Solving 3-Momentum Integral: Seeking Assistance

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SUMMARY

The integral for 3-momentum, represented as d^3p, is calculated using spherical coordinates, specifically through the equation d^3p = 4π∫p² dp. The integration process involves transforming the Cartesian coordinates into spherical coordinates, where the volume element is expressed as ∫dxdydz = ∫r²sin(θ)dr dθ dφ. By integrating over the angles θ and φ, which do not depend on the function being integrated, an additional factor of 4π is introduced, simplifying the integral to 4π∫r²dr.

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

But I don't know what how is this integtal done.

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

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

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