Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I know how transform an integral below,

$$ \iint f(\mathbf v_{1})f(\mathbf v_{2})d^3\mathbf v_{1}d^3\mathbf v_{2}, $$

using relative speed coordinates: we just use $$ m_{1} \mathbf v_{1} + m_{2}\mathbf v_{2} = M\mathbf V, \quad \mathbf v = \mathbf v_{1} - \mathbf v_{2} , $$

and then we may use spherical coordinates.

But if I have an integral like

$$ \iint f(\mathbf r_{1})f(\mathbf r_{2})d^3\mathbf r_{1}d^3\mathbf r_{2}, $$

I don't know how to transform it by using a spherical coordinates of center of masses. In Pathria's book called "Statistical Mechanics" I saw a transform that I need, but I don't understand how it was made. Can you help me?


And it's not a homework!

share|improve this question
I presume that you mean there are two integrals over $v_1$ and $v_2$ in the first case, and over $r_1$ and $r_2$ in the second case. –  Vijay Murthy May 5 '12 at 17:23
Oh yes, sorry for my misinterpretation. –  PhysiXxx May 5 '12 at 17:25
add comment

2 Answers

up vote 1 down vote accepted

You can do the transformation to the relative coordinate $\mathbf{r} = \mathbf{r}_1-\mathbf{r}_2$ and center-of-mass coordinates $M\mathbf{R} = m_1\mathbf{r}_1+m_2\mathbf{r}_2$ and do one of the integrals trivially provided the two functions inside the integrals depend only on $\mathbf{r}$ (or only on $\mathbf{R}$). Otherwise you will still be left with two integrals one over $\mathbf{r}$ and the other over $\mathbf{R}$.

In the part that you refer to in Pathria's book the two functions $$f(\mathbf{r}_1) = r \frac{\partial u(r)}{\partial r}; \qquad f(\mathbf{r}_2)= g(\mathbf{r}_2 - \mathbf{r}_1)$$ depend only on $\mathbf{r}$. A transformation to $\mathbf{r}$ and $\mathbf{R}$ coordinates then decouples the $\mathbf{R}$ integral which has given a factor of the system volume (equation (16) in the image). Further if the functions inside the integrand depend only on $r=|\mathbf{r}|$, then one can transform to spherical coordinates as is done in Pathria..

share|improve this answer
Thank's. You answered on my two questions, even on the question which I didn't ask! –  PhysiXxx May 5 '12 at 18:48
OK. But what did you NOT ask that I DID answer? Curious to know :) –  Vijay Murthy May 5 '12 at 18:50
"...coordinates then decouples the R integral which has given a factor of the system volume (equation (16) in the image)..." –  PhysiXxx May 6 '12 at 19:54
add comment

An integral like $$\iint f(\mathbf{r}_1,\mathbf{r}_2)d^3\mathbf{r}_1d^3\mathbf{r}_2,$$ will not in general simplify nicely without additional assumptions about $f$. If you know it to be a product, then $$\iint f(\mathbf{r}_1)f(\mathbf{r}_2)d^3\mathbf{r}_1d^3\mathbf{r}_2=\left(\int f(\mathbf{r})d^3\mathbf{r}\right)^2.$$ If you know $f(\mathbf{r}_1,\mathbf{r}_2)$ to be invariant under translations, then it's a function of the centre of mass and relative coordinates, so the change of variables you need is $$\mathrm{R}=\frac{m_1\mathbf{r}_1+m_2\mathbf{r}_2}{m_1+m_2},$$ $$\mathbf{r}=\mathbf{r}_2-\mathbf{r}_1.$$

share|improve this answer
Thank's! You helped me. –  PhysiXxx May 5 '12 at 18:49
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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