Part B · the action · spine

Rigid bodies, and what a tensor actually is

Angular momentum is not parallel to angular velocity, and the object that relates them cannot be described by nine numbers — only by nine numbers per basis, related to one another by a law that is a theorem, not a definition.

needs 12, 05, 03~3 h slides + 6 h problemsproblems: ./mr problems 16

This unit does two jobs at once, and they are the same job.

The physics job: a rigid body has six degrees of freedom, three of them rotational, and the rotational three do not behave like the translational three. Momentum is $m\vec v$ with $m$ a number; angular momentum is not $(\text{number})\times\vec\omega$. The relation between $\vec L$ and $\vec\omega$ needs an object with nine components, and the consequences are strange enough that a rotating body can flip over without anything touching it.

The mathematics job: that object is the first tensor in the course, and this is where the word gets its definition. A tensor is a multilinear map. It is not "an array that transforms in a certain way" — the array, and the way it transforms, are both consequences of choosing a basis, and the transformation law is a theorem with a three-line proof. Getting this straight now is what makes units 28 to 35 possible: the electromagnetic field tensor, the stress tensor, the metric, the connection and the Riemann tensor are all built on exactly this definition.

How to read this deck

One body runs through all six sections: a uniform rectangular block, a hardback book, $0.24 \times 0.16 \times 0.04$ m and $0.60$ kg. Section 2 computes its inertia tensor at the centre and at a corner; section 3 verifies the transformation law on it component by component, and exhibits nine numbers per basis that are not a tensor; section 4 diagonalises it; section 5 predicts that spinning it about one particular axis will make it flip, with a growth rate and a time in seconds; section 6 puts it in Euler angles and then does the same for a top.

Proofs are on the slides where they are one line and in these notes under Full proof where they are not; results imported from elsewhere are marked cited. Terms with a dotted underline are in the glossary (G); a solid underline is a piece of the running example, shown on hover.

16 · overview

One body, six sections

the physics

A rigid body is a mass distribution whose internal distances are frozen. Its configuration space is $SO(3)\times\mathbb R^3$: six numbers. Three of them rotate, and rotation is where the surprises are.

$\vec L = \mathsf I\vec\omega$, with $\mathsf I$ the inertia tensor of §2, and $\vec L \nparallel \vec\omega$ except along three special directions. Everything else in the unit follows from that one fact.

the mathematics

A tensor of type $(p,q)$ is a multilinear map $(V^*)^p\times V^q\to\mathbb R$, where $V^*$, the dual space, is the set of linear maps $V\to\mathbb R$ (§3). Pick a basis and you get an array; pick another and you get a different array. The rule connecting them is a theorem.

Nine numbers are not a tensor. Nine numbers per basis, obeying that rule, are.

the route to the flip · a sketch

1. Rotational kinetic energy is quadratic in $\vec\omega$. Polarising it gives a symmetric bilinear form: $\mathsf I$ (§2).

2. A symmetric form has three perpendicular principal axes, with moments $I_1\lt I_2\lt I_3$ for the book (§4).

3. In axes carried by the body $\mathsf I$ is constant, and $\dot{\vec L}=0$ becomes three ODEs, Euler's equations (§5).

4. Spin at rate $\Omega$ about axis 2 and linearise: a small wobble grows like $e^{\lambda t}$, with $\lambda = 0.58\,\Omega$ for the book (§5).

Dotted underline: a glossary term (G index, N notes and proofs). Solid: the running example on hover — one block, a $0.60$ kg hardback book.

The order is: what a rigid body is (§1), what the inertia tensor is and how to compute it (§2), what a tensor is in general (§3), the spectral theorem and principal axes (§4), Euler's equations and the instability (§5), Euler angles and tops (§6).

The route, slide by slide. Step 1 is From a kinetic energy to a bilinear form; the book's moments come in The block, about its centre. Step 2 is The spectral theorem, proved and Principal axes, and the ellipsoid. Step 3 is Euler's equations, which uses unit 03's transport theorem to move $\dot{\vec L}=\vec\tau$ into the body's own axes. Step 4 is Linearising about a principal axis, and the number is in The intermediate-axis theorem, with a number: at two turns a second, $\lambda = 7.32\ \mathrm{s^{-1}}$, and a $1\%$ wobble becomes a flip in $0.63$ s. Section 3 is not on the route because it does not change any number; it says what kind of object $\mathsf I$ is, which is the part later units need.

§3 is the one that later units cash in. If you are short of time, §3 and §4 are the spine; §6 is the most classical and the least load-bearing, though the gyroscope is on the problem sheet and the Euler-angle singularity is the thing unit 31 explains properly.

Everything in §1 and §2 is Newtonian and could have appeared in Part A. It is here because Part B is where the course builds the two objects relativity needs — an action principle, which units 11 to 15 built, and an honest definition of a tensor, which is this unit. The inertia tensor is the first place in physics where nine numbers per basis will not do.

16 · overview

The book, and the axis that will not hold

The block has three axes through its centre that the whole unit is about: along the spine, across the cover, and normal to the cover. They are its principal axes, and the numbers beside them are its moments of inertia about each, in $\mathrm{kg\,m^2}$.

The smallest moment goes with the longest edge: spinning about the spine keeps the mass closest to the axis.

the prediction

Toss the book spinning about axis 1 or axis 3 and it wobbles steadily. Toss it about axis 2, the middle moment, and it turns over as well, within about $0.6$ s: the flip. The route sketched on the previous slide ends in that number.

The book lying flat, with its three principal axes through the centre. Only the middle one, across the cover, is unstable; the curved arrow is a spin about it.

section 1

A body that cannot deform

Six degrees of freedom, three of them living in a group.

16 · the body

What a rigid body is, and where it lives

definition

A rigid body is a mass distribution — a density $\rho$ on a compact set $B\subset\mathbb R^3$, or finitely many point masses $m_\alpha$ — together with the constraint that $\lVert\vec x_\alpha(t)-\vec x_\beta(t)\rVert$ is independent of $t$ for every pair.

So the body moves as a whole: whatever carries it from one instant to the next preserves every distance between its points. The theorem says what such a map can be.

theorem · rigid motions are affine

If $\Phi:\mathbb R^3\to\mathbb R^3$ preserves all distances, then $\Phi(\vec x) = Q\vec x + \vec c$ for some $Q\in O(3)$ and $\vec c\in\mathbb R^3$.

Why: put $\vec c = \Phi(0)$ and $\Psi = \Phi-\vec c$. Then $\Psi$ fixes $0$ and preserves distances, so it preserves $\lVert\Psi(\vec x)\rVert$ and hence, by polarisation, all inner products; a map preserving inner products sends an orthonormal basis to an orthonormal basis and is therefore linear.

full proof in the notes (N)

Full proof

Let $\Psi$ fix the origin and preserve distances. Then $\lVert\Psi(\vec x)\rVert = \lVert\Psi(\vec x)-\Psi(0)\rVert = \lVert\vec x\rVert$, so norms are preserved. For any $\vec x,\vec y$, $$\lVert\Psi(\vec x)-\Psi(\vec y)\rVert^2 = \lVert\Psi(\vec x)\rVert^2 - 2\,\Psi(\vec x)\cdot\Psi(\vec y) + \lVert\Psi(\vec y)\rVert^2,$$ and the same identity holds without $\Psi$. Since the left side and the two norm terms are unchanged by $\Psi$, so is $\Psi(\vec x)\cdot\Psi(\vec y) = \vec x\cdot\vec y$. Now let $\{\vec e_i\}$ be orthonormal; $\{\Psi(\vec e_i)\}$ is orthonormal too, hence a basis. For any $\vec x$, the components of $\Psi(\vec x)$ in that basis are $\Psi(\vec x)\cdot\Psi(\vec e_i) = \vec x\cdot\vec e_i = x_i$, so $\Psi(\vec x) = \sum_i x_i\,\Psi(\vec e_i)$, which is linear in $\vec x$. Its matrix has orthonormal columns, so it is in $O(3)$.

Note what the proof did not assume: neither linearity nor continuity nor surjectivity. Distance preservation alone is enough, because the Euclidean distance comes from an inner product. For a general real normed space the analogous statement needs surjectivity, and is the Mazur–Ulam theorem (1932).

16 · the body

A configuration is a rotation and a shift

Continuity in $t$ and $\det Q = \pm1$ force $\det Q = +1$: a body cannot reflect itself. So a configuration of the body is a pair

$$\big(R,\ \vec X\big)\in SO(3)\times\mathbb R^3,$$

with $\vec X$ the centre of mass and $R$ the rotation carrying a reference copy of the body onto its present position. That is the configuration space, and it has six dimensions — unit 12's generalised coordinates, for the first time on a space that is not $\mathbb R^n$.

not a vector space

$SO(3)$ is a group and a smooth surface in $\mathbb R^9$, and it is curved: you cannot add two rotations. Unit 31 says what kind of object it is; section 6 shows what goes wrong if you insist on coordinates for it.

Put a reference copy with its centre of mass at $O$. The body now is that copy rotated by $R$, then moved by $\vec X$: the material point at $\vec r$ in the copy is at $\vec x = \vec X + R\vec r$. Schematic, drawn in a plane.

Reading the figure. The affine theorem says the map from the reference copy to the present body is $\vec r\mapsto Q\vec r+\vec c$; taking the reference copy's centre of mass at the origin makes $\vec c$ the present centre of mass $\vec X$, because a rigid motion carries the centre of mass to the centre of mass. The body axes drawn in the present copy are the columns of $R$: the images of the reference copy's axes, which coincide with the space axes. The next slide differentiates $\vec x = \vec X + R\vec r$ in time.

Why $\det = +1$. $t\mapsto\det Q(t)$ is continuous with values in $\{+1,-1\}$ and equals $+1$ at $t=0$ (where $\Phi$ is the identity), so it is $+1$ for all $t$. This is the same connectedness argument that made $O(3)$ split into two pieces in unit 03.

Six, not nine. $SO(3)$ sits inside the nine-dimensional space of $3\times3$ matrices, cut out by $R^{\mathsf T}R = \mathbb 1$. That is six independent equations (the matrix equation is symmetric), so $9-6=3$ dimensions, plus three for $\vec X$. Unit 03 derived $\dot R = R\Omega$ with $\Omega$ antisymmetric from the same constraint; antisymmetric $3\times3$ matrices form a three-dimensional space, which is the same count seen from the tangent space.

16 · the body

Angular velocity, recalled

A point of the body fixed at $\vec r$ in the body frame sits at $\vec x(t) = \vec X(t) + R(t)\vec r$ in the space frame. Unit 03 differentiated $R^{\mathsf T}R = \mathbb 1$ to get

$$\Omega := R^{\mathsf T}\dot R \quad\text{antisymmetric},\qquad \Omega\vec u = \vec\omega\times\vec u,$$

which defines the angular velocity $\vec\omega$ — as a vector only because $3\times3$ antisymmetric matrices happen to form a three-dimensional space. Hence

$$\dot{\vec x} = \dot{\vec X} + R\,(\vec\omega\times\vec r) = \vec V + \vec\omega_{\text{sp}}\times(\vec x-\vec X).$$

Every velocity in the body is one translation plus one rotation. Two readings of the same $\vec\omega$: components in the body frame, $\vec\omega$, and in the space frame, $\vec\omega_{\text{sp}} = R\vec\omega$.

which frame

$\Omega = R^{\mathsf T}\dot R$ gives body components; $\tilde\Omega = \dot R R^{\mathsf T}$ gives space components, and $\tilde\Omega = R\Omega R^{\mathsf T}$. Sign and frame errors here account for most wrong answers in rigid-body dynamics. This unit works in the body frame throughout, and says so each time.

The transport theorem of unit 03, $\left(\tfrac{\mathrm d}{\mathrm dt}\right)_{\text{sp}} = \left(\tfrac{\mathrm d}{\mathrm dt}\right)_{\text{body}} + \vec\omega\times$, is the same statement, and it is what turns $\dot{\vec L}=\vec\tau$ into Euler's equations in §5.

Where $\Omega$ comes from. Differentiate $R^{\mathsf T}R = \mathbb 1$: $\dot R^{\mathsf T}R + R^{\mathsf T}\dot R = 0$, so $\Omega = R^{\mathsf T}\dot R$ satisfies $\Omega^{\mathsf T} = -\Omega$. Write $\Omega = \begin{pmatrix}0&-\omega_3&\omega_2\\ \omega_3&0&-\omega_1\\ -\omega_2&\omega_1&0\end{pmatrix}$ and check on basis vectors that $\Omega\vec u = \vec\omega\times\vec u$.

The accident of three dimensions. Antisymmetric $n\times n$ matrices form a space of dimension $n(n-1)/2$, which equals $n$ only when $n=3$. In four dimensions angular velocity has six components and is not a vector at all; it is a $2$-form, which unit 34 will name. Every time this course writes $\vec\omega$, it is using a coincidence. The coincidence also explains why the cross product exists only in three dimensions.

Space versus body. From $\dot R = R\Omega$, $\dot R R^{\mathsf T} = R\Omega R^{\mathsf T}$, which is $\tilde\Omega$; in §3's language, $\tilde\Omega$ and $\Omega$ are the component arrays of one linear map in two bases, and $R\Omega R^{\mathsf T}$ is the transformation law for a $(1,1)$-tensor when the basis change is orthogonal.

16 · the body

$\mathfrak{so}(3)$, and the exponential that gets you back

definition

so(3) is the vector space of antisymmetric $3\times3$ real matrices. It is the tangent space to $SO(3)$ at the identity: every curve $R(t)$ in $SO(3)$ with $R(0)=\mathbb 1$ has $\dot R(0)$ antisymmetric, and conversely.

the exponential map

$\displaystyle \exp A := \sum_{n\ge0}\frac{A^n}{n!}$, absolutely convergent for every matrix since $\lVert A^n\rVert\le\lVert A\rVert^n$ in the operator norm.

theorem

$A^{\mathsf T}=-A \implies \exp A\in SO(3)$.

Why: $(\exp A)^{\mathsf T}\exp A = \exp(A^{\mathsf T})\exp A = \exp(-A)\exp(A) = \mathbb 1$, using $\exp(X)\exp(-X)=\mathbb 1$ for commuting arguments; and $t\mapsto\det\exp(tA)$ is continuous, equals 1 at $t=0$, never 0.

So $\exp$ carries the three-dimensional space $\mathfrak{so}(3)$ into the group. The next slide says what it does to a given antisymmetric matrix, and draws it.

Full proof of the two facts the theorem used. (i) $(\exp A)^{\mathsf T} = \exp(A^{\mathsf T})$: transpose the partial sums term by term, since $(A^n)^{\mathsf T} = (A^{\mathsf T})^n$, and pass to the limit, transposition being continuous. (ii) If $XY = YX$ then $\exp X\exp Y = \exp(X+Y)$: both series converge absolutely, so their Cauchy product converges to the product of the sums, and its $n$th term is $\sum_{k=0}^n\frac{X^k}{k!}\frac{Y^{n-k}}{(n-k)!} = \frac{(X+Y)^n}{n!}$ by the binomial theorem — which holds for commuting matrices and fails otherwise. Put $Y=-X$: $\exp X\exp(-X) = \exp 0 = \mathbb 1$, so $\exp X$ is always invertible and $\det\exp(tA)\ne0$.

Full proof that $\dot R(0)$ is antisymmetric and conversely. Forward: differentiate $R^{\mathsf T}R=\mathbb 1$ at $t=0$. Backward: given antisymmetric $A$, the curve $R(t)=\exp(tA)$ lies in $SO(3)$ by the theorem and has $\dot R(0)=A$. So the tangent space is exactly $\mathfrak{so}(3)$, and its dimension 3 confirms the count on the previous slide.

Why the course needs this. $\exp$ is the bridge between a Lie group and its Lie algebra, and the structure recurs: the Lorentz group and its algebra in unit 26, and the exponential of a connection along a curve — parallel transport — in unit 32. Here it also supplies the cleanest description of a rotation: an axis and an angle, three numbers, valid for every rotation, and singular only where $\theta$ wraps.

16 · the body

Rodrigues' formula, drawn

Rodrigues' formula

For a unit vector $\hat n$ with matrix $K$ ($K\vec u = \hat n\times\vec u$),

$$\exp(\theta K) = \mathbb 1 + \sin\theta\,K + (1-\cos\theta)\,K^2,$$

the rotation by $\theta$ about $\hat n$.

Why: $K^3=-K$, so the series splits into the $K$ terms $\theta-\theta^3/3!+\dots=\sin\theta$ and the $K^2$ terms $\theta^2/2!-\theta^4/4!+\dots=1-\cos\theta$.

consequence

$\exp:\mathfrak{so}(3)\to SO(3)$ is onto: every rotation is a rotation by some angle about some axis (Euler, 1775). Not injective — $\theta$ and $\theta+2\pi$ agree — which is the first hint that $SO(3)$ is not $\mathbb R^3$.

Split $\vec u$ into a part along $\hat n$ and a part $\vec u_\perp$ across it. The rotation leaves the first alone and turns the second in the plane spanned by $\vec u_\perp = -K^2\vec u$ and $\hat n\times\vec u = K\vec u$: the tip runs round the dashed circle.

