3, first line: the count for a symmetric \((0,2)\) is the number of unordered pairs with repetition, \(\binom{n+1}{2}\).

Worked solution

Idea. Each part is the same question — which features of an array are properties of the tensor — answered by pushing the law through and seeing what cancels.

1. \(\delta'^a_b = A_{ai}S_{jb}\delta^i_j = A_{ai}S_{ib} = (AS)_{ab} = \delta_{ab}\): the identity array in every basis. For the \((0,2)\) version, \(\delta'_{ab} = S_{ia}S_{ib} = (S^{\mathsf T}S)_{ab}\), and with \(S = \mathrm{diag}(2,1,1)\) this is \(\mathrm{diag}(4,1,1) \ne \mathbb 1\). The first is the identity map, the second is the inner product; they are different objects that coincide numerically in orthonormal bases.

2. \(T' = S^{\mathsf T}TS\), so \((T')^{\mathsf T} = S^{\mathsf T}T^{\mathsf T}S\): the transpose transforms the same way. Hence \((T')_{(ab)} = (T_{(ij)})'\) and \((T')_{[ab]} = (T_{[ij]})'\), and in particular a symmetric tensor is symmetric in every basis. For a \((1,1)\)-tensor the law is \(T' = ATS = S^{-1}TS\), a similarity, and transposition does not commute with it. Explicitly, with \(T = \begin{pmatrix}0&1\\1&0\end{pmatrix}\) and \(S = \mathrm{diag}(1,2)\), \(T' = \begin{pmatrix}0&2\\ \tfrac12&0\end{pmatrix}\), which is not symmetric.

3. Symmetric \((0,2)\): entries with \(i\le j\), i.e. \(\binom{n+1}{2} = \tfrac{n(n+1)}2\). Antisymmetric \((0,2)\): entries with \(i\lt j\), i.e. \(\binom n2 = \tfrac{n(n-1)}2\). Totally antisymmetric \((0,3)\): entries with \(i\lt j\lt k\), i.e. \(\binom n3 = \tfrac{n(n-1)(n-2)}6\).

\(n=3\)\(n=4\)
symmetric \((0,2)\)610
antisymmetric \((0,2)\)36
antisymmetric \((0,3)\)14

4. Vector components: \(\vec v = v^i\vec e_i = v'^a\vec f_a = v'^aS_{ia}\vec e_i\), so \(v^i = S_{ia}v'^a\), i.e. \(v' = S^{-1}v = Av\). Covector components: \(\alpha'_a = \alpha(\vec f_a) = S_{ia}\alpha_i\), i.e. \(\alpha' = S^{\mathsf T}\alpha\). These matrices agree exactly when \(S^{\mathsf T} = S^{-1}\), i.e. \(S\) orthogonal — and then both are \(R = S^{\mathsf T}\). \(\;\blacksquare\)

Check. Part 3 against the slides: a symmetric \((0,2)\)-tensor in \(n=3\) has 6 components, which is the count for an inertia tensor ✓; an antisymmetric one has 3, which is why \(\vec\omega\) can masquerade as a vector in three dimensions and cannot in four, where the count is 6 ✓. Part 3 at \(n=4\), antisymmetric \((0,2)\): 6 — the six components of the electromagnetic field tensor of unit 28, three electric and three magnetic ✓. Part 1, limit: for \(S\to\mathbb 1\) both arrays are \(\mathbb 1\) and nothing distinguishes them, which is the trap ✓.

Principal axes

P7 An L-shaped lamina medium

exercises: assembling a body from pieces with the parallel-axis theorem, diagonalising, the perpendicular-axis check

Three unit squares, each of mass 1 kg and side 1 m, are glued edge to edge into an L: their centres are at \((\tfrac12,\tfrac12)\), \((\tfrac12,\tfrac32)\) and \((\tfrac32,\tfrac12)\) in the plane \(z=0\).

  1. Find the centre of mass.
  2. Compute \(\mathsf I\) about the centre of mass, in the frame with axes along the square edges. (A single square of mass \(\mu\) and side \(\ell\) has \(I_{11}=I_{22}=\mu\ell^2/12\), \(I_{12}=0\) about its own centre.)
  3. Find the principal moments and the principal axes, and say which axis the L is easiest to spin about.
  4. Check your three principal moments against the perpendicular-axis theorem of P2.
Hints · 3 rungs
Rung 1

Part 2 is a \(3\times3\) array of numbers; part 3 is three numbers and two directions. The body is flat, so the array is block diagonal and only a \(2\times2\) eigenvalue problem appears.

Rung 2

Compute each square's contribution about the common centre of mass using the parallel-axis theorem, then add. For the off-diagonal entry remember \(I_{12} = -\sum\mu\,\Delta x\,\Delta y\) with \(\Delta\) measured from the common centre of mass, and the individual squares contribute nothing of their own to it.

Rung 3

First line: the centre of mass is at \((\tfrac56,\tfrac56)\), so the three displacements are \((-\tfrac13,-\tfrac13)\), \((-\tfrac13,\tfrac23)\) and \((\tfrac23,-\tfrac13)\). Then \(I_{12} = -\big[(-\tfrac13)(-\tfrac13)+(-\tfrac13)(\tfrac23)+(\tfrac23)(-\tfrac13)\big] = \tfrac13\).

Worked solution

Idea. Build the tensor as a sum of Steiner-shifted pieces. The symmetry of the L across the line \(y = x\) then guarantees the principal axes without any algebra, and the eigenvalues come out of a \(2\times2\) whose off-diagonal entry is all that matters.

1. \(\vec X = \tfrac13\big[(\tfrac12,\tfrac12)+(\tfrac12,\tfrac32)+(\tfrac32,\tfrac12)\big] = \big(\tfrac56,\tfrac56\big)\) m.

