3
$\begingroup$

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.

$\endgroup$

1 Answer 1

4
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$ Sep 5, 2021 at 14:21
  • 1
    $\begingroup$ @BrandonBocklund You're right, that was a typo. Now it's fixed. Thanks $\endgroup$
    – wzkchem5
    Sep 5, 2021 at 14:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.