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In
statistics the
Maxwell–Boltzmann distribution is a particular
probability distribution named after
James Clerk Maxwell and
Ludwig Boltzmann. It was first defined and used in
physics (in particular in
statistical mechanics) for describing particle speeds in idealized
gases where the particles move freely inside a stationary container without interacting with one another, except for very brief
collisions
in which they exchange energy and momentum with each other or with
their thermal environment. Particle in this context refers to gaseous
particles (
atoms or
molecules), and the system of particles is assumed to have reached
thermodynamic equilibrium.
^{[1]} While the distribution was first derived by Maxwell in 1860 on heuristic grounds,
^{[2]} Boltzmann later carried out significant investigations into the physical origins of this distribution.
A particle speed probability distribution indicates which speeds are
more likely: a particle will have a speed selected randomly from the
distribution, and is more likely to be within one range of speeds than
another. The distribution depends on the
temperature of the system and the mass of the particle.
^{[3]} The Maxwell–Boltzmann distribution applies to the classical
ideal gas, which is an idealization of real gases. In real gases, there are various effects (e.g.,
van der Waals interactions,
vortical flow,
relativistic speed limits, and
quantum exchange interactions) that can make their speed distribution different from the Maxwell–Boltzmann form. However,
rarefied
gases at ordinary temperatures behave very nearly like an ideal gas and
the Maxwell speed distribution is an excellent approximation for such
gases. Thus, it forms the basis of the
Kinetic theory of gases, which provides a simplified explanation of many fundamental gaseous properties, including
pressure and
diffusion.
^{[4]}
Distribution function
The speed probability density functions of the speeds of a few
noble gases at a temperature of 298.15 K (25 °C). The
yaxis
is in s/m so that the area under any section of the curve (which
represents the probability of the speed being in that range) is
dimensionless.
The Maxwell–Boltzmann distribution is the function
^{[5]}
 $f(v)={\sqrt {\left({\frac {m}{2\pi kT}}\right)^{3}}}\,4\pi v^{2}e^{{\frac {mv^{2}}{2kT}}},$
where
$m$ is the particle mass and
$kT$ is the product of
Boltzmann's constant and
thermodynamic temperature.
An interesting point to be noted is that the MaxwellBoltzmann
distribution will not vary with the value of m/T i.e the ratio of mass
of the molecule to its absolute temperature; mathematically (Derivative
of f(v)/derivative of (m/T))=0. This
probability density function gives the probability, per unit speed, of finding the particle with a speed near
$v$. This equation is simply the Maxwell distribution (given in the infobox) with distribution parameter
$a={\sqrt {kT/m}}$. In probability theory the Maxwell–Boltzmann distribution is a
chi distribution with three degrees of freedom and
scale parameter $a={\sqrt {kT/m}}$.
The simplest
ordinary differential equation satisfied by the distribution is:
 $kTvf'(v)+f(v)(mv^{2}2kT)=0,$
 $f(1)={\sqrt {\frac {2}{\pi }}}e^{{\frac {m}{2kT}}}\left({\frac {m}{kT}}\right)^{3/2}$
or in unitless presentation:
 $a^{2}xf'(x)+\left(x^{2}2a^{2}\right)f(x)=0,$
 $f(1)={\frac {{\sqrt {\frac {2}{\pi }}}e^{{\frac {1}{2a^{2}}}}}{a^{3}}}.$
Note that a distribution (function) is not the same as the
probability. The distribution (function) stands for an average number,
as in all three kinds of statistics (Maxwell–Boltzmann,
Bose–Einstein,
Fermi–Dirac). With the
Darwin–Fowler method of mean values the Maxwell–Boltzmann distribution is obtained as an exact result.
Typical speeds
The mean speed, most probable speed (mode), and rootmeansquare can be obtained from properties of the Maxwell distribution.
 The most probable speed, v_{p}, is the speed most likely to be possessed by any molecule (of the same mass m) in the system and corresponds to the maximum value or mode of f(v). To find it, we calculate the derivative df/dv, set it to zero and solve for v:
 ${\frac {df(v)}{dv}}=0$
which yields:
 $v_{p}={\sqrt {\frac {2kT}{m}}}={\sqrt {\frac {2RT}{M}}}$
where R is the gas constant and M = N_{A} m is the molar mass of the substance.
For diatomic nitrogen (N_{2}, the primary component of air) at room temperature (300 K), this gives $v_{p}=422$ m/s
 The mean speed is the expected value of the speed distribution
 $\langle
v\rangle =\int _{0}^{\infty }v\,f(v)\,dv={\sqrt {\frac {8kT}{\pi
m}}}={\sqrt {\frac {8RT}{\pi M}}}={\frac {2}{\sqrt {\pi }}}v_{p}$
 The root mean square speed is the secondorder moment of speed:
 ${\sqrt
{\langle v^{2}\rangle }}=\left(\int _{0}^{\infty
}v^{2}\,f(v)\,dv\right)^{1/2}={\sqrt {\frac {3kT}{m}}}={\sqrt {\frac
{3RT}{M}}}={\sqrt {\frac {3}{2}}}v_{p}$
The typical speeds are related as follows:
 $0.886\langle v\rangle =v_{p}<\langle v\rangle <{\sqrt {\langle v^{2}\rangle }}=1.085\langle v\rangle .$
Derivation and related distributions
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.
^{[2]}^{[6]} After Maxwell,
Ludwig Boltzmann in 1872
^{[7]}
also derived the distribution on mechanical grounds and argued that
gases should over time tend toward this distribution, due to collisions
(see
Htheorem). He later (1877)
^{[8]} 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 singleparticle
microstate, under certain assumptions:
^{[1]}^{[9]}

