[Physics] Calculation of the partition function for a classical 2D gas lying on the surface of a sphere of constant radius $R$

classical-mechanicscoordinate systemspartition functionstatistical mechanics

I'm kind of confused with this system. My first question is. Is the Hamiltonian of one particle of this gas the following?
$$H(x,y,z,p_{x},p_{y},p_z)=\frac{1}{2m}\left(p_{x}^{2}+p_{y}^{2}+p_{z}^{2}\right)\to \frac{1}{2m}\left(\frac{p_{\theta}^{2}}{R^{2}}+\frac{p_{\phi}^{2}}{R^{2}\sin^{2}\theta}\right)$$
with $p_{\phi}=$constant.

If so, then my second question is: in order to obtain the classical partition function of one particle, should I integrate over the 6-dimensional phase space (Considering in spherical coordinates $r=$constant, $p_r=0$ and $p_{\phi}=$constant)? i.e.
$$Z=\int \exp\left[-\frac{\beta}{2m}\left(\frac{p_{\theta}^{2}}{R^{2}}+\frac{p_{\phi}^{2}}{R^{2}\sin^{2}\theta}\right)\right] J\, drd\theta d\phi dp_rdp_{\theta}dp_{\phi} \qquad (1)$$

or should I just integrate over the 4-D phase space, since $r=$constant, $p_r=0$?

$$Z=\int \exp\left[-\frac{\beta}{2m}\left(\frac{p_{\theta}^{2}}{R^{2}}+\frac{p_{\phi}^{2}}{R^{2}\sin^{2}\theta}\right)\right] J\, d\theta d\phi dp_{\theta}dp_{\phi} \qquad (2)$$

And my final question is: in the expressions (1) and (2), is $J=1$ or $J=R^2\sin \theta $ ?

Best Answer

  1. Yes. The Hamiltonian is your $H$. $p_r=0$ because $\frac{\partial L}{\partial \dot r}=0$ with $L:=\frac{m}{2}(R^2\dot\theta^2+R^2\sin^2\theta\dot\phi^2)$.

  2. The integral is over the 4-dimensional phase space: $(\theta,\phi,p_{\theta},p_{\phi})$, because the particles just move over the 2D surface of the sphere.

  3. $J=1$ regardless of the coordinate system you are expressing the Hamiltonian. This is because of the Liouville's theorem that states that the phase volume is conserved under canonical transformations. In particular a coordinate transformation is included. So, in 3 dimensions:

$$Z=\frac{1}{h^3}\int d^3q d^3p \, \, \exp[-\beta H(\bar q,\bar p)] \qquad (*)$$

being $\bar q = (x,y,z)$ or $q=(r,\theta,\phi)$, etc.

*Note. The constant $h$ introduced in the integral (*) in order to maintain $Z$ dimensionless. So $h$ is just a constant (yet unknown), with units of action, i.e. units of angular momentum.

Related Question