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For a rigid body with three distinct positive principal moments of inertia $0<I_1<I_2<I_3$ and zero external torque about its centre of mass, steady rotation about the first or third principal axis is Lyapunov stable in angular velocity, while rotation about the intermediate principal axis is unstable. Here steady rotation means $\mathbf{\omega }=\Omega {\hat{e}}_i$ in the principal body frame, with constant $\Omega \ne 0$. A small sideways perturbation remains small about the extreme principal moments of inertia, but generically grows about the intermediate one. Stable does not mean that the wobble decays: without damping, there is no asymptotic stability. Coincident principal moments of inertia and a stationary body are excluded.^[ref]

No external torque is needed for a body to tumble. Conservation of angular momentum fixes $\mathbf{L}$ in the fixed space frame, but the body can turn relative to $\mathbf{L}$, so its angular velocity components in the principal body frame change. The torque-free rigid-body invariants constrain this motion. A supported gyroscope with gravitational torque has different governing assumptions.

Let $e_1,e_2,e_3$ be a principal body frame, let $I_1,I_2,I_3$ be the corresponding principal moments of inertia, and write the angular velocity as $\mathbf{\omega }={\sum }_i{\omega }_i{e}_i$. The body-frame components of the angular momentum areNo external torque acts, so $\mathbf{L}$ is constant in the fixed space frame. Since a body-fixed basis vector satisfies ${\dot{e}}_i=\mathbf{\omega }\times e_i$,Taking body-frame components gives the Euler equations for a torque-free rigid body:

A fixed space frame is an inertial orthonormal frame whose axes remain fixed in space. A principal body frame is attached to the rigid body and aligned with the principal axes of its inertia tensor, so that the tensor is diagonal with entries $I_1,I_2,I_3$.

If $e_i(t)$ are the body axes and $\omega$ is the angular velocity, then the derivative of a body-fixed basis vector isIn the principal frame,For torque-free motion, the angular momentum has zero space derivative. ThereforeTaking body-frame components gives the Euler equations for a torque-free rigid body:

For an axisymmetric body, put $I_1=I_2=I$ and let $e_3$ be its symmetry axis. Decompose $\omega ={\omega }_1{e}_1+{\omega }_2{e}_2+{\omega }_3{e}_3$. ThenThus $L$ is a linear combination of $\omega$ and $e_3$, proving that the angular momentum, angular velocity, and symmetry axis are always coplanar. This is the coplanarity in a torque-free axisymmetric rigid body.

A useful readout of a tumble is the alignment of the intermediate principal axis ${\hat{e}}_2$ with the conserved angular momentum direction $\hat{L}$. In the principal body frame, write $L_i=I_i{\omega }_i$ and $L=|\mathbf{L}|$. ThenA change from $c_2\simeq +1$ to $c_2\simeq -1$ means that the directed body axis has turned nearly $180^{\circ }$ relative to $\mathbf{L}$. It does not mean that $\mathbf{L}$ reversed in the fixed space frame. The unsigned line through that principal axis hides this distinction, so a marked end matters in an actual demonstration.

For the normalized intermediate-axis model with $\mathbf{\omega }(0)=(0.01,1,0)$, the rotational kinetic energy and squared angular momentum are $E=1.00005$ and $L^2=4.0001$. Hence $L^2-2E{I}_2=-0.0001$: this initial condition lies close to, but off, the intermediate-axis separatrix. An independent calculation evolves ${\dot{\mathbf{L}}}_{\mathrm{body}}={\mathbf{L}}_{\mathrm{body}}\times I^{-1}{\mathbf{L}}_{\mathrm{body}}$ and a rotation matrix taking body components to space components. The intermediate principal axis crosses $c_2=0$ at normalized times approximately $10.3773$ and $31.1319$, alternating long nearly aligned intervals with quicker reversals.

Both calculations use step size $0.005$ without rescaling either invariant. Over $0\le t\le 50$, the tumbling calculation changes rotational kinetic energy and squared angular momentum by less than $4\times 10^{-13}$ relative to their initial values. Reconstructing $\mathbf{L}$ in the fixed space frame gives relative error below $3\times 10^{-12}$. Halving the step changes the body components and rotation matrix entries by less than $3\times 10^{-10}$. These checks distinguish physical tumbling from numerical drift. On the exact intermediate-axis separatrix, approach to an intermediate-axis rotation takes infinite time; the recurrent off-separatrix example here is different.