${\frac {N_{i}}{N}}={\frac {\exp(E_{i}/kT)}{\sum _{j}\exp(E_{j}/kT)}}$


(1)

where:
 i and j are indices (or labels) of the singleparticle micro states,
 N_{i} is the average number of particles in the singleparticle microstate i,
 N is the total number of particles in the system,
 E_{i} is the energy of microstate i,
 T is the equilibrium temperature of the system,
 k is the Boltzmann constant.
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.
The denominator in Equation (
1) is simply a normalizing factor so that the
N_{i}/
N add up to 1 — in other words it is a kind of
partition function (for the singleparticle system, not the usual partition function of the entire system).
Because velocity and speed are related to energy, Equation (
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 and momentum for massive non
relativistic particles is

$E={\frac {p^{2}}{2m}}$


(2)

where
p^{2} is the square of the momentum vector
p = [
p_{x},
p_{y},
p_{z}]. We may therefore rewrite Equation (
1) as:

${\frac {N_{i}}{N}}={\frac {1}{Z}}\exp \left[{\frac {p_{i,x}^{2}+p_{i,y}^{2}+p_{i,z}^{2}}{2mkT}}\right]$


(3)

where
Z is the
partition function, corresponding to the denominator in Equation (
1). Here
m is the molecular mass of the gas,
T is the thermodynamic temperature and
k is the
Boltzmann constant. This distribution of
N_{i}/
N is
proportional to the
probability density function f_{p} for finding a molecule with these values of momentum components, so:

$f_{\mathbf
{p} }(p_{x},p_{y},p_{z})={\frac {c}{Z}}\exp \left[{\frac
{p_{x}^{2}+p_{y}^{2}+p_{z}^{2}}{2mkT}}\right]$


(4)

The
normalizing constant c, can be determined by recognizing that the probability of a molecule having
some momentum must be 1. Therefore the integral of equation (
4) over all
p_{x},
p_{y}, and
p_{z} must be 1.
It can be shown that:

$c={\frac {Z}{(2\pi mkT)^{3/2}}}$


(5)

Substituting Equation (
5) into Equation (
4) gives:

$f_{\mathbf
{p} }(p_{x},p_{y},p_{z})=\left(2\pi mkT\right)^{3/2}\exp \left[{\frac
{p_{x}^{2}+p_{y}^{2}+p_{z}^{2}}{2mkT}}\right]$ (6)

The distribution is seen to be the product of three independent
normally distributed variables
$p_{x}$,
$p_{y}$, and
$p_{z}$, with variance
$mkT$. Additionally, it can be seen that the magnitude of momentum will be distributed as a Maxwell–Boltzmann distribution, with
$a={\sqrt {mkT}}$.
The Maxwell–Boltzmann distribution for the momentum (or equally for the
velocities) can be obtained more fundamentally using the
Htheorem at equilibrium within the
Kinetic theory of gases framework.
Distribution for the energy
The energy distribution is found imposing

$f_{E}(E)dE=f_{p}({\textbf {p}})d^{3}{\textbf {p}},$


(7)

where
$d^{3}{\textbf {p}}$ is the infinitesimal phasespace volume of momenta corresponding to the energy interval
$dE$. Making use of the spherical symmetry of the energymomentum dispersion relation
$E={\textbf {p}}^{2}/2m$, this can be expressed in terms of
$dE$ as

$d^{3}{\textbf {p}}=4\pi {\textbf {p}}^{2}d{\textbf {p}}=4\pi m{\sqrt {2mE}}dE.$


(8)

