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Why $d^3\mathbf{p}=p^2\;dp \; d\Omega$ ?

where $d\Omega$ is the solid angle that covers a particle with 3-momentum $\mathbf{p}$...

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Does spherical coordinates not help? Specifically the volume element & Jacobian –  Michael Brown Mar 18 '13 at 10:25
    
@MichaelBrown: It could be, but I don't know how...Only if I regard the momentum being the radial component, but I don't know why I can do that....Btw, thank's for your comment! –  Thanos Mar 18 '13 at 10:36
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yes, $p$ denotes the magnitude of the vector $\mathbf{p}$ and the angles give the direction. This is just as valid to do for momentum vectors as for position vectors. :) –  Michael Brown Mar 18 '13 at 10:40
    
@MichaelBrown: OK!!!Now it makes sence! Thank you very much for your help! Would you mind giving an answer so that I can vote it up? The minimum I can do to thank you! –  Thanos Mar 18 '13 at 10:43
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1 Answer 1

up vote 4 down vote accepted

This is just the standard measure for integration given in spherical coordinates. If you write the momentum vector $\mathbf{p}$ in terms of the magnitude $p$ and polar angles you can write

$$ \begin{array}{lcl} \int\mathrm{d}^{3}\mathbf{p} &\equiv& \int_{-\infty}^{\infty}\mathrm{d}p_{x}\int_{-\infty}^{\infty}\mathrm{d}p_{y}\int_{-\infty}^{\infty}\mathrm{d}p_{z}\\ &=& \int_{0}^{\infty}\mathrm{d}p\ p^{2}\int_{-1}^{1}\mathrm{d}\left(\cos\theta\right)\int_{0}^{2\pi}\mathrm{d}\phi\\ &\equiv&\int_{0}^{\infty}\mathrm{d}p\ p^{2}\int\mathrm{d}\Omega. \end{array} $$

Strip off the integrals to get what you want. These manipulations hold for any 3-vector integration variable - position, momentum, whatever.

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