# Center-of-mass and relative coordinate integral transformation

Assuming I have a function $$f(r_s,r)$$ where $$\vec{r_s}$$ is the center-of-mass- and $$\vec{r}$$ the relative-coordinate of a two particle system and I perform the following integral in spherical coordinates (assuming the function is only dependent on the absolute value of the coordinates, so that angle dependencies get trivially integrated over) $$16\pi^2\int_0^\infty\int_0^\infty f(r_s,r)r_s^2r^2\,dr_s\,dr$$ I now want to rewrite the integral but in terms of the individual particle coordinates $$r_1$$ and $$r_2$$. The transformation rules are given by $$\vec{r_s}=\frac{m_1\vec{r_1}+m_2\vec{r_2}}{m_1m_2}\hspace{1em}\text{and}\hspace{1em}\vec{r}=\vec{r_2}-\vec{r_1}$$ I am unsure as on how to write the proper measure of the integral so that the result of both ends up being the same.

• If $\vec{r}_s$ and $\vec{r}$ are vectors what is the meaning of ${\rm d}r_s$ and ${\rm d} r$? Commented Jul 7, 2021 at 6:27

It looks like you have a pair of 3D integration variables in spherical coordinates, albeit ones in which you've integrated out the angular coordinates. You might have to undo that and put those back in to the integral. Once you do, you can use a Jacobian (or a pair of Jacobians, depending on how you look at it) to convert to integrals in terms of the Cartesian components of $$\vec{r}_s, \vec{r}$$. From there, you can again use a Jacobian transformation using the equations you give above to convert to an integral in terms of $$\vec{r}_1, \vec{r}_2$$.