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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?

Ivo Terek
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    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 $M$. – childofsaturn Jul 06 '18 at 04:02
  • My goal is to get a feeling for Marsden-Weinstein reduction, and how to reduce Hamiltonians in practice. I just took a course in symplectic geometry, and I wanted to see it in action (no pun intended). Perhaps I misunderstood the generality of $H$... So I can take any $V\colon \Bbb S^2 \to \Bbb R$, assume ${\rm SO}(2)$-invariance and go for it? – Ivo Terek Jul 06 '18 at 04:06
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    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 $S^2$. – childofsaturn Jul 06 '18 at 04:14
  • Ah, now I get what you're saying. Just to confirm (and you can point me to some reference if you don't want to explain everything), any physically significant Hamiltonian will always have that $\frac{1}{2}g^{ij}p_ip_j$ term because this would be the Legendre transform of $L\colon TM \to \Bbb R$ given by $L(v) = \langle v,v\rangle/2$? I know from differential geometry that the Euler-Lagrange equations for this Lagrangian $L$ produce the geodesic equations, but I never played around with Legendre transforms and such (just know that they should appear here somewhere)... – Ivo Terek Jul 06 '18 at 04:19
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    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. – childofsaturn Jul 06 '18 at 04:25
  • Then I'll look it up. Thank you very much for the comments, they were very helpful! :) – Ivo Terek Jul 06 '18 at 04:27
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    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. – childofsaturn Jul 06 '18 at 04:33
  • That makes sense... since $\dim T^*\Bbb S^1 = 2$ is "too low", it doesn't seem that you can reduce $H$ at all. I guess setting up the Hamilton equations should be easy (just $\dot{q} =\cdots$ and $\dot{p}=\cdots$), but the resulting system must be a mess. – Ivo Terek Jul 06 '18 at 04:38
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    @IvoTerek Perhaps some interesting examples in this: http://wwwf.imperial.ac.uk/~dholm/classnotes/GeomMech2-2nd.pdf – RedPen Jul 06 '18 at 06:35
  • Related: https://physics.stackexchange.com/q/111216/2451 – Qmechanic Jul 06 '18 at 08:15

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