Simplifying double integrals of isotropic functions

Question

Cross-posted from physics SE. I am studying statistical mechanics, and I am studying ideas surrounding potential of mean force and n-body density functions.

In a derivation, they mention that $$-\left\langle \frac{\partial U}{\partial r_1} \right\rangle=k_BT\frac{d}{dr_1}\ln g(|r_1,r_2|)$$ and then they say, "since integrating over the average force yields work, we get", $$w(r)= -k_BT\ln g(r)$$

But I feel like I don't understand how the integration works here. I understand $\langle F\rangle = -k_BT\frac{d}{dr_1}\ln g(r_1,r_2)$ But now what I am doing is I am dragging particle 2 from $\infty$ to some distance $r$ from particle 1. So, the integration should look like: $$w(r) = \int_{\infty} ^{r_1+r} \int_{0}^R \langle F \rangle \cdot r dr_1dr_2 = -\int_{\infty}^{r_1+r} \int_{0}^{\infty} k_BT\frac{d}{dr_1}\ln g(|r_1-r_2|) dr_1 dr_2$$

I am not sure how to safely reduce these integrals into the nice statement $$w(r) = -k_BT\ln g(r)$$ How does one go about reducing such an integral? Or if there is an error in the way I have set this up, I would appreciate knowing that too.

I would appreciate any advice you have for me.

Answer 1

The work is a single integral over $|r_1-r_2|$, not a double integral over $r_1$ and $r_2$. As you are fixing particle 1, you shouldn't integrate over particle 1. Moreover, the work is $w(r) = \int Fdr$, not $w(r) = \int Frdr$, as you can see from dimensional analysis ($dr$ has the dimension of length, too). Therefore, you treat $r_1$ as constant, integrate over $r_2$ and define $r$ as $|r_1-r_2|$, and you get the result trivially.