Phase space integral in noninteracting dipole system

In summary: The Lagrangian and Hamiltonian for a system of dipoles is given by:L=-T-V=\sum_{i=1}^N\left\{\frac{m\dot{\vec{r}_i}^2}{2}+\vec{\mu}_i\cdot\vec{E}\right\}andH=-\sum_{i=1}^N\left\{\frac{\vec{p}_i^2}{2m}-E\mu\cos\theta_i\right\}
  • #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]
 
Last edited:
  • #3
raisins said:
But ##p_{\theta_i},p_{\phi_i}\in[0,\infty)## s
From where this comes from?
 

Related to Phase space integral in noninteracting dipole system

1. What is a phase space integral?

A phase space integral is a mathematical concept used in physics to calculate the total probability of a system's state based on its position and momentum. It involves integrating over all possible values of position and momentum to determine the likelihood of a particular state.

2. How is a phase space integral different from a regular integral?

A phase space integral involves integrating over multiple variables (position and momentum) whereas a regular integral typically only involves one variable. Additionally, the limits of integration for a phase space integral are determined by the physical constraints of the system, rather than being explicitly given.

3. What is the significance of phase space in physics?

Phase space is a mathematical representation of all possible states of a physical system. It allows scientists to analyze and predict the behavior of a system by considering all possible combinations of position and momentum.

4. How is phase space integral used in statistical mechanics?

In statistical mechanics, phase space integrals are used to calculate the partition function, which is a fundamental quantity that describes the thermodynamic properties of a system. This allows scientists to make predictions about the behavior of a system at the microscopic level.

5. Can phase space integrals be applied to all physical systems?

Phase space integrals can be applied to any physical system that can be described in terms of position and momentum. However, the complexity of the integral may vary depending on the number of particles and the interactions between them.

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