I Phase space integral in noninteracting dipole system

[raisins]

Hi all,

Consider a system of $N$ noninteracting, identical electric point dipoles (dipole moment $\vec{\mu}$) subjected to an external field $\vec{E}=E\hat{z}$. The Lagrangian for this system is
$$L=T-V=\sum_{i=1}^N\left\{\frac{m\dot{\vec{r}_i}^2}{2}+\vec{\mu}_i\cdot\vec{E}\right\}=\sum_{i=1}^N\left\{\frac{m\dot{\vec{r}_i}^2}{2}+E\mu\cos\theta_i\right\}$$
and the Hamiltonian is
$$H=\sum_{i=1}^N\left\{\frac{\vec{p}_i^2}{2m}-E\mu\cos\theta_i\right\}$$
As I see it, 5 sets of generalized coordinates $\left(\left\{\vec{r}_i\right\},\left\{\theta_i\right\},\left\{\phi_i\right\}\right)$, where $\theta_i,\phi_i$ are the usual spherical angles, are needed to describe this system. Now, the momenta conjugate to $\theta_i,\phi_i$ (call them $p_{\theta_i},p_{\phi_i}$) are both cyclic, since $\dot{\theta}_i,\dot{\phi}_i$ appear nowhere in the Lagrangian. But do we not still have to integrate over them to find the partition function; ie.
$$Z=\frac{1}{N!}\int \prod_{i=1}^N\frac{d^3\vec{r}_id^3\vec{p}_i}{h^3}\,\frac{d\theta_idp_{\theta_i}}{h}\,\frac{d\phi_idp_{\phi_i}}{h}e^{-\beta H}$$
But $p_{\theta_i},p_{\phi_i}\in[0,\infty)$ so doesn't that integral blow up? Am I wrong in thinking $p_{\theta_i},p_{\phi_i}$ can take any real, positive value? Or, because they're cyclic, do we just omit them from the integration?

Any help would be appreciated, thank you!

 

[anuttarasammyak]

I am not familiar with momenta conjugate. Taking polar coordinate for both coordinate and momenta spaces
dv=r^2 sin^2\theta dr d\theta d\phi
dV=R^2 sin^2\Theta dR d\Theta d\Phi
where I noted small r for coordinate space and capital R for momentum space.
So in total number of states in infinitesimal phase space elements is
\frac{\prod_{i=1}^N dv_idV_i}{h^{3N}}

 

[andresB]

But $p_{\theta_i},p_{\phi_i}\in[0,\infty)$ s

From where this comes from?