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What is the canonical momentum in rigid body systems? It appears to me that it can't be the angular momentum as these have non-vanishing Poisson brackets, but any canonical coordinates have $\left\{p_i,p_j\right\}=0$.

If I'm right, what is the relation between canonical momentum and angular momentum?

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  • $\begingroup$ Well, that's just different Poisson structures. $\endgroup$
    – Qmechanic ♦
    Commented Dec 20, 2019 at 16:57
  • $\begingroup$ I think you will have to consider two new Hamiltonian variables $\theta_i$ and $l_i$ and the Poisson brackets should be written in terms of these variables. Here $l_i$ will be the canonical momentum $\frac{\partial L}{\partial \dot{\theta}}$. $\endgroup$
    – Manvendra Somvanshi
    Commented Dec 20, 2019 at 17:27
  • $\begingroup$ @Qmechanic, not sure what you mean. Can you address my argument? $\endgroup$
    – Joshua Tilley
    Commented Dec 20, 2019 at 20:52

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The rigid body, though Hamiltonian, is not a symplectic dynamical system, hence does not follow the same pattern as textbook mechanics. Instead it is described by a Lie-Poisson bracket with nontrivial Casimirs. These are functions with zero Poissson bracket with everything, hence constants of the motion. In general, fixing an independent set of Casimirs at a fixed value produces a foliation of the Poisson manifold into symplectic leaves. This is discussed in detail in the Book Mechanics and symmetry by Marsden and Ratiu.

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