Rigid body dynamics — Euler angles, Euler's equations of motion: Visual walkthrough

Step 1 — What is spinning, and how do we measure it?

WHAT. Picture a solid brick tumbling in empty space. At any instant it is turning about some line through its center — an axis — at some speed. We bundle "which line" and "how fast" into a single arrow called the angular velocity, written $ω$ (Greek "omega").

WHY an arrow? Because a spin needs two pieces of information: a direction (the axis it turns about) and a size (how fast). An arrow carries exactly those two things — its direction is the axis, its length is the turning rate in radians per second. We point the arrow along the axis using the right-hand rule: curl your right fingers the way the body turns, your thumb points along $ω$ .

PICTURE. The amber arrow is the spin axis; the cyan circle is the body sweeping around it.

Figure — Rigid body dynamics — Euler angles, Euler's equations of motion

Step 2 — Two ways to look: the lab and the bug

WHAT. There are two viewpoints on the tumbling brick:

The space frame $(X, Y, Z)$ — you, standing still in the lab, watching it tumble.
The body frame $(x, y, z)$ — a bug glued to the brick, riding along, turning with it.

WHY two frames? Because the brick's mass distribution looks constant to the bug and constantly changing to you. A moment we will meet — the inertia tensor — is frozen for the bug and a shifting nightmare for you. So the bug's frame makes the algebra clean. The catch: the bug's frame is rotating, and we will have to pay a price for that in Step 5.

PICTURE. Same brick, two coordinate crosses: white $(X, Y, Z)$ fixed, cyan $(x, y, z)$ tilted with the body.

Step 3 — The inertia tensor, and why we align to principal axes

WHAT. Spinning has a "reluctance": some bodies resist being spun about one axis more than another. That reluctance is the inertia tensor $I$ . It is the machine that turns the spin arrow $ω$ into the angular momentum arrow $L$ (the "amount of spinning-motion stored"): $L = I ω .$

WHY does $L$ point a different way than $ω$ ? For a lopsided body, $I$ tilts the output: you spin about one line but the stored angular momentum leans toward a heavier direction. In general $I$ is a full 3×3 table of nine numbers (messy). But every rigid body has three special perpendicular axes — the principal axes — along which spinning produces $L$ parallel to $ω$ . Choose the body frame $(x, y, z)$ to lie along them and the table collapses to just three numbers on its diagonal: $I = I_{1} 00 0 I_{2} 0 00 I_{3}, L_{i} = I_{i} ω_{i} .$

WHY bother? Because $L_{i} = I_{i} ω_{i}$ (three clean scalar products) beats a full matrix multiply every time. This one choice is the reason Euler's equations are short. More detail in Inertia tensor and principal axes.

PICTURE. Left: a general body — $ω$ (amber) and $L$ (cyan) point different ways. Right: spin about a principal axis — the two arrows line up.

Step 4 — Newton's law for rotation

WHAT. A twisting push is a torque, written $τ$ (Greek "tau"). The fundamental law of rotation — the rotational cousin of $F = m a$ — says: torque equals the rate of change of angular momentum, measured in the fixed lab frame. $τ = (\frac{d L}{d t})_{space} .$

WHY the "space" subscript? Because this law is only true in an inertial (non-accelerating) frame — the lab. The little symbol $(\frac{d}{d t})_{space}$ reads "the rate of change as the lab observer sees it." That distinction is about to matter enormously.

PICTURE. A wrench applies torque $τ$ ; the angular-momentum arrow $L$ grows in the direction of $τ$ over a small time.

Step 5 — The price of riding the body: the transport theorem

WHAT. We want to work in the bug's frame (where the $I_{i}$ are constant), but the law lives in the lab frame. So we need the bridge between "how the lab sees a vector change" and "how the bug sees it change." That bridge is the transport theorem: $(\frac{d A}{d t})_{space} = (\frac{d A}{d t})_{body} + ω \times A .$

WHY the extra $ω \times A$ term? Imagine an arrow $A$ painted on the brick and not moving at all for the bug. The lab observer still sees it swinging around, because the whole brick is turning. That swinging is a tiny sideways shift, perpendicular to both the spin axis $ω$ and the arrow $A$ — exactly what the cross product $ω \times A$ produces.

