Our professor's notes read, "In general, in Hamiltonian dynamics a constant of motion will reduce the dimension of the phase space by two dimensions, not just one as it does in Lagrangian dynamics." To demonstrate this, he uses the central force Hamiltonian,
$$H=\frac{P_r^2}{2m}+\frac{p_{\theta}^2}{2mr^2}+ \frac{p_{\phi}}{2mr^2 sin^2 \theta} + V(r).$$
By Hamilton's equation $\dot{p_{\phi}}=0$, this is a constant of the motion. As a result, specifying $p_{\phi}=\mu$ gives us a 5 dimensional manifold. The notes go on to state, "Furthermore, on each invariant submanifold the Hamiltonian can be written:
$$H=\frac{P_r^2}{2m}+\frac{p_{\theta}^2}{2mr^2}+ \frac{\mu}{2mr^2 sin^2 \theta} + V(r),$$
which is a Hamiltonian involving only two freedoms $r$ and $\theta$. Therefore the motion actually occurs on a 4-dimensional submanifold of the 5-dimensional submanifold of $T^*Q$ . . ."
However, to me it looks like we still have five degrees of freedom: $p_{\theta},$ $p_r,$ $r,$ $\theta,$ and $\phi$. Thus, I'm not sure what he means when he says that the presence of a constant of motion reduces the dimension of the cotangent manifold by 2. Is he saying that if we specify a numerical value for H, then the dimension is reduced from 5 to 4? Or does just the presence of a cyclic coordinate reduce the dimension from 6 to 4?
Best Answer
Recall that $$ \dot{\mathbf{p}}=-\frac{\partial H}{\partial\mathbf{q}} $$ Since $H$ does not actually depend on $\phi$, then $$ \dot{p}_\phi=0=-\frac{\partial H}{\partial\phi} $$
This will eliminate $p_\phi$ and $\phi$ from your coordinates: $H(p_r,p_\theta,p_\phi,r,\theta,\phi)\to H(p_r,p_\theta,r,\theta)$.