2. Displacements from \(\vec X\): \(\vec d_1 = (-\tfrac13,-\tfrac13)\), \(\vec d_2 = (-\tfrac13,\tfrac23)\), \(\vec d_3 = (\tfrac23,-\tfrac13)\). Each square contributes \(\tfrac1{12}\) to \(I_{11}\) about its own centre, plus \(\mu(\Delta y)^2\):

\[ I_{11} = 3\cdot\tfrac1{12} + \Big[\tfrac19+\tfrac49+\tfrac19\Big] = \tfrac14+\tfrac23 = \tfrac{11}{12},\qquad I_{22} = \tfrac{11}{12} \]

by the same sum with \(\Delta x\). And

\[ I_{12} = -\sum_\alpha\mu\,\Delta x_\alpha\,\Delta y_\alpha = -\Big[\tfrac19-\tfrac29-\tfrac29\Big] = \tfrac13 . \] \[ \mathsf I \;\hat=\; \begin{pmatrix}\tfrac{11}{12}&\tfrac13&0\\[2pt] \tfrac13&\tfrac{11}{12}&0\\[2pt] 0&0&\tfrac{11}6\end{pmatrix}\ \mathrm{kg\,m^2} \]

with \(I_{33} = I_{11}+I_{22}\) by P2, since the body is a lamina.

3. For \(\begin{pmatrix}A&B\\B&A\end{pmatrix}\) the eigenvectors are \((1,1)/\sqrt2\) and \((1,-1)/\sqrt2\) with eigenvalues \(A\pm B\):

\[ I_1 = \tfrac{11}{12}-\tfrac13 = \tfrac7{12} = 0.58\overline3,\qquad I_2 = \tfrac{11}{12}+\tfrac13 = \tfrac54 = 1.25,\qquad I_3 = \tfrac{11}6 = 1.8\overline3\ \mathrm{kg\,m^2}. \]

The smallest belongs to \((1,-1)/\sqrt2\), the anti-diagonal running between the tips of the two arms. Mass lying along an axis contributes nothing to the moment about it, and both arms lie close to that line — so this is the axis the L is easiest to spin about. The equal diagonal entries are forced by the mirror symmetry of the L in the line \(y=x\), which swaps the two coordinates.

4. \(I_1+I_2 = \tfrac7{12}+\tfrac{15}{12} = \tfrac{22}{12} = \tfrac{11}6 = I_3\) ✓ exactly, as the perpendicular-axis theorem demands for a lamina.

Check. Dimensions \(\mathrm{kg\,m^2}\) ✓. Trace: \(\tfrac{11}{12}+\tfrac{11}{12}+\tfrac{11}6 = \tfrac{11}3 = 3.6\overline6\), and \(I_1+I_2+I_3 = \tfrac7{12}+\tfrac{15}{12}+\tfrac{22}{12} = \tfrac{44}{12} = \tfrac{11}3\) ✓. Limit: gluing a fourth square at \((\tfrac32,\tfrac32)\) makes a \(2\times2\) square of mass 4, for which the symmetry of P8 forces \(I_1 = I_2 = \tfrac{4\cdot4}{12} = \tfrac43\) and \(I_{12}=0\) — and indeed the fourth displacement \((\tfrac12,\tfrac12)\) about the new centre \((1,1)\) cancels the off-diagonal sum ✓.

P8 Proof: the cube, and a body that only looks symmetric hard

exercises: symmetry forcing degeneracy, a converse that fails, constructing a counterexample

  1. Let a body be invariant under the rotation \(C\) by \(2\pi/n\) about \(\hat k\) through \(O\). Prove \(\mathsf I_O(C\vec u,C\vec w) = \mathsf I_O(\vec u,\vec w)\), and deduce that the array \(M\) of \(\mathsf I_O\) commutes with \(C\).
  2. Using only the two four-fold axes through opposite face centres, prove that a uniform cube of side \(a\) about its centre has \(\mathsf I = \tfrac16Ma^2\,\delta\), so that every axis through the centre is principal and a cube can never exhibit the intermediate-axis instability.
  3. Show that the converse fails: construct a body with \(\mathsf I = \lambda\delta\) about its centre of mass which has no rotation axis of order three or more, so that nothing in part 1 forces the degeneracy. Six point masses on the coordinate axes will do.
  4. What, then, does \(\mathsf I\) remember about a mass distribution, and what does it forget? Answer in terms of the moments \(\int\rho\,x_{i_1}x_{i_2}\cdots x_{i_k}\,\mathrm dV\) of the mass distribution, \(k = 0,1,2,\dots\).
Hints · 3 rungs
Rung 1

Parts 1 and 2 are proofs; part 3 wants an explicit list of masses and positions; part 4 is two sentences. For part 3, ask what has to be true of \(\sum m_\alpha x_{\alpha i}x_{\alpha j}\) and note that six free masses give you more parameters than constraints.

Rung 2

Part 1: change variables \(\vec r = C\vec s\) in the defining integral and use that \(C\) is orthogonal with unit Jacobian. Part 2: a matrix commuting with a rotation by \(\pi/2\) about \(\vec e_3\) is forced into a particular shape; write it out. Part 3: with masses on the axes, \(\mathsf I\) is automatically diagonal, so you need only make the three diagonal entries equal while keeping the centre of mass at the origin.

Rung 3

Part 2, first line: let \(C_3\) be the rotation by \(\pi/2\) about \(\vec e_3\). \(MC_3 = C_3M\) forces \(M_{11}=M_{22}\), \(M_{12}=-M_{21}=0\) (by symmetry of \(M\)) and \(M_{13}=M_{23}=0\). Now repeat with the rotation about \(\vec e_1\). Part 3, first line: put masses \(m_i\) at \(\pm\ell_i\vec e_i\), which fixes the centre of mass at the origin automatically.

Worked solution

Idea. Symmetry constrains \(\mathsf I\) through a commutation relation, and commuting with enough rotations leaves only a multiple of the identity. The converse fails because \(\mathsf I\) sees only second moments, and second moments can be made isotropic by hand.

