Simplifying double integrals of isotropic functions

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.

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.
• Good answer, but do you mean to say in the second sentence that $r_1$ is fixed and not to integrate, but then in the last sentence that $r_2$ is constant and $r_1$ is integrated? Sep 5 '21 at 14:21