Editing Maxwell–Boltzmann distribution (section)

==Derivation and related distributions==
===Maxwell–Boltzmann statistics===
{{main|Maxwell–Boltzmann statistics#Derivations|Boltzmann distribution}}
The original derivation in 1860 by [[James Clerk Maxwell]] was an argument based on molecular collisions of the [[Kinetic theory of gases]] as well as certain symmetries in the speed distribution function; Maxwell also gave an early argument that these molecular collisions entail a tendency towards equilibrium.<ref name=MaxwellA/><ref name=MaxwellB/><ref>{{Cite journal | last1 = Gyenis | first1 = Balazs | doi = 10.1016/j.shpsb.2017.01.001 | title = Maxwell and the normal distribution: A colored story of probability, independence, and tendency towards equilibrium | journal = Studies in History and Philosophy of Modern Physics | volume = 57 | pages = 53–65 | year = 2017| arxiv = 1702.01411 | bibcode = 2017SHPMP..57...53G | s2cid = 38272381 }}</ref> After Maxwell, [[Ludwig Boltzmann]] in 1872<ref>Boltzmann, L., "Weitere studien über das Wärmegleichgewicht unter Gasmolekülen." ''Sitzungsberichte der Kaiserlichen Akademie der Wissenschaften in Wien, mathematisch-naturwissenschaftliche Classe'', '''66''', 1872, pp. 275–370.</ref> also derived the distribution on mechanical grounds and argued that gases should over time tend toward this distribution, due to collisions (see [[H-theorem]]). He later (1877)<ref>Boltzmann, L., "Über die Beziehung zwischen dem zweiten Hauptsatz der mechanischen Wärmetheorie und der Wahrscheinlichkeitsrechnung respektive den Sätzen über das Wärmegleichgewicht." ''Sitzungsberichte der Kaiserlichen Akademie der Wissenschaften in Wien, Mathematisch-Naturwissenschaftliche Classe''. Abt. II, '''76''', 1877, pp. 373–435. Reprinted in ''Wissenschaftliche Abhandlungen'', Vol. II, pp. 164–223, Leipzig: Barth, 1909. '''Translation available at''': http://crystal.med.upenn.edu/sharp-lab-pdfs/2015SharpMatschinsky_Boltz1877_Entropy17.pdf {{Webarchive|url=https://web.archive.org/web/20210305005604/http://crystal.med.upenn.edu/sharp-lab-pdfs/2015SharpMatschinsky_Boltz1877_Entropy17.pdf |date=2021-03-05 }}</ref> derived the distribution again under the framework of [[statistical thermodynamics]]. The derivations in this section are along the lines of Boltzmann's 1877 derivation, starting with result known as [[Maxwell–Boltzmann statistics]] (from statistical thermodynamics). Maxwell–Boltzmann statistics gives the average number of particles found in a given single-particle [[Microstate (statistical mechanics)|microstate]]. Under certain assumptions, the logarithm of the fraction of particles in a given microstate is linear in the ratio of the energy of that state to the temperature of the system: there are constants <math>k</math> and <math>C</math> such that, for all <math>i</math>,
<math display="block">-\log \left(\frac{N_i}{N}\right) = \frac{1}{k}\cdot\frac{E_i}{T} + C.</math>
The assumptions of this equation are that the particles do not interact, and that they are classical; this means that each particle's state can be considered independently from the other particles' states. Additionally, the particles are assumed to be in thermal equilibrium.<ref name="StatisticalPhysics" /><ref>{{ cite book | last = Parker | first = Sybil P. | title = McGraw-Hill Encyclopedia of Physics | date = 1993 | publisher = McGraw-Hill | isbn = 978-0-07-051400-3 | edition = 2nd}}</ref>

This relation can be written as an equation by introducing a normalizing factor:
{{NumBlk||<math display="block">
\frac{N_i} N =
\frac{
    \exp\left(-\frac{E_i}{k_\text{B}T}\right)
}{
    \displaystyle \sum_j \exp\left(-\tfrac{E_j}{k_\text{B}T}\right)
}</math>|{{EquationRef|1}}}}

where:
* {{mvar|N<sub>i</sub>}} is the expected number of particles in the single-particle microstate {{mvar|i}},
* {{mvar|N}} is the total number of particles in the system,
* {{mvar|E<sub>i</sub>}}  is the energy of microstate {{mvar|i}},
* the sum over index {{mvar|j}} takes into account all microstates, 
* {{mvar|T}} is the equilibrium temperature of the system,
* {{math|''k''<sub>B</sub>}} is the [[Boltzmann constant]].
The denominator in {{EquationNote|1|equation 1}} is a normalizing factor so that the ratios <math>N_i:N</math>  add up to unity — in other words it is a kind of [[partition function (statistical mechanics)|partition function]] (for the single-particle system, not the usual partition function of the entire system).

Because velocity and speed are related to energy, Equation ({{EquationNote|1}}) can be used to derive relationships between temperature and the speeds of gas particles. All that is needed is to discover the density of microstates in energy, which is determined by dividing up momentum space into equal sized regions.

===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.

===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.

===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 &times; 3}} identity matrix.

===Distribution for the speed===

The Maxwell–Boltzmann distribution for the speed follows immediately from the distribution of the velocity vector, above. Note that the speed is
<math display="block">v = \sqrt{v_x^2 + v_y^2 + v_z^2}</math>
and the [[volume element]] in [[spherical coordinates]]
<math display="block"> dv_x\, dv_y\, dv_z = v^2 \sin \theta\, dv\, d\theta\, d\phi = v^2 \, dv \, d\Omega</math>
where <math>\phi</math> and <math>\theta</math> are the [[Spherical coordinate system|spherical coordinate]] angles of the velocity vector. [[Spherical coordinate system#Integration and differentiation in spherical coordinates|Integration]] of the probability density function of the velocity over the solid angles <math>d\Omega</math> yields an additional factor of <math>4\pi</math>.
The speed distribution with substitution of the speed for the sum of the squares of the vector components:
{{Equation box 1 |indent=: |equation=<math>
f (v) =
\sqrt{\frac{2}{\pi}} \, \biggl[\frac{m}{k_\text{B}T}\biggr]^{3/2} v^2 \exp\left(-\frac{mv^2}{2k_\text{B}T}\right).
</math>}}