Integral of d3p

  • #1

Main Question or Discussion Point

I was recently told that the integral for a 3-momentum d^3p = 4∏∫p^2 dp

But I dunno what how is this integtal done.

Any help?
 

Answers and Replies

  • #2
jfizzix
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This integral was done in spherical coordinates
[itex]\int dxdydz = \int r^{2}sin(\theta)dr d\theta d\phi[/itex]
if what you're integrating over does not depend on [itex]\theta[/itex] or [itex]\phi[/itex], then you can integrate over those variables giving you an additional factor of [itex]4 \pi[/itex]. Thus
[itex]\int dxdydz = 4\pi \int r^{2}d r[/itex].

I expect d^3p is shorthand for dp_{x}dp_{y}dp_{z}.
 
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