1. Substituting \(\vec r = C\vec s\) in

\[ \mathsf I_O(C\vec u,C\vec w) = \int_B\rho(\vec r)\Big[(C\vec u\cdot C\vec w)\lVert\vec r\rVert^2-(C\vec u\cdot\vec r)(C\vec w\cdot\vec r)\Big]\mathrm dV \]

uses \(|\det C| = 1\) (so \(\mathrm dV\) is unchanged), \(C(B) = B\) and \(\rho(C\vec s)=\rho(\vec s)\) (the symmetry hypothesis), and \(C\vec u\cdot C\vec s = \vec u\cdot\vec s\), \(\lVert C\vec s\rVert = \lVert\vec s\rVert\) (orthogonality). The result is \(\mathsf I_O(\vec u,\vec w)\). In components, \(\mathsf I(C\vec u,C\vec w) = \vec u^{\mathsf T}C^{\mathsf T}MC\vec w\), so \(C^{\mathsf T}MC = M\), i.e. \(MC = CM\). \(\;\blacksquare\)

2. Let \(C_3\) be the rotation by \(\pi/2\) about \(\vec e_3\): \(C_3\vec e_1 = \vec e_2\), \(C_3\vec e_2 = -\vec e_1\), \(C_3\vec e_3 = \vec e_3\). Writing out \(MC_3 = C_3M\) entry by entry gives \(M_{11} = M_{22}\), \(M_{12} = -M_{21}\) — which with \(M_{12}=M_{21}\) forces \(M_{12}=0\) — and \(M_{13} = M_{23} = -M_{13}\), so \(M_{13}=M_{23}=0\). Thus \(M = \mathrm{diag}(\mu,\mu,\nu)\). Repeating with the four-fold rotation about \(\vec e_1\) forces \(M_{22} = M_{33}\), i.e. \(\mu = \nu\). So \(M = \mu\,\delta\), and \(\mu = I_{11} = \tfrac{M}{12}(a^2+a^2) = \tfrac16Ma^2\) from the block formula. Every unit vector is then an eigenvector; \(\vec L = \mu\vec\omega\) always; and Euler's equations lose their quadratic terms entirely, so there is no unstable axis. \(\;\blacksquare\)

3. Put masses \(m_1\) at \(\pm\ell_1\vec e_1\), \(m_2\) at \(\pm\ell_2\vec e_2\), \(m_3\) at \(\pm\ell_3\vec e_3\) — six masses, centre of mass at the origin by construction, and \(\mathsf I\) diagonal because every \(\sum m x_ix_j\) with \(i\ne j\) vanishes. Writing \(q_i := 2m_i\ell_i^2\), the diagonal entries are \(I_{11} = q_2+q_3\), \(I_{22} = q_3+q_1\), \(I_{33} = q_1+q_2\). Equality of all three requires \(q_1=q_2=q_3 =: q\), giving \(\mathsf I = 2q\,\delta\). Now choose \(m_1 = 1\), \(\ell_1 = 1\); \(m_2 = 4\), \(\ell_2 = \tfrac12\); \(m_3 = 9\), \(\ell_3 = \tfrac13\) (all in kg and m): then \(q_i = 2\) for each, and \(\mathsf I = 4\,\delta\ \mathrm{kg\,m^2}\). The three pairs have different masses at different distances, so no rotation carries the body to itself except the identity and the three two-fold axis rotations — certainly no three-fold or four-fold axis. \(\;\blacksquare\)

4. \(\mathsf I\) is built entirely from the second moments \(Q_{ij} := \int\rho\,x_ix_j\,\mathrm dV\): by the definition, \(I_{ij} = (\operatorname{tr}Q)\,\delta_{ij}-Q_{ij}\), and conversely \(Q_{ij} = \tfrac12(\operatorname{tr}\mathsf I)\,\delta_{ij}-I_{ij}\), so \(\mathsf I\) and the second-moment array carry exactly the same six numbers. The zeroth moment (the mass) and the first moments (zero about the centre of mass) enter the dynamics separately; every moment with \(k\ge3\) — the difference between a cube and a sphere with the same mass and the same second moments — is invisible to \(\mathsf I\). That is why a cube rotates dynamically like a sphere while being obviously not one. (In the language of multipoles, which unit 10 develops for gravity, the trace-free part of \(Q\) is the quadrupole moment.)

Check. Part 2 against example 1: \(\tfrac{M}{12}(a^2+a^2) = \tfrac16Ma^2\), and the trace gives \(3\times\tfrac16Ma^2 = \tfrac12Ma^2 = 2\int\rho\lVert\vec r\rVert^2\mathrm dV = 2M\cdot\tfrac{a^2}{4}\) ✓. Part 3: the trace of \(4\delta\) is \(12\), and \(2\sum m_\alpha\lVert\vec r_\alpha\rVert^2 = 2[2(1)(1)+2(4)(\tfrac14)+2(9)(\tfrac19)] = 2[2+2+2] = 12\) ✓. Triangle inequality: \(4+4\ge4\), strict, so the body is not a lamina ✓ — as it is not.

Euler's equations, tops and instability

P9 The Earth wobbles, but not at 305 days routine

exercises: solving Euler's equations for a symmetric top, a real number, identifying the failing hypothesis

The Earth is very nearly a symmetric top with \(I_1 = I_2\lt I_3\), spinning about an axis close to \(\vec u_3\), and to a first approximation torque-free.

  1. From Euler's equations with \(I_1 = I_2\) and \(\vec\tau = 0\), show that \(\omega_3\) is constant and that \((\omega_1,\omega_2)\) rotates in the body frame at the rate \(\Omega_b = (I_3-I_1)\omega_3/I_1\).
  2. Given the Earth's dynamical ellipticity \((I_3-I_1)/I_3 = 3.2737949\times10^{-3}\), compute \(\Omega_b\) and the period \(2\pi/\Omega_b\) in days.
  3. The measured period of the Earth's polar wobble is 433 days. State which hypothesis of part 1 is false, and in which direction it must push the period.
  4. What would the period be if the Earth were an exact sphere? Reconcile that with your answer to 3.
Hints · 3 rungs
Rung 1

Part 1 is a two-line derivation; part 2 is one number in days; parts 3 and 4 are one sentence each. Note that \(\omega_3\) is one revolution per day, so the period you want is \(1\) day divided by a small number.

Rung 2

With \(I_1=I_2\), Euler's third equation has \((I_1-I_2)\omega_1\omega_2 = 0\) on the right. The remaining two are linear with constant coefficients — unit 06's territory — and the substitution \(\zeta = \omega_1+i\omega_2\) collapses them into one. For part 2, convert the given \((I_3-I_1)/I_3\) into \((I_3-I_1)/I_1\) before dividing.

Rung 3

First line: \(I_1\dot\omega_1 = (I_1-I_3)\omega_2\omega_3\) and \(I_1\dot\omega_2 = (I_3-I_1)\omega_3\omega_1\), so with \(\zeta = \omega_1+i\omega_2\), \(\dot\zeta = i\Omega_b\zeta\). For part 2, if \(H := (I_3-I_1)/I_3\) then \((I_3-I_1)/I_1 = H/(1-H)\).