Using then (
8) in (
7), and expressing everything in terms of the energy
$E$, we get
 $f_{E}(E)dE={\frac {1}{(2\pi mkT)^{3/2}}}e^{E/kT}4\pi m{\sqrt
{2mE}}dE=2{\sqrt {\frac {E}{\pi }}}\left({\frac
{1}{kT}}\right)^{3/2}\exp \left({\frac {E}{kT}}\right)dE$
and finally

$f_{E}(E)=2{\sqrt {\frac {E}{\pi }}}\left({\frac {1}{kT}}\right)^{3/2}\exp \left({\frac {E}{kT}}\right)$ (9)

Since the energy is proportional to the sum of the squares of the
three normally distributed momentum components, this distribution is a
gamma distribution; in particular, it is a
chisquared distribution with three degrees of freedom.
By the
equipartition theorem,
this energy is evenly distributed among all three degrees of freedom,
so that the energy per degree of freedom is distributed as a chisquared
distribution with one degree of freedom:
^{[10]}
 $f_{\epsilon
}(\epsilon )\,d\epsilon ={\sqrt {\frac {1}{\pi \epsilon kT}}}~\exp
\left[{\frac {\epsilon }{kT}}\right]\,d\epsilon$
where
$\epsilon$
is the energy per degree of freedom. 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 chisquared distribution with one
degree of freedom, and the total energy will be distributed according to
a chisquared distribution with five degrees of freedom. This has
implications in the theory of the
specific heat of a gas.
The Maxwell–Boltzmann distribution can also be obtained by considering the gas to be a type of
quantum gas for which the approximation
ε >> k T may be made.
Distribution for the velocity vector
Recognizing that the velocity probability density
f_{v} is proportional to the momentum probability density function by
 $f_{\mathbf {v} }d^{3}v=f_{\mathbf {p} }\left({\frac {dp}{dv}}\right)^{3}d^{3}v$
and using
p = m
v we get

$f_{\mathbf
{v} }(v_{x},v_{y},v_{z})=\left({\frac {m}{2\pi kT}}\right)^{3/2}\exp
\left[{\frac {m(v_{x}^{2}+v_{y}^{2}+v_{z}^{2})}{2kT}}\right]$

which is the Maxwell–Boltzmann velocity distribution. The probability
of finding a particle with velocity in the infinitesimal element [
dv_{x},
dv_{y},
dv_{z}] about velocity
v = [
v_{x},
v_{y},
v_{z}] is
 $f_{\mathbf {v} }\left(v_{x},v_{y},v_{z}\right)\,dv_{x}\,dv_{y}\,dv_{z}.$
Like the momentum, this distribution is seen to be the product of three independent
normally distributed variables
$v_{x}$,
$v_{y}$, and
$v_{z}$, but with variance
${\frac {kT}{m}}$. It can also be seen that the Maxwell–Boltzmann velocity distribution for the vector velocity [
v_{x},
v_{y},
v_{z}] is the product of the distributions for each of the three directions:
 $f_{v}\left(v_{x},v_{y},v_{z}\right)=f_{v}(v_{x})f_{v}(v_{y})f_{v}(v_{z})$
where the distribution for a single direction is
 $f_{v}(v_{i})={\sqrt {\frac {m}{2\pi kT}}}\exp \left[{\frac {mv_{i}^{2}}{2kT}}\right].$
Each component of the velocity vector has a
normal distribution with mean
$\mu _{v_{x}}=\mu _{v_{y}}=\mu _{v_{z}}=0$ and standard deviation
$\sigma _{v_{x}}=\sigma _{v_{y}}=\sigma _{v_{z}}={\sqrt {\frac {kT}{m}}}$, so the vector has a 3dimensional normal distribution, a particular kind of
multivariate normal distribution, with mean
$\mu _{\mathbf {v} }={\mathbf {0} }$ and standard deviation
$\sigma _{\mathbf {v} }={\sqrt {\frac {3kT}{m}}}$.
The Maxwell–Boltzmann distribution for the speed follows immediately
from the distribution of the velocity vector, above. Note that the speed
is
 $v={\sqrt {v_{x}^{2}+v_{y}^{2}+v_{z}^{2}}}$
and the
volume element in
spherical coordinates
 $dv_{x}\,dv_{y}\,dv_{z}=v^{2}\sin \theta \,dv\,d\theta \,d\phi$
where
$\phi$ and
$\theta$
are the "course" (azimuth of the velocity vector) and "path angle"
(elevation angle of the velocity vector). Integration of the normal
probability density function of the velocity, above, over the course
(from 0 to
$2\pi$) and path angle (from 0 to
$\pi$), with substitution of the speed for the sum of the squares of the vector components, yields the speed distribution.