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## Velocity volume element to momentum volume element

Posted by peeterjoot on April 9, 2013

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Motivation

One of the problems I attempted had integrals over velocity space with volume element $d^3\mathbf{u}$. Initially I thought that I’d need a change of variables to momentum space, and calculated the corresponding momentum space volume element. Here’s that calculation.

# Guts

We are working with a Hamiltonian

\begin{aligned}\epsilon = \sqrt{ (p c)^2 + \epsilon_0^2 },\end{aligned} \hspace{\stretch{1}}(1.1)

where the rest energy is

\begin{aligned}\epsilon_0 = m c^2.\end{aligned} \hspace{\stretch{1}}(1.2)

Hamilton’s equations give us

\begin{aligned}u_\alpha = \frac{ p_\alpha/c^2 }{\epsilon},\end{aligned} \hspace{\stretch{1}}(1.3)

or

\begin{aligned}p_\alpha = \frac{ m u_\alpha }{\sqrt{1 - \mathbf{u}^2/c^2}}.\end{aligned} \hspace{\stretch{1}}(1.4)

This is enough to calculate the Jacobian for our volume element change of variables

\begin{aligned}du_x \wedge du_y \wedge du_z &= \frac{\partial(u_x, u_y, u_z)}{\partial(p_x, p_y, p_z)}dp_x \wedge dp_y \wedge dp_z \\ &= \frac{1}{{c^6 \left( { m^2 + (\mathbf{p}/c)^2 } \right)^{9/2}}}\begin{vmatrix}m^2 c^2 + p_y^2 + p_z^2 & - p_y p_x & - p_z p_x \\ -p_x p_y & m^2 c^2 + p_x^2 + p_z^2 & - p_z p_y \\ -p_x p_z & -p_y p_z & m^2 c^2 + p_x^2 + p_y^2\end{vmatrix}dp_x \wedge dp_y \wedge dp_z \\ &= m^2 \left( { m^2 + \mathbf{p}^2/c^2 } \right)^{-5/2}dp_x \wedge dp_y \wedge dp_z.\end{aligned} \hspace{\stretch{1}}(1.5)

That final simplification of the determinant was a little hairy, but yielded nicely to Mathematica.

Our final result for the velocity volume element in momentum space, in terms of the particle energy is

\begin{aligned}d^3 \mathbf{u} = \frac{c^6 \epsilon_0^2 } {\epsilon^5} d^3 \mathbf{p}.\end{aligned} \hspace{\stretch{1}}(1.6)