The figure is the formula. Since $K^2\vec u = (\hat n\cdot\vec u)\hat n-\vec u = -\vec u_\perp$ (next paragraph), Rodrigues reads $\exp(\theta K)\vec u = \vec u + \sin\theta\,K\vec u - (1-\cos\theta)\vec u_\perp = \vec u_\parallel + \cos\theta\,\vec u_\perp + \sin\theta\,(\hat n\times\vec u)$. The two vectors $\vec u_\perp$ and $\hat n\times\vec u$ are perpendicular and of equal length, so the last two terms are a point on a circle of radius $\lVert\vec u_\perp\rVert$, at angle $\theta$ from $\vec u_\perp$: plane rotation, done in the plane across the axis.

$K^3=-K$. With $\hat n$ a unit vector, $K^2\vec u = \hat n\times(\hat n\times\vec u) = (\hat n\cdot\vec u)\hat n - \vec u$ by the BAC–CAB identity of unit 08, so $K^2 = \hat n\hat n^{\mathsf T}-\mathbb 1$ and $K^3\vec u = K\big((\hat n\cdot\vec u)\hat n-\vec u\big) = -K\vec u$ since $K\hat n=0$.

Why $\exp$ is onto here. Every $R\in SO(3)$ has a real eigenvalue: its characteristic polynomial is a real cubic, so it has a real root, and $\det R=1$ with $|{\lambda}|=1$ for every eigenvalue forces one eigenvalue to be exactly $+1$. Its eigenvector $\hat n$ is the axis; on $\hat n^{\perp}$, $R$ is a plane rotation by some $\theta$; Rodrigues with that $\hat n,\theta$ reproduces it. This is Euler's rotation theorem.

16 · the body

Kinetic energy splits in two

König's theorem

For a rigid body with centre of mass $\vec X$, total mass $M$ and angular velocity $\vec\omega$,

$$T \;=\; \underbrace{\tfrac12 M\lVert\dot{\vec X}\rVert^2}_{\text{translation}} \;+\; \underbrace{\tfrac12\int_B \rho\,\lVert\vec\omega\times\vec r\rVert^2\,\mathrm dV}_{\text{rotation}},\qquad \vec r = \text{position from }\vec X .$$

proof

$\dot{\vec x} = \dot{\vec X} + \vec\omega\times\vec r$ (previous slide, in whichever frame), so

$$\lVert\dot{\vec x}\rVert^2 = \lVert\dot{\vec X}\rVert^2 + 2\,\dot{\vec X}\cdot(\vec\omega\times\vec r) + \lVert\vec\omega\times\vec r\rVert^2 .$$

Integrate $\tfrac12\rho\,(\cdot)$ over $B$. The cross term is $\dot{\vec X}\cdot\big(\vec\omega\times\int_B\rho\,\vec r\,\mathrm dV\big)$, and $\int_B\rho\,\vec r\,\mathrm dV = 0$ because $\vec r$ is measured from the centre of mass. $\;\blacksquare$

The same split works about any point that is fixed in space (a pivot): then $\dot{\vec X}$-terms are absent and $T$ is purely rotational about that point.

only two points

The split is clean only about the centre of mass or a fixed pivot. About any other point the cross term survives, and every "$I$" you write is wrong unless you say which point it is about. The block gets two: its centre, and a corner.

Why $\int\rho\vec r = 0$. That is the definition of the centre of mass, $\vec X = \frac1M\int_B\rho\,\vec x\,\mathrm dV$, rearranged: with $\vec r = \vec x-\vec X$, $\int\rho\vec r = \int\rho\vec x - M\vec X = 0$. Unit 08 used the same identity to reduce the two-body problem.

A fixed pivot. If some point $\vec P$ of the body is held fixed, write positions from $\vec P$: then $\dot{\vec x} = \vec\omega\times\vec r$ exactly, with no translation term at all, and $T = \frac12\int\rho\lVert\vec\omega\times\vec r\rVert^2$ with $\vec r$ from $\vec P$. The heavy top of §6 is this case.

Where this is going. The rotational term is a quadratic function of $\vec\omega$. Any quadratic form comes from a symmetric bilinear form by polarisation, and that bilinear form is the inertia tensor. Section 2 does exactly that, and nothing more: the tensor is not an extra postulate, it is what $T_{\text{rot}}$ already is.

König's theorem is also how unit 12's Lagrangian for a rigid body is written: $L = T - V$ with $T$ split this way, which is why a rolling body's kinetic energy is $\frac12 M v^2 + \frac12 I\omega^2$ and not one or the other.

section 2

The inertia tensor

A quadratic form was there all along; give it its proper name.

16 · inertia

From a kinetic energy to a bilinear form

Lagrange's identity (unit 08): $\lVert\vec u\times\vec r\rVert^2 = \lVert\vec u\rVert^2\lVert\vec r\rVert^2 - (\vec u\cdot\vec r)^2$. So the rotational energy is

$$T_{\text{rot}} = \tfrac12\int_B\rho\Big(\lVert\vec\omega\rVert^2\lVert\vec r\rVert^2-(\vec\omega\cdot\vec r)^2\Big)\mathrm dV,$$

a quadratic form in $\vec\omega$. Polarise it — replace one $\vec\omega$ by $\vec u$ and the other by $\vec w$ — and the object underneath appears.

definition · the inertia tensor

About a chosen origin $O$, $\ \mathsf I_O: \mathbb R^3\times\mathbb R^3\to\mathbb R$,

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

Bilinear and symmetric by inspection, and $T_{\text{rot}} = \tfrac12\mathsf I_O(\vec\omega,\vec\omega)$.

positivity

$\mathsf I_O(\vec u,\vec u) = \int\rho\lVert\vec u\times\vec r\rVert^2\ge0$, with equality iff $\vec r\parallel\vec u$ for almost every point of $B$ — that is, iff the whole body lies on the line through $O$ along $\vec u$.

So $\mathsf I_O$ is positive definite for any body that is not a straight line of mass. Dimensions: $[\rho]\,[\mathrm dV]\,[r^2] = \mathrm{kg\,m^2}$, and $\tfrac12\mathsf I(\vec\omega,\vec\omega)$ is $\mathrm{kg\,m^2\,s^{-2}} = \mathrm J$ ✓.

Nothing has been assumed about bases. $\mathsf I_O$ is defined by an integral over the body and two input vectors, and no coordinate system appears in its definition — which is exactly what will make §3's transformation law a theorem.

Polarisation. A symmetric bilinear form is determined by its quadratic form: $B(\vec u,\vec w) = \frac12\big[Q(\vec u+\vec w)-Q(\vec u)-Q(\vec w)\big]$. So writing down $T_{\text{rot}}$ as a quadratic form in $\vec\omega$ already determines $\mathsf I_O$ uniquely; the displayed formula is what that recipe produces. Check it: expanding $\lVert\vec u+\vec w\rVert^2\lVert\vec r\rVert^2-((\vec u+\vec w)\cdot\vec r)^2$ and subtracting the two pure terms leaves twice the bracket in the definition.

Lagrange's identity. $\lVert\vec u\times\vec r\rVert^2 = \lVert\vec u\rVert^2\lVert\vec r\rVert^2\sin^2\theta = \lVert\vec u\rVert^2\lVert\vec r\rVert^2(1-\cos^2\theta)$, and $\vec u\cdot\vec r = \lVert\vec u\rVert\lVert\vec r\rVert\cos\theta$. Or expand both sides in components.

Point masses. For $N$ point masses the integral is a sum: $\mathsf I_O(\vec u,\vec w) = \sum_\alpha m_\alpha\big[(\vec u\cdot\vec w)\lVert\vec r_\alpha\rVert^2-(\vec u\cdot\vec r_\alpha)(\vec w\cdot\vec r_\alpha)\big]$. Every statement in this unit holds verbatim for either. The problem sheet uses both.

The moment of inertia about an axis is the special case $\mathsf I_O(\hat n,\hat n)$ for a unit vector $\hat n$: the number that appears in $T = \frac12 I\omega^2$ for rotation about a fixed axis. The elementary treatment of rigid bodies never gets beyond that diagonal slice, which is why it never has to face the fact that $\vec L\nparallel\vec\omega$.

16 · inertia

Angular momentum is a covector

$\displaystyle \vec L_O = \int_B\rho\,\vec r\times(\vec\omega\times\vec r)\,\mathrm dV$. Pair it with a test vector $\vec u$ and use $\vec u\cdot(\vec r\times\vec s) = \vec s\cdot(\vec u\times\vec r)$:

$$\vec L_O\cdot\vec u = \int_B\rho\,\big(\vec\omega\times\vec r\big)\cdot(\vec u\times\vec r)\,\mathrm dV = \mathsf I_O(\vec\omega,\vec u).$$

so, properly

Angular momentum is the linear functional $\ \vec u\mapsto\mathsf I_O(\vec\omega,\vec u)$ — an element of the dual space $V^*$, written $\mathsf I_O(\vec\omega,\cdot)$. The Euclidean inner product then turns it into the arrow we draw. In components, $L_i = I_{ij}\omega_j$.

Unit 04 planted this: the derivative of a function is a covector, and the gradient is what you get after using an inner product. Same distinction, now structural. Units 28 and 32 raise and lower indices with a metric that is not the identity, and then the difference is unavoidable.

The vector identity. $\vec u\cdot(\vec r\times\vec s) = \det[\vec u,\vec r,\vec s] = \det[\vec s,\vec u,\vec r] = \vec s\cdot(\vec u\times\vec r)$: the scalar triple product is invariant under cyclic permutation (unit 08). Applying it with $\vec s = \vec\omega\times\vec r$ gives the displayed line. Then expand $(\vec\omega\times\vec r)\cdot(\vec u\times\vec r) = (\vec\omega\cdot\vec u)\lVert\vec r\rVert^2-(\vec\omega\cdot\vec r)(\vec u\cdot\vec r)$ — Binet–Cauchy, or BAC–CAB — which is exactly the bracket in the definition of $\mathsf I_O$.

Why "covector" is the honest word. $\mathsf I_O$ eats two vectors and returns a number. Feed it one and you are left with a machine that eats one vector and returns a number: a linear functional. Nothing in that says which vector it is; to convert it you need an inner product, and the answer depends on which one you choose. In Euclidean space with an orthonormal basis the conversion is invisible — the components are numerically the same — and that is precisely why it is worth being careful now rather than in unit 32, where it is not invisible.

Energy from the pairing. $T_{\text{rot}} = \frac12\vec L\cdot\vec\omega$, which is the covector eating the vector: a pairing needing no inner product at all. That is the cleanest statement of rotational kinetic energy available.

16 · inertia

The block's $\vec L$, tilted $22.8^\circ$ off $\vec\omega$

the fact this unit is about

$\vec L$ is not parallel to $\vec\omega$. It is, exactly when $\vec\omega$ is an eigenvector of $\mathsf I_O$ — three directions out of infinitely many.

Spin the block about its body diagonal through a corner, $\hat\omega = (0.8242,0.5494,0.1374)$: $\mathsf I_O$ returns an $\vec L$ tilted $\mathbf{22.8^\circ}$ away. $\vec L$ is fixed in space when no torque acts, so $\vec\omega$ is not — and the body wobbles.

In the corner frame $L_i = I_{ij}\omega_j$. The diagonal entries are unequal — $I_{22}$ is more than twice $I_{11}$ — and the off-diagonal ones are not zero, so $\mathsf I_O$ stretches and shears $\vec\omega$ instead of scaling it. Only along an eigenvector do the two effects leave the direction alone.

The book seen along the normal to the plane of $\vec\omega$ and $\vec L$, so the angle between them is shown true. $\vec\omega$ runs along the body diagonal from the corner $O$; $\vec L$ leans $22.8^\circ$ towards the cover's short edge $b$. Arrow lengths are not to scale.

The $22.8^\circ$. $\mathsf I_O\hat\omega = (1.1209,\,1.6263,\,0.5714)\times10^{-3}$ in $\mathrm{kg\,m^2\,s^{-1}}$ per unit $\lVert\vec\omega\rVert$; its angle with $\hat\omega$ is $\arccos\big(\mathsf I_O\hat\omega\cdot\hat\omega/\lVert \mathsf I_O\hat\omega\rVert\big) = 22.776^\circ$. The tilt would be zero only if $\hat\omega$ were an eigenvector. It nearly is — §4 finds the smallest-moment axis just $3.376^\circ$ from the diagonal — but "nearly an eigenvector" buys nothing here, because the three eigenvalues differ by a factor of nine and a small misalignment is amplified by that ratio. For a cube the diagonal is exactly an eigenvector, and the tilt is exactly zero; that is a problem on the sheet.

How the figure was drawn. The viewing direction is $\hat\omega\times\mathsf I_O\hat\omega$, normalised: $(0.114,-0.398,0.910)$ in the corner frame, mostly along the cover normal, so the book is seen nearly from above and its thickness almost vanishes. Both arrows lie in the drawing plane, which is why the angle can be measured off the page; any other view would distort it.

16 · inertia · example 1

The block, about its centre

worked example · the block

Take $\vec e_1,\vec e_2,\vec e_3$ along the edges $a,b,c$, origin at the centre, so $B = [-\tfrac a2,\tfrac a2]\times[-\tfrac b2,\tfrac b2]\times[-\tfrac c2,\tfrac c2]$ and $\rho = m/abc$ is constant. Components are $I_{ij} = \mathsf I(\vec e_i,\vec e_j)$:

$$I_{11} = \int_B\rho\,(y^2+z^2)\,\mathrm dV,\qquad I_{12} = -\int_B\rho\,xy\,\mathrm dV .$$

Off-diagonal, first. By Fubini (unit 05) the triple integral factorises, and $\int_{-b/2}^{b/2}y\,\mathrm dy = 0$: every off-diagonal entry vanishes.

Diagonal. $\int_B\rho\,y^2\mathrm dV = \rho\,a c\int_{-b/2}^{b/2}y^2\mathrm dy = \rho\,ac\cdot\tfrac{b^3}{12} = \tfrac{m}{12}b^2$, and likewise for $z$. So $I_{11} = \tfrac m{12}(b^2+c^2)$, and cyclically.

With $m = 0.60$ kg, $a=0.24$, $b=0.16$, $c=0.04$ m:

axis	formula	$\mathrm{kg\,m^2}$
$1$ (spine)	$\tfrac m{12}(b^2+c^2)$	0.00136
$2$ (across)	$\tfrac m{12}(a^2+c^2)$	0.00296
$3$ (thickness)	$\tfrac m{12}(a^2+b^2)$	0.00416

Already diagonal, and $I_1\lt I_2\lt I_3$ strictly: three different numbers, which is what §5 needs.

checks

Dimensions $\mathrm{kg\,m^2}$ throughout ✓. Limit $b,c\to0$ (a thin rod along $\vec e_1$): $I_1\to0$ and $I_2=I_3\to ma^2/12$, the standard rod ✓. Limit $a=b=c$: all three equal $ma^2/6$, so every axis through the centre of a cube is equivalent ✓. Trace $= \tfrac m6(a^2+b^2+c^2) = 2\int\rho\lVert\vec r\rVert^2\mathrm dV$, which §3 shows is basis-independent: $0.00848\ \mathrm{kg\,m^2}$.

Why the off-diagonal terms vanish. Not an accident of the block: each coordinate plane is a mirror plane of the body, and $\rho(x,y,z)=\rho(-x,y,z)$ makes $\int\rho\,xy$ an integral of an odd function over a symmetric interval. Any body with two orthogonal mirror planes through the chosen origin has those axes principal, which §4 states as a theorem.

The factorisation, in full. $\int_B\rho\,y^2\,\mathrm dV = \rho\int_{-a/2}^{a/2}\mathrm dx\int_{-c/2}^{c/2}\mathrm dz\int_{-b/2}^{b/2}y^2\,\mathrm dy = \rho\,a\,c\,\big[\tfrac{y^3}3\big]_{-b/2}^{b/2} = \rho\,ac\,\tfrac{b^3}{12}$, and $\rho\,ac = m/b$, giving $mb^2/12$. Fubini applies because the integrand is continuous on a compact box (unit 05).

Which axis is which. The smallest moment belongs to the axis along the longest edge, because $I_{11}$ involves $b$ and $c$ — the other two. That inversion catches people out. For the book: spinning about the spine is easiest, about the axis perpendicular to the cover is hardest, and across the cover is in between. Axis 2 — across the cover, in its plane — is the intermediate one, and it is the one that will not hold.

Trace check, numerically. $0.00136+0.00296+0.00416 = 0.00848$, and $\frac{0.60}{6}(0.0576+0.0256+0.0016) = 0.1\times0.0848 = 0.00848$ ✓.

