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I was pondering rigid body mechanics and the non-vanishing Poisson brackets $\left\{J_x,J_y\right\}=J_z$ etc. However in arbitrary coordinates $q^i$ with conjugate momenta $p_i$ we are supposed to have $\left\{q^i,q^j\right\}=\left\{p_i,p_j\right\}=0$ and $\left\{p_i,q^j\right\}=\delta_i^j$, so what is going on here? Can we express the poisson bracket in terms of $\phi^i,J_j$ coordinates?

Qmechanic
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Joshua Tilley
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1 Answers1

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The symplectic manifold phase space for the classical version of the angular momentum algebra is the two-sphere $S^2$. The symplectic form , in terms of spherical polar angles $\theta$, $\phi$, is $$ \omega = \sin\theta\, d\theta\wedge d \phi, $$ i.e. the area 2-form for $S^2$. Then we take $$ J_x= \sin\theta \cos \phi\\ J_y= \sin\theta \sin\phi\\ J_z= \cos\theta $$ as the moment maps. Using hamilton's equations $dH= -\omega(V_H,-)$ with $H= J_x$, etc gives us $$ V_{J_x}=\sin\theta \partial_\theta+\cos\phi \cot \theta \partial_\phi\\ V_{J_y}=-\cos\theta \partial_\theta +\sin\phi \cot \theta \partial_\phi\\ V_{J_z}= \partial_\phi. $$ So from $$ \{J_x,J_y\}= \omega(V_{J_x},V_{J_y}) $$ we read read off (after a bit of algebra) that $$ \{J_x,J_y\}= \cos\theta =J_z. $$ The other Poisson brackets work the same.

Incidentaly, there is a discussion of the Lagrangian formalism for the Euler equations of a rigid body at Lagrangian of the Euler equations

mike stone
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  • Isn't the configuration space $SO\left(3\right)$? Also, I didn't phrase very well, but what I was really wondering is whether angular momentum is canonical momentum after a particular choice of coordinates, and if not, what is? – Joshua Tilley Dec 20 '19 at 16:03
  • @Joshu Tilley. The configuration space of a rigid body top is $SO(3)$ and the associated phase space is 6 dimensional and has infinite volume. Both classical and the quantum version of the top can have any value of total angular momentum $J$. The two dimensional $S^2$ is the phase space for a spin with a definite angular momentum $J$. The spin phase space must have finite volume becsue the quatum Hilbert space is finite dimensional. I discussed the spin case in my answer because if you just want the Poisson brackets its symplectic geometry is simpler than the top. – mike stone Dec 20 '19 at 17:01