Worked solution

Idea. A symmetric top makes Euler's equations linear, and the answer is a single rotation rate in the body frame. Comparing it with the measured value is a measurement of the one hypothesis the derivation made about the Earth.

1. With \(I_1 = I_2\) and \(\vec\tau=0\), the third equation is \(I_3\dot\omega_3 = (I_1-I_2)\omega_1\omega_2 = 0\), so \(\omega_3\) is constant. The first two are

\[ I_1\dot\omega_1 = (I_1-I_3)\omega_3\,\omega_2,\qquad I_1\dot\omega_2 = (I_3-I_1)\omega_3\,\omega_1 . \]

Put \(\Omega_b := (I_3-I_1)\omega_3/I_1\) and \(\zeta := \omega_1+i\omega_2\); then \(\dot\zeta = -\Omega_b\omega_2+i\Omega_b\omega_1 = i\Omega_b\zeta\), so \(\zeta(t) = \zeta_0e^{i\Omega_bt}\). The transverse part of \(\vec\omega\) has constant magnitude and rotates at \(\Omega_b\) in the body frame. \(\;\blacksquare\)

2. With \(H := (I_3-I_1)/I_3 = 3.2737949\times10^{-3}\),

\[ \frac{I_3-I_1}{I_1} = \frac{H}{1-H} = 3.2845478\times10^{-3},\qquad \frac{2\pi}{\Omega_b} = \frac{I_1}{I_3-I_1}\times\frac{2\pi}{\omega_3} = \frac{1\ \text{day}}{3.2845478\times10^{-3}} . \]

That is \(\mathbf{304.456}\) sidereal days, or \(2.6233\times10^{7}\) s, which is \(303.62\) days of \(86400\) s — about ten months. This is Euler's 1765 prediction, and the number is fixed entirely by the Earth's oblateness.

3. The false hypothesis is rigidity. The Earth yields elastically as the wobble stresses it, so the mass distribution is not carried unchanged by the rotation, and \(\mathsf I\) is not constant in the body frame — which is the one thing the derivation of Euler's equations needed. A body that yields responds as though it were rounder, i.e. with a smaller effective \((I_3-I_1)\), so the period must be longer than 304 days. It is: \(433\) days, a factor \(433/303.62 = 1.43\) — the Chandler wobble, found by S. C. Chandler in 1891.

4. For an exact sphere \(I_3 = I_1\), so \(\Omega_b = 0\) and the period is infinite: \(\vec\omega\) never moves in the body frame, because every axis is principal. There is no contradiction with part 3 — elasticity pushes the Earth's response towards the spherical case, i.e. towards a longer period, and the observed 433 days sits between 304 days and infinity.

Check. Dimensions: \(\Omega_b\) is (dimensionless)\(\times\omega_3\), a frequency ✓. Limit \(I_3\to I_1\): \(\Omega_b\to0\) ✓, part 4. Sign: \(I_3\gt I_1\) for an oblate body, so \(\Omega_b\gt0\) and the wobble runs the same way as the spin ✓. Magnitude: a wobble amplitude of \(0.1''\) of arc corresponds to a few metres of polar displacement, which is what geodesy measures ✓.

P10 Flipping a phone medium

exercises: principal moments of a block, the intermediate-axis growth rate, turning a growth rate into a time

A smartphone is a uniform rectangular block of mass \(m = 0.171\) kg with edges \(a = 0.1467\) m (long), \(b = 0.0715\) m (wide), \(c = 0.0078\) m (thick).

  1. Compute the three principal moments about its centre, and say which physical axis is the intermediate one.
  2. Compute the dimensionless ratio \(\lambda/\Omega\) for rotation about that axis, and then \(\lambda\) for a toss at \(1.5\) turns per second.
  3. If the toss imparts a \(5\%\) transverse wobble, how long until the phone has flipped? Compare with the flight time of a toss to a height of \(0.5\) m, and say whether the flip is avoidable.
  4. Compare \(\lambda/\Omega\) here with the slides' \(0.58255\) for the hardback block, and explain in one sentence why the phone is worse.
Hints · 3 rungs
Rung 1

Three numbers in \(\mathrm{kg\,m^2}\), then a dimensionless ratio, a rate in \(\mathrm{s^{-1}}\), and two times in seconds. No new theory: the slides' formulas, used carefully.

Rung 2

\(I_k = \tfrac m{12}(\text{sum of the squares of the other two edges})\) — so the longest edge carries the smallest moment. The growth rate is \(\lambda = \Omega\sqrt{(I_3-I_2)(I_2-I_1)/(I_1I_3)}\), and a perturbation \(\varepsilon\) reaches order one after \(\ln(1/\varepsilon)/\lambda\). For part 3, a toss to height \(h\) is in the air for \(2\sqrt{2h/g}\).

Rung 3

First line: \(I_1 = \tfrac{0.171}{12}(0.0715^2+0.0078^2) = \tfrac{0.171}{12}(5.1729\times10^{-3}) = 7.3717\times10^{-5}\ \mathrm{kg\,m^2}\).

Worked solution

Idea. Everything is the slides' formula; the work is arithmetic done carefully and then converted into a time that can be compared with an experiment you can perform.

1. \(a^2 = 2.15209\times10^{-2}\), \(b^2 = 5.11225\times10^{-3}\), \(c^2 = 6.084\times10^{-5}\), and \(m/12 = 1.425\times10^{-2}\):

axis alongformula\(\mathrm{kg\,m^2}\)
\(a\) (the long edge)\(\tfrac m{12}(b^2+c^2)\)\(7.3717\times10^{-5}\)
\(b\) (the width)\(\tfrac m{12}(a^2+c^2)\)\(3.0754\times10^{-4}\)
\(c\) (the thickness)\(\tfrac m{12}(a^2+b^2)\)\(3.7952\times10^{-4}\)

So \(I_1 = 7.3717\times10^{-5}\), \(I_2 = 3.0754\times10^{-4}\), \(I_3 = 3.7952\times10^{-4}\), and the intermediate axis is the one along the width — the horizontal axis lying in the screen, perpendicular to the long edge.