16 · inertia · example 1, continued

Moving the origin: the parallel-axis theorem

parallel-axis theorem (Steiner)

If the centre of mass is displaced from $O$ by $\vec d$ (so $\vec r_O = \vec r_{\text{cm}}+\vec d$), then for all $\vec u,\vec w$

$$\mathsf I_O(\vec u,\vec w) = \mathsf I_{\text{cm}}(\vec u,\vec w) + M\Big[(\vec u\cdot\vec w)\lVert\vec d\rVert^2-(\vec u\cdot\vec d)(\vec w\cdot\vec d)\Big].$$

proof

Substitute $\vec r_O = \vec r+\vec d$ into the definition. The bracket is bilinear in $\vec r_O$, so it expands into three groups: the one quadratic in $\vec r$ gives $\mathsf I_{\text{cm}}$, the one quadratic in $\vec d$ gives $M[\dots]$ since $\int\rho\,\mathrm dV = M$, and the two cross groups each carry exactly one factor $\int_B\rho\,\vec r\,\mathrm dV = \vec 0$. $\;\blacksquare$

checks

Dimensions $\mathrm{kg\,m^2}$ ✓. Limit $\vec d\to0$: $\mathsf I_O\to\mathsf I_{\text{cm}}$ ✓. Sign: the added form is $M\lVert\vec u\times\vec d\rVert^2\ge0$, so moving off the centre of mass only increases every moment ✓.

The expansion, in full. Write $\vec s = \vec r+\vec d$. Then $(\vec u\cdot\vec w)\lVert\vec s\rVert^2-(\vec u\cdot\vec s)(\vec w\cdot\vec s)$ equals $$\Big[(\vec u\cdot\vec w)\lVert\vec r\rVert^2-(\vec u\cdot\vec r)(\vec w\cdot\vec r)\Big] +\Big[(\vec u\cdot\vec w)\lVert\vec d\rVert^2-(\vec u\cdot\vec d)(\vec w\cdot\vec d)\Big] +\Big[2(\vec u\cdot\vec w)(\vec r\cdot\vec d)-(\vec u\cdot\vec r)(\vec w\cdot\vec d)-(\vec u\cdot\vec d)(\vec w\cdot\vec r)\Big].$$ Multiply by $\rho$ and integrate. The third bracket is linear in $\vec r$, so it is a linear function of $\int_B\rho\,\vec r\,\mathrm dV$, which vanishes about the centre of mass.

Only from the centre of mass. The theorem does not let you jump between two arbitrary points in one step — go via the centre of mass, subtracting one Steiner term and adding another. Forgetting this is the most common error in inertia calculations.

The familiar scalar version is the diagonal slice: putting $\vec u=\vec w=\hat n$ with $\hat n\perp\vec d$ gives $I_O = I_{\text{cm}}+Md^2$. The tensor version also tells you what the off-diagonal entries become, which the scalar version cannot.

16 · inertia · example 1, continued

Steiner at the block's corner

On the block, with $\vec d = (\tfrac a2,\tfrac b2,\tfrac c2)$ from a corner to the centre, the added form contributes $\tfrac m4(b^2+c^2)$ to $I_{11}$ and $-\tfrac m4ab$ to $I_{12}$:

$$I^O_{11}=\tfrac m3(b^2+c^2),\quad I^O_{12}=-\tfrac m4ab,$$

giving $\mathsf I_O$, with every off-diagonal entry now nonzero: a corner is not a symmetry point.

numbers

$I^O_{11} = 0.00544$, $I^O_{12} = -0.00576$, and for the axis drawn, $I^O_{33} = I_3 + \tfrac M4(a^2+b^2) = 0.00416 + 0.01248 = 0.01664$. Trace $0.03392 = 4\times0.00848$.

From the corner $O$ to the centre is $\vec d = (\tfrac a2,\tfrac b2,\tfrac c2)$. Two vertical axes, one through each, stand $\tfrac12\sqrt{a^2+b^2} = 0.144$ m apart, and the moment about the corner's axis is the central one plus $M$ times that distance squared.

What the figure shows. The scalar parallel-axis theorem, as the special case $\vec u=\vec w=\vec e_3$ of the tensor version: $(\vec e_3\cdot\vec e_3)\lVert\vec d\rVert^2-(\vec e_3\cdot\vec d)^2 = d_1^2+d_2^2$, the squared distance between the two vertical axes. So $I^O_{33} = \tfrac m{12}(a^2+b^2)+\tfrac m4(a^2+b^2) = \tfrac m3(a^2+b^2) = 0.2\times0.0832 = 0.01664$, the bottom-right entry of $\mathsf I_O$. The off-diagonal entries have no such picture; they come from the $-(\vec u\cdot\vec d)(\vec w\cdot\vec d)$ term with $\vec u\ne\vec w$.

The numbers. $\frac{0.60}{3}(0.0256+0.0016) = 0.2\times0.0272 = 0.00544$; $-\frac{0.60}{4}(0.24)(0.16) = -0.15\times0.0384 = -0.00576$. The trace is $\frac{m}{3}\cdot2(a^2+b^2+c^2) = 0.4\times0.0848 = 0.03392$, four times the central trace, because $M\lVert\vec d\rVert^2\times2 = 2\times0.6\times\frac{0.0848}{4} = 0.02544$ is added.

section 3

What a tensor actually is

A multilinear map. The array is a shadow; the transformation law is a theorem.

16 · tensors

The dual space, and why it is not $V$

definition

For a real vector space $V$ of dimension $n$, the dual space $V^*$ is the space of linear maps $V\to\mathbb R$. Its elements are covectors.

the dual basis

Given a basis $\{\vec e_i\}$ of $V$, there are unique $\{\varepsilon^j\}\subset V^*$ with $\varepsilon^j(\vec e_i)=\delta^j_i$, and they are a basis of $V^*$. Hence $\dim V^* = n$.

Why: a linear map is determined by its values on a basis, so the condition defines each $\varepsilon^j$ uniquely; any $\alpha\in V^*$ equals $\sum_j\alpha(\vec e_j)\varepsilon^j$, since both sides agree on every $\vec e_i$; and $\sum_jc_j\varepsilon^j=0$ evaluated at $\vec e_i$ gives $c_i=0$.

$V$ and $V^*$ have the same dimension, so they are isomorphic — but no isomorphism is singled out by $V$ alone. The next slide shows why, with a picture.

What does give an isomorphism is an inner product: $\vec v\mapsto\langle\vec v,\cdot\rangle$. Unit 04's gradient is that map applied to the derivative; §2's $\vec L$ is that map inverted. Unit 32 calls it lowering an index.

$V^{**}$, by contrast, is canonically $V$ in finite dimensions: the map $\vec v\mapsto(\alpha\mapsto\alpha(\vec v))$ is linear, injective (if $\alpha(\vec v)=0$ for all $\alpha$ then $\vec v=0$, by taking $\alpha$ to be a dual basis element), and both spaces have dimension $n$. No choices were made. That is why $(1,0)$-tensors — multilinear maps $V^*\to\mathbb R$ — may be called vectors without apology on the next slide.

Where this bites, later. In unit 28 the electromagnetic field is an antisymmetric $(0,2)$-tensor $F_{\mu\nu}$, and $F^{\mu\nu}$ is a genuinely different array obtained through the Minkowski metric, which has a minus sign: raising an index flips signs on some components. In unit 32 the metric varies from point to point, and the two arrays differ everywhere. Here, with the Euclidean inner product and orthonormal bases, the conversion is numerically the identity — which is why the distinction can only be planted in this unit, not forced.

Notation. Upper indices label components of vectors and basis covectors; lower indices label components of covectors and basis vectors. The pattern is: a basis object and a component carry opposite index positions, which is what makes sums like $\vec v = v^i\vec e_i$ come out basis-independent.

16 · tensors

A covector, drawn as a stack of lines

no canonical isomorphism

$V\cong V^*$ — both have dimension $n$ — but not in any way determined by $V$ alone. In $\mathbb R^2$ take $\vec f_1 = \vec e_1$, $\vec f_2 = \vec e_1+\vec e_2$. Then

$$\varphi^1 = \varepsilon^1-\varepsilon^2 \ne \varepsilon^1,$$

although $\vec f_1 = \vec e_1$ did not move. The dual of a basis vector depends on all the other basis vectors. Any rule "$\vec e_i\mapsto\varepsilon^i$" is a choice of basis in disguise.

A covector is drawn by its level lines: $\alpha(\vec v)$ counts how many lines $\vec v$ crosses. $\varepsilon^1$ must be $0$ on $\vec e_2$, $\varphi^1$ must be $0$ on $\vec f_2$, so their lines tilt differently — though both read $1$ on the same vector $\vec e_1=\vec f_1$.

Reading the figure. A nonzero linear map $\alpha:\mathbb R^2\to\mathbb R$ is fixed by the family of parallel lines $\alpha = 0, 1, 2,\dots$; the value on a vector is read off by where its tip lands. The dual-basis condition $\varepsilon^1(\vec e_1)=1,\ \varepsilon^1(\vec e_2)=0$ says: the $0$-line contains $\vec e_2$ and the $1$-line passes through the tip of $\vec e_1$. Change $\vec e_2$ to $\vec f_2=\vec e_1+\vec e_2$ and the $0$-line must turn to contain it, which tilts every line, while the $1$-line still passes through the tip of the unchanged $\vec f_1$. That is the whole content of "the dual of a basis vector depends on all the others".

Check the $\mathbb R^2$ example. $\varphi^1$ must satisfy $\varphi^1(\vec f_1)=1,\ \varphi^1(\vec f_2)=0$. Try $\varphi^1 = \varepsilon^1-\varepsilon^2$: on $\vec f_1=\vec e_1$ it gives $1-0=1$ ✓; on $\vec f_2=\vec e_1+\vec e_2$ it gives $1-1=0$ ✓. So changing $\vec e_2$ alone changed $\varepsilon^1$. Nothing local about it.

16 · tensors

The definition

definition · tensor

A tensor of type $(p,q)$ on $V$ is a multilinear map

$$T:\underbrace{V^*\times\dots\times V^*}_{p}\times\underbrace{V\times\dots\times V}_{q}\longrightarrow\mathbb R .$$

"Multilinear" means linear in each slot with the others held fixed. The set of them is a vector space $T^p_q(V)$ under pointwise operations.

$(0,1)$

a covector. E.g. $\vec L_O$.

$(1,0)$

a map $V^*\to\mathbb R$, i.e. an element of $V^{**}=V$: a vector.

$(0,2)$

a bilinear form. The inertia tensor. Later: the metric.

$(1,1)$

$T(\alpha,\vec v)$: equivalently a linear map $V\to V$, by $\vec v\mapsto T(\cdot,\vec v)$. E.g. $\Omega$.

the tensor product

$(S\otimes T)(\dots) = S(\text{its slots})\,T(\text{its slots})$, with the arguments split in order. Bilinear in $S$ and $T$, associative, not commutative.

Why multilinear and not just "an array". Three reasons, all of which pay off later. (i) It is defined without reference to a basis, so any statement proved about $T$ is automatically coordinate-free. (ii) It says what you may do with a tensor — feed it vectors — which is how every physical statement using one is actually phrased. (iii) It makes the transformation law a theorem with a two-line proof, instead of an axiom that has to be remembered.

The $(1,1)$ correspondence. Given multilinear $T:V^*\times V\to\mathbb R$, define $\hat T:V\to V$ by requiring $\alpha(\hat T\vec v)=T(\alpha,\vec v)$ for all $\alpha$; this determines $\hat T\vec v$ because a vector is determined by all its pairings (that is $V^{**}=V$ again). Conversely any linear map gives such a $T$. So "$(1,1)$-tensor" and "linear map $V\to V$" are the same thing, which is why the trace and the determinant of a $(1,1)$-tensor make sense with no extra structure, while the trace of a $(0,2)$-tensor does not.

The inertia tensor is $(0,2)$ and not a linear map. Physics writes $\vec L = \mathsf I\vec\omega$ as though $\mathsf I$ were $(1,1)$. What it really does is take the $(0,2)$-tensor $\mathsf I(\vec\omega,\cdot)\in V^*$ and then raise its index with the Euclidean inner product. Harmless here; fatal in unit 32 if done silently.

Tensor product, concretely. If $\alpha,\beta\in V^*$ then $(\alpha\otimes\beta)(\vec u,\vec w)=\alpha(\vec u)\beta(\vec w)$, a $(0,2)$-tensor. Not symmetric: $(\alpha\otimes\beta)(\vec u,\vec w)\ne(\alpha\otimes\beta)(\vec w,\vec u)$ in general. Not every $(0,2)$-tensor is of this form either — those that are are called simple, and they are a measure-zero subset. The inertia tensor of a single point mass at $\vec r$ is $m(\lVert\vec r\rVert^2 g - \vec r^\flat\otimes\vec r^\flat)$, writing $g$ for the Euclidean inner product read as a $(0,2)$-tensor and $\vec r^\flat := g(\vec r,\cdot)$ for the covector it sends $\vec r$ to. The second term is simple; $g = \sum_i\varepsilon^i\otimes\varepsilon^i$ is not, since a simple tensor's array has rank at most 1 and $g$'s has rank 3. The point mass's inertia tensor has rank 2 — its array kills $\vec r$ and is $m\lVert\vec r\rVert^2$ times the identity on $\vec r^{\perp}$ — so even this smallest of bodies needs two simple tensors to write down, and no fewer.

16 · tensors

Components, and why there are $n^{p+q}$ of them

definition

Given a basis $\{\vec e_i\}$ with dual $\{\varepsilon^j\}$, the components of $T\in T^p_q(V)$ are the numbers

$$T^{i_1\dots i_p}{}_{j_1\dots j_q} := T\big(\varepsilon^{i_1},\dots,\varepsilon^{i_p},\vec e_{j_1},\dots,\vec e_{j_q}\big).$$

theorem

The $n^{p+q}$ tensors $\vec e_{i_1}\otimes\dots\otimes\varepsilon^{j_q}$ form a basis of $T^p_q(V)$, and

$$T = T^{i_1\dots}{}_{\dots j_q}\ \vec e_{i_1}\otimes\dots\otimes\varepsilon^{j_q}.$$

So $\dim T^p_q(V) = n^{p+q}$, and a tensor is determined by its components in one basis.

proof

Spanning. Call the right-hand side $S$. Evaluate $S$ on $(\varepsilon^{k_1},\dots,\vec e_{l_q})$: every factor is a Kronecker delta, so only the term with matching indices survives and $S$ agrees with $T$ there. Two multilinear maps agreeing on all tuples of basis elements agree everywhere, by expanding each argument.

Independence. If a combination vanishes, evaluate it on a tuple of basis elements: the same delta collapse leaves exactly one coefficient, which is therefore zero. $\;\blacksquare$

For the inertia tensor, $n=3$, $(p,q)=(0,2)$: nine components, reduced to six by symmetry. For the Riemann tensor of unit 33, $n=4$, $(p,q)=(1,3)$: $256$ components, reduced to $20$ by its symmetries — the same argument, run once.

"Determined by its components in one basis" is the sentence the whole unit turns on. A tensor is not the array; but an array in a stated basis does fix the tensor, and therefore fixes the array in every other basis. That is why the transformation law is forced, and why a rule that assigns arrays to bases freely cannot come from a tensor.

Expanding the arguments. If $\vec v = v^i\vec e_i$ then multilinearity gives $T(\dots,\vec v,\dots) = v^iT(\dots,\vec e_i,\dots)$, and doing this in every slot writes $T(\alpha,\dots,\vec w)$ as a sum of components times products of the arguments' components. It is the same argument that says a linear map is determined by its matrix.

Index placement. The upper indices of the components sit where the $V^*$ arguments go, and the lower indices where the $V$ arguments go. The mnemonic that survives into curved space: a component index is "up" when it pairs with a basis vector written "down", so that a repeated index summed once up and once down produces something basis-independent.

Symmetry cuts the count. A symmetric $(0,2)$-tensor in $n$ dimensions has $n(n+1)/2$ independent components — six for $n=3$, which is what makes an inertia tensor a six-number object once symmetry is used. The next slides make the symmetry itself basis-independent, which is what allows that count to mean anything.

16 · tensors

The transformation law is a theorem

theorem

Let $\vec f_a = S_{ia}\vec e_i$ be a second basis, $S$ invertible, $A = S^{-1}$. Then $\varphi^a = A_{ai}\varepsilon^i$, and for any tensor

$$T'^{a\dots}{}_{b\dots} \;=\; A_{ai}\cdots S_{jb}\cdots\ T^{i\dots}{}_{\dots j} :$$

each upper index contracts with $A = S^{-1}$, each lower with $S$.

proof

$T'^{a\dots}{}_{\dots b} = T(\varphi^a,\dots,\vec f_b,\dots)$ by the definition of components in the new basis. Substitute $\varphi^a = A_{ai}\varepsilon^i$, $\vec f_b = S_{jb}\vec e_j$, pull the scalars out by multilinearity, and what is left is $T^{i\dots}{}_{\dots j}$. $\;\blacksquare$

There is no content in the law beyond "$T$ does not know about bases".

For a $(0,2)$-tensor, $T'_{ab} = S_{ia}S_{jb}T_{ij}$: that is $T' = S^{\mathsf T}TS$.

orthonormal case

Both bases orthonormal makes $S$ orthogonal. Put $R := S^{\mathsf T} = S^{-1}$, whose rows are the new basis vectors, $R_{ai} = \vec f_a\cdot\vec e_i$:

$$\mathsf I' = R\,\mathsf I\,R^{\mathsf T} .$$

Vectors and covectors now transform by the same $R$ — which is why Euclidean physics can ignore the distinction.

