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Maxwell–Boltzmann distribution
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===Distribution for the momentum vector=== The potential energy is taken to be zero, so that all energy is in the form of kinetic energy. The relationship between [[Kinetic energy#Kinetic energy of rigid bodies|kinetic energy and momentum]] for massive non-[[special relativity|relativistic]] particles is {{NumBlk||<math display="block">E=\frac{p^2}{2m}</math>|{{EquationRef|2}}}} where {{math|''p''<sup>2</sup>}} is the square of the momentum vector {{math|1='''p''' = [''p<sub>x</sub>'', ''p<sub>y</sub>'', ''p<sub>z</sub>'']}}. We may therefore rewrite Equation ({{EquationNote|1}}) as: {{NumBlk||<math display="block"> \frac{N_i}{N} = \frac{1}{Z} \exp \left(-\frac{p_{i, x}^2 + p_{i, y}^2 + p_{i, z}^2}{2m k_\text{B}T}\right)</math> |{{EquationRef|3}}}} where: * {{mvar|Z}} is the [[partition function (statistical mechanics)|partition function]], corresponding to the denominator in {{EquationNote|1|equation 1}}; * {{mvar|m}} is the molecular mass of the gas; * {{mvar|T}} is the thermodynamic temperature; * {{math|''k''<sub>B</sub>}} is the [[Boltzmann constant]]. This distribution of {{math|''N{{sub|i}}'' : ''N''}} is [[Proportionality (mathematics)|proportional]] to the [[probability density function]] {{mvar|''f''<sub>'''p'''</sub>}} for finding a molecule with these values of momentum components, so: {{NumBlk||<math display="block"> f_\mathbf{p} (p_x, p_y, p_z) \propto \exp \left(-\frac{p_x^2 + p_y^2 + p_z^2}{2m k_\text{B}T}\right)</math>|{{EquationRef|4}}}} The [[normalizing constant]] can be determined by recognizing that the probability of a molecule having ''some'' momentum must be 1. Integrating the exponential in {{EquationNote|4|equation 4}} over all {{mvar|p<sub>x</sub>}}, {{mvar|p<sub>y</sub>}}, and {{mvar|p<sub>z</sub>}} yields a factor of <math display="block">\iiint_{-\infty}^{+\infty} \exp\left(-\frac{p_x^2 + p_y^2 + p_z^2}{2m k_\text{B}T}\right) dp_x\, dp_y\, dp_z = \Bigl[ \sqrt{\pi} \sqrt{2m k_\text{B}T} \Bigr]^3</math> So that the normalized distribution function is: {{Equation box 1 |indent=: |equation= <math> f_\mathbf{p} (p_x, p_y, p_z) = \left[\frac{1}{2\pi m k_\text{B}T}\right]^{3/2} \exp\left(-\frac{p_x^2 + p_y^2 + p_z^2}{2m k_\text{B}T}\right)</math> |cellpadding |border |border colour = #50C878 |background colour = #ECFCF4|ref=6}} The distribution is seen to be the product of three independent [[normal distribution|normally distributed]] variables <math>p_x</math>, <math>p_y</math>, and <math>p_z</math>, with variance <math>m k_\text{B}T</math>. Additionally, it can be seen that the magnitude of momentum will be distributed as a Maxwell–Boltzmann distribution, with <math display="inline">a = \sqrt{m k_\text{B}T}</math>. The Maxwell–Boltzmann distribution for the momentum (or equally for the velocities) can be obtained more fundamentally using the [[H-theorem]] at equilibrium within the [[Kinetic theory of gases]] framework.
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