Editing Maxwell–Boltzmann statistics (section)

===Derivation===

We can deduce the Maxwell–Boltzmann distribution from Maxwell–Boltzmann statistics, starting with the Maxwell-Boltzmann probability for energy states and substituting the kinetic energy <math>E=\frac{1}{2} m v^2</math> to express the probability in terms of velocity:

:<math>
\begin{align}
P(E) &=\frac{1}{Z}~\exp \left(\frac{-E }{ k_B T}\right)\\
\rightarrow P(v)&= \frac{1}{Z}~\exp \left(\frac{- m v^2 }{ 2k_B T}\right)
\end{align}
</math>

In 3D, this is proportional to the surface area of a sphere, <math>4\pi v^2</math>. Thus, the probability density function (PDF) for speed <math>v</math> becomes:

:<math>
f(v) = C \cdot 4\pi v^2 \exp\left(-\frac{m v^2}{2 k_B T}\right)
</math>

To find the normalization constant <math>C</math>, we require the integral of the probability density function over all possible speeds to be unity:

:<math>
\begin{align}
\int_0^\infty f(v) dv &= 1\\
\rightarrow C \int_0^\infty 4\pi v^2 \exp\left(-\frac{m v^2}{2 k_B T}\right) dv &= 1
\end{align}
</math>

Evaluating the integral using the known result <math>\int_0^\infty v^2 e^{-a v^2} dv = \frac{\sqrt{\pi}}{4 a^{3/2}}</math>, with <math> a = \frac{m}{2 k_B T}</math>, we obtain:

:<math>
\begin{align}
C \cdot 4\pi \cdot \frac{\sqrt{\pi}}{4\left(\frac{m}{2 k_B T}\right)^{3/2}} = 1
\quad \rightarrow C = \left(\frac{m}{2 \pi k_B T}\right)^{3/2}
\end{align}
</math>

Therefore, the Maxwell–Boltzmann speed distribution is:

:<math>
f(v) = \left(\frac{m}{2 \pi k_B T}\right)^{3/2} 4\pi v^2 \exp\left(-\frac{m v^2}{2 k_B T}\right)
</math>