Why $\varphi^a = A_{ai}\varepsilon^i$. Check the defining property: $\varphi^a(\vec f_b) = A_{ai}\varepsilon^i(S_{jb}\vec e_j) = A_{ai}S_{jb}\delta^i_j = A_{ai}S_{ib} = (AS)_{ab} = \delta_{ab}$ ✓, and uniqueness of the dual basis does the rest.

The two matrices really are different. 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. $v' = Av$. Covector components: $\alpha_a' = \alpha(\vec f_a) = S_{ia}\alpha_i$, i.e. $\alpha' = S^{\mathsf T}\alpha$. These agree only when $S^{\mathsf T} = S^{-1}$. Take $S = \mathrm{diag}(2,1,1)$ — stretch one axis — and a vector's first component halves while a covector's doubles. That is the whole covector story in one example, and it is why the distinction is invisible in this unit's orthonormal frames and unavoidable in unit 32's.

Read the boxed formula carefully. $R$ here is the change of basis, with rows $\vec f_a$. Some books use the matrix whose columns are the new basis vectors, and then the law reads $R^{\mathsf T}\mathsf IR$; some rotate the body instead of the frame, which is the inverse operation again. The formula is not the thing to memorise — the two-line proof is, since it regenerates whichever convention you are in.

Historical note. Ricci and Levi-Civita's absolute differential calculus (1900) defined tensors exactly by the transformation law, and that is how a generation of physicists, Einstein included, learnt them. The modern definition is due to the algebraists of the 1930s and 1940s; it is the same objects, with the theorem and the definition swapped, and swapping them is what makes units 31 to 35 tractable on a page rather than on twenty.

16 · tensors

Index notation, and the convention that earns its keep

the Einstein summation convention

An index repeated once up and once down in a single term is summed over $1,\dots,n$; the summation sign is not written. An index appearing once is free and must appear, in the same position, in every term of the equation.

$T'_{ab} = S_{ia}S_{jb}T_{ij}$ is nine equations, each a double sum; $L_i = I_{ij}\omega_j$, all indices down, is three — legitimate only in orthonormal frames, as the right-hand column explains.

what it buys

A wrong equation usually announces itself: if the free indices do not match on both sides, or an index is repeated three times, or twice in the same position, something is wrong before any number is computed. It is dimensional analysis for index expressions.

$\delta$ twice over

$\delta^i_j$ — the components of the identity $(1,1)$-tensor — is a tensor: $A_{ai}S_{jb}\delta^i_j = (AS)_{ab} = \delta^a_b$, the same array in every basis.

$\delta_{ij}$ — the components of the Euclidean inner product, a $(0,2)$-tensor — is a tensor too, but its array is not the same in every basis: $S^{\mathsf T}\mathbb 1 S = S^{\mathsf T}S$, which is $\mathbb 1$ only for orthogonal $S$. The identity matrix and the metric are different objects that look alike in orthonormal frames.

This unit writes all inertia indices down and keeps every basis orthonormal, so $\sum$ over a repeated pair of lower indices is safe here and only here. Unit 28 starts obeying the up–down rule strictly, and unit 32 has no choice.

Free versus dummy. In $T'_{ab} = S_{ia}S_{jb}T_{ij}$ the free indices are $a,b$ and the dummies are $i,j$. Renaming a dummy changes nothing; renaming a free index changes which equation you are looking at. A dummy may never be reused for two different sums in the same term — $S_{ia}S_{ib}T_{ii}$ is meaningless.

Why up and down, and not just "repeated". Under a general change of basis, a sum like $v^i\alpha_i$ is invariant — $A_{ai}v^i\cdot S_{jb}\alpha_j$ contracted over $a=b$ gives $(SA)_{ji}v^i\alpha_j = v^i\alpha_i$ — whereas $v^iv^i$ is not. The up–down rule is precisely the syntactic condition that guarantees the result is a tensor. In orthonormal Euclidean frames both kinds of sum survive, which is why physics books written before 1905 never distinguished them.

The computation for $\delta^i_j$, spelled out. Its transformed components are $\delta'^a_b = A_{ai}S_{jb}\delta^i_j = A_{ai}S_{ib} = (AS)_{ab} = \delta_{ab}$ since $A=S^{-1}$. So the identity tensor has the identity array in every basis — the one array that may safely be memorised. Unit 32's Kronecker delta is this one, and its metric is the other.

A habit worth forming now. When an index expression confuses you, write out one component in full with explicit sums. Three minutes of that beats an hour of staring, and every index identity in units 28 to 35 was checked that way at least once.

16 · tensors

Contraction, and the trace that survives

contraction

For $T$ of type $(p,q)$ with $p,q\ge1$, contracting the $r$th upper against the $s$th lower slot gives a tensor of type $(p-1,q-1)$ with components $\;(CT)^{\dots}{}_{\dots} = T^{\dots i\dots}{}_{\dots i\dots}$, summed.

theorem · it is a tensor

For $(1,1)$, since $SA=\mathbb 1$:

$$T'^a{}_a = A_{ai}S_{ja}T^i{}_j = T^i{}_i .$$

The trace of a $(1,1)$-tensor is a number attached to the tensor, not to a basis. $\;\blacksquare$

not for $(0,2)$

$\sum_iT_{ii}$ is not basis-independent: under $T'=S^{\mathsf T}TS$ it becomes $\operatorname{tr}(S^{\mathsf T}TS)$, equal to $\operatorname{tr}(T)$ only when $S^{\mathsf T}S=\mathbb 1$. Contraction needs one index up and one down; two down needs a metric to raise one first.

So $\operatorname{tr}\mathsf I$ is meaningful — because this unit only ever changes between orthonormal bases, which is the same as saying the metric was used silently. With $\sum_i\delta_{ii} = 3$ and $\sum_ix_ix_i = \lVert\vec r\rVert^2$,

$$\operatorname{tr}\mathsf I_O = 2\int_B\rho\,\lVert\vec r\rVert^2\,\mathrm dV$$

For the block at its corner, $0.03392\ \mathrm{kg\,m^2}$ in every orthonormal frame — a cheap and merciless check on any inertia computation.

The general contraction. Define it basis-free: on a simple tensor $\vec v\otimes\alpha\otimes(\text{rest})$, the contraction of that vector against that covector is $\alpha(\vec v)\times(\text{rest})$; extend by linearity, which is legitimate because simple tensors span. Checking in components reproduces the displayed formula, and the theorem above generalises verbatim: the $A\cdots S$ pair on the contracted indices collapses to a delta.

Why "trace of a matrix is basis-independent" is the same statement. A linear map is a $(1,1)$-tensor; its array changes by $T'=ATS = S^{-1}TS$, a similarity, and trace is invariant under similarity. A bilinear form's array changes by $T'=S^{\mathsf T}TS$, a congruence, and trace is not invariant under congruence. Two different transformation rules, two different invariants: for congruence the invariants are the rank and the signature, not the eigenvalues — Sylvester's law of inertia, cited, proved in any linear-algebra treatment of quadratic forms and used again in unit 26 to fix the Minkowski signature.

Then how can §4 talk about the eigenvalues of $\mathsf I$? Only because the bases are orthonormal: there congruence and similarity coincide, and the eigenvalues of the array are genuine invariants of the pair (tensor, inner product). The principal moments are properties of $\mathsf I$ and the Euclidean metric together. In unit 32 exactly this is what $g^{\mu\alpha}T_{\alpha\nu}$ does: raise one index so a trace can be taken.

The number. $2\int\rho\lVert\vec r\rVert^2\mathrm dV$ for the block about a corner is $\frac{2m}{3}(a^2+b^2+c^2) = 0.4\times0.0848 = 0.03392$ ✓, matching the sum of the diagonal entries $0.00544+0.01184+0.01664$.

16 · tensors

Symmetric and antisymmetric parts

definition

For a $(0,2)$-tensor, $\ T_{(ij)} := \tfrac12(T_{ij}+T_{ji})$ is the symmetric part and $\ T_{[ij]} := \tfrac12(T_{ij}-T_{ji})$ the antisymmetric part, with $T = T_{(ij)}+T_{[ij]}$.

theorem

The splitting is basis-independent: if $T$ is symmetric in one basis it is symmetric in all.

Why: $T'_{ba} = S_{ib}S_{ja}T_{ij} = S_{ja}S_{ib}T_{ij} = S_{ia}S_{jb}T_{ji}$ after renaming dummies, which is $T'_{ab}$ exactly when $T_{ji}=T_{ij}$. Or without indices: $(S^{\mathsf T}TS)^{\mathsf T} = S^{\mathsf T}T^{\mathsf T}S$.

Dimension count: $\tfrac{n(n+1)}2 + \tfrac{n(n-1)}2 = n^2$. In $n=3$: $6+3$. $\mathsf I$ lives entirely in the first summand.

not for $(1,1)$

"$T^i{}_j$ is symmetric" is meaningless as a statement about a tensor: the law is $T' = S^{-1}TS$, and a similarity does not preserve symmetry. Take

$$T = \begin{pmatrix}0&1\\1&0\end{pmatrix},\ \ S=\begin{pmatrix}1&0\\0&2\end{pmatrix}$$ $$\Rightarrow\ T' = \begin{pmatrix}0&2\\ \tfrac12&0\end{pmatrix}.$$

Antisymmetry of a $(0,2)$-tensor, by contrast, is intrinsic — which is why the field tensor of unit 28 is an antisymmetric $(0,2)$-tensor, and why $\Omega$ is honestly antisymmetric only once an index is lowered with the metric.

Unit 34 builds the whole exterior algebra out of the antisymmetric summand; unit 32's metric and unit 22's stress tensor live in the symmetric one, alongside $\mathsf I$.

Check the counterexample. $S^{-1} = \mathrm{diag}(1,\tfrac12)$, so $S^{-1}TS = \begin{pmatrix}1&0\\0&\frac12\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix} \begin{pmatrix}1&0\\0&2\end{pmatrix} = \begin{pmatrix}0&1\\ \frac12&0\end{pmatrix} \begin{pmatrix}1&0\\0&2\end{pmatrix} = \begin{pmatrix}0&2\\ \frac12&0\end{pmatrix}$, not symmetric ✓. Note the change of basis is not orthogonal; restricted to orthogonal $S$ the symmetry of a $(1,1)$ array is preserved, which is another face of the same coincidence that lets this unit be careless.

Why $\mathsf I$ is symmetric. Directly from its definition: the bracket $(\vec u\cdot\vec w)\lVert\vec r\rVert^2-(\vec u\cdot\vec r)(\vec w\cdot\vec r)$ is unchanged by swapping $\vec u$ and $\vec w$. It is not a property of a clever choice of basis, and §4 needs that: the spectral theorem applies to symmetric bilinear forms, and if symmetry were basis-dependent the principal axes would be too.

The general pattern. For $(0,k)$-tensors the group of permutations of the $k$ slots acts, and the decomposition into irreducible pieces under that action is what "symmetry type" means. For $k=2$ there are exactly two pieces. For $k=3$ there are more than two, which is why the Riemann tensor of unit 33 needs its symmetries listed rather than named.

16 · tensors · example 2

The law, verified component by component

worked example · the block at a corner, rotated $45^\circ$ about $\vec e_3$

New basis $\vec f_1 = \tfrac1{\sqrt2}(\vec e_1+\vec e_2)$, $\vec f_2 = \tfrac1{\sqrt2}(-\vec e_1+\vec e_2)$, $\vec f_3=\vec e_3$, so $R$ is the $45^\circ$ rotation. Route A, the law:

$$I'_{11} = \tfrac12\big(I_{11}+2I_{12}+I_{22}\big)$$ $$= \tfrac12(0.00544-0.01152+0.01184),$$

which is $0.00288$. Likewise $I'_{12} = \tfrac12(I_{22}-I_{11}) = 0.00320$, $I'_{22} = \tfrac12(I_{11}-2I_{12}+I_{22}) = 0.01440$, $I'_{13} = \tfrac1{\sqrt2}(I_{13}+I_{23}) = -0.00169706$, $I'_{23} = \tfrac1{\sqrt2}(I_{23}-I_{13}) = 0.00033941$, $I'_{33} = 0.01664$.

Route B, the definition — integrate again in the new coordinates $x' = \tfrac{x+y}{\sqrt2},\ y' = \tfrac{-x+y}{\sqrt2}$, over the same block:

$$I'_{11} = \int_B\rho\,(y'^2+z^2)\,\mathrm dV,$$ $$y'^2 = \tfrac12(x^2-2xy+y^2).$$

With $\int\rho x^2 = \tfrac m3a^2$ and $\int\rho xy = \tfrac m4ab$ at the corner:

$$I'_{11} = \tfrac m6(a^2+b^2)-\tfrac m4\,ab+\tfrac m3c^2$$ $$= 0.00832-0.00576+0.00032 = 0.00288 .$$

Same number, from an integral that never mentioned $R$. All six agree.

checks

Trace: $0.00288+0.01440+0.01664 = 0.03392$, unchanged ✓. Symmetry preserved ✓. Special case $a=b$: $I'_{12}=I'_{23}=0$, so $\vec f_2$ — the direction the mirror $x\leftrightarrow y$ reverses — is principal ✓; $\vec f_1$ and $\vec e_3$ lie in the mirror, and $I'_{13} = -\tfrac{m}{2\sqrt2}ac$ survives.

The point of doing it twice. Route B is what "the tensor does not know about bases" means operationally: $\mathsf I(\vec f_1,\vec f_1)$ is an integral over the body with a fixed pair of input vectors, and it can be evaluated without ever writing a transformation law. Route A is the law. They agree because the law is a theorem about Route B, not an extra rule. If you ever doubt an index gymnastics result, Route B is always available.

Route A in full, by the two-line proof. $I'_{ab} = R_{ai}R_{bj}I_{ij}$. With $c=s=1/\sqrt2$ and $R = \begin{pmatrix}c&s&0\\-s&c&0\\0&0&1\end{pmatrix}$: $I'_{11}=c^2I_{11}+2csI_{12}+s^2I_{22}$; $I'_{12}=-csI_{11}+(c^2-s^2)I_{12}+csI_{22}$, and $c^2-s^2=0$ at $45^\circ$, which is why the middle term drops and $I'_{12}=\frac12(I_{22}-I_{11})$; $I'_{13}=cI_{13}+sI_{23}$; $I'_{23}=-sI_{13}+cI_{23}$; $I'_{33}=I_{33}$.

Route B's three integrals. Over $[0,a]\times[0,b]\times[0,c]$ with $\rho = m/abc$: $\int\rho x^2 = \rho bc\frac{a^3}{3} = \frac m3a^2$; $\int\rho xy = \rho c\frac{a^2}{2}\frac{b^2}{2} = \frac m4ab$. Then $\int\rho y'^2 = \frac12\big(\frac m3a^2 - 2\cdot\frac m4ab + \frac m3b^2\big) = \frac m6(a^2+b^2)-\frac m4ab$. Numerically $0.1\times0.0832 = 0.00832$ and $0.15\times0.0384 = 0.00576$, and $\int\rho z^2 = \frac m3c^2 = 0.00032$. Total $0.00288$ ✓.

The remaining five, by Route B, are a good hour on paper and are the first problem of the sheet in a smaller case. The two that are easiest to get wrong are $I'_{13}$ and $I'_{23}$, because the $1/\sqrt2$ appears once, not twice.

16 · tensors

Nine numbers per basis that are not a tensor

Define a rule $N$: in each orthonormal basis, return the array $\mathrm{diag}(I_{11},I_{22},I_{33})$ — "the moments of inertia about the three coordinate axes". Nine numbers per basis, computed from the body by an honest integral, one array for every basis. It is not a tensor.

the falsification, on the block

In $\{\vec e_i\}$: $N = \mathrm{diag}(0.00544,\ 0.01184,\ 0.01664)$.

Transport it by the law: $\big(RNR^{\mathsf T}\big)_{11} = \tfrac12(0.00544+0.01184) = \mathbf{0.00864}$.

Apply the rule in $\{\vec f_a\}$: $N'_{11} = I'_{11} = \mathbf{0.00288}$.

A factor of three apart. One rule, two answers, so no tensor produces them.

what exactly fails

A rule assigning an array to each basis comes from a tensor iff it obeys the transformation law.

Why ($\Leftarrow$): define $T$ by the array in one basis via the component formula; in any other basis its components are given by the law, hence equal the rule's array. ($\Rightarrow$) is the previous theorem.

So the one-sentence answer: the numbers are fine, the rule is not equivariant — it does not commute with change of basis, so there is no single multilinear map behind it. $N$ discards $I_{12}$, and $I_{12}$ is exactly the information that $I'_{11}$ needs.

Why this matters physically. Each entry of $N$ is a real, measurable moment of inertia: hang the block from each axis in turn and time the oscillations. The numbers are not wrong. What is wrong is treating the triple as a description of the body's rotational inertia, because it cannot answer "what is the moment about this axis" for an axis you did not measure. The tensor can, from six numbers: $\mathsf I(\hat n,\hat n) = I_{ij}n_in_j$.