WHY the cross product specifically? We need an operation that (a) gives zero when $A$ lies along the spin axis (a point on the axis doesn't move) and (b) is largest when $A$ sticks straight out (that point swings fastest). The cross product does precisely this: $∣ ω \times A ∣ = ∣ ω ∣∣ A ∣ sin α$ , zero when the angle $α = 0$ , maximal at $α = 9 0^{\circ}$ , and pointing sideways (perpendicular to both).

PICTURE. An arrow fixed to the spinning body; the bug sees it still, but the lab sees its tip trace a circle — the velocity of the tip is $ω \times A$ (amber), perpendicular to the grey radius.

Step 6 — Feed $L$ through the bridge

WHAT. Apply the transport theorem with $A = L$ , then substitute into Newton's rotation law from Step 4: $τ = (\frac{d L}{d t})_{body} + ω \times L .$

WHY. This is the whole trick in one line. The left side is the lab law. On the right, the first term is now measured in the bug's frame, where $L = I ω$ with constant $I_{i}$ — so it will differentiate cleanly. The second term is the unavoidable rotating-frame surcharge.

PICTURE. A flow diagram: lab rate splits into "bug rate" + "gyroscopic swing"; below, $L$ (cyan) and its swing $ω \times L$ (amber) drawn as arrows.

Step 7 — Read off the three components

WHAT. Write Step 6 out along the three principal axes. Two ingredients:

In the body frame, $L_{i} = I_{i} ω_{i}$ with each $I_{i}$ constant, so $(\frac{d L _{i}}{d t})_{body} = I_{i} \overset{ω}{˙}_{i},$ where $\overset{ω}{˙}_{i}$ (omega with a dot) means "the rate at which the number $ω_{i}$ changes."
The cross product, first component: $(ω \times L)_{1} = ω_{2} L_{3} - ω_{3} L_{2} = ω_{2} (I_{3} ω_{3}) - ω_{3} (I_{2} ω_{2}) = (I_{3} - I_{2}) ω_{2} ω_{3} .$

WHY $ω_{2} L_{3} - ω_{3} L_{2}$ ? That is the definition of the first slot of a cross product: it mixes the other two axes (2 and 3), never its own. Cycling $1 \to 2 \to 3 \to 1$ gives the other two slots.

PICTURE. The cyclic wheel $1 \to 2 \to 3 \to 1$ showing which indices feed each component, with the sign pattern.

Assembling — putting each component of $τ = (d L / d t)_{body} + ω \times L$ together:

