# Integral over momentum space of the distribution function

I have $$\int \frac{d^3p}{(2\pi)^3}p\frac{\partial f}{\partial p} = \int \frac{d^2\hat{p}}{(2\pi)^3}\int^\infty_0 p^2dp\ p \frac{\partial f}{\partial p} ,$$ where $$f$$ is the distribution function, $$\hat{p}$$ is the unit momentum vector, $$p$$ is the magnitude of the momentum vector.

How do I get from the left hand side to the right hand side of the above equation?

• It is not clear what the rhs means here (what is $\hat{p}$?): shouldn't it be just a transformation to spherical coordinates? Jul 7 at 9:54
• I updated the question. Jul 7 at 10:04
• In other words, it is spherical coordinates - the double integral is over a unit sphere with constant density. Jul 7 at 10:07

It looks like we've written $${\rm d}^3p$$ in spherical coordinates. The volume element in 3D momentum space is $${\rm d}^3p=p^2\sin\theta\;{\rm d}p\;{\rm d}\theta\;{\rm d}\phi$$ and it looks like they've condensed the angular part into $$\sin\theta\;{\rm d}\theta\;{\rm d}\phi\equiv{\rm d}^2\hat{p}$$