= Solution
Assume the velocity-to-momentum Legendre map is locally invertible. Differentiate the <Hamiltonian> while regarding $q,p,t$ as independent:
$$
dH=\sum_i\dot q_i\,dp_i+\sum_i p_i\,d\dot q_i-dL=\sum_i\dot q_i\,dp_i-\sum_i\frac{\partial L}{\partial q_i}dq_i-\frac{\partial L}{\partial t}dt.
$$
The velocity variations cancel because $p_i=\partial L/\partial\dot q_i$. Combining the remaining coefficients with the <Euler-Lagrange equations> gives <Hamilton's equations>
$$
\boxed{\dot q_i=\frac{\partial H}{\partial p_i},\qquad\dot p_i=-\frac{\partial H}{\partial q_i}.}
$$
An <ignorable coordinate> does not occur explicitly in the Lagrangian, so its conjugate momentum is conserved. For the spherical motion, $\phi$ is ignorable and the <Hamiltonian> has no explicit time dependence. The two constants are \b[$p_\phi$ and $H$].
The polar <Hamilton equations> are
$$
\dot\theta=\frac{p_\theta}{ma^2},\qquad\dot p_\theta=\frac{p_\phi^2\cos\theta}{ma^2\sin^3\theta}-V'(\theta).
$$
A constant polar angle therefore requires $p_\theta=0$ and vanishing of the second right-hand side. For $0<\theta<\pi$ with $\cos\theta\ne0$, this is
$$
\boxed{p_\phi^2=\frac{ma^2\sin^3\theta}{\cos\theta}V'(\theta).}
$$
The expression must be nonnegative to represent a real momentum. Once it holds, $\phi$ advances uniformly at $p_\phi/(ma^2\sin^2\theta)$, so it is also sufficient with those initial data. At the equator the undivided equation instead requires $V'(\pi/2)=0$, with any $p_\phi$; dividing by cosine would wrongly discard this special case. The polar coordinate chart itself excludes the poles.
Back to article page
