Context: I know that if we have a particle (say, with unit mass) moving in the plane $\Bbb R^2$ subject to a spherically symmetric potential $V\colon \Bbb R^2 \to \Bbb R$, it will move along the integral curves of the Hamiltonian vector field of $H\colon T^*\Bbb R^2 \cong \Bbb R^4 \to \Bbb R$ given by $$H(q^1,q^2,p_1,p_2) = \frac{p_1^2+p_2^2}{2} + V(r),$$great. Since $V$ and $H$ are ${\rm SO}(2)$-invariant, we have the moment map of the induced action ${\rm SO}(2)\circlearrowright \Bbb R^4$ given in polar coordinates by $\mu(r,\theta,p_r,p_\theta) = p_\theta$, so we can reduce the Hamiltonian at a level $\xi \neq 0$ to $$H_\xi(r,p_r) = \frac{p_r^2}{2} + V_{\rm eff}(r),$$where $V_{\rm eff}(r) = V(r) + \xi^2/2r^2$ is the effective potential.

Question: I wanted, as a self-posed exercise, see what happens if we look at a similar situation in the unit sphere $\Bbb S^2$, with ${\rm SO}(2)$ acting by rotation in the $z$-axis. We can write $$T^*\Bbb S^2 = \{(q^1,q^2, q^3, p_1,p_2,p_3)\in \Bbb R^6 \mid (q^1)^2+(q^2)^2+(q^3)^2 = 1\mbox{ and } q^1p_1+q^2p_2+q^3p_3=0\}.$$Calling spherical coordinates $$q^1 = \cos\theta\cos\phi, \quad q^2 = \sin\theta\cos\phi, \quad q^3 = \sin \phi,$$I went through the hassle of computing $p_\theta$ and $p_\phi$ in terms of $p_1$, $p_2$ and $p_3$, and I checked that the moment map of the induced action ${\rm SO}(2)\circlearrowright T^*\Bbb S^2$ is given in these coordinates by $\mu(\theta,\phi,p_\theta,p_\phi) = p_\theta$. So I'd like to know

what is the Hamiltonian that governs the motion of a particle in the sphere? As in, what should I reduce? Of course, I can take anything ${\rm SO}(2)$-invariant and nice enough, but I want something physically meaningful.

I google around a bit, but found only things about quantum mechanics, which is not the case. And I'm also not so quick on my feet with the classical references since I'm a mathematician with no training in physics whatsoever. I thought about adding some potential to the "kinectic energy" $\|p\|^2/2$, but spherical symmetry is sort of a given here, since we're constrained to the sphere, so I got lost.

As a side question, how much harder this gets if we look at the "full" action ${\rm SO}(3)\circlearrowright \Bbb S^2$? Is there a way to escape using Rodrigues' rotation formula and stuff like that?

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    $\begingroup$ What's your goal? To compute/solve for motion on the sphere? The general Hamiltonian on a manifold $M$ just reads $H = \frac{1}{2}g^{ij}p_i p_j + V$ where $g$ is the Riemannian metric on $M$, $p$ are coordinates for the fiber directions in $T^*M$ and $V: M \rightarrow \mathbb{R}$ is the potential. If $V=0$ this is just the geodesic flow on $M$. $\endgroup$ – childofsaturn Jul 6 '18 at 4:02
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    $\begingroup$ I think even the case $V=0$ is quite interesting to work out on its own, in which case you recover many of the standard aspects of Riemannian geometry such as the geodesic flow, isometry groups, Killing vector fields and so on. But otherwise the $V$ you choose depends on the physical situation you are considering. For instance if you want to consider a particle moving on the sphere but under the influence of gravity, $V$ would be the height function on $S^2$. $\endgroup$ – childofsaturn Jul 6 '18 at 4:14
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    $\begingroup$ That's the correct picture, yes. I am not sure of a place where this type of thing is worked out in satisfactory detail but chapter 1 of Takhtajan's book "Quantum Mechanics for Mathematicians" contains some good stuff (the first chapter is actually on classical mechanics) including the Legendre transform that you mention. $\endgroup$ – childofsaturn Jul 6 '18 at 4:25
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    $\begingroup$ One last thing I should add is that adding some simple $V$ can make a very simple problem quite non-trivial. For example take a particle moving on circle. It's very trivial to solve the motion for $V=0$ but turning on $V$ to be the height function (imagine a vertical circle) i.e $V = \text{sin}(\theta)$, you need to invoke elliptic functions to solve it exactly. $\endgroup$ – childofsaturn Jul 6 '18 at 4:33
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    $\begingroup$ @IvoTerek Perhaps some interesting examples in this: wwwf.imperial.ac.uk/~dholm/classnotes/GeomMech2-2nd.pdf $\endgroup$ – RedPen Jul 6 '18 at 6:35

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