Phase space integral in noninteracting dipole system

  • #1
raisins
3
1
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!
 
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  • #2
I am not familiar with momenta conjugate. Taking polar coordinate for both coordinate and momenta spaces
[tex]dv=r^2 sin^2\theta dr d\theta d\phi[/tex]
[tex]dV=R^2 sin^2\Theta dR d\Theta d\Phi[/tex]
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
[tex]\frac{\prod_{i=1}^N dv_idV_i}{h^{3N}}[/tex]
 
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  • #3
raisins said:
But ##p_{\theta_i},p_{\phi_i}\in[0,\infty)## s
From where this comes from?
 
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