# Maxwell Boltzmann distribution: Going from momentum to energy

I am learning about the Maxwell Boltzmann distribution, and am trying to convert the equation from momentum into energy, but I'm stuck on changing $d^n p$ into $dE$. I have the equation: $$E=\frac{|\textbf{p}|^2}{2m}$$ And when I looked on wikipedia, I see that whoever wrote it changed it as: $$d^3 \textbf{p} = 4\pi |\textbf{p}|^2d|\textbf{p}|= 4\pi m\sqrt{2mE}dE$$ I understand how $4\pi |\textbf{p}|^2d|\textbf{p}|= 4\pi m\sqrt{2mE}dE$, but I don't understand how they got: $$d^3 p = 4\pi |\textbf{p}|^2d|\textbf{p}|$$ Is there a certain identity or something that I need to make use of? Also how doe this vary for different dimensions? Could someone point me in the right direction as to how to solve this please?

• I don't know very much about this field but this reminds of saying $d^3\textbf{x}=4\pi |\textbf{x}|^2d|\textbf{x}|=4\pi r^2 dr$ in position space, when you assume spherical symmetry. This looks like the same thing but in momentum space Commented Dec 3, 2015 at 18:20
• @rbncruise - Note that the $d^{3}p \rightarrow 4\pi \ p^{2} \ dp$ approximation only works if the distribution is one-dimensional (i.e., if there is spherical symmetry as tmwilson26 pointed out). Commented Dec 3, 2015 at 19:45

$$d^3p = p^2\sin(\theta_p)d\theta_p d\phi_p dp$$
$$\int_{\theta,\phi}d^3p = \int_{\theta,\phi}p^2\sin(\theta_p)d\theta_p d\phi_p dp$$
Integrating over the angular coordinates gives $4\pi p^2 dp$
• I see, and do you know how to perform this for an arbitrary number of dimensions $d^n \textbf{p}$? Commented Dec 3, 2015 at 18:38