# Partition function in spherical coordinates

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.

• Do you have a guess as to what $J$ might be? – jacob1729 Feb 7 '19 at 17:43
• @jacob1729 To me it is 1, but I do not know why. – Mauricio Feb 7 '19 at 17:44
• 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. – jacob1729 Feb 7 '19 at 17:56
• Related. Liouville’s theorem preserves phase space volumes under canonical transformations. – Cosmas Zachos Feb 8 '19 at 4:08
• peeterjoot.wordpress.com/2013/02/11/… Might help. – Cosmas Zachos Feb 8 '19 at 4:14

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.