Visual walkthrough — Torque-free rotation — Euler's equations, asymmetric top

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Step 1 — What is the "spin arrow" $\vec{\omega }$?

WHAT. When something rotates, at each instant it turns about one line in space — an axis. We draw an arrow along that axis. Its length is how fast we spin (radians per second), its direction is the axis, pointed by the right-hand rule (curl fingers the way it turns, thumb gives the arrow). We call this arrow $\vec{\omega }$ ("omega"). The little arrow on top just means "this is a thing with a direction," not just a number.

WHY this object. A rotation needs three facts: which axis, how fast, which sense. A single arrow packs all three. That is why we use a vector instead of an angle.

PICTURE. Below, a flat spinning disc. The purple arrow $\vec{\omega }$ points straight up out of the disc. Spin twice as fast → arrow twice as long.

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Step 2 — The body has three stiffness numbers: the principal moments

WHAT. A rigid body resists being spun. How much it resists depends on which axis you spin it about. For a nicely shaped body there are three special perpendicular axes — the principal axes — with three "resistance-to-spin" numbers $I_1,I_2,I_3$, the principal moments of inertia. Big $I$ = mass spread far from that axis = hard to spin up.

WHY these axes. Along any other axis the resistance mixes the three together (an ugly $3\times 3$ table). Along the principal axes it separates cleanly, one number per axis. That is the whole reason we glue our coordinate frame to the body — see Inertia tensor and principal axes.

PICTURE. A brick (matchbox) with its three axes drawn: the long-thin axis (small $I$), the flat axis (large $I$), and the in-between axis. The arrows are sized to hint at each $I$.

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Step 3 — $\vec{L}$ and $\vec{\omega }$ point in different directions

WHAT. Multiply each component of $\vec{\omega }$ by a different number $I_i$ and the result is a new arrow $\vec{L}$ pointing somewhere else. Only if all $I_i$ are equal (a sphere) do the two arrows line up.

WHY it matters. In empty space with no torque, it is $\vec{L}$ that stays frozen in direction — not $\vec{\omega }$. So if $\vec{\omega }\ne \vec{L}$, then keeping $\vec{L}$ fixed forces $\vec{\omega }$ to move. That moving-$\vec{\omega }$ is the wobble. This is exactly Mistake A in the parent note.

PICTURE. Same brick, spinning about an axis that is not one of the three special ones. The purple $\vec{\omega }$ and the magenta $\vec{L}$ split apart because the stretch factors $I_1,I_2,I_3$ differ.

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Step 4 — Two reasons an arrow seems to change: the transport theorem

WHAT. We keep our axes glued to the tumbling body. Watching any arrow $\vec{A}$ from inside that spinning frame, it can appear to change for two separate reasons.

WHY. Reason one: the arrow genuinely changes (grows, tilts). Reason two: even if the arrow is frozen in real space, our own axes swing past it, so it looks like it moved. We must add both.

PICTURE. A frozen arrow (grey) sits still in the room. Our rotating axes (violet) sweep by. From our seat the arrow appears to swing backwards — that apparent swing is the $\vec{\omega }\times \vec{A}$ term.

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Step 5 — Feed $\vec{L}$ into the theorem

WHAT. Space-frame Newton for rotation says ${\left(\frac{d\vec{L}}{dt}\right)}_{\text{space}}=\vec{N}$, the external torque. Torque-free means $\vec{N}=\vec{0}$. Now swap the awkward space-derivative for the body-derivative using Step 4.

WHY. In the body frame the stiffness numbers $I_i$ are constant, so ${\left(\frac{d\vec{L}}{dt}\right)}_{\text{body}}=(I_1{\dot{\omega }}_1,I_2{\dot{\omega }}_2,I_3{\dot{\omega }}_3)$ — clean. All the mess moves into the tidy cross-product term.

PICTURE. A flow box: real-space law → apply transport → set $\vec{N}=0$ → the body-frame law that couples the components.

$\underset{\text{body-frame change of }\vec{L}}{\underbrace{(I_1{\dot{\omega }}_1,I_2{\dot{\omega }}_2,I_3{\dot{\omega }}_3)}}+\underset{\text{frame-swing term}}{\underbrace{\vec{\omega }\times \vec{L}}}=\underset{\text{no torque}}{\underbrace{\vec{0}}}$

Each dotted symbol ${\dot{\omega }}_i$ means "rate of change of that spin component." We now just need the cross product.

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Step 6 — Work out the cross product, term by term

WHAT. The cross product $\vec{\omega }\times \vec{L}$ has three components. Using $\vec{\omega }=({\omega }_1,{\omega }_2,{\omega }_3)$ and $\vec{L}=(I_1{\omega }_1,I_2{\omega }_2,I_3{\omega }_3)$, the first component is $(\vec{\omega }\times \vec{L}{)}_1=\underset{{\omega }_2{I}_3{\omega }_3}{\underbrace{{\omega }_2{L}_3}}-\underset{{\omega }_3{I}_2{\omega }_2}{\underbrace{{\omega }_3{L}_2}}=(I_3-I_2){\omega }_2{\omega }_3.$

WHY. The cross product's first slot always uses the other two slots (2 and 3), never itself. That is why axis 1's fate is decided by ${\omega }_2$ and ${\omega }_3$ — the coupling. The difference $I_3-I_2$ appears because both terms carry an ${\omega }_2{\omega }_3$; subtracting leaves the gap in stiffness.

