Editing Maxwell–Boltzmann distribution (section)

== In ''n''-dimensional space ==
In {{mvar|n}}-dimensional space, Maxwell–Boltzmann distribution becomes:
<math display="block"> f(\mathbf{v}) ~ d^n\mathbf{v} = \biggl[\frac{m}{2 \pi k_\text{B}T}\biggr]^{n/2} \exp\left(-\frac{m|\mathbf{v}|^2}{2k_\text{B}T}\right) ~d^n\mathbf{v} </math>

Speed distribution becomes:
<math display="block"> f(v) ~ dv = A \exp\left(-\frac{mv^2}{2k_\text{B} T}\right)  v^{n-1} ~ dv </math>
where <math> A </math> is a normalizing constant.

The following integral result is useful:
<math display="block">\begin{align}
\int_{0}^{\infty} v^a \exp\left(-\frac{mv^2}{2k_\text{B} T}\right) dv  
&= \left[\frac{2k_\text{B} T}{m}\right]^\frac{a+1}{2} \int_{0}^{\infty} e^{-x}x^{a/2} \, dx^{1/2} \\[2pt]
&= \left[\frac{2k_\text{B} T}{m}\right]^\frac{a+1}{2} \int_{0}^{\infty} e^{-x}x^{a/2}\frac{x^{-1/2}}{2} \, dx \\[2pt]
&= \left[\frac{2k_\text{B} T}{m}\right]^\frac{a+1}{2} \frac{\Gamma{\left(\frac{a+1}{2}\right)}}{2}
\end{align}</math>
where <math> \Gamma(z)</math> is the [[Gamma function]]. This result can be used to calculate the [[Moment (mathematics)|moments]] of speed distribution function:
<math display="block"> \langle v \rangle 
= \frac
{\displaystyle \int_{0}^{\infty} v \cdot v^{n-1} \exp\left(-\tfrac{mv^2}{2k_\text{B} T}\right) \, dv}
{\displaystyle \int_{0}^{\infty} v^{n-1} \exp\left(-\tfrac{mv^2}{2k_\text{B} T}\right) \, dv}
= \sqrt{\frac{2k_\text{B} T}{m}} ~~ \frac{\Gamma{\left(\frac{n+1}{2}\right)}}{\Gamma{\left(\frac{n}{2}\right)}}</math>
which is the [[expectation value|mean]] speed itself <math display="inline">v_\mathrm{avg} = \langle v \rangle = \sqrt{\frac{2k_\text{B} T}{m}} \  \frac{\Gamma \left(\frac{n+1}{2}\right)}{\Gamma \left(\frac{n}{2}\right)}.</math>

<math display="block"> \begin{align}
\langle v^2 \rangle 
&= \frac
{\displaystyle\int_{0}^{\infty} v^2 \cdot v^{n-1} \exp\left(-\tfrac{mv^2}{2k_\text{B} T}\right) \, dv}
{\displaystyle\int_{0}^{\infty} v^{n-1} \exp\left(-\tfrac{mv^2}{2k_\text{B}T}\right) \, dv} \\[1ex]
&= \left[\frac{2k_\text{B}T}{m}\right] \frac{\Gamma {\left(\frac{n+2}{2}\right)}}{\Gamma {\left(\frac{n}{2}\right)}} \\[1.2ex]
&= \left[\frac{2k_\text{B}T}{m}\right] \frac{n}{2} = \frac{n k_\text{B}T}{m}
\end{align}</math>
which gives root-mean-square speed  <math display="inline">v_\text{rms} = \sqrt{\langle v^2 \rangle} = \sqrt{\frac{n k_\text{B}T}{m}}.</math>

The derivative of speed distribution function:
<math display="block">\frac{df(v)}{dv} = A \exp\left(-\frac{mv^2}{2k_\text{B}T}\right) \biggl[-\frac{mv}{k_\text{B}T} v^{n-1}+(n-1)v^{n-2}\biggr] = 0 </math>

This yields the most probable speed ([[Mode (statistics)|mode]]) <math display="inline">v_\text{p} = \sqrt{\left(n-1\right) k_\text{B}T/m}.</math>