2.

\[ \frac{(I_3-I_2)(I_2-I_1)}{I_1I_3} = \frac{(7.1983\times10^{-5})(2.3382\times10^{-4})}{(7.3717\times10^{-5})(3.7952\times10^{-4})} = 0.60161,\qquad \frac{\lambda}{\Omega} = 0.77563 . \]

At \(1.5\) turns per second, \(\Omega = 3\pi = 9.4248\ \mathrm{s^{-1}}\), so \(\lambda = \mathbf{7.3102}\ \mathrm{s^{-1}}\) and the e-folding time is \(0.1368\) s.

3. From \(\varepsilon = 0.05\) to order one takes \(\ln(20)/\lambda = 2.9957/7.3102 = \mathbf{0.410}\) s. A toss to \(h = 0.5\) m is airborne for \(2\sqrt{2h/g} = 2\sqrt{0.10194} = 0.639\) s. Since \(0.410\lt0.639\), the flip happens with well over a third of the flight to spare: not avoidable. Reducing the initial wobble to \(1\%\) only buys \(\ln5/\lambda = 0.22\) s, still short of the flight time — the logarithm is a weak defence.

4. \(0.77563\) against \(0.58255\): the phone is worse because its three moments are more unequal in the right way. For a body thin enough to count as a lamina, \(I_3\simeq I_1+I_2\) and the quantity under the root collapses to \((I_2-I_1)/(I_1+I_2)\), which grows as \(I_1/I_2\) shrinks — and for a plate \(I_1/I_2 \simeq (b/a)^2\), the squared aspect ratio of the face. The phone's face is the narrower one: \(b/a = 0.0715/0.1467 = 0.487\) against the book's \(0.16/0.24 = 0.667\), so \(I_1/I_2 = 0.2397\) against \(0.4595\). Thinness helps too, but only by pushing the body closer to that lamina limit: giving the book the phone's \(c/a\) while keeping its face would raise \(0.58255\) only to \(0.61624\), whereas giving it the phone's \(b/a\) raises it to \(0.69792\).

Check. Dimensions: the ratio is dimensionless, \(\lambda\) is \(\mathrm{s^{-1}}\), the times are seconds ✓. Bound: \(\lambda/\Omega\lt1\) ✓, as the triangle inequality guarantees. Triangle inequality itself: \(I_1+I_2-I_3 = 1.7339\times10^{-6}\), and \(\tfrac m6c^2 = 1.7339\times10^{-6}\) ✓ — small, because the phone is nearly a lamina. Experiment: toss a phone over a bed, spinning about the width axis, and watch it turn over; spin it about the long edge or about the thickness axis instead and it does not.

P11 Proof: the extreme axes are Lyapunov stable hard

exercises: building a Lyapunov function from conserved quantities, the gap between linear and nonlinear stability, what "stable" is a statement about

A free rigid body with \(I_1\lt I_2\lt I_3\) conserves \(2T = \sum_kI_k\omega_k^2\) and \(\lVert\vec L\rVert^2 = \sum_kI_k^2\omega_k^2\).

  1. Show that the linearisation about steady rotation of rate \(\Omega\) about \(\vec u_1\) has eigenvalues \(\{0,\pm i\lambda_1\}\) with \(\lambda_1\) real, so that unit 02's classification calls it a centre and Hartman–Grobman is silent.
  2. Prove that \(F := \lVert\vec L\rVert^2-2I_1T\) is conserved, equals \(I_2(I_2-I_1)\omega_2^2+I_3(I_3-I_1)\omega_3^2\), and is positive definite in \((\omega_2,\omega_3)\). Deduce Lyapunov stability of the \(\vec u_1\) axis, giving the \(\delta(\varepsilon)\) explicitly.
  3. Do the same for \(\vec u_3\), and show that at \(\vec u_2\) both candidate functions are indefinite — as they must be.
  4. State precisely what has been proved: is it that \(\vec\omega(t)\) stays near \(\Omega\vec u_1\), or something weaker? Give the sharpest true statement.
Hints · 3 rungs
Rung 1

Parts 2, 3 and 4 are proofs and a careful statement; part 1 is a \(3\times3\) Jacobian. A Lyapunov function is a conserved (or decreasing) quantity that is positive definite about the equilibrium; the work is to find a combination of the two conserved quantities that has a minimum there.

Rung 2

Any combination \(\alpha\lVert\vec L\rVert^2+\beta\,2T\) is conserved. Choose \(\alpha,\beta\) so that the \(\omega_1^2\) term drops out and the other two have the same sign. For part 4, notice that nothing in \(F\) constrains \(\omega_1\) at all.

Rung 3

First line: \(\lVert\vec L\rVert^2-2I_1T = \sum_kI_k^2\omega_k^2-I_1\sum_kI_k\omega_k^2 = \sum_kI_k(I_k-I_1)\omega_k^2\), and the \(k=1\) term vanishes identically.

Worked solution

Idea. The two conserved quantities are two ellipsoids; the right combination is one whose level sets near the axis are small ellipses in the transverse variables alone. Then conservation traps the trajectory with no approximation at all.

1. Write \(\vec\omega = (\Omega+v_1,v_2,v_3)\). The linearised Euler system is \(\dot v_1 = 0\), \(\dot v_2 = \tfrac{(I_3-I_1)\Omega}{I_2}v_3\), \(\dot v_3 = \tfrac{(I_1-I_2)\Omega}{I_3}v_2\). Its matrix is block diagonal, with a \(1\times1\) block \(0\) — the line of equilibria, one for each \(\Omega\) — and a \(2\times2\) block whose determinant is \(-\tfrac{(I_3-I_1)(I_1-I_2)}{I_2I_3}\Omega^2 = +\lambda_1^2\) with \(\lambda_1^2 := \tfrac{(I_3-I_1)(I_2-I_1)}{I_2I_3}\Omega^2\gt0\), and trace 0. Eigenvalues of a traceless \(2\times2\) with positive determinant are \(\pm i\lambda_1\). A centre: unit 02 warned that linearisation decides nothing here.

