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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.

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The $p$ in the differential form $d^3p$ is a vector, and stands for the $N$ degrees of freedom ... so it's $\sum^{3N}dp_i$. – Chris Gerig Jul 18 '12 at 22:34
@ChrisGerig: +1 but na, something's weird with the units. Is the power of $h$ really only 3? – NikolajK Jul 18 '12 at 22:35
My mistake, you are probably right... factor is $\frac{1}{h^{3N}}$ – physicsgrad Jul 18 '12 at 22:42
@physicsgrad: Still unnice, as the volume (propably a constant but dimensionful factor) is just tropped, but Chris explaination is the right one here. – NikolajK Jul 18 '12 at 22:54
Yes, I omitted it because I didn't think it relevant but you are right. – physicsgrad Jul 18 '12 at 22:56

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.

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


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

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.

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Its the $N$th power of your $I$, and the $3N$th power of a 1D version of $I$. – Arnold Neumaier Jul 19 '12 at 12:32
@ArnoldNeumaier: Oh, I didn't see the blooper, fixed. – Ron Maimon Jul 20 '12 at 3:22
You fixed only half; it still says $I^{3N}$ instead of $I^N$. – Arnold Neumaier Jul 20 '12 at 10:30
@ArnoldNeumaier: Thanks. I think I am starting to appreciate the nature of the difficulties in writing a technical book. – Ron Maimon Jul 20 '12 at 17:40
Yes. Books with on the average less than 1 mistakes per technical page are not frequent. – Arnold Neumaier Jul 21 '12 at 20:10

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