Integration of partition-function over many momentum variables

Question

My integral looks like

$$Z(\beta) = \frac{1}{h^3}\int d^3p\ \exp{\left(-\frac{\beta}{2m}\sum^{3N}_{i=1}p_i^2\right)}.$$

I'm confused about how to integrate over seemingly 3N variables in only a 3-dimensional integral.

Answer 1

It's a 3N dimensional integral, but it reduces to the N-th power of a 3-dimensional integral (and ultimately to the 3N-th power of a 1 dimensional integral), so you probably have a sloppy source.

$$ Z = \int (\prod_i d^3p_i) e^{-\beta \sum_i {p_i^2\over 2m}} = \prod_i (\int e^{-\beta {p^2\over 2m}} d^3p) = I^N $$

Where

$$ I = \int e^{-\beta {p^2\over 2m}} d^3 p$$

The integral I is really the product of three independent gaussians in $p_x$,$p_y$,$p_z$, so the answer is

$$ I = ({\sqrt{m}\over \sqrt{2\pi \beta}})^3$$

Which is the cube of the integral of each Gaussian separately. So that

$$ Z= {m^{3N\over 2} \over (2\pi \beta)^{3N\over 2}}$$

and taking the log gives the free energy of the ideal gas:

$$ \beta F = {3N\over 2}\log(T)$$

and you can read off the specific heat of the ideal gas from this formula--- ${3N\over 2}$. This works for any quadratic variables in H, and this is the equipartition theorem.