2. \(F\) is a fixed linear combination of two conserved quantities, hence conserved. Expanding,

\[ F = \sum_kI_k^2\omega_k^2-I_1\sum_kI_k\omega_k^2 = \sum_kI_k(I_k-I_1)\omega_k^2 = I_2(I_2-I_1)\omega_2^2+I_3(I_3-I_1)\omega_3^2, \]

the \(k=1\) term vanishing identically. Both coefficients are strictly positive because \(I_1\lt I_2\lt I_3\), so \(F\ge c_1(\omega_2^2+\omega_3^2)\) with \(c_1 := \min\{I_2(I_2-I_1),\,I_3(I_3-I_1)\}\gt0\), and also \(F\le c_2(\omega_2^2+\omega_3^2)\) with \(c_2\) the corresponding maximum.

Given \(\varepsilon\gt0\), choose \(\delta := \varepsilon\sqrt{c_1/c_2}\). If \(\omega_2(0)^2+\omega_3(0)^2\lt\delta^2\) then \(F\lt c_2\delta^2 = c_1\varepsilon^2\) initially, hence for all time, hence \(\omega_2(t)^2+\omega_3(t)^2\lt\varepsilon^2\) for all \(t\). \(\;\blacksquare\)

3. Take \(G := 2I_3T-\lVert\vec L\rVert^2 = \sum_kI_k(I_3-I_k)\omega_k^2 = I_1(I_3-I_1)\omega_1^2+I_2(I_3-I_2)\omega_2^2\), again conserved, again positive definite in the two transverse variables, and the same argument applies at \(\vec u_3\).

At \(\vec u_2\), the analogous \(\lVert\vec L\rVert^2-2I_2T = \sum_kI_k(I_k-I_2)\omega_k^2 = I_1(I_1-I_2)\omega_1^2+I_3(I_3-I_2)\omega_3^2\) has coefficients of opposite signs: its level sets near the axis are hyperbolas, not ellipses, and it traps nothing. Likewise \(2I_2T-\lVert\vec L\rVert^2\) is its negative. This had to happen — the slides' intermediate-axis theorem makes that axis genuinely unstable, and a Lyapunov function would prove otherwise.

4. What is proved is that the transverse components stay small: for every \(\varepsilon\) there is \(\delta\) with \(\omega_2^2+\omega_3^2\lt\varepsilon^2\) for all time. It is not proved that \(\vec\omega(t)\) stays near the particular vector \(\Omega\vec u_1\): nothing above pins \(\omega_1\) to \(\Omega\). The sharpest true statement is that the axis \(\mathbb R\vec u_1\) is stable, with \(\omega_1\) controlled only indirectly, through \(I_1\omega_1^2 = 2T-I_2\omega_2^2-I_3\omega_3^2\), which fixes it to within \(O(\varepsilon^2)\) of its initial value. That is enough: it says the spin axis does not wander away from the body's long axis, which is the physical question.

Check. Dimensions: \(F\) is \(\mathrm{kg^2m^4s^{-2}}\), the same as \(\lVert\vec L\rVert^2\) ✓. Limits: as \(I_2\to I_1\), \(c_1\to0\) and the bound degenerates — correct, since a symmetric top's transverse components are then unconstrained by \(F\), and indeed \(\vec\omega\) precesses freely on a cone (P9) without the axis being unstable ✓. Consistency: the strength of the argument is that it uses no approximation, so it holds for large perturbations too — which the picture of nested closed polhodes on the slides confirms ✓.

P12 The gyroscope, exactly hard

exercises: a Lagrangian with two cyclic coordinates, an effective potential, steady precession and nutation, naming a silent approximation

A bicycle wheel is idealised as a uniform hoop of mass \(M = 1.2\) kg and radius \(R = 0.33\) m, mounted on a light axle pivoted at a point \(l = 0.10\) m from the wheel's centre, and spun at \(\omega_3 = 40\ \mathrm{s^{-1}}\). Take \(g = 9.81\ \mathrm{m\,s^{-2}}\), and write \(L_3 = I_3\omega_3\).

  1. Write the Lagrangian in Euler angles, identify the two cyclic coordinates and their conserved momenta, and reduce to a one-dimensional problem in \(\theta\) with an effective potential.
  2. Derive the condition for steady precession at a fixed tilt \(\theta_0\), and show it is a quadratic in \(\dot\phi\). State when it has real roots.
  3. Numbers: with \(\theta_0 = 60^\circ\), compute \(I_1\), \(I_3\), both roots, the precession period of the slow one, and the elementary estimate \(Mgl/L_3\). How large is the error? Compute the nutation frequency both exactly and as \(L_3/I_1\).
  4. The elementary argument writes \(\vec\tau = \mathrm d\vec L/\mathrm dt\) with \(\vec L = L_3\hat u_3\) and gets \(\dot\phi = Mgl/L_3\). Name the approximation it makes silently, show it is the one responsible for your error in part 3, and name a second phenomenon it misses entirely.
Hints · 3 rungs
Rung 1

Parts 1 and 2 are derivations; part 3 is six numbers; part 4 is a diagnosis in two or three sentences. The wheel is a symmetric top about a fixed pivot, so everything on the slides' heavy-top slide applies with \(I_1\) computed about the pivot.

Rung 2

For a hoop about its own centre, \(I_3 = MR^2\) and \(I_1 = I_2 = \tfrac12MR^2\); shift \(I_1\) to the pivot with the parallel-axis theorem. In Euler angles with \(I_1=I_2\), \(T = \tfrac12I_1(\dot\theta^2+\dot\phi^2\sin^2\theta)+\tfrac12I_3(\dot\psi+\dot\phi\cos\theta)^2\) and \(V = Mgl\cos\theta\). Steady precession comes from the \(\theta\) Euler–Lagrange equation with \(\ddot\theta = 0\).

Rung 3

First line: \(I_3 = MR^2 = 1.2\times0.1089 = 0.13068\ \mathrm{kg\,m^2}\) and \(I_1 = \tfrac12MR^2+Ml^2 = 0.06534+0.012 = 0.07734\ \mathrm{kg\,m^2}\), so \(L_3 = 5.2272\ \mathrm{kg\,m^2\,s^{-1}}\) and \(Mgl = 1.1772\) N m.

Worked solution

Idea. Two cyclic coordinates reduce three degrees of freedom to one, and steady precession is the statement that the remaining one sits at a minimum of its effective potential. The elementary formula is the leading term of one of the two roots.

