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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!

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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

2 Answers 2

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..

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Thank's. You answered on my two questions, even on the question which I didn't ask! –  PhysiXxx May 5 '12 at 18:48
1  
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

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 $$\mathrm{R}=\frac{m_1\mathbf{r}_1+m_2\mathbf{r}_2}{m_1+m_2},$$ $$\mathbf{r}=\mathbf{r}_2-\mathbf{r}_1.$$

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Thank's! You helped me. –  PhysiXxx May 5 '12 at 18:49

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