A second non-tensor, in the same spirit: the rule $M_{ij} = |v_i|\,|v_j|$ built from a fixed vector $\vec v$. Without the bars, $v_iv_j$ are the components of the tensor $\vec v^\flat\otimes\vec v^\flat$; the bars throw away signs, and which signs they throw away depends on the basis, so the rule cannot commute with a rotation that mixes positive and negative components. A third: let the rule return the identity array in every basis, general ones included, read as the components of a bilinear form. Read as a $(1,1)$-tensor it is a tensor — $\delta^i_j$; read as $(0,2)$ it is not, since the law demands $S^{\mathsf T}S$. That is the $\delta_{ij}$ trap of the index-notation slide, and it is the reason "the metric is $\delta_{ij}$" is a statement about a choice of basis, not about space.

The quotient theorem, which the problem sheet touches: if $N_{ij}v_j$ are the components of a vector for every vector $\vec v$, then $N_{ij}$ are the components of a tensor. That is the practical converse used in continuum mechanics to prove Cauchy's stress tensor is a tensor (unit 22) rather than assuming it.

The numbers. $(RNR^{\mathsf T})_{11} = c^2N_{11}+2csN_{12}+s^2N_{22}$ with $N_{12}=0$ and $c^2=s^2=\frac12$: $\frac12(0.00544+0.01184) = \frac12(0.01728) = 0.00864$. Against $I'_{11}=0.00288$ from the previous slide. The discrepancy is $-I_{12} = +0.00576$, precisely the term $N$ threw away.

16 · tensors

What the diagonal rule throws away

A moment of inertia about an axis measures how much mass lies far from it. The $45^\circ$ axis $\vec f_1$ through the corner cuts across the middle of the cover, so most of the book is close to it: $I'_{11} = 0.00288$.

The rule $N$ kept only the moments about $\vec e_1$ and $\vec e_2$, and carried to $\vec f_1$ the law can only average them: $0.00864$. What is missing is $I_{12} = -\int\rho\,xy\,\mathrm dV$, which is large because all the mass sits in the quadrant between $\vec e_1$ and $\vec e_2$ — exactly where $\vec f_1$ points.

The far corner's distance to $\vec f_1$ (dotted) is $(a-b)/\sqrt2 = 0.057$ m; to the $\vec e_2$ axis it is $a = 0.24$ m.

The cover seen from above, corner $O$ at the origin. The moments about four axes through $O$, in $\mathrm{kg\,m^2}$: the tensor gives $0.00288$ about $\vec f_1$; the diagonal rule, transported, claims three times as much.

The same numbers, read off the picture. For a unit vector $\hat n$ in the cover plane, $\mathsf I_O(\hat n,\hat n) = I_{11}n_1^2 + 2I_{12}n_1n_2 + I_{22}n_2^2$. With $\hat n = \vec f_1 = (1,1,0)/\sqrt2$ that is $\tfrac12(0.00544 - 0.01152 + 0.01184) = 0.00288$; with $\hat n = \vec f_2 = (-1,1,0)/\sqrt2$ the cross term changes sign, $\tfrac12(0.00544+0.01152+0.01184) = 0.01440$. The cross term is what tells the two diagonals apart, and the diagonal rule has nothing that could: it would give $0.00864$ for both, though one axis runs through the book and the other points away from it.

Why "the mass between the axes" makes $I_{12}$ negative. $I_{12} = -\int\rho\,xy$, and from the corner every point of the block has $x\ge0$, $y\ge0$. Put the origin at the centre instead and the four quadrants cancel — which is why the central array is diagonal and the corner array is not.

section 4

Principal axes

Three directions in every body where $\vec L$ and $\vec\omega$ finally agree.

16 · principal axes

The spectral theorem, proved

theorem

Let $B$ be a symmetric bilinear form on a Euclidean space $V$ of dimension $n$. Then $V$ has an orthonormal basis $\{\vec u_k\}$ and reals $\lambda_1\le\dots\le\lambda_n$ with $B(\vec u_j,\vec u_k) = \lambda_k\delta_{jk}$: the array of $B$ is diagonal in that basis.

proof · induction on $n$

$Q(\vec u) = B(\vec u,\vec u)$ is continuous on the unit sphere $S$, which is compact, so it attains a maximum at some $\vec u_n\in S$. By Lagrange multipliers (unit 04) applied to $Q$ under $\lVert\vec u\rVert^2=1$, at that point $\nabla Q = \lambda\nabla g$, i.e. $2B_{ij}u_j = 2\lambda u_i$: $\vec u_n$ is an eigenvector with eigenvalue $\lambda=Q(\vec u_n)$.

Let $W = \vec u_n^{\perp}$. For $\vec w\in W$, $B(\vec u_n,\vec w) = \lambda\,\vec u_n\cdot\vec w = 0$, so $B$ restricts to a symmetric form on $W$, of dimension $n-1$. Apply the hypothesis to $W$. $\;\blacksquare$

Why the theorem is about a form and a metric together. "Diagonal in an orthonormal basis" uses both $B$ and the inner product; the previous section explained why. Without a metric the right statement is Sylvester's law of inertia — a symmetric form can be brought to $\mathrm{diag}(1,\dots,1,-1,\dots,-1,0,\dots)$ by some basis, and only the signature is invariant. Unit 26 uses exactly that version to define Minkowski signature, where no orthonormal basis in the Euclidean sense exists.

Uniqueness. The eigenvalues are unique; the eigenvectors are not, when eigenvalues repeat. If $\lambda_1=\lambda_2$ every vector in their plane is an eigenvector, and the body has a continuum of principal axes there. That is the symmetric top of §6.

Alternative proof, the one most books give: the characteristic polynomial has a complex root $\lambda$ with eigenvector $\vec z$, and $\bar{\vec z}^{\mathsf T}B\vec z$ computed two ways forces $\lambda=\bar\lambda$. It is shorter but tells you nothing about what the eigenvalues are; the extremal proof hands you the variational characterisation, and that is what the problem sheet's Rayleigh-style bound uses.

16 · principal axes

Why the maximum is an eigenvector

The proof's first step, drawn in the plane. On the unit circle, $Q$ is largest where the level curve $Q = \lambda_{\max}$ just touches the circle from outside. There the two curves share a normal, so $\nabla Q = 2B\vec u$ is parallel to $\vec u$: $B\vec u = \lambda\vec u$.

Two things the proof gives free of charge:

the eigenvalues are real and the eigenvectors orthogonal, without ever touching a characteristic polynomial or a complex number;
$\lambda_{\max} = \max_{\lVert\vec u\rVert=1}Q(\vec u)$ and $\lambda_{\min}=\min$ — a variational characterisation, which is how one bounds a principal moment without diagonalising anything.

The same theorem, in the guise of the generalised eigenvalue problem $K\vec v = \omega^2M\vec v$, is what produced normal modes in unit 07. Here the second form is the Euclidean inner product, and the eigenvalues are moments of inertia.

Level curves of $Q = 3u_1^2 + 2u_1u_2 + 2u_2^2$ on the unit circle. Where $\nabla Q$ has a part along the circle, $Q$ still grows; only where it is radial — the eigenvectors — can $Q$ be extreme.

What the figure is. The form has array $\begin{pmatrix}3&1\\1&2\end{pmatrix}$, chosen only for legibility; the book's arrays have eigenvalues too far apart to draw on one page. Its eigenvalues are $(5\pm\sqrt5)/2 = 3.618,\ 1.382$, with the top eigenvector at $31.7^\circ$ to the first axis. The ellipse $Q=\lambda_{\max}$ has semi-axes $1$ and $\sqrt{\lambda_{\max}/\lambda_{\min}} = 1.618$, so it touches the circle at $\pm\vec u_{\max}$ and lies outside it elsewhere; $Q=\lambda_{\min}$ lies inside and touches at $\pm\vec u_{\min}$. The dashed curve is the level through a generic point, which crosses the circle instead of touching it.

The Lagrange-multiplier step, carefully. $Q(\vec u) = B_{ij}u_iu_j$ with $B_{ij}=B_{ji}$, so $\partial Q/\partial u_i = 2B_{ij}u_j$; the constraint $g=\lVert\vec u\rVert^2$ has $\partial g/\partial u_i = 2u_i$. The constraint's gradient never vanishes on $S$, so the multiplier theorem of unit 04 applies at the maximiser. Symmetry of $B$ is used exactly once, in differentiating $Q$ — and without it the conclusion is false.

16 · principal axes

Principal axes, and the ellipsoid

definition

The eigenvectors of $\mathsf I_O$ are the principal axes at $O$; the eigenvalues $I_1\le I_2\le I_3$ are the principal moments. In a principal basis $\ \vec L = (I_1\omega_1,I_2\omega_2,I_3\omega_3)$ and $T = \tfrac12\sum_kI_k\omega_k^2$.

$\vec L\parallel\vec\omega$ exactly when $\vec\omega$ is along a principal axis. Every body has such an orthonormal triple about every point — and a whole plane or sphere of principal axes where the moments coincide.

the triangle inequality

$I_1+I_2\ge I_3$, with equality iff the body is a lamina in the $1$–$2$ plane.

Why: $I_1+I_2-I_3 = \int\rho\big[(x_2^2+x_3^2)+(x_1^2+x_3^2)-(x_1^2+x_2^2)\big] = 2\int\rho\,x_3^2\ \ge 0$.

the inertia ellipsoid

$\mathcal E = \{\vec u:\mathsf I_O(\vec u,\vec u)=1\}$ — an ellipsoid because $\mathsf I_O$ is positive definite, with semi-axes $1/\sqrt{I_k}$ along the principal axes. It is the tensor, drawn.

Fattest semi-axis $\leftrightarrow$ smallest moment: the block's ellipsoid is longest along the spine.

check on the block

$I_1+I_2-I_3 = 0.00136+0.00296-0.00416 = 0.00016\ \mathrm{kg\,m^2}$, and $2\int\rho z^2 = \tfrac m6c^2 = 0.1\times0.0016 = 0.00016$ ✓. As $c\to0$ the block becomes a lamina and the slack closes.

The lamina case is the perpendicular-axis theorem: for a flat body in the $1$–$2$ plane, $I_3 = I_1+I_2$. It has no three-dimensional analogue, and it is on the problem sheet. Note it holds only for the axis perpendicular to the lamina and only for the lamina's own plane.

Reading the ellipsoid. For a unit vector $\hat n$, the moment about that axis is $\mathsf I(\hat n,\hat n)$, and the ellipsoid's radius in direction $\hat n$ is $1/\sqrt{\mathsf I(\hat n,\hat n)}$. So directions where the ellipsoid is long are directions that are easy to spin. Poinsot's construction in §5 puts this ellipsoid to work: the free motion is the ellipsoid rolling without slipping on a fixed plane.

Why the eigenvalues deserve the name "moments". $I_k = \mathsf I(\vec u_k,\vec u_k)$ is literally the moment of inertia about the $k$th principal axis, so the spectral theorem says: among all axes through $O$, the largest and smallest moments are attained along principal axes, and the intermediate one is a saddle of $\hat n\mapsto\mathsf I(\hat n,\hat n)$ on the sphere. That saddle is the whole of §5.

Limit check on the triangle inequality. For a sphere $I_1=I_2=I_3$ and the inequality is strict ($2I\gt I$). For a rod along axis 3, $I_3=0$ and $I_1=I_2$, so $I_1+I_2 = 2I_1 \gt 0 = I_3$: satisfied, and the "ellipsoid" degenerates to an infinite cylinder, since $\mathsf I$ is only semi-definite there.

16 · principal axes

The block's inertia ellipsoid, drawn

The block's ellipsoid at its centre has semi-axes $1/\sqrt{I_k}$, in $\mathrm{kg^{-1/2}\,m^{-1}}$: $27.1$ along the spine, $18.4$ across the cover, $15.5$ through it. Its section in the plane of the cover is drawn. Long where the moment is small: the ellipsoid is stretched along the spine, as the book is.

Any other axis through the centre is read off the same way: its moment is $1/r^2$, with $r$ the radius in that direction. In the direction $\hat n$ drawn, $38^\circ$ from the spine, $r = 22.6$ and $\mathsf I(\hat n,\hat n) = 0.00197\ \mathrm{kg\,m^2}$.

The ellipsoid $\mathsf I(\vec u,\vec u)=1$ cut by the plane of the cover, over the cover's outline (drawn at its own scale). Neither is round, and both are longest along the spine.

The numbers. $1/\sqrt{0.00136} = 27.12$, $1/\sqrt{0.00296} = 18.38$, $1/\sqrt{0.00416} = 15.50$. For $\hat n = (\cos38^\circ,\sin38^\circ,0)$, $\mathsf I(\hat n,\hat n) = 0.00136\cos^238^\circ + 0.00296\sin^238^\circ = 0.001966$, and $1/\sqrt{0.001966} = 22.55$: the point $r\hat n$ lies on the ellipse, as the definition says it must.

Why the ellipsoid copies the body's shape, roughly. A long direction of the body is one with little mass far from the axis along it, so a small moment, so a long semi-axis. The match is not exact — the ellipsoid's axis ratio here is $27.1/18.4 = 1.48$ against the cover's $0.24/0.16 = 1.5$ — and for a thin plate it fails badly, since $I_1+I_2\ge I_3$ stops the ellipsoid from being as flat as the plate. It is a picture of the second moments, not of the body.

16 · principal axes · example 3

Diagonalising the corner

worked example · $\mathsf I_O$ at the block's corner

The array is symmetric with every off-diagonal entry nonzero, so the corner frame is not principal. Its characteristic polynomial is a real cubic; solving it numerically:

$I_k\ (\mathrm{kg\,m^2})$	principal axis $\vec u_k$ in the corner frame
0.0018489	$(0.85562,\ 0.50443,\ 0.11604)$
0.0152235	$(-0.51290,\ 0.85642,\ 0.05901)$
0.0168476	$(-0.06961,\ -0.11000,\ 0.99149)$

The easiest axis through the corner points nearly, but not exactly, along the body diagonal $(a,b,c)/\lVert\cdot\rVert = (0.82416,0.54944,0.13736)$: the angle between them is $\arccos(0.99826) = 3.376^\circ$.

Why not exactly? The diagonal is a symmetry axis only for a cube. Set $a=b=c$: then $I_{ii}=\tfrac23ma^2$, $I_{ij}=-\tfrac14ma^2$, and

$$\mathsf I_O\,\vec n = \big(\tfrac23-\tfrac14-\tfrac14\big)ma^2\,\vec n$$

for $\vec n=(1,1,1)$: the diagonal is exactly an eigenvector, with moment $\tfrac16ma^2$ — the smallest, as the near-alignment suggested.

checks

Trace: $0.0018489+0.0152235+0.0168476 = 0.033920$, the invariant computed twice already ✓. Positivity: all three $\gt0$, as required for a body that is not a line ✓. Triangle: $0.0018489+0.0152235-0.0168476 = 0.0002248\gt0$ ✓. Order of magnitude: the largest, $0.01685$, exceeds the central $I_3=0.00416$ by roughly $M\lVert\vec d\rVert^2$ projected — Steiner, again ✓.

How the numbers were obtained. The characteristic polynomial of the $3\times3$ array was solved numerically; a symmetric eigenvalue routine gives the same to twelve digits. Doing it by hand is an afternoon and not instructive; what is instructive is the two checks — trace and positivity — which catch almost every arithmetic slip, and both pass here.

The cube computation, in full. With $a=b=c$ each row of $\mathsf I_O$ reads $(\frac23,-\frac14,-\frac14)ma^2$ up to cyclic order, so acting on $(1,1,1)$ gives $(\frac23-\frac14-\frac14)ma^2 = \frac16ma^2$ in each slot. The other two eigenvalues are equal, by the symmetry argument of the next slide, and equal $\frac23ma^2+\frac14ma^2 = \frac{11}{12}ma^2$ — check the trace: $\frac16+2\cdot\frac{11}{12} = \frac16+\frac{11}6 = 2$, against $\operatorname{tr} = 3\cdot\frac23 = 2$ ✓.

Why $3.4^\circ$ and not more. The corner tensor is dominated by the Steiner term $M(\lVert\vec d\rVert^2\delta-\vec d\otimes\vec d)$, whose smallest eigenvector is exactly $\vec d$, i.e. the body diagonal, with eigenvalue 0. The central tensor perturbs it, and the perturbation is small because $M\lVert\vec d\rVert^2 = 0.0127\ \mathrm{kg\,m^2}$ is several times the central moments. The tilt measures how far the block is from a cube.

16 · principal axes

Symmetry forces degeneracy

theorem

Suppose the body is carried onto itself, with density preserved, by the rotation $C$ through $2\pi/n$ about a unit vector $\hat k$ through $O$. Then $\hat k$ is a principal axis; and if $n\ge3$, the other two principal moments are equal.

proof

Change of variables (unit 05) with $|\det C|=1$ gives $\mathsf I_O(C\vec u,C\vec w) = \mathsf I_O(\vec u,\vec w)$: the form is $C$-invariant. Hence the array $M$ of $\mathsf I_O$ satisfies $C^{\mathsf T}MC = M$, i.e. $MC = CM$.

