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Statistical mechanics

by arnesmeets
Tags: mechanics, statistical
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May22-06, 05:38 PM
P: 18
This shouldn't be too hard but I'm struggling.

Consider N identical particles in a box of volume V. The relation between their linear momentum and kinetic energy is given by [tex]E = c\left|\overrightarrow{p}\right|[/tex], where c is the speed of light. So, the Hamiltonian of the system is

[tex]H_N\left(\overrightarrow{q_1},\,\cdots,\overrightarrow{q_N},\overrighta rrow{p_1},\,\cdots,\overrightarrow{p_N}\right) = c \sum_{i = 1}^N \left|p_i\right|[/tex]
Now, I'm trying to find the average energy of the system. So, I need to get Z:

[tex]Z = \int \exp\left(-\beta \cdot c \cdot \sum_i \left|p_i\right|\right) d\overrightarrow{q_1}\cdots d\overrightarrow{q_N} d\overrightarrow{p_1}\cdots d\overrightarrow{p_N} = V^{N} \cdot \left(\int \exp\left(-\beta \cdot c \cdot \left|\overrightarrow{p}\right|\right) d\overrightarrow{p}\right)^N[/tex]

Now, I've got two questions:

1) Is there any mistake in the calculations for Z above?
2) How to evaluate the integral??
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May25-06, 07:18 AM
P: 18
Does nobody have a clue?
May25-06, 08:59 AM
P: 76
You can solve the integral quite easily by transforming to spherical coordinates (r,\theta, \phi). So the integral becomes:

[tex] \int_0^{2\pi} \int_0^{\pi} \int_0^{\infty} e^{-\beta r} r^2\sin{\theta} dr d\theta d\phi = 4\pi \int_0^{\infty} r^2 e^{-\beta r}dr[/tex]

I think you can now solve the rest on your own

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