Is the free 3D rigid body/Euler top an integrable system?

Question

Or is there chaos?

The base of the configuration space is 3D. There are 4 constants of motion, namely, the energy, and the 3 components of the angular momentum.

So there are still 2 degrees of freedom.

Answer 1

Ref. 1 argues that the 3D free rigid body/Euler top (which is well-known from e.g. the intermediate axis theorem/tennis racket theorem/Dzhanibekov effect) is an integrable system$^1$ as follows. Let us only discuss the generic case where the principal moments of inertia $(I_1,I_2,I_3)$ are not all equal$^2$.

1. On one hand, the 6D configuration manifold in the laboratory/inertial frame has 3 constants of motion$^3$ (COM): the angular momentum vector $\vec{L}\neq\vec{0}$; and 1 integral of motion (IOM): the kinetic energy $$T~\equiv~\sum_{i=1}^3\frac{L_i^2}{2I_i}~>~0,\tag{1}$$ which leaves a 2D surface, as OP already mentions. Ref. 1 gives topological arguments why it must be a 2-torus $\mathbb{S}^1\times \mathbb{S}^1$. (For constant $T$ the 2D surface is bounded and closed, and hence compact, for starters.)

2. On the other hand, Euler's equations for a free rigid body in the body/rotating frame closes on the angular momentum vector $\vec{L}\in\mathbb{R}^3\backslash \{\vec{0}\}$: $$\begin{align}\forall i ~\in~\mathbb{Z}_3:~~ \dot{L}_i ~\equiv~& I_i \dot{\Omega}_i ~=~\Omega_{i+1}(I_{i+1}-I_{i-1}) \Omega_{i-1}\cr ~\equiv~&L_{i-1}(I_{i-1}^{-1}-I_{i+1}^{-1}) L_{i+1} ~\stackrel{(1)+(3)}{=}~\{L_i,T\}. \end{align}\tag{2}$$ In fact it is a Hamiltonian system on a 3D Poisson manifold with $so(3)$ Poisson bracket $$ \forall i,j ~\in~\mathbb{Z}_3:~~ \{L_i,L_j\}~=~-\sum_{k=1}^3\epsilon_{ijk} L_k.\tag{3}$$ The angular momentum square $\vec{L}^2$ is a Casimir (and hence an IOM), so we can restrict to a 2D symplectic leaf $\mathbb{S}^2$, which is Liouville integrable. The Hamiltonian $T$ serves as an action variable, while the angle variable (and hence $\vec{L}$, and in turn $\vec{\Omega}$) parametrizes a circle $\mathbb{S}^1$.

3. Comparing section 1 & 2, we conclude that the second circle $\mathbb{S}^1$ only lives in the basemanifold of the 6D configuration manifold. It is possibly incommensurate with the first circle $\mathbb{S}^1$.

References:

1. V.I. Arnold, Mathematical methods of Classical Mechanics, 2nd eds., 1989; $\S$28B + $\S$29.

2. L.D. Landau & E.M. Lifshitz, Mechanics, vol. 1 (1976); $\S$37.

3. H. Goldstein, Classical Mechanics; Section 5.6.

4. W.B. Heard, Rigid Body Mechanics, (2006); Section 5.1.

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$^1$ In particular, there is no chaos.

$^2$ If some of the principal moments of inertia $(I_1,I_2,I_3)$ are equal, then at least 1 of the angular momentum components $L_i$ is a 3rd IOM (besides $T$ and $\vec{L}^2$), so that the symplectic leaf $\mathbb{S}^2$ becomes superintegrable. The symmetric top performs a regular precession. However, the 2-torus $\mathbb{S}^1\times \mathbb{S}^1$ may still be incommensurate.

$^3$ In the spirit of this Phys.SE post, we can realize an arbitrary moment of inertia tensor of a rigid body by e.g. using 6 point particles constrained to be pairwise opposite, and span an orthonormal frame. Then the Hamiltonian becomes $$ H ~=~\sum_{i=1}^3 \frac{{\bf p}_i^2}{m_i} ~-~ \sum_{i=1}^3 \lambda_i({\bf r}_i^2-1) ~-~ \frac{1}{2}\sum_{k=1}^3 \epsilon_{ijk}\mu_k(|{\bf r}_i-{\bf r}_j|^2-2),\tag{4}$$ with $3\times 3=9$ position variables $({\bf r}_1,{\bf r}_2,{\bf r}_3)$ and where $(\lambda_1,\lambda_2,\lambda_3,\mu_1,\mu_2,\mu_3)$ are 6 Lagrange multipliers. In the laboratory/inertial frame the angular momentum vector $$\vec{L}~=~2\sum_{i=1}^3{\bf r}_i\times {\bf p}_i\tag{5}$$ is formally 3 IOM in this constrained system. We cannot eliminate the constraints in a global manner, so it is not immediately clear from this construction (4) that the system is Liouville integrable. In contrast, we know from the description in the body/rotating frame, that $T$ and $\vec{L}^2$ are 2 genuine IOM.

Answer 2

Yes - rigid body motion in 3 space can be modeled by the Euler equations, which because of cross-axis coupling are nonlinear - but they can be integrated. A complete reference on how to model rigid body dynamics is given in these MIT lecture Notes. Solve equations (9), (10) and (11) for the rate of change of angular velocity around each principle axis, and integrate each to solve for the angular velocities. In this form you would have the set of equations needed for simulation, and you would need to specify the initial conditions of angular momentum and velocity - as well as the body's moments of inertia.

The lecture notes also provide a good discussion on the stability of these equations which are dependent on the ratios of the moments of inertia.