$C$ fixes $\hat k$, so $M\hat k$ is fixed by $C$ too; the fixed space of $C$ is the line $\mathbb R\hat k$ (for $n\ge2$), so $M\hat k\parallel\hat k$: principal.

On $\hat k^\perp$, $M$ commutes with the plane rotation by $2\pi/n$. Distinct eigenvalues would give two eigenlines, each carried into itself by it — impossible for $n\ge3$, as the notes check. $\;\blacksquare$

consequences

Cylinder, cone, square prism about their axis: $I_1=I_2$, a symmetric top, and every axis perpendicular to $\hat k$ is principal.

Cube about its centre: three four-fold axes force $I_1=I_2=I_3=\tfrac16ma^2$, so $\mathsf I = \tfrac16ma^2\,\delta$ and every direction is principal. A cube has the inertia tensor of a sphere.

The block has only two-fold axes, so nothing is forced: $I_1\lt I_2\lt I_3$, all distinct — which is why it is the interesting case.

The change of variables. $\mathsf I_O(C\vec u,C\vec w) = \int_B\rho(\vec r) [(C\vec u\cdot C\vec w)\lVert\vec r\rVert^2-(C\vec u\cdot\vec r)(C\vec w\cdot\vec r)]\mathrm dV$. Substitute $\vec r = C\vec s$; $C$ orthogonal preserves dot products and $\lVert\cdot\rVert$, the Jacobian is 1, and the symmetry hypothesis says $\rho(C\vec s)=\rho(\vec s)$ with $C(B)=B$. What is left is $\mathsf I_O(\vec u,\vec w)$.

The cube, carefully. Take the three four-fold axes through opposite face centres. The first forces the other two moments equal; the second forces another pair equal; together all three are equal. Then $\mathsf I = \lambda\delta$, and the trace gives $3\lambda = 2\int\rho\lVert\vec r\rVert^2 = 3\cdot\frac{ma^2}{6}\cdot\ldots$ — more simply, $\lambda = \frac m{12}(a^2+a^2) = \frac16ma^2$ from example 1 ✓. A cube therefore spins stably about every axis, has $\vec L\parallel\vec\omega$ always, and never flips. Unit 16's whole drama needs an asymmetric body.

The $n\ge3$ step, in full. Because $MC=CM$, $C$ maps every eigenspace of $M$ into itself: if $M\vec v = \mu\vec v$ then $M(C\vec v) = CM\vec v = \mu\,C\vec v$. Suppose the two eigenvalues on $\hat k^\perp$ differ. Then each of the two eigenlines is carried into itself by $C$, so $C$ sends some nonzero $\vec v\in\hat k^\perp$ to $\pm\vec v$. A plane rotation by $2\pi/n$ does that only if $2\pi/n$ is a multiple of $\pi$, i.e. $n\le2$. So for $n\ge3$ the two eigenvalues are equal.

Why $n=2$ is not enough. A two-fold axis gives $C=\mathrm{diag}(-1,-1,1)$, which commutes with any matrix diagonal in that frame; it forces $\hat k$ principal but says nothing about the other two. The block has exactly three two-fold axes, and gets exactly that: a principal frame, with three unrelated moments.

The converse is false. Equal moments do not imply symmetry of the body. One can build a deliberately lumpy body with $\mathsf I\propto\delta$; it is sometimes called an inertially spherical body, and it shows that $\mathsf I$ remembers only the second moments of the mass distribution — everything the multipole expansion of unit 10 called higher order is invisible to it.

section 5

Euler's equations, and the axis that will not hold

Three coupled nonlinear ODEs, and a prediction you can test by throwing a book.

16 · Euler

Euler's equations

Angular momentum obeys $\big(\tfrac{\mathrm d\vec L}{\mathrm dt}\big)_{\text{space}}=\vec\tau$ about the centre of mass or a fixed pivot. Unit 03's transport theorem converts it to the body frame, where $\mathsf I$ is constant:

$$\Big(\tfrac{\mathrm d\vec L}{\mathrm dt}\Big)_{\text{body}}+\vec\omega\times\vec L=\vec\tau .$$

Take the body frame principal, so $L_k=I_k\omega_k$ with $I_k$ constants:

Euler's equations (1758) $$\begin{aligned}I_1\dot\omega_1&=(I_2-I_3)\,\omega_2\omega_3+\tau_1\\ I_2\dot\omega_2&=(I_3-I_1)\,\omega_3\omega_1+\tau_2\\ I_3\dot\omega_3&=(I_1-I_2)\,\omega_1\omega_2+\tau_3\end{aligned}$$

what the body frame costs

These are equations for the components of $\vec\omega$ in a frame that is itself rotating with the body. Solving them gives $\vec\omega(t)$ in the body; to get the body's orientation in space you must then integrate $\dot R = R\Omega$, which is §6's business and is the harder half.

Why bother: in the space frame $\mathsf I$ changes with time, and $\vec L = \mathsf I(t)\vec\omega$ is useless. The quadratic terms are the price of a constant $\mathsf I$, and they are also where all the physics is — set $I_1=I_2=I_3$ and they vanish.

Dimensions: each term is $\mathrm{kg\,m^2\,s^{-2}} = \mathrm{N\,m}$ ✓. Limit $\mathsf I\propto\delta$: $\dot{\vec\omega}=\vec\tau/I$, the rotational $\vec F=m\vec a$ ✓.

Full derivation. $\vec L$ is one covector; its space-frame and body-frame component arrays are related by $R$. The transport theorem (unit 03) says that for any vector-valued function, $\left(\frac{\mathrm d}{\mathrm dt}\right)_{\text{space}}$ applied to it equals $\left(\frac{\mathrm d}{\mathrm dt}\right)_{\text{body}}+\vec\omega\times$ acting on its body components. Applied to $\vec L$ and set equal to $\vec\tau$ this gives the displayed equation. In the principal body frame $L_k = I_k\omega_k$ with $I_k$ time-independent — that is the whole reason for working there — so $(\mathrm d\vec L/\mathrm dt)_{\text{body}}$ has components $I_k\dot\omega_k$, and the cross product $\vec\omega\times\vec L$ has first component $\omega_2L_3-\omega_3L_2 = (I_3-I_2)\omega_2\omega_3$. Move it to the right.

The same equations from unit 12. They are the Euler–Lagrange equations of $L = \frac12\sum I_k\omega_k^2 - V$ on the configuration space $SO(3)$, with the $\omega_k$ written in Euler angles. Deriving them that way is two pages and produces the same three lines, which is a good demonstration that the Lagrangian formalism does not always shorten the work — it shortens the thinking, by never requiring a constraint force.

Torque about which point. The same two points as König: the centre of mass, or a point of the body fixed in space. About any other moving point there are extra terms. The free top below uses the centre of mass, the heavy top of §6 uses the pivot.

History. Euler presented these to the Berlin Academy in 1758 (printed 1765), nearly a century before Hamilton coined the word "tensor" (1846, in a different sense) and seventy years before Cauchy proved (1829) that a real symmetric matrix has real eigenvalues and orthogonal eigenvectors. The principal axes themselves had been found by Segner in 1755, as three perpendicular axes about which the products of inertia vanish.

16 · Euler

The free top: two surfaces, one curve

With $\vec\tau=0$, two quantities are conserved, both immediately from the equations:

$$2T=\sum_kI_k\omega_k^2,\qquad \lVert\vec L\rVert^2=\sum_kI_k^2\omega_k^2 .$$

Why: multiply the $k$th equation by $\omega_k$ and add — the right-hand side telescopes to zero. Multiply by $I_k\omega_k$ and add — same.

In $\vec\omega$-space these are two ellipsoids, and the motion runs along their intersection — generically a closed curve, the polhode.

Poinsot, 1834

Equivalently: the inertia ellipsoid, carried by the body, rolls without slipping on a plane fixed in space and perpendicular to $\vec L$. The polhode is its track on the ellipsoid.

Polhodes on the energy ellipsoid. Closed loops encircle axes 1 and 3; the two curves through axis 2 cross there — a saddle, and the whole story of the next three slides.

The two conservation laws, in full. Multiply Euler's first equation by $\omega_1$: $I_1\omega_1\dot\omega_1 = (I_2-I_3)\omega_1\omega_2\omega_3$. Do the same for the other two and add: the right-hand side is $[(I_2-I_3)+(I_3-I_1)+(I_1-I_2)]\omega_1\omega_2\omega_3 = 0$, and the left is $\frac{\mathrm d}{\mathrm dt}\frac12\sum I_k\omega_k^2$. Multiplying instead by $I_k\omega_k$ gives $\sum I_k^2\omega_k\dot\omega_k$ on the left and $[I_1(I_2-I_3)+I_2(I_3-I_1)+I_3(I_1-I_2)]\omega_1\omega_2\omega_3 = 0$ on the right.

Why an intersection of two ellipsoids is generically a closed curve. Both are compact; their intersection is compact, and for generic values it is a one-dimensional manifold, hence a finite union of circles. The exceptional values are those where the surfaces are tangent — which happens exactly at the principal axes, and at $2TI_2=\lVert\vec L\rVert^2$, where the intersection degenerates into two crossing curves through $\pm\vec u_2$. Those are the separatrices in the figure.

Poinsot's construction, stated precisely (cited: Arnold §29). Let $\vec\rho = \vec\omega/\sqrt{2T}$, so $\mathsf I(\vec\rho,\vec\rho)=1$: $\vec\rho$ lies on the inertia ellipsoid. The outward normal there is $\nabla_{\vec\rho}\mathsf I(\vec\rho,\vec\rho) = 2\mathsf I\vec\rho\propto\vec L$, fixed in space, and $\vec\rho\cdot\vec L = 2T/\sqrt{2T}$ is constant — so the tangent plane at the contact point is fixed. The contact point has velocity $\vec\omega\times\vec\rho = 0$, so the rolling is without slipping.

The name. Polhode, from Greek for "pole path"; its track on the fixed plane is the herpolhode. The Earth's polhode is the Chandler wobble, §6.

16 · Euler

Linearising about a principal axis

$\vec\omega = \Omega\vec u_2 + \vec v$ with $\lVert\vec v\rVert\ll\Omega$ — steady spin about axis 2, slightly disturbed. (From here on $\Omega$ is a spin rate, a number in $\mathrm{s^{-1}}$, not the matrix $R^{\mathsf T}\dot R$ of §1.) To first order in $\vec v$, the second Euler equation gives $I_2\dot v_2 = (I_3-I_1)v_3v_1 = O(v^2)$, so $\omega_2=\Omega$ is constant at this order, and the other two decouple from it:

$$I_1\dot v_1=(I_2-I_3)\,\Omega\,v_3,\qquad I_3\dot v_3=(I_1-I_2)\,\Omega\,v_1 .$$

Differentiate the first and substitute the second — a linear constant-coefficient system, unit 06:

$$\ddot v_1 = \frac{(I_2-I_3)(I_1-I_2)}{I_1I_3}\,\Omega^2\,v_1 \;=\; \frac{(I_3-I_2)(I_2-I_1)}{I_1I_3}\,\Omega^2\,v_1 .$$

the sign decides everything

Spin about the axis with moment $I_\alpha$ and call the other two $I_\beta,I_\gamma$. The same two lines give $\ddot v_\beta=\sigma\Omega^2v_\beta$ with

$$\sigma \;=\; -\,\frac{(I_\alpha-I_\beta)(I_\alpha-I_\gamma)}{I_\beta I_\gamma}.$$

The product $(I_\alpha-I_\beta)(I_\alpha-I_\gamma)$ is positive when $I_\alpha$ is the largest or the smallest, and negative when it lies strictly between. So $\sigma\lt0$ — oscillation — at the extreme axes, and $\sigma\gt0$ — exponential growth — at the intermediate one, and nowhere else.

Do the bookkeeping slowly; this is where sign errors breed. Perturb about axis 1: $\vec\omega = (\Omega,v_2,v_3)$. Euler 2 and 3 linearise to $I_2\dot v_2 = (I_3-I_1)\Omega v_3$ and $I_3\dot v_3 = (I_1-I_2)\Omega v_2$, so $\ddot v_2 = \frac{(I_3-I_1)(I_1-I_2)}{I_2I_3}\Omega^2v_2$. With $I_1\lt I_2\lt I_3$ the numerator is $(+)(-)\lt0$: oscillation. Perturb about axis 3: $\ddot v_1 = \frac{(I_2-I_3)(I_3-I_1)}{I_1I_2}\Omega^2v_1$, numerator $(-)(+)\lt0$: oscillation. Perturb about axis 2, as on the slide: numerator $(I_2-I_3)(I_1-I_2) = (-)(-)\gt0$: growth. Three explicit computations, one line each, and they are what the box's $\sigma$ summarises. When in doubt, redo them rather than recall the formula.

Why $\omega_\alpha$ stays constant. Its equation has the product of the two small components on the right, so $\dot\omega_\alpha = O(v^2)$: at linear order the spin rate about the chosen axis does not change. At second order it does, slowly, and that is what makes the full nonlinear motion periodic rather than unbounded.

This is unit 02's programme. Steady rotation about a principal axis is an equilibrium of the Euler system on $\mathbb R^3$ — in fact a whole line of equilibria, one for each $\Omega$. Linearise, read the eigenvalues of the Jacobian: $\pm\lambda$ real is a saddle, $\pm i\lambda$ is a centre. Unit 02 also warned that a centre in the linearisation proves nothing about the nonlinear system. That warning is cashed two slides on.

16 · Euler · example 4

The intermediate-axis theorem, with a number

intermediate-axis theorem

For $I_1\lt I_2\lt I_3$, steady rotation about the intermediate principal axis is linearly unstable, with growth rate

$$\lambda=\Omega\sqrt{\frac{(I_3-I_2)(I_2-I_1)}{I_1I_3}} ,$$

while rotation about either extreme axis is a centre, not a saddle.

checks

Dimensions: the ratio is dimensionless, so $[\lambda]=\mathrm{s^{-1}}$ ✓. Limit $I_2\to I_1$ or $I_2\to I_3$: $\lambda\to0$, and a symmetric top has no intermediate axis ✓. Bound: $\lambda\lt\Omega$ always, from $I_1+I_2\ge I_3$ ✓.

worked example · the block, thrown

$I_1=0.00136$, $I_2=0.00296$, $I_3=0.00416\ \mathrm{kg\,m^2}$:

$$\frac{(0.00120)(0.00160)}{(0.00136)(0.00416)}=0.33937,$$

whose root is $0.58255$. At two turns a second, $\Omega=4\pi=12.566\ \mathrm{s^{-1}}$ and $\lambda=7.3206\ \mathrm{s^{-1}}$.

A $1\%$ wobble grows like $e^{\lambda t}$, reaching order one after $\ln(100)/\lambda = \mathbf{0.63}$ s — the length of a throw, which is why the flip is always seen and never avoidable.

About the spine, instead, a steady wobble of period $0.83$ s; about the cover normal, $0.55$ s.

The prediction is falsifiable in your kitchen. Put a rubber band round a hardback so it cannot open. Toss it spinning about the axis lying in the cover and perpendicular to the spine. At two turns a second a toss of about half a second is a turn or so about that axis, and the book arrives back in your hand having flipped over as well — about one flip per toss. Tossed about the spine, or about the axis through the cover, it does not.

Why the flip time is only logarithmically sensitive. The time to grow from an initial wobble $\varepsilon$ to order one is $\ln(1/\varepsilon)/\lambda$. Halving $\varepsilon$ adds only $\ln2/\lambda = 0.095$ s. That is why the interval between successive flips of a freely tumbling body looks almost regular even though nothing enforces a period — the same logarithm that makes the wing nut Dzhanibekov filmed on Salyut 7 in 1985 flip at a nearly steady rate, every few seconds, for as long as the camera runs. The exact interval requires the full nonlinear solution in Jacobi elliptic functions, which this course does not build; what is derived here is the growth rate, and that is the falsifiable part.

The extreme-axis frequencies. About axis 1: $\Omega\sqrt{(I_3-I_1)(I_2-I_1)/(I_2I_3)} = 12.566\times0.6032 = 7.580\ \mathrm{s^{-1}}$, period $0.829$ s. About axis 3: $\Omega\sqrt{(I_3-I_1)(I_3-I_2)/(I_1I_2)} = 12.566\times0.9136 = 11.481\ \mathrm{s^{-1}}$, period $0.547$ s. These are the wobbles you see superposed on a stable toss.

The bound $\lambda\lt\Omega$. The triangle inequality $I_1+I_2\ge I_3$ gives $I_3-I_2\le I_1$, and $I_2\lt I_3$ gives $I_2-I_1\lt I_3$. Multiplying, $(I_3-I_2)(I_2-I_1)\lt I_1I_3$, so the ratio under the root is below 1 and $\lambda\lt\Omega$. The instability never outruns the spin itself; for the block it reaches $58\%$ of it.