1. With \(I_1=I_2\) about the pivot,

\[ \mathcal L = \tfrac12I_1\big(\dot\theta^2+\dot\phi^2\sin^2\theta\big)+\tfrac12I_3\big(\dot\psi+\dot\phi\cos\theta\big)^2-Mgl\cos\theta . \]

Neither \(\phi\) nor \(\psi\) appears, so

\[ p_\psi = I_3(\dot\psi+\dot\phi\cos\theta) = L_3,\qquad p_\phi = I_1\dot\phi\sin^2\theta+L_3\cos\theta = L_z \]

are conserved. Solving the second for \(\dot\phi\) and substituting into the energy gives \(E' = \tfrac12I_1\dot\theta^2+V_{\text{eff}}(\theta)\) with

\[ V_{\text{eff}}(\theta) = \frac{(L_z-L_3\cos\theta)^2}{2I_1\sin^2\theta}+Mgl\cos\theta . \]

2. The \(\theta\) equation is \(I_1\ddot\theta = I_1\dot\phi^2\sin\theta\cos\theta-L_3\dot\phi\sin\theta+Mgl\sin\theta\). Setting \(\ddot\theta=0\) and dividing by \(\sin\theta_0\ne0\),

\[ I_1\cos\theta_0\,\dot\phi^2-L_3\,\dot\phi+Mgl = 0,\qquad \dot\phi_\pm = \frac{L_3\pm\sqrt{L_3^2-4I_1Mgl\cos\theta_0}}{2I_1\cos\theta_0}, \]

real precisely when \(L_3^2\ge4I_1Mgl\cos\theta_0\): spin the top too slowly and no steady precession at that tilt exists at all.

3. \(I_3 = 0.13068\), \(I_1 = 0.07734\ \mathrm{kg\,m^2}\), \(L_3 = 5.2272\ \mathrm{kg\,m^2\,s^{-1}}\), \(Mgl = 1.1772\) N m, \(I_1\cos\theta_0 = 0.03867\). Discriminant \(5.2272^2-4(0.03867)(1.1772) = 27.32363-0.18210 = 27.14153\), root \(5.209753\). Hence

\[ \dot\phi_- = \frac{5.2272-5.209753}{0.07734} = \mathbf{0.225583}\ \mathrm{s^{-1}},\qquad \dot\phi_+ = \frac{5.2272+5.209753}{0.07734} = 134.95\ \mathrm{s^{-1}} . \]

Period of the slow precession \(2\pi/\dot\phi_- = \mathbf{27.853}\) s. The elementary estimate is \(Mgl/L_3 = 1.1772/5.2272 = 0.225207\ \mathrm{s^{-1}}\), period \(27.900\) s — low by \(\mathbf{0.167\%}\). Nutation: \(V_{\text{eff}}''(\theta_0) = 350.94\) J with \(L_z = 2.62668\ \mathrm{kg\,m^2\,s^{-1}}\), so \(\omega_{\text{nut}} = \sqrt{350.94/0.07734} = 67.362\ \mathrm{s^{-1}}\) (period \(0.09327\) s), against \(L_3/I_1 = 67.587\ \mathrm{s^{-1}}\) — high by \(0.33\%\).

4. The silent approximation is that \(\vec L\) points along the symmetry axis. In truth \(\vec L = I_1\dot\phi\sin\theta\,\hat e_\perp + L_3\hat u_3\): the precession itself carries transverse angular momentum, and dropping it requires \(I_1\dot\phi\ll L_3\). Here \(I_1\dot\phi_-/L_3 = 3.34\times10^{-3}\). Expanding the exact root, \(\dot\phi_- = \tfrac{Mgl}{L_3}\big(1+\tfrac{I_1Mgl\cos\theta_0}{L_3^2}+\dots\big)\), and the correction is \(I_1Mgl\cos\theta_0/L_3^2 = 1.67\times10^{-3}\) — exactly the \(0.167\%\) error, and exactly \(\cos\theta_0\) times the neglected ratio. The phenomenon missed entirely is nutation: the elementary argument produces one rate, not two roots, and no oscillation in \(\theta\) whatever. It also cannot tell you that steady precession ceases to exist below a critical spin.

Check. Dimensions: \(Mgl/L_3\) is \(\mathrm{N\,m}/(\mathrm{kg\,m^2\,s^{-1}}) = \mathrm{s^{-1}}\) ✓; \(V_{\text{eff}}''\) is \(\mathrm{J}\) and \(V_{\text{eff}}''/I_1\) is \(\mathrm{s^{-2}}\) ✓. Special case \(\theta_0 = 90^\circ\): the quadratic degenerates to \(-L_3\dot\phi+Mgl=0\), so \(\dot\phi = Mgl/L_3 = 0.225207\ \mathrm{s^{-1}}\) exactly, and the fast root runs off to infinity ✓. Existence: \(L_3^2 = 27.32\) against \(4I_1Mgl\cos\theta_0 = 0.182\), a factor 150, so this wheel is very fast and the errors are correspondingly tiny ✓. Limit \(\omega_3\to\infty\): \(\dot\phi_-\simeq Mgl/L_3\) dies like \(1/\omega_3\), and the relative error \(I_1Mgl\cos\theta_0/L_3^2\) like \(1/\omega_3^2\), so the elementary formula becomes exact faster than the precession disappears ✓.

Another way

Part 2 can be had from \(V_{\text{eff}}'(\theta_0) = 0\) instead of from the equation of motion. Differentiating and writing \(\dot\phi = (L_z-L_3\cos\theta)/(I_1\sin^2\theta)\) reproduces the same quadratic; the route through \(V_{\text{eff}}\) has the advantage that \(V_{\text{eff}}''\) — needed for the nutation frequency — is then already in hand.