PICTURE. The determinant grid for $\vec{\omega }\times \vec{L}$, with the row-1 entries highlighted and the cancellation of the common ${\omega }_2{\omega }_3$ shown.

Moving that term to the right side ($I_1{\dot{\omega }}_1=-(\vec{\omega }\times \vec{L}{)}_1=(I_2-I_3){\omega }_2{\omega }_3$) and cycling $1\to 2\to 3\to 1$:

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Step 7 — Degenerate cases: sphere and symmetric top

WHAT. Test the machine on simple bodies.

• Sphere $I_1=I_2=I_3$: every right side has an $(I_a-I_b)=0$ factor, so ${\dot{\omega }}_i=0$. The spin arrow is frozen in the body — no wobble ever.
• Symmetric top $I_1=I_2=I_{\perp }\ne I_3$: Euler-3 gives $I_3{\dot{\omega }}_3=(I_1-I_2){\omega }_1{\omega }_2=0$, so ${\omega }_3$ is constant. The other two obey ${\dot{\omega }}_1=-{\Omega }_b{\omega }_2,{\dot{\omega }}_2=+{\Omega }_b{\omega }_1$ with ${\Omega }_b=\frac{I_3-I_{\perp }}{I_{\perp }}{\omega }_3$ — pure circular motion, so $\vec{\omega }$ sweeps a cone about axis 3.

WHY show these. A formula you cannot break on the easy cases is untrustworthy. These also connect to Symmetric top and gyroscopic precession (free precession, Earth's Chandler wobble).

PICTURE. Left: sphere, arrows all zero on the RHS, $\vec{\omega }$ still. Right: symmetric top, $({\omega }_1,{\omega }_2)$ tracing a circle so $\vec{\omega }$ traces a cone.

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Step 8 — The intermediate-axis instability, in one sign

WHAT. Asymmetric top $I_1<I_2<I_3$. Spin nearly about one axis, call its big spin $\Omega$, and let the other two be tiny. Combining two Euler equations gives a single equation ${\ddot{\omega }}_1=k{\omega }_1$ with $k=\frac{(I_2-I_3)(I_3-I_1)}{I_1{I}_2}{\Omega }^2(\text{for spin about axis 3}).$

WHY. The sign of $k$ is destiny. If $k<0$ the solution is $\cos ,\sin$ — a bounded wobble (stable). If $k>0$ the solution is $e^{\pm \sqrt{k}t}$ — it grows (unstable).

• Axis 3 (largest): $(I_2-I_3)<0$, $(I_3-I_1)>0$ → $k<0$ → stable.
• Axis 1 (smallest): both gaps negative → product positive → $k<0$ → stable.
• Axis 2 (middle): $(I_3-I_2)>0$ and $(I_2-I_1)>0$ → $k>0$ → unstable — it tumbles.

PICTURE. Three phase pictures: closed loops (stable) for axes 1 and 3, and a saddle (runaway) for axis 2 — the Dzhanibekov / tennis-racket effect. See Stability analysis and linearization.

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The one-picture summary

Two frozen scalars — energy $2T=I_1{\omega }_1^2+I_2{\omega }_2^2+I_3{\omega }_3^2$ and $L^2=I_1^2{\omega }_1^2+I_2^2{\omega }_2^2+I_3^2{\omega }_3^2$ — each carve a surface in $\omega$-space (two nested ellipsoids). The spin arrow tip is trapped on their intersection curve, the polhode. Near the long/short axes the intersections are tiny closed loops (stable wobble); near the middle axis they become the two great saddle-curves that carry $\vec{\omega }$ all the way around (the tumble). One figure holds the whole story.

Recall Feynman retelling of the whole walkthrough

A spinning thing carries a "spin arrow" pointing along its axis (Step 1). Because the object is heavier to spin one way than another, there are three special stiffness numbers (Step 2), and multiplying the spin arrow by these unequal numbers makes a second arrow — the momentum arrow — that points a different way (Step 3). Out in space, with nothing pushing, it is the momentum arrow that must hold still. But since it doesn't line up with the spin arrow, keeping momentum still forces the spin arrow to keep swinging. To do the bookkeeping we sit on the tumbling body; from that seat an arrow seems to move both because it really moves and because our seat is turning (Step 4). Writing "no push" in that language (Step 5) and grinding out the cross product (Step 6) gives Euler's three equations: each axis is spun up by the gap in the other two stiffnesses times the other two spins. On a ball there are no gaps, so it never wobbles; on a symmetric top the spin arrow just circles in a cone (Step 7). And for a lopsided body, one little sign decides all: spin about the biggest or smallest axis and disturbances circle harmlessly, but spin about the middle axis and they blow up — the phone flips itself end over end (Step 8). All of it lives on the crossing of two energy-and-momentum ellipsoids (summary).