Editing Maxwell–Boltzmann distribution (section)

===Distribution for the energy===

The energy distribution is found imposing
{{NumBlk||<math display="block"> f_E(E) \, dE = f_p(\mathbf p) \, d^3 \mathbf p,</math>|{{EquationRef|7}}}}
where <math>d^3 \mathbf p</math> is the infinitesimal phase-space volume of momenta corresponding to the energy interval {{mvar|dE}}.
Making use of the spherical symmetry of the energy-momentum dispersion relation <math>E = \tfrac{| \mathbf p|^2}{2m},</math> this can be expressed in terms of {{mvar|dE}} as
{{NumBlk|:|<math>  d^3 \mathbf p = 4 \pi | \mathbf p |^2 d |\mathbf p| = 4 \pi m \sqrt{2mE} \ dE.</math>|{{EquationRef|8}}}}
Using then ({{EquationNote|8}}) in ({{EquationNote|7}}), and expressing everything in terms of the energy {{mvar|E}}, we get
<math display="block">\begin{align}
f_E(E) dE &= \left[\frac{1}{2\pi m k_\text{B}T}\right]^{3/2} \exp\left(-\frac{E}{k_\text{B}T}\right) 4 \pi m \sqrt{2mE} \ dE \\[1ex]
 &= 2 \sqrt{\frac{E}{\pi}} \, \left[\frac{1}{k_\text{B}T}\right]^{3/2} \exp\left(-\frac{E}{k_\text{B}T}\right) \, dE
\end{align}</math>
and finally

{{Equation box 1 |indent=: |equation=
<math>f_E(E) = 2 \sqrt{\frac{E}{\pi}} \, \left[\frac{1}{k_\text{B}T}\right]^{3/2} \exp\left(-\frac{E}{k_\text{B}T} \right)</math>
|cellpadding |border |border colour = #50C878 |background colour = #ECFCF4|ref=9}}

Since the energy is proportional to the sum of the squares of the three normally distributed momentum components, this energy distribution can be written equivalently as a [[gamma distribution]], using a shape parameter, <math>k_\text{shape} = 3/2</math> and a scale parameter, <math>\theta_\text{scale} = k_\text{B}T.</math>

Using the [[equipartition theorem]], given that the energy is evenly distributed among all three degrees of freedom in equilibrium, we can also split <math>f_E(E) dE</math> into a set of [[chi-squared distribution]]s, where the energy per degree of freedom, {{mvar|ε}} is distributed as a chi-squared distribution with one degree of freedom,<ref>{{ cite book | title = Statistical Thermodynamics: Fundamentals and Applications | first1 = Normand M. | last1 = Laurendeau | publisher = Cambridge University Press | year = 2005 | isbn = 0-521-84635-8 | page = [https://books.google.com/books?id=QF6iMewh4KMC&pg=PA434 434] | url = https://books.google.com/books?id=QF6iMewh4KMC}}</ref>
<math display="block">f_\varepsilon(\varepsilon)\,d\varepsilon
= \sqrt{\frac{1}{\pi\varepsilon k_\text{B}T}} ~ \exp\left(-\frac{\varepsilon}{k_\text{B}T}\right)\,d\varepsilon</math>

At equilibrium, this distribution will hold true for any number of degrees of freedom. For example, if the particles are rigid mass dipoles of fixed dipole moment, they will have three translational degrees of freedom and two additional rotational degrees of freedom. The energy in each degree of freedom will be described according to the above chi-squared distribution with one degree of freedom, and the total energy will be distributed according to a chi-squared distribution with five degrees of freedom. This has implications in the theory of the [[specific heat]] of a gas.