16 · Euler · example 4, continued

The flip, integrated

The linearisation only says the wobble starts to grow. To see what it grows into, integrate Euler's three equations as they stand, for the book at two turns a second about axis 2, started with a $1\%$ wobble along the growing direction.

Until about $0.5$ s the wobble follows $1\%\cdot e^{\lambda t}$ (dashed), as linearised.
Around $0.7$ s the spin component $\omega_2$ runs from $+\Omega$ to $-\Omega$: the book has turned over. The midpoint is at $0.72$ s, just after the estimate $0.63$ s.
Then it is calm again, spinning the other way up — for about five seconds, before the next flip.

Body components of $\vec\omega$, in units of $\Omega = 4\pi\ \mathrm{s^{-1}}$, from a numerical solution of Euler's equations for the book. The flip is a short event between long quiet stretches, which is why a tossed book seems to turn over all at once.

How the curves were computed. Euler's equations with $\vec\tau = 0$ and the book's $I_1, I_2, I_3$, integrated by fourth-order Runge–Kutta with step $0.2$ ms (unit 06's methods would do; this is only bookkeeping). Initial condition $\vec\omega(0) = \Omega\,(0.01,\ 1,\ -0.0066)$: the second and third components of the linear growing mode, $v_3/v_1 = I_1\lambda/\big((I_2-I_3)\Omega\big) = -0.66$. The energy $\tfrac12\sum I_k\omega_k^2$ is conserved by the integration to better than one part in $10^{13}$ over six seconds, a check that the step is small enough.

Why the flip is quick and the rest is slow. Near $\pm\Omega\vec u_2$ the motion is exponentially slow to leave — it starts from a $1\%$ wobble and needs $\ln(100)/\lambda$ to get anywhere — and once away it crosses over in about $0.4$ s ($\omega_2$ from $0.9\,\Omega$ to $-0.9\,\Omega$). The next flip, from the other side, again starts from a small wobble, so the interval between flips is set by the logarithm of how small. For this start it is $5.0$ s; the trajectory is one of the curves through the saddle of the polhode figure, run near the separatrix.

16 · Euler

The stable axes need more than a linearisation

unit 02's warning, cashed

At axes 1 and 3 the linearisation has eigenvalues $\pm i\lambda$: a centre. Hartman–Grobman says nothing about centres, and unit 02 exhibited systems with a linear centre that spiral out. Linear stability is not stability. Something else is needed.

Lyapunov stability of the extreme axes

$F:=\lVert\vec L\rVert^2-2I_1T=\sum_kI_k(I_k-I_1)\omega_k^2$ is conserved, and equals $I_2(I_2-I_1)\omega_2^2+I_3(I_3-I_1)\omega_3^2$ — a positive definite function of $(\omega_2,\omega_3)$ when $I_1$ is strictly smallest. So $F$ is a Lyapunov function: rotation about axis 1 is stable.

$G:=2I_3T-\lVert\vec L\rVert^2=\sum_kI_k(I_3-I_k)\omega_k^2$ does the same for the largest. $\;\blacksquare$

Read the figure of the free top again: the level sets of $F$ near axis 1 are the small closed polhodes. A trajectory starting on a small one stays on it forever, because $F$ cannot change — not merely for a long time.

what is proved, precisely

For every $\varepsilon\gt0$ there is $\delta\gt0$ such that $\lVert\vec\omega(0)-\Omega\vec u_1\rVert\lt\delta$ implies $\lVert\vec\omega(t)-\Omega'\vec u_1\rVert\lt\varepsilon$ for all $t$, for some nearby $\Omega'$. Stability of the axis, not of a particular $\Omega$: the spin rate may settle slightly different, and does.

And for axis 2, both $F$ and $G$ are indefinite: neither works, as it must be, since the axis is genuinely unstable.

Full proof

$F = \lVert\vec L\rVert^2-2I_1T = \sum_kI_k^2\omega_k^2-I_1\sum_kI_k\omega_k^2 = \sum_kI_k(I_k-I_1)\omega_k^2$. The $k=1$ term vanishes identically, leaving $F = I_2(I_2-I_1)\omega_2^2+I_3(I_3-I_1)\omega_3^2$, with both coefficients strictly positive when $I_1\lt I_2\le I_3$. Both $\lVert\vec L\rVert^2$ and $T$ are conserved along the free motion, so $F$ is. Given $\varepsilon$, let $c = \min\{I_2(I_2-I_1),I_3(I_3-I_1)\}\,\varepsilon^2$ and pick $\delta$ small enough that $F(\vec\omega(0))\lt c$; then $\omega_2^2+\omega_3^2\lt\varepsilon^2$ for all $t$. That is Lyapunov stability of the $\vec u_1$ axis. It says nothing about $\omega_1$, which is why the statement is about the axis: $\omega_1$ is pinned only by $T$, to within $O(\varepsilon^2)$.

Why this is the right kind of argument. Unit 02 introduced Lyapunov functions exactly for the situation where linearisation is silent, and the free rigid body is the classical example. Note that no approximation was made anywhere: $F$ is exactly conserved and exactly positive definite, so the conclusion holds for perturbations of any size, not just small ones — stronger than anything a linearisation could give.

The indefinite case. With $\alpha=2$: $\lVert\vec L\rVert^2-2I_2T = I_1(I_1-I_2)\omega_1^2+I_3(I_3-I_2)\omega_3^2$, whose coefficients have opposite signs. Its level sets near the axis are hyperbolas, not ellipses — the crossing curves in the figure — and the conserved quantity, far from trapping the trajectory, is precisely what channels it away.

Energy dissipation changes the answer. A real body flexes and loses energy at fixed $\lVert\vec L\rVert$, so it drifts to the smallest $T$ compatible with its $\vec L$, which is rotation about the axis of largest moment. Explorer 1 (1958) was spin-stabilised about its long, small-moment axis, and its flexible wire antennas dissipated enough energy that it was soon found tumbling end over end instead. Nothing in this unit's conservative analysis predicts that, and nothing in it contradicts it: the Lyapunov argument assumes $T$ is constant.

section 6

Coordinates for a group, and the tops they describe

Three angles that almost work, and the gyroscope they explain.

16 · Euler angles

Three angles, and where they fail

Euler angles ($z$–$x$–$z$)

$R(\phi,\theta,\psi)=R_z(\phi)\,R_x(\theta)\,R_z(\psi)$: precess by $\phi$ about the space $z$, tilt by $\theta$, spin by $\psi$ about the body $3$-axis. In body components,

$$\begin{aligned}\omega_1&=\dot\phi\sin\theta\sin\psi+\dot\theta\cos\psi\\ \omega_2&=\dot\phi\sin\theta\cos\psi-\dot\theta\sin\psi\\ \omega_3&=\dot\phi\cos\theta+\dot\psi\end{aligned}$$

The Jacobian $\partial(\omega_1,\omega_2,\omega_3)/\partial(\dot\phi,\dot\theta,\dot\psi)$ has determinant $\;-\sin\theta$. At $\theta=0$ or $\pi$ it drops rank: only $\dot\phi+\dot\psi$ is determined, and the map from angles to rotations is not locally invertible.

$\phi$ turns $x$ to the line of nodes, about $z$; $\theta$ tilts $z$ to the body $3$-axis, about that line; $\psi$ turns the line to the body $1$-axis, about axis $3$. The line of nodes is where the body's $1$–$2$ plane (dashed) cuts the $x$–$y$ plane.

Where the $\omega$ formulas come from. $\vec\omega$ is the sum of three contributions: $\dot\phi$ about the space $z$-axis, $\dot\theta$ about the line of nodes, $\dot\psi$ about the body $3$-axis. Express each in body components. The body $3$-axis is $(0,0,1)$; the line of nodes is $(\cos\psi,-\sin\psi,0)$; the space $z$-axis is $(\sin\theta\sin\psi,\ \sin\theta\cos\psi,\ \cos\theta)$. Add, weighted by $\dot\psi,\dot\theta,\dot\phi$, and the three displayed lines appear. Checking the middle one: the space $z$ contributes $\dot\phi\sin\theta\cos\psi$ and the nodes contribute $-\dot\theta\sin\psi$ ✓.

The determinant. Expanding along the third row of the coefficient matrix and using $\sin^2\psi+\cos^2\psi=1$ gives $-\sin\theta$. At $\theta=0$ the first two rows become proportional: the $\dot\phi$ column is $(0,0,1)$, identical to the $\dot\psi$ column, so a change in $\phi$ and an equal, opposite change in $\psi$ produce no motion at all. That is gimbal lock: two of the three axes have become parallel and one degree of freedom is invisible.

16 · Euler angles

Gimbal lock: no single chart covers the rotations

theorem · no global chart

There is no homeomorphism from $SO(3)$ onto an open subset of $\mathbb R^3$.

Why: $SO(3)$ is closed and bounded in $\mathbb R^9$, so compact; its image would be compact, hence closed, and open, so $\varnothing$ or $\mathbb R^3$ — and $\mathbb R^3$ is not compact. $\;\blacksquare$

gimbal lock

Not sloppiness, and not fixable by cleverer angles: by the theorem, every three-parameter description of rotations is singular somewhere. Unit 31 covers the space with several charts instead — the idea general relativity cannot do without.

On the block, with the space $z$ along the conserved $\vec L$, $\theta$ is the tilt of the cover normal from $\vec L$. Toss the book about the cover normal and $\theta\approx0$: the chart is singular just where the motion is calmest.

At $\theta = 0$ the body $3$-axis lies along $z$ and the two planes coincide, so the line of nodes is not defined. $\phi$ and $\psi$ are then turns about the same axis, and only $\phi+\psi$ is seen.

The figure and the determinant. The Jacobian of the previous slide has $\dot\phi$-column $(\sin\theta\sin\psi,\sin\theta\cos\psi,\cos\theta)$ and $\dot\psi$-column $(0,0,1)$; at $\theta=0$ they are equal, which is the picture: two of the three turning axes have become one, and the three angles span only two directions of motion.

The block's $\theta$, in full. The Euler angle $\theta$ is the angle between the space $z$-axis and the body $3$-axis, because the space $z$-axis has body components $(\sin\theta\sin\psi,\sin\theta\cos\psi,\cos\theta)$ and the body $3$-axis is $(0,0,1)$. Choose the space $z$ along the conserved $\vec L$; its body components are $\vec L/\lVert\vec L\rVert$, whose third entry is $I_3\omega_3/\lVert\vec L\rVert$, so $\cos\theta = I_3\omega_3/\lVert\vec L\rVert$. For the block that is the cover normal's tilt from $\vec L$, and integrating $\dot R = R\Omega$ for the thrown book is exactly the job of finding $\phi,\theta,\psi$ as functions of time — the harder half promised in §5, and the half that needs Jacobi elliptic functions. In a flip about axis 2, $\theta$ starts at $\pi/2$, nods towards $\vec L$ and back, while $\phi$ and $\psi$ carry the half-turn — so the flip itself is far from the singular points; it is the calm toss about the cover normal that sits on one.

The compactness argument, in detail. $SO(3)\subset\mathbb R^9$ is bounded (every entry has modulus at most 1, since columns are unit vectors) and closed (it is the preimage of $(\mathbb 1,1)$ under the continuous map $R\mapsto(R^{\mathsf T}R,\det R)$), hence compact by Heine–Borel. A homeomorphism carries compact to compact; $\mathbb R^3$ is connected, so its only clopen subsets are $\varnothing$ and itself; and $\mathbb R^3$ is not compact. Hence no such homeomorphism exists. The learner's metric-space topology is all this needs.

What $SO(3)$ actually is. The closed ball of radius $\pi$ in $\mathbb R^3$ — the axis-angle vector $\theta\hat n$ of Rodrigues — with antipodal boundary points identified, since a rotation by $\pi$ about $\hat n$ equals one by $\pi$ about $-\hat n$. That is real projective space $\mathbb{RP}^3$, which unit 31 names. Quaternions give a two-to-one covering by the 3-sphere and are how flight software avoids gimbal lock — not by escaping the theorem, but by using four numbers with one constraint instead of three free ones.

Apollo 11 carried a gimbal-lock warning for its three-gimbal inertial platform, and Michael Collins, near it in lunar orbit, asked Mission Control half in jest for a fourth gimbal for Christmas. The problem was real hardware built to the shape of this theorem.

16 · tops

The symmetric top, free: two cones

Let $I_1=I_2\ne I_3$ — a symmetric top. Euler's third equation gives $\dot\omega_3=0$, and the first two become

$$\dot\omega_1=-\Omega_b\omega_2,\ \ \dot\omega_2=\Omega_b\omega_1,$$ $$\Omega_b:=(I_3-I_1)\omega_3/I_1,$$

a rotation of $(\omega_1,\omega_2)$ at constant rate (unit 06). So in the body frame $\vec\omega$ sweeps a cone about the symmetry axis at $\Omega_b$; in the space frame $\vec L$ is fixed and $\vec\omega$ sweeps a second cone about it. The body cone rolls on the space cone.

check

$I_3\to I_1$: $\Omega_b\to0$, free rotation about a fixed axis — correct for a sphere ✓. Dimensions: $\mathrm{s^{-1}}$ ✓.

Schematic, for an oblate top ($I_3\gt I_1$). The body cone, fixed in the body around $\hat u_3$, rolls without slipping on the space cone around the fixed $\vec L$; they touch along $\vec\omega$, the one line of the body momentarily at rest.

Reading the figure. It is drawn for $I_3 = 1.6\,I_1$ with $\vec\omega$ at $35^\circ$ to the symmetry axis, so that the cones are visible; the Earth's are the same shape with angles a million times smaller. From $\vec L = I_1\vec\omega + (I_3-I_1)\omega_3\hat u_3$, $\vec L$ is a positive combination of $\vec\omega$ and $\hat u_3$ when $I_3\gt I_1$, so it lies between them: $\tan$ of its angle from $\hat u_3$ is $(I_1/I_3)\tan35^\circ$, giving $23.6^\circ$, and the space cone's half-angle is $35^\circ-23.6^\circ = 11.4^\circ$. The wide body cone therefore encloses the narrow space cone and rolls round the inside of it. For a prolate body ($I_3\lt I_1$, a rugby ball) $\vec\omega$ lies between $\vec L$ and $\hat u_3$ instead, and the body cone rolls on the outside.

Solving the two equations. Set $\zeta=\omega_1+i\omega_2$; then $\dot\zeta = i\Omega_b\zeta$, so $\zeta=\zeta_0e^{i\Omega_bt}$ — unit 06's complex exponential doing exactly the job it was introduced for. $|\zeta|$ is constant, so $\vec\omega$ keeps a constant angle with the symmetry axis and a constant magnitude.

Why the space cone. $\vec L$, $\vec\omega$ and the symmetry axis $\vec u_3$ are always coplanar: $\vec L = I_1(\omega_1,\omega_2,0)+I_3\omega_3\vec u_3 = I_1\vec\omega+(I_3-I_1)\omega_3\vec u_3$. With $\vec L$ fixed in space, that identity forces $\vec\omega$ and $\vec u_3$ to precess together about $\vec L$, at the rate $\lVert\vec L\rVert/I_1$. Two rates, then: $\Omega_b$ in the body and $\lVert\vec L\rVert/I_1$ in space. The rolling-cones picture is the two of them drawn at once.

A hand-sized version: a spiral pass in American football is, to a first approximation, a free symmetric top. On a wobbly pass the nose circles the fixed $\vec L$ at the space-cone rate $\lVert\vec L\rVert/I_1$, which is what a spectator sees; $\Omega_b$ is the rate at which $\vec\omega$ circles the ball's own axis, seen only from on board.

16 · tops

The Earth, a symmetric top that is not quite rigid

the Earth, which is a symmetric top

The Earth's dynamical ellipticity is $(I_3-I_1)/I_3=3.2738\times10^{-3}$ (cited: IERS Conventions 2010), so $(I_3-I_1)/I_1=3.2845\times10^{-3}$ and

$$\frac{2\pi}{\Omega_b}=\frac{I_1}{I_3-I_1}\,T_{\text{sid}}=\mathbf{303.6}\ \text{d},$$

with $T_{\text{sid}} = 86164$ s the sidereal day.

The pole should wander in a circle with that period — Euler, 1765. Chandler found the wobble in 1891; its period is 433 days.

Not a failed prediction of this theory but of one of its hypotheses: the Earth is not rigid. Its yielding lengthens the period by $43\%$ — which Newcomb (1892) read at once as evidence of an elastic Earth.

In the figure of the previous slide, the Earth's $\hat u_3$ is its figure axis through the geographic poles, and $\vec\omega$ is the axis the sky turns about. They are a few tenths of an arcsecond apart — some metres at the pole. In space the body cone rolls round $\vec L$ once a day; seen from the ground, $\vec\omega$ circles the figure axis at $\Omega_b$, once in $304$ days for a rigid Earth.

