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Suppose I write the Hamiltonian/energy of my system in spherical coordinates ($r,\theta,\varphi$) with conjugated momentums($p_r,p_\theta,p_\varphi$).

How do I calculate the partition function?

If

$$Z=\int e^{-\beta H}d^3r\;d^3p=\int e^{-\beta H}r^2\sin(\theta)J(p_r,p_\theta,p_\varphi)drd\theta d\varphi\;dp_rdp_\theta dp_\varphi,$$

what should $J$ be?

Edit: to add to my question, I tried to write the momentums as functions of $p_x,p_y,p_z$ (and thus as a function of $x,y,z$ also) but it is a mess and I do not think that's the good approach.

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  • $\begingroup$ Do you have a guess as to what $J$ might be? $\endgroup$
    – jacob1729
    Feb 7, 2019 at 17:43
  • $\begingroup$ @jacob1729 To me it is 1, but I do not know why. $\endgroup$
    – Mauricio
    Feb 7, 2019 at 17:44
  • $\begingroup$ sorry I initially misread this (thought you were imposing spherical polars on $p$ space). I suspect the answer is to consider the change of variables as a canonical transform, but am not actually that sure. $\endgroup$
    – jacob1729
    Feb 7, 2019 at 17:56
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    $\begingroup$ Related. Liouville’s theorem preserves phase space volumes under canonical transformations. $\endgroup$ Feb 8, 2019 at 4:08
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    $\begingroup$ peeterjoot.wordpress.com/2013/02/11/… Might help. $\endgroup$ Feb 8, 2019 at 4:14

1 Answer 1

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I'll spare you the cotangent bundles and differential geometrese and just summarize the takeaway that, in fact, $J=1/r^2 \sin \theta$ so that the entire phase-space Jacobian is 1, $$ dx dy dz ~ dp_x dp_y dp_z = dr d\theta d\phi ~dp_r dp_\theta dp_\phi . $$

A direct (blood, sweat and tears) derivation is available in Peter Joot's Blog.

The reason is that Cartesian to spherical is a point canonical transformation, so it preserves phase-space volumes (Liouville's theorem―which also holds for motion, since that is also a canonical transformation generated by the Hamiltonian).

To rationalize this, consider a free particle of mass m =1. The Hamiltonian is then $\vec p ^2/2$, generating $$ \frac{d\vec r}{dt} = \{\vec r , \vec p ^2 \}/2 = \vec p. $$ It is simple in Cartesian coordinates, but in spherical coordinates, given the line element $$ d\vec r= \hat r ~ dr +\hat \theta ~ r d\theta + \hat \phi ~ r \sin\theta d\phi, $$ you have $$ \dot{\vec r}= \hat r ~ \dot{r} +\hat \theta ~ r \dot{\theta} + \hat \phi ~ r \sin\theta ~\dot{\phi} \\ =\vec p= \hat r ~ p_r +\hat \theta ~ \frac{1}{r} p_\theta + \hat \phi ~\frac{1}{ r \sin\theta} p_\phi ~~. $$

These are the canonical conjugate momenta gotten from the canonical procedure and, e.g., $p_\phi$ is not the projection of $\vec p$ in the direction $\phi$!

You've seen this covariant bit before in the gradient expressed in spherical coordinates, $$ \nabla = \hat r ~ \partial_r +\hat \theta ~ \frac{1}{r} \partial_\theta + \hat \phi ~\frac{1}{ r \sin\theta} \partial_\phi , $$ not coincidentally, as it is proportional to the quantization of the momentum when you transcend classical mechanics.

Twice the Hamiltonian is, in this language, $$ 2H= \vec p^2 = p_r^2 + \frac{p_\theta^2}{r^2} + \frac{p_\phi^2}{r^2\sin^2\theta }, $$ (and the Liouville one-form would be $\vec p \cdot d\vec r = p_r dr + p_\theta d\theta +p_\phi d\phi $. In components, $p_r=\dot{r}, \quad p_\theta /r = r \dot{\theta}, \quad p_\phi/ r \sin\theta= r\sin \theta ~ \dot{\phi} $. )

The volume element in phase space, then, by above, is $$ d^3 \vec r ~ d^3 \vec p= r^2 \sin \theta ~ dr d\theta d\phi ~ \frac{1}{r^2 \sin \theta} dp_r dp_\theta dp_\phi, $$ collapsing to the top line.

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