P13 Euler angles and the rank that drops medium

exercises: composing rotations, a Jacobian determinant, a topological obstruction from compactness

With \(R(\phi,\theta,\psi) = R_z(\phi)R_x(\theta)R_z(\psi)\), the body components of \(\vec\omega\) are

\[ \omega_1 = \dot\phi\sin\theta\sin\psi+\dot\theta\cos\psi,\quad \omega_2 = \dot\phi\sin\theta\cos\psi-\dot\theta\sin\psi,\quad \omega_3 = \dot\phi\cos\theta+\dot\psi . \]
  1. Derive these by writing \(\vec\omega\) as the sum of three contributions — about the space \(z\)-axis, about the line of nodes, and about the body \(3\)-axis — expressed in body components.
  2. Compute \(\det\big(\partial(\omega_1,\omega_2,\omega_3)/\partial(\dot\phi,\dot\theta,\dot\psi)\big)\) and identify where it vanishes. At such a point, exhibit a nonzero \((\dot\phi,\dot\theta,\dot\psi)\) giving \(\vec\omega = 0\), and say what that means physically.
  3. Verify that \(\omega_1^2+\omega_2^2 = \dot\theta^2+\dot\phi^2\sin^2\theta\), and hence write the kinetic energy of a symmetric top in Euler angles.
  4. Prove that no three coordinates can cover \(SO(3)\) without such a degeneracy: there is no homeomorphism from \(SO(3)\) onto an open subset of \(\mathbb R^3\).
Hints · 3 rungs
Rung 1

Parts 1, 3 and 4 are short proofs; part 2 is one determinant and one explicit vector. Part 4 needs only metric-space topology: compactness, connectedness, and the fact that a homeomorphism preserves both.

Rung 2

Part 1: the body \(3\)-axis has body components \((0,0,1)\); the line of nodes is the \(x\)-axis after the \(R_x(\theta)\) step, so in body components it is \((\cos\psi,-\sin\psi,0)\); the space \(z\)-axis is what \(R^{\mathsf T}\) does to \((0,0,1)\). Part 4: \(SO(3)\) is closed and bounded in \(\mathbb R^9\), and \(\mathbb R^3\) is connected.

Rung 3

Part 2, first line: expand the determinant along the third column, which is \((0,0,1)^{\mathsf T}\). Only one minor survives, and it is \(\det\begin{pmatrix}\sin\theta\sin\psi&\cos\psi\\ \sin\theta\cos\psi&-\sin\psi\end{pmatrix}\); collapse it with \(\sin^2\psi+\cos^2\psi=1\).

Worked solution

Idea. Euler angles are three rotations about three different axes, so \(\vec\omega\) is a sum of three vectors — each easy once written in the body frame. The degeneracy appears when two of the three axes coincide, and no choice of angles can avoid it because the obstruction is topological, not algebraic.

1. In body components: the \(\dot\psi\) rotation is about \((0,0,1)\); the \(\dot\theta\) rotation is about the line of nodes, which is the body \(1\)-axis rotated back by \(\psi\), i.e. \((\cos\psi,-\sin\psi,0)\); the \(\dot\phi\) rotation is about the space \(z\)-axis, whose body components are \((\sin\theta\sin\psi,\ \sin\theta\cos\psi,\ \cos\theta)\). Adding \(\dot\phi\), \(\dot\theta\), \(\dot\psi\) times these three gives the three displayed lines.

2. The matrix is

\[ J = \begin{pmatrix}\sin\theta\sin\psi & \cos\psi & 0\\ \sin\theta\cos\psi & -\sin\psi & 0\\ \cos\theta & 0 & 1\end{pmatrix},\qquad \det J = -\sin\theta\,(\sin^2\psi+\cos^2\psi) = -\sin\theta . \]

It vanishes at \(\theta = 0\) and \(\theta = \pi\). At \(\theta = 0\) the first and third columns both read \((0,0,1)^{\mathsf T}\), so \((\dot\phi,\dot\theta,\dot\psi) = (1,0,-1)\) gives \(\vec\omega = 0\): changing \(\phi\) and \(\psi\) oppositely produces no rotation at all. Physically, the precession and spin axes have become the same line, one degree of freedom has become invisible, and the angular velocity cannot be solved for from the angle rates — gimbal lock.

3. \(\omega_1^2+\omega_2^2 = (\dot\phi\sin\theta\sin\psi+\dot\theta\cos\psi)^2+(\dot\phi\sin\theta\cos\psi-\dot\theta\sin\psi)^2\). The cross terms are \(+2\dot\phi\dot\theta\sin\theta\sin\psi\cos\psi\) and \(-2\dot\phi\dot\theta\sin\theta\cos\psi\sin\psi\), which cancel; the squares leave \(\dot\phi^2\sin^2\theta(\sin^2\psi+\cos^2\psi)+\dot\theta^2 = \dot\phi^2\sin^2\theta+\dot\theta^2\) ✓. Hence for \(I_1=I_2\),

\[ T = \tfrac12I_1(\omega_1^2+\omega_2^2)+\tfrac12I_3\omega_3^2 = \tfrac12I_1\big(\dot\theta^2+\dot\phi^2\sin^2\theta\big)+\tfrac12I_3\big(\dot\psi+\dot\phi\cos\theta\big)^2 , \]

which is P12's Lagrangian kinetic term.

4. \(SO(3)\subset\mathbb R^9\) is bounded — every entry of an orthogonal matrix has modulus at most 1 — and closed, being the preimage of \((\mathbb 1,1)\) under the continuous map \(R\mapsto(R^{\mathsf T}R,\det R)\). By Heine–Borel it is compact. Suppose \(h:SO(3)\to U\) were a homeomorphism onto an open \(U\subseteq\mathbb R^3\). Then \(U\) is compact, hence closed in \(\mathbb R^3\); being open and closed in the connected space \(\mathbb R^3\), it is \(\varnothing\) or \(\mathbb R^3\); it is not empty, and \(\mathbb R^3\) is not compact. Contradiction. \(\;\blacksquare\)

Check. Part 2 sanity: at a generic \(\theta\), \(|\det J| = |\sin\theta|\le1\), and it is largest at \(\theta = \pi/2\) where the three axes are as close to mutually perpendicular as Euler angles allow ✓. Part 3, special case \(\theta\) constant and \(\dot\psi=0\): \(T\) reduces to \(\tfrac12(I_1\sin^2\theta+I_3\cos^2\theta)\dot\phi^2\), which is \(\tfrac12\mathsf I(\hat n,\hat n)\dot\phi^2\) for rotation about the fixed space \(z\)-axis ✓. Part 4 consistency: the axis–angle description \(\theta\hat n\) of Rodrigues also uses three numbers, and it too is singular — at \(\theta = \pi\), where \(\hat n\) and \(-\hat n\) give the same rotation ✓.