Hamilton's principle/Applications/The rigid body

The rigid body is the first system in this volume whose configuration space is not a piece of Euclidean space. Its orientations form a curved, three-dimensional group manifold, and that fact — rather than any difficulty in the forces — is what makes rigid-body motion instructive. It also makes the variational treatment worth doing carefully, because the naive coordinate approach hides the structure and introduces singularities that are not physical.

The configuration space

Stability of rotation about the three principal axes.

A rigid body is a system of particles whose mutual distances are fixed. Those constraints are holonomic in the sense of Hamilton's principle/Mathematics/Holonomic constraints — relations among coordinates, not velocities — so elimination is legitimate.

Eliminating them leaves six degrees of freedom: three for the position of the centre of mass and three for the orientation. The kinetic energy separates,

$T=\frac{1}{2}M|\dot{R}{|}^2+T_{\text{rot}}\mathrm{(}\mathrm{1}\mathrm{)}$

exactly as Hamilton separates the motion of the centre of gravity from the relative motion in the two-body problem of Hamilton's principle/Applications/Central force and the Kepler problem. The translational part is a free particle. Everything of interest is in $T_{\text{rot}}$.

The orientation is an element of the rotation group. That space is compact and curved, and it is not covered by any single coordinate patch — a fact that will reappear below as a defect of the Euler angles rather than of the body.

Rotational kinetic energy

The effective potential of a central-force problem.

In a frame fixed in the body, with axes along the principal axes of inertia, the angular velocity $\omega =({\omega }_1,{\omega }_2,{\omega }_3)$ gives

$T_{\text{rot}}=\frac{1}{2}(I_1{\omega }_1^2+I_2{\omega }_2^2+I_3{\omega }_3^2)\mathrm{(}\mathrm{2}\mathrm{)}$

with $I_1,I_2,I_3$ the principal moments of inertia. This is a metric on the orientation space, and it is a left-invariant one: the kinetic energy of a motion does not change if the whole body is first rotated. That invariance is a symmetry of the action, so Hamilton's principle/Mathematics/Noether's first theorem applies and the conserved quantity is the angular momentum in the space frame.

Euler angles, and their defect

The usual coordinates are the Euler angles $(\varphi ,\vartheta ,\psi )$, in terms of which

${\omega }_1=\dot{\varphi }\sin \vartheta \sin \psi +\dot{\vartheta }\cos \psi ,{\omega }_2=\dot{\varphi }\sin \vartheta \cos \psi -\dot{\vartheta }\sin \psi ,{\omega }_3=\dot{\varphi }\cos \vartheta +\dot{\psi }\mathrm{(}\mathrm{3}\mathrm{)}$

Substituting (3) into (2) gives a Lagrangian in three coordinates and their velocities, and the Euler-Lagrange equations of Hamilton's principle/Mathematics/Several dependent variables apply directly.

But the coordinates are singular at $\vartheta =0$. There $\varphi$ and $\psi$ generate the same rotation, the map from angles to orientations degenerates, and the equations blow up. Nothing happens to the body at that configuration; the singularity is in the chart. It is the same gimbal lock that afflicts mechanical gyroscope mountings, and it is a consequence of trying to cover a curved compact space with three angles.

This is worth stating plainly because it is a general lesson about the variational method: the Euler-Lagrange equations are only as good as the coordinates, and a singular chart produces singular equations for a non-singular motion.

The torque-free case

Take the body free, so $V=0$ and $L=T_{\text{rot}}$. Two conservation laws follow without solving anything.

$L$ carries no explicit time, so by Hamilton's principle/Mathematics/The Beltrami identity the energy is conserved:

$2T=I_1{\omega }_1^2+I_2{\omega }_2^2+I_3{\omega }_3^2=\text{const}\mathrm{(}\mathrm{4}\mathrm{)}$

$L$ is invariant under rotations of the body in space, so by Noether the space-frame angular momentum is conserved, and so is its magnitude:

$|𝐋{|}^2=I_1^2{\omega }_1^2+I_2^2{\omega }_2^2+I_3^2{\omega }_3^2=\text{const}\mathrm{(}\mathrm{5}\mathrm{)}$

Equations (4) and (5) are two surfaces in the space of body-frame angular velocity: an ellipsoid and a second ellipsoid with squared semi-axes. The motion is confined to their intersection, a closed curve, and the entire qualitative behaviour can be read off without integrating anything.

The full equations of motion, obtained from (2) and (3), are Euler's:

$I_1{\dot{\omega }}_1=(I_2-I_3){\omega }_2{\omega }_3\mathrm{(}\mathrm{6}\mathrm{)}$

with the other two by cyclic permutation. Note that (6) is written in the body frame, where the moments of inertia are constant; in the space frame the inertia tensor varies and the equations are worse.

The intermediate axis theorem

The intersection curves of (4) and (5) supply a result that is easy to state and easy to demonstrate.

Take $I_1<I_2<I_3$ and consider rotation about a principal axis, perturbed slightly. Linearising (6) about rotation near axis 1 gives a perturbation equation with growth rate squared

${\lambda }^2=\frac{(I_2-I_1)(I_3-I_1)}{I_2{I}_3}{\omega }_1^2\mathrm{(}\mathrm{7}\mathrm{)}$

up to sign, and the analogous expressions for the other axes. For the smallest moment, both factors in the numerator are positive; for the largest, both are negative; in each case the product of the corresponding rates is negative, $\lambda$ is imaginary, and the perturbation oscillates. Stable.

For the intermediate axis the two factors have opposite signs, $\lambda$ is real, and the perturbation grows exponentially. Unstable.

This is the tennis-racket theorem, and it is visible in the geometry as well: the intersection curves near the intermediate axis are hyperbolic rather than closed loops around the axis. A book or a phone spun about its intermediate axis tumbles; about either of the others it does not.

The result required no integration. It came from two conserved quantities and the shape of their intersection, which is the characteristic payoff of the conservation-law machinery.

The heavy symmetric top

Add gravity, with the body supported at a point on its symmetry axis. Then $I_1=I_2$, and

$L=\frac{1}{2}{I}_1({\dot{\vartheta }}^2+{\dot{\varphi }}^2{\sin }^2\vartheta )+\frac{1}{2}{I}_3(\dot{\varphi }\cos \vartheta +\dot{\psi }{)}^2-M{g}_0\lambda \cos \vartheta \mathrm{(}\mathrm{8}\mathrm{)}$

with $\lambda$ the distance from support to centre of mass.

Both $\varphi$ and $\psi$ are cyclic, so two momenta are conserved immediately, and the energy is a third constant. Three conserved quantities for three degrees of freedom: the system is integrable, and the problem reduces to a single first-order equation in $\vartheta$,

$\frac{1}{2}{I}_1{\dot{\vartheta }}^2+V_{\text{eff}}(\vartheta )=E^{\prime }\mathrm{(}\mathrm{9}\mathrm{)}$

which is a one-dimensional problem in an effective potential, exactly as the Kepler problem reduced on Hamilton's principle/Applications/Central force and the Kepler problem.

$V_{\text{eff}}$ rises to infinity at $\vartheta =0$ and $\pi$ when the two conserved momenta differ, so $\vartheta$ oscillates between turning points: the axis nutates while it precesses. The qualitative motion follows from the shape of a one-dimensional potential, without solving (9), which requires elliptic functions.

Provenance: every equation on this page is derived here or is standard. Euler's equations (6) are named for their author; no primary source for them has been verified for this volume, and none is quoted. See Hamilton's principle/Sources.

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Hamilton's principle/Applications/Continuum systems: the vibrating string takes the step from finitely many degrees of freedom to infinitely many.