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}}$$ > [!formula] Euler's equations, term by term (take equation 1) > $$\underbrace{I_1}_{\text{reluctance about axis 1}}\ \underbrace{\dot\omega_1}_{\text{how fast the axis-1 spin changes}}\ -\ \underbrace{(I_2-I_3)}_{\text{shape asymmetry}}\ \underbrace{\omega_2\omega_3}_{\text{the other two spins, multiplied}}\ =\ \underbrace{\tau_1}_{\text{applied twist about axis 1}}$$ > The moved-over cross-product piece became $-(I_2-I_3)\omega_2\omega_3$ (sign flips crossing the equals sign). We hunted this box since the top of the page — it is now fully earned. --- ## Step 8 — The degenerate & limiting cases (never leave a gap) **WHAT & WHY.** A derivation you can trust must survive its edge cases. Here are all of them. **Case A — sphere, all equal ($I_1=I_2=I_3$).** Every difference $I_i-I_j=0$, so all coupling terms die: $$I\dot\omega_i = \tau_i.$$ With no torque, $\boldsymbol\omega$ is frozen — **any** axis is a stable spin axis. This is the one case where the naive "$\boldsymbol\tau=\mathbf I\dot{\boldsymbol\omega}$" is actually right. **Case B — spin exactly on a principal axis, no torque.** Say $\omega_2=\omega_3=0$. Then equation 1 gives $I_1\dot\omega_1 = 0$: the spin holds forever. A body spun perfectly about a principal axis never wanders. **Case C — symmetric top, torque-free ($I_1=I_2\ne I_3$, $\boldsymbol\tau=0$).** Equation 3 loses its coupling ($I_1-I_2=0$), so $\omega_3=$ const. Equations 1,2 become, with the constant $\Omega \equiv \dfrac{I_3-I_1}{I_1}\omega_3$, $$\dot\omega_1 = -\Omega\,\omega_2,\qquad \dot\omega_2 = +\Omega\,\omega_1,$$ whose solution $\omega_1=A\cos\Omega t,\ \omega_2=A\sin\Omega t$ makes the transverse spin **circle** the symmetry axis at rate $\Omega$ — free precession, the [[Chandler wobble]] of the Earth. **Case D — intermediate-axis (tennis-racket) instability.** For $I_1<I_2<I_3$, spin nearly about the middle axis 2 and linearize: the small deviations obey $\ddot q = +k^2 q$ with $k^2>0$, so they **grow exponentially** — the flipping-phone effect. About the largest or smallest axis the sign flips to $-k^2$ → stable oscillation. **PICTURE.** Four mini-panels, one per case: sphere (arrows anywhere), locked principal spin, circling transverse spin, and the tumbling middle-axis flip. ![[deepdives/dd-physics-2.1.21-d2-s08.png]] > [!mistake] Don't skip the sign when crossing the equals sign > $\boldsymbol\omega\times\mathbf L$ sits on the *right* with $+(I_3-I_2)\omega_2\omega_3$. Moving it to the left with the $I_1\dot\omega_1$ flips it to $-(I_2-I_3)\omega_2\omega_3$. Both boxed forms are equal — just watch which side the term lives on. --- ## The one-picture summary ![[deepdives/dd-physics-2.1.21-d2-s09.png]] Everything on one blueprint: the spin arrow $\boldsymbol\omega$ (Step 1) feeds the inertia machine on principal axes to make $\mathbf L$ (Step 3); the lab law $\boldsymbol\tau=(d\mathbf L/dt)_{\text{space}}$ (Step 4) passes through the transport bridge (Step 5–6) and splits into the bug's clean rate plus the gyroscopic swing (Step 7), landing on Euler's boxed equations. > [!recall]- Feynman retelling — the whole walk in plain words > Picture a brick tumbling in space. To describe its spin we use one arrow, $\boldsymbol\omega$: it points along the axis and its length is how fast (Step 1). Now, two people watch: you in the lab, and a bug glued to the brick (Step 2). The bug's life is easier, because to the bug the brick's "heaviness in each direction" — the inertia — never changes, and along three special axes it becomes just three plain numbers $I_1,I_2,I_3$ that turn spin into stored spinning-motion $\mathbf L$ (Step 3). The rulebook of rotation says a twist $\boldsymbol\tau$ changes $\mathbf L$ — but that rule only works for the lab observer (Step 4). To move the rule onto the bug's desk we need a bridge: whenever you switch to a spinning viewpoint, any arrow gains an extra swing $\boldsymbol\omega\times\mathbf L$, because the whole frame is turning under it (Step 5). Push $\mathbf L$ across that bridge (Step 6), read off the three axes using the fact that a cross product always mixes the *other two* directions (Step 7), and out drop Euler's three equations. The surprise inside them: even with nobody pushing, the multiplied spins $\omega_j\omega_k$ times the shape-difference $I_j-I_k$ keep changing the spin — that's why a tumbling phone flips about its middle axis but not its ends (Step 8). > [!recall]- Quick self-check > Why does the body frame make the algebra clean? ::: Because the inertia tensor is constant (diagonal, $I_i$ fixed) there, unlike in the lab where it changes every instant. > Where does the $\boldsymbol\omega\times\mathbf L$ term come from? ::: The transport theorem — a vector fixed in a rotating frame still swings for the lab observer, and that swing is $\boldsymbol\omega\times$ the vector. > What kills all the coupling terms? ::: Equal moments of inertia ($I_1=I_2=I_3$, a sphere) — every $I_i-I_j=0$. > In equation 1, what does $(I_2-I_3)\omega_2\omega_3$ represent physically? ::: Gyroscopic coupling — the spins about axes 2 and 3 driving a change about axis 1, present even with zero torque.