16 · tops

The heavy top, from a Lagrangian

A symmetric top of mass $M$ on a fixed pivot, in gravity $g$, with its centre of mass a distance $l$ up the symmetry axis $\hat u_3$, tilted $\theta$ from the vertical. Using the Euler-angle expressions and $I_1=I_2$,

$$\begin{aligned}\mathcal L&=\tfrac12I_1\big(\dot\theta^2+\dot\phi^2\sin^2\theta\big)\\&\quad+\tfrac12I_3\big(\dot\psi+\dot\phi\cos\theta\big)^2-Mgl\cos\theta .\end{aligned}$$

$\phi$ and $\psi$ are cyclic (unit 12), so

$$\begin{aligned}p_\psi&=I_3\omega_3=:L_3,\\ p_\phi&=I_1\dot\phi\sin^2\theta+L_3\cos\theta=:L_z\end{aligned}$$

are conserved, and $\theta$ alone is left: one degree of freedom, moving in an effective potential that the next slide writes down and draws.

Schematic. Gravity $Mg$ acts at the centre of mass, a distance $l$ up the axis $\hat u_3$; steady precession carries it round the dashed circle.

Reading the figure. The two angles that stay are the tilt $\theta$, between the vertical and $\hat u_3$, and the precession angle $\phi$, the azimuth of $\hat u_3$ round the vertical; the third, $\psi$, is the spin of the disc about its own axis and is not drawn. The unit vector $\hat e_\perp$ is perpendicular to $\hat u_3$ in the vertical plane through it, on the upper side; the gyroscope example uses it. With $\theta$ held fixed, the centre of mass runs round the dashed circle at the rate $\dot\phi$.

Getting $V_{\text{eff}}$. The energy is $E = \frac12I_1(\dot\theta^2+\dot\phi^2\sin^2\theta)+\frac{L_3^2}{2I_3}+Mgl\cos\theta$. Solve $p_\phi = L_z$ for $\dot\phi = (L_z-L_3\cos\theta)/(I_1\sin^2\theta)$ and substitute; the constant $L_3^2/2I_3$ is dropped. What remains is $E' = \frac12I_1\dot\theta^2+V_{\text{eff}}(\theta)$: one degree of freedom, solvable by quadrature exactly as unit 00 and unit 08 did. $V_{\text{eff}}\to+\infty$ at $\theta\to0,\pi$ (unless $L_z=\pm L_3$), so $\theta$ is trapped between two turning points — the top nods.

Released from rest, a top does not precess steadily: it starts at $\dot\phi=0$, falls, picks up precession, rises again, and traces cusps or loops in $\theta$–$\phi$. Steady precession is the special initial condition that lands exactly on the minimum of $V_{\text{eff}}$.

16 · tops

Steady precession, and the well it sits in

Eliminating $\dot\phi$ with $L_z$ leaves $\theta$ in the effective potential

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

steady precession

$\theta\equiv\theta_0$ requires $V_{\text{eff}}'(\theta_0)=0$, which in terms of $\dot\phi$ is the quadratic

$$I_1\cos\theta_0\,\dot\phi^2-L_3\,\dot\phi+Mgl=0 .$$

Two roots when $L_3^2\ge4I_1Mgl\cos\theta_0$: a slow and a fast precession. Spin the top slower than that and steady precession at that angle is impossible.

For $L_3$ large the slow root is $\dot\phi\simeq Mgl/L_3$, independent of $\theta_0$ — the textbook answer. Small oscillations of $\theta$ about $\theta_0$ are nutation, at $\omega_{\text{nut}}=\sqrt{V_{\text{eff}}''(\theta_0)/I_1}\simeq L_3/I_1$.

$V_{\text{eff}}(\theta)$ for the toy gyroscope of the next slide, with $L_z$ set by its slow precession at $60^\circ$. Steady precession sits at the bottom of the well; nutation is the oscillation across it, $71\ \mathrm{s^{-1}}$ for this well.

Reading the plot. The dashed line is the gravitational term $Mgl\cos\theta$ alone, which has no minimum at all: by itself the top would fall. The first term of $V_{\text{eff}}$, the kinetic energy of precession at fixed $L_z$, is zero exactly where $L_z = L_3\cos\theta$ — close to $60^\circ$ here — and rises steeply on both sides; it is what walls the well in. At the bottom the two curves almost touch, because for a fast top the precession carries very little energy: $V_{\text{eff}}(60^\circ) = 0.0303$ J, of which $Mgl\cos60^\circ$ is $0.0294$. The curvature there is $V_{\text{eff}}'' = 1.382$ J, and $\sqrt{1.382/I_1} = 71.05\ \mathrm{s^{-1}}$ is the nutation frequency of the next slide.

The quadratic, derived. The Euler–Lagrange equation for $\theta$ is $I_1\ddot\theta = I_1\dot\phi^2\sin\theta\cos\theta-L_3\dot\phi\sin\theta+Mgl\sin\theta$ (using $I_3(\dot\psi+\dot\phi\cos\theta)=L_3$). Set $\ddot\theta=0$ and divide by $\sin\theta\ne0$. Note it is a quadratic in $\dot\phi$, not a formula for it: two precession rates are compatible with the same tilt and spin.

The two roots. $\dot\phi_\pm = \frac{L_3\pm\sqrt{L_3^2-4I_1Mgl\cos\theta_0}}{2I_1\cos\theta_0}$. For large $L_3$, expand the root: $\dot\phi_-\simeq Mgl/L_3$ and $\dot\phi_+\simeq L_3/(I_1\cos\theta_0)$. The fast one exists mathematically and is almost never what a released top does, because releasing it from rest selects the slow branch. At $\theta_0=\pi/2$ the quadratic degenerates to a linear equation and the fast root goes to infinity: the slow root $Mgl/L_3$ is then exact.

Nutation frequency. Perturb $\theta=\theta_0+\eta$ in $I_1\ddot\theta=-V_{\text{eff}}'(\theta)$: $I_1\ddot\eta=-V_{\text{eff}}''(\theta_0)\eta$, which is unit 06's harmonic oscillator with $\omega_{\text{nut}}^2=V_{\text{eff}}''(\theta_0)/I_1$. For a fast top the dominant term in $V_{\text{eff}}''$ is $L_3^2/I_1$, giving $\omega_{\text{nut}}\simeq L_3/I_1$ — much faster than the precession, whose rate is $Mgl/L_3$: their ratio is $L_3^2/(I_1Mgl)$, the same large parameter that made the expansion valid.

16 · tops · example 5

A gyroscope, and the approximation nobody mentions

worked example · a toy gyroscope

Disc $M=0.15$ kg, radius $R=0.03$ m, pivot distance $l=0.04$ m, spun at $\omega_3=300\ \mathrm{s^{-1}}$, held at $\theta_0=60^\circ$.

$I_3=\tfrac12MR^2=6.75\times10^{-5}$, $I_1=\tfrac14MR^2+Ml^2=2.7375\times10^{-4}\ \mathrm{kg\,m^2}$, $L_3=2.025\times10^{-2}\ \mathrm{kg\,m^2\,s^{-1}}$, $Mgl=5.886\times10^{-2}$ N m.

Exact: $1.36875\times10^{-4}\dot\phi^2-2.025\times10^{-2}\dot\phi+5.886\times10^{-2}=0$, discriminant $3.7784\times10^{-4}$, roots

$$\dot\phi_-=\mathbf{2.9661}\ \mathrm{s^{-1}},\qquad \dot\phi_+=145.0\ \mathrm{s^{-1}} .$$

Approximate: $Mgl/L_3=2.9067\ \mathrm{s^{-1}}$ — low by $2.0\%$. Precession period $2.12$ s exactly, $2.16$ s approximately.

Nutation: $71.05\ \mathrm{s^{-1}}$ exactly, against $L_3/I_1=73.97$ — high by $4.1\%$. Period $0.088$ s: a fast shiver on the slow sweep.

what $\vec\tau=\mathrm d\vec L/\mathrm dt$ hides

The elementary argument sets $\vec L=L_3\hat u_3$, but $\vec L=I_1\dot\phi\sin\theta\,\hat e_\perp+L_3\hat u_3$ ($\hat e_\perp$ as in the heavy-top figure): precession adds angular momentum. Dropping it needs $I_1\dot\phi\ll L_3$: here $8.12\times10^{-4}$ against $2.025\times10^{-2}$, or $4.01\%$, and the error in $\dot\phi$ is $\cos\theta_0$ times that — $2.0\%$, to the digit.

checks

Dimensions: $Mgl/L_3$ is $\mathrm{N\,m}/(\mathrm{kg\,m^2\,s^{-1}})=\mathrm{s^{-1}}$ ✓. Limit $\omega_3\to\infty$: $\dot\phi_-\to0$ and the discrepancy $\to0$ ✓. Existence: $L_3^2=4.10\times10^{-4}\gg4I_1Mgl\cos\theta_0=3.22\times10^{-5}$: two real roots, so the top is comfortably fast ✓.

Every number, recomputed. $I_3 = 0.5\times0.15\times0.0009 = 6.75\times10^{-5}$; $I_1 = 0.25\times0.15\times0.0009+0.15\times0.0016 = 3.375\times10^{-5}+2.4\times10^{-4} = 2.7375\times10^{-4}$; $L_3 = 6.75\times10^{-5}\times300 = 0.02025$; $Mgl = 0.15\times9.81\times0.04 = 0.05886$. Quadratic coefficient $I_1\cos60^\circ = 1.36875\times10^{-4}$. Discriminant $0.02025^2-4(1.36875\times10^{-4})(0.05886) = 4.100625\times10^{-4}-3.2226\times10^{-5} = 3.77837\times10^{-4}$, square root $0.0194380$. Slow root $(0.02025-0.0194380)/(2.7375\times10^{-4}) = 2.96613$; fast root $(0.02025+0.0194380)/(2.7375\times10^{-4}) = 144.979$.

The nutation number. With $\dot\phi=\dot\phi_-$, $L_z = I_1\dot\phi\sin^2\theta_0+L_3\cos\theta_0 = 1.0734\times10^{-2}$; then $V_{\text{eff}}''(\theta_0) = 1.3820\ \mathrm{J}$ (evaluated from the formula on the previous slide), and $\sqrt{1.3820/2.7375\times10^{-4}} = 71.05\ \mathrm{s^{-1}}$. The approximation $L_3/I_1 = 73.97$ overshoots because it drops the gravitational and geometric corrections to $V_{\text{eff}}''$.

The two percentages, related. Expanding the square root, $\dot\phi_- = \frac{Mgl}{L_3}\Big(1+\frac{I_1Mgl\cos\theta_0}{L_3^2}+\dots\Big)$, so the elementary formula is low by the relative amount $\epsilon := I_1Mgl\cos\theta_0/L_3^2 = (2.7375\times10^{-4})(0.05886)(0.5)/(4.1006\times10^{-4}) = 0.01965$, i.e. $2.0\%$ ✓. Meanwhile the neglected transverse angular momentum has $I_1\dot\phi_-/L_3 \simeq I_1Mgl/L_3^2 = 2\epsilon = 3.93\%$, against the computed $4.01\%$. So the elementary formula errs at exactly the order of the term it discarded, and the factor between the two is $\cos\theta_0$. That is the honest version of "the gyroscope precesses because torque changes angular momentum".

What else the elementary argument hides: nutation entirely. It produces one rate, not two roots, and no oscillation in $\theta$. A real gyroscope released at $\theta_0$ nutates visibly until friction damps it onto the steady branch, which is why the elementary formula seems to work in the lecture-room demonstration — you watch it after the nutation has gone.

16 · closing

The block, closed

section	what the block gave
1 · the body	six degrees of freedom; $\vec\omega$ from $R^{\mathsf T}\dot R$
2 · inertia	$\mathsf I_{\text{cm}}$ by integration; $\mathsf I_O$ by Steiner
3 · tensors	$\mathsf I'=R\mathsf IR^{\mathsf T}$ checked against a fresh integral; $\mathrm{diag}(I_{kk})$ shown not to be a tensor — $0.00288$ against $0.00864$
4 · principal axes	three moments and three axes at the corner; the easiest axis $3.38^\circ$ off the diagonal
5 · Euler	$\lambda=0.58255\,\Omega$: a flip in $0.63$ s
6 · angles	Euler angles for its orientation, and why they must fail somewhere

what this unit installed for later

22 the stress tensor, another symmetric $(0,2)$-tensor, this time a field. 27, 28 four-vectors and $F_{\mu\nu}$, where raising and lowering stop being invisible. 31–33 the metric, the connection and the curvature: same definition, on a manifold. 35 the field equations, an identity between two $(0,2)$-tensors.

And one sentence to keep: a tensor is a multilinear map; the transformation law is what its components must do because the map does not move.

If one thing survives from this unit, make it the two-line proof of the transformation law. Every later index manipulation in the course is that proof, applied. When unit 32 asks why $\partial_\mu V^\nu$ is not a tensor, the answer is the same shape as this unit's $\mathrm{diag}(I_{kk})$: a rule that assigns arrays to bases, which fails to be equivariant, and fails in a way you can exhibit by computing the same thing twice.

The physics that will not be needed again, and is worth having anyway: Poinsot's construction, the heavy top's two precession rates, gimbal lock. The physics that will: the intermediate-axis argument is the prototype of a linear stability analysis around a relative equilibrium, which unit 17 generalises and unit 39 uses on the universe.

Two habits to take forward. First, compute every invariant twice — the trace check caught nothing in this unit because everything was right, and it will catch a great deal later. Second, when an index expression is in doubt, evaluate the tensor on basis vectors directly; that is Route B of example 2, and it is always available.

16 · checkpoint

Done when

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.

Reading

Landau & Lifshitz, Mechanics, §§31–37 — terse and complete, and it uses nothing this course has not reached.
Arnold, Mathematical Methods, ch. 6 (§§26–30) for the geometry; skip §31, which needs manifolds (unit 31).
Goldstein ch. 4–5 for Euler angles and the heavy top in full detail.
For the algebra: any linear algebra text's chapter on multilinear maps — the definition matters more than the source.

17 integrability and chaos, which takes the free top as an example of an integrable system · 22 the stress tensor, this unit's algebra applied to a continuum · 27 relativistic dynamics, the first place the index positions matter · and eventually 31–35, which are this unit with $\delta$ replaced by $g_{\mu\nu}$.

./mr problems 16 — thirteen problems, about six hours, including the gyroscope and a collection of nine numbers per basis that you must convict.

Test the checkpoint on a body that is not in this deck, so that no number can be recalled. A good one: point masses of $1$, $2$ and $3$ kg at $(1,0,0)$, $(0,1,0)$, $(0,0,1)$ m. About the origin $\mathsf I = \mathrm{diag}(5,4,3)\ \mathrm{kg\,m^2}$ — check that. Then rotate to a frame whose first axis is $(1,1,1)/\sqrt3$, verify the law component by component, and confirm that the trace is still $12$ and the eigenvalues still $5,4,3$. Predict which axis is unstable: the intermediate moment is $4$, so the $\vec e_2$ axis. If your two frames disagree, the error is almost always a transposed $R$.

A second, sharper test of §4: make the three masses equal. Then $\mathsf I = 2\delta$ exactly, every axis is principal, and no axis is unstable. The symmetry theorem predicts only part of this: $(1,1,1)$ is a three-fold axis of that configuration, which forces the two moments perpendicular to it to be equal; that the third equals them too is a property of these particular positions, not of the symmetry.

The "one sentence" of the checkpoint should contain the word basis twice: nine numbers do not determine an object until you say which basis they are in, and they do not determine a tensor until the numbers in every other basis are determined too. If your sentence mentions only "the transformation law", press further: why must there be one?

Then ./mr done 16.

Rigid bodies, and what a tensor actually is

One body, six sections

The book, and the axis that will not hold

A body that cannot deform

What a rigid body is, and where it lives

A configuration is a rotation and a shift

Angular velocity, recalled

$\mathfrak{so}(3)$, and the exponential that gets you back

Rodrigues' formula, drawn

Kinetic energy splits in two

The inertia tensor

From a kinetic energy to a bilinear form

Angular momentum is a covector

The block's $\vec L$, tilted $22.8^\circ$ off $\vec\omega$

The block, about its centre

Moving the origin: the parallel-axis theorem

Steiner at the block's corner

What a tensor actually is

The dual space, and why it is not $V$

A covector, drawn as a stack of lines

The definition

Components, and why there are $n^{p+q}$ of them

The transformation law is a theorem

Index notation, and the convention that earns its keep

Contraction, and the trace that survives

Symmetric and antisymmetric parts

The law, verified component by component

Nine numbers per basis that are not a tensor

What the diagonal rule throws away

Principal axes

The spectral theorem, proved

Why the maximum is an eigenvector

Principal axes, and the ellipsoid

The block's inertia ellipsoid, drawn

Diagonalising the corner

Symmetry forces degeneracy

Euler's equations, and the axis that will not hold

Euler's equations

The free top: two surfaces, one curve

Linearising about a principal axis

The intermediate-axis theorem, with a number

The flip, integrated

The stable axes need more than a linearisation

Coordinates for a group, and the tops they describe

Three angles, and where they fail

Gimbal lock: no single chart covers the rotations

The symmetric top, free: two cones

The Earth, a symmetric top that is not quite rigid

The heavy top, from a Lagrangian

Steady precession, and the well it sits in

A gyroscope, and the approximation nobody mentions

The block, closed

Done when

Reading

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