Maxwell–Boltzmann distribution

Maxwell–Boltzmann

Probability density function
Cumulative distribution function
Parameters	${\displaystyle a>0}$
Support	${\displaystyle x\in (0;\infty )}$
PDF	${\displaystyle {\sqrt {\frac {2}{\pi }}}{\frac {x^{2}e^{-x^{2}/\left(2a^{2}\right)}}{a^{3}}}}$
CDF	${\displaystyle \operatorname {erf} \left({\frac {x}{{\sqrt {2}}a}}\right)-{\sqrt {\frac {2}{\pi }}}{\frac {xe^{-x^{2}/\left(2a^{2}\right)}}{a}}}$ where erf is the error function
Mean	${\displaystyle \mu =2a{\sqrt {\frac {2}{\pi }}}}$
Mode	${\displaystyle {\sqrt {2}}a}$
Variance	${\displaystyle \sigma ^{2}={\frac {a^{2}(3\pi -8)}{\pi }}}$
Skewness	${\displaystyle \gamma _{1}={\frac {2{\sqrt {2}}(16-5\pi )}{(3\pi -8)^{3/2}}}}$
Ex. kurtosis	${\displaystyle \gamma _{2}=4{\frac {\left(-96+40\pi -3\pi ^{2}\right)}{(3\pi -8)^{2}}}}$
Entropy	${\displaystyle \ln \left(a{\sqrt {2\pi }}\right)+\gamma -{\frac {1}{2}}}$

In physics (in particular in statistical mechanics), the Maxwell–Boltzmann distribution is a particular probability distribution named after James Clerk Maxwell and Ludwig Boltzmann.

It was first defined and used 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. The term "particle" in this context refers to gaseous particles only (atoms or molecules), and the system of particles is assumed to have reached thermodynamic equilibrium.^[1] The energies of such particles follow what is known as Maxwell-Boltzmann statistics, and the statistical distribution of speeds is derived by equating particle energies with kinetic energy.

Mathematically, the Maxwell–Boltzmann distribution is the chi distribution with three degrees of freedom (the components of the velocity vector in Euclidean space), with a scale parameter measuring speeds in units proportional to the square root of ${\displaystyle T/m}$ (the ratio of temperature and particle mass).^[2]

The Maxwell–Boltzmann distribution is a result of the kinetic theory of gases, which provides a simplified explanation of many fundamental gaseous properties, including pressure and diffusion.^[3] The Maxwell–Boltzmann distribution applies fundamentally to particle velocities in three dimensions, but turns out to depend only on the speed (the magnitude of the velocity) of the particles. 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 kinetic theory of gases 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. Ideal plasmas, which are ionized gases of sufficiently low density, frequently also have particle distributions that are partially or entirely Maxwellian.^[4]

The distribution was first derived by Maxwell in 1860 on heuristic grounds.^[5] Boltzmann later, in the 1870s, carried out significant investigations into the physical origins of this distribution.

Distribution function[edit]

Assuming the system of interest contains a large number of particles, the fraction of the particles within an infinitesimal element of three-dimensional velocity space, ${\displaystyle d^{3}v}$, centered on a velocity vector of magnitude ${\displaystyle v}$, is ${\displaystyle f(v)d^{3}v}$, in which

${\displaystyle f(v)~\mathrm {d} ^{3}v=\left({\frac {m}{2\pi kT}}\right)^{3/2}\,e^{-{\frac {m|v|^{2}}{2kT}}}~\mathrm {d} ^{3}v,}$

where ${\displaystyle m}$ is the particle mass and ${\displaystyle kT}$ is the product of Boltzmann's constant and thermodynamic temperature.

The speed probability density functions of the speeds of a few noble gases at a temperature of 298.15 K (25 °C). The y-axis 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.

One can write the element of velocity space as d${\displaystyle ^{3}v}$ = d${\displaystyle v_{x}}$d${\displaystyle v_{y}}$d${\displaystyle v_{z}}$, for velocities in a standard Cartesian coordinