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Maxwell–Boltzmann distribution
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===Distribution for the velocity vector=== Recognizing that the velocity probability density {{math|''f''<sub>'''v'''</sub>}} is proportional to the momentum probability density function by <math display="block">f_\mathbf{v} d^3\mathbf{v} = f_\mathbf{p} \left(\frac{dp}{dv}\right)^3 d^3\mathbf{v}</math> and using {{math|1='''p''' = ''m'''''v'''}} we get {{Equation box 1 |indent=: |equation= <math> f_\mathbf{v} (v_x, v_y, v_z) = \biggl[\frac{m}{2\pi k_\text{B}T} \biggr]^{3/2} \exp\left(-\frac{m\left(v_x^2 + v_y^2 + v_z^2\right)}{2 k_\text{B}T}\right) </math> |cellpadding |border |border colour = #50C878 |background colour = #ECFCF4}} which is the Maxwell–Boltzmann velocity distribution. The probability of finding a particle with velocity in the infinitesimal element {{math|[''dv<sub>x</sub>'', ''dv<sub>y</sub>'', ''dv<sub>z</sub>'']}} about velocity {{math|1='''v''' = [''v<sub>x</sub>'', ''v<sub>y</sub>'', ''v<sub>z</sub>'']}} is <math display="block">f_\mathbf{v}{\left(v_x, v_y, v_z\right)}\, dv_x\, dv_y\, dv_z.</math> Like the momentum, this distribution is seen to be the product of three independent [[normal distribution|normally distributed]] variables <math>v_x</math>, <math>v_y</math>, and <math>v_z</math>, but with variance <math display="inline">k_\text{B}T / m</math>. It can also be seen that the Maxwell–Boltzmann velocity distribution for the vector velocity {{math|[''v<sub>x</sub>'', ''v<sub>y</sub>'', ''v<sub>z</sub>'']}} is the product of the distributions for each of the three directions: <math display="block">f_\mathbf{v}{\left(v_x, v_y, v_z\right)} = f_v (v_x)f_v (v_y)f_v (v_z)</math> where the distribution for a single direction is <math display="block"> f_v (v_i) = \sqrt{\frac{m}{2 \pi k_\text{B}T}} \exp \left(-\frac{mv_i^2}{2k_\text{B}T}\right).</math> Each component of the velocity vector has a [[normal distribution]] with mean <math>\mu_{v_x} = \mu_{v_y} = \mu_{v_z} = 0</math> and standard deviation <math display="inline">\sigma_{v_x} = \sigma_{v_y} = \sigma_{v_z} = \sqrt{k_\text{B}T / m}</math>, so the vector has a 3-dimensional normal distribution, a particular kind of [[multivariate normal distribution]], with mean <math> \mu_{\mathbf{v}} = \mathbf{0} </math> and covariance <math display="inline">\Sigma_{\mathbf{v}} = \left(\frac{k_\text{B}T}{m}\right)I</math>, where <math>I</math> is the {{nowrap|3 × 3}} identity matrix.
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