- #1

raisins

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- 1

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!