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?

  • $\begingroup$ It is not clear what the rhs means here (what is $\hat{p}$?): shouldn't it be just a transformation to spherical coordinates? $\endgroup$ Jul 7 at 9:54
  • $\begingroup$ I updated the question. $\endgroup$
    – IanDsouza
    Jul 7 at 10:04
  • $\begingroup$ In other words, it is spherical coordinates - the double integral is over a unit sphere with constant density. $\endgroup$ 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 $$\sin\theta\;{\rm d}\theta\;{\rm d}\phi\equiv{\rm d}^2\hat{p}$$


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