# Partition function for weakly interacting gases

I'm studying the Susskind lecture on statistical mechanics. A potential energy of pairwise interactions has been defined:

$$\sum_{n>m}U(|x_n -x_m|)$$

We want to calculate the partition function, and we start by evaluating the potential energy term by considering the interaction of a single pair, through the integral of potential energy between two particles over all possible positions:

$$\int dx_1^3 dx_2^3 U(|x_1-x_2|)$$

Now,there are 2 steps,

1. Assuming that $$U$$ goes to zero at large distances, hold particle $$1$$ fixed and integrate over all possible positions of particle $$2$$: $$\int dx^3 U(|x|)=U_0$$
2. Integrating over the position of particle $$1$$ gives a factor of volume, $$V$$.

Conclusion: $$\int dx_1^3 dx_2^3 U(|x_1-x_2|)=VU_0$$

I cannot understand steps 1 and 2. It looks pure magic to me. Is there a formal way to explain how the integral is broken up in two parts, why we can consider the position fixed, and why the results are equivalent?

For simplicity, let us first assume that the potential $$U$$ has finite range: $$U(|x|)=0$$ whenever $$|x|>R$$. Then, for any $$x_1$$ located at a distance at least $$R$$ from the boundary of the vessel enclosing your system, $$\int_V\mathrm{d}x_2\, U(|x_1-x_2|) = \int_{\mathbb{R}^3}\mathrm{d}y\, U(|y|) = U_0,$$ where the first identity follows from the change of variables $$y=x_2-x_1$$, (which has Jacobian $$1$$). Then, splitting the integral over $$x_1$$ into two parts according to whether $$x_1$$ is at a distance at least $$R$$ from the boundary or not, we obtain $$\int_V\mathrm{d}x_1\int_V\mathrm{d}x_2\, U(|x_1-x_2|) = U_0 V + O(R\partial V),$$ where the error term is of the order of the area $$\partial V$$ of the vessel.
The situation is entirely similar if $$U$$ does not have finite range (but is, of course, integrable). First, we can fix some $$\epsilon>0$$ and observe that there exists $$R(\epsilon)$$ such that $$\Bigl| \int_{|y|\leq R(\epsilon)}\mathrm{d}y\, U(|y|) - U_0 \Bigr| \leq \epsilon.$$ From this, we get $$\int_V\mathrm{d}x_1\int_V\mathrm{d}x_2\, U(|x_1-x_2|) = U_0 V + O(\epsilon V) + O(R(\epsilon)\partial V).$$ Letting $$V\to\infty$$ (say, along a sequence of cubes or other reasonable shapes) and then letting $$\epsilon\to 0$$, we obtain $$\lim_{V\to\infty} \frac1V \int_V\mathrm{d}x_1\int_V \mathrm{d}x_2\, U(|x_1-x_2|) = U_0,$$ which is a precise way of stating the identity.
• Dear Prof. Yvan Velenik, I'd have a question regarding your answer: In e.g. Wolfgang Nolting. Theoretical Physics 9. Fundamentals of Many-body Physics. Springer. Second Edition chapter 2, p.45, right after equation 2.48, the last equality you've derived but without the limit for the function $e^{-\alpha |x-y|}/(|x-y|)$ is given. Is this correct or meant to be in the limit $V\to\infty$? If you have time, I'd really appreciate if you could take a look on my related math SE question. Commented Jun 12, 2022 at 7:08
• I looked at your question on Math.SE. The answer is basically the one I've given above: to make precise sense of the identity you want, divide by $V$ and let $V\to\infty$. This is necessary to get rid of $o(V)$ error terms, as explained in the answer above. In Nolting's book, he's also explicitly taking the $V\to\infty$ limit (two lines below (2.48)), although he's being very cavalier with it, since there's still a $V$ remaining in his formula after having taken the limit (in the equation preceding equation (2.49))! Commented Jun 19, 2022 at 10:33