Editing Equipartition theorem (section)

===Kinetic energies and the Maxwell–Boltzmann distribution===
The original formulation of the equipartition theorem states that, in any physical system in [[thermal equilibrium]], every particle has exactly the same average translational [[kinetic energy]], {{math|{{sfrac|3|2}}''k''<sub>B</sub>''T''}}.<ref name="mcquarrie_2000a">{{cite book | last = McQuarrie | first = DA | year = 2000 | title = Statistical Mechanics | edition = revised 2nd | publisher = University Science Books | isbn = 978-1-891389-15-3 | pages = [https://archive.org/details/statisticalmecha00mcqu_0/page/121 121–128] | url = https://archive.org/details/statisticalmecha00mcqu_0/page/121 }}</ref> However, this is true only for [[ideal gas]], and the same result can be derived from the [[Maxwell–Boltzmann distribution]]. First, we choose to consider only the Maxwell–Boltzmann distribution of velocity of the z-component
<math display="block">f (v_z) = \sqrt{\dfrac{m}{2\pi k_\text{B}T}}\;e^{\frac{-m{v_z}^2}{2k_\text{B}T}}</math>

with this equation, we can calculate the mean square velocity of the {{mvar|z}}-component
<math display="block">\langle {v_z}^2 \rangle = \int_{-\infty}^{\infty} f (v_z){v_z}^2 dv_z = \dfrac{k_\text{B}T}{m}</math>

Since different components of velocity are independent of each other, the average translational kinetic energy is given by
<math display="block">\langle E_k \rangle = \dfrac 3 2 m \langle {v_z}^2 \rangle = \dfrac 3 2 k_\text{B}T</math>

Notice, the [[Maxwell–Boltzmann distribution]] should not be confused with the [[Boltzmann distribution]], which the former can be derived from the latter by assuming the energy of a particle is equal to its translational kinetic energy.

As stated by the equipartition theorem. The same result can also be obtained by averaging the particle energy using the probability of finding the particle in certain quantum energy state.<ref name="configint"/>