I was trying to do an n-dimensional Maxwell-Boltzmann distribution for velocity, but, I can't figure out what is happening in the following passage:
$$f(v){d}^{n}v = \left( \frac{m}{2\pi kT} \right)^{n/2} e^{-\frac{m |v|^{2}}{2kT}} d^{n}v \ \ \ \rightarrow \ \ \ f(v) dv = {(\operatorname{const})} \ \ e^{-\frac{mv^{2}}{2kT}} \ \ v^{n-1} dv \ \ ,$$
where:
$$ (d^n v) = (dv_{1})(dv_{2}) (…) (dv_{n}) \ \ ,$$
and
$$ v^2 = v_{1}^2 + v_{2}^2 + … + v_{n}^2 \ \ .$$
Source:
Maxwell-Boltzmann distribution in n-dimensional space
I couldn't find it anywhere else to double check…
Best Answer
This is a transition in velocity space from Cartesian coordinates to polar ones, with $v$ the polar radius. The constant absorbs the $\left(\frac{m}{2\pi kT}\right)^{n/2}$, as well as a solid angle factor due to integration over the non-polar angles. You don't need to worry so much about that, because of the spherical symmetry. You do need the constant to calculate the probability in a region of velocity space, but you can get it by normalizing the PDF. This is equivalent to computing the $n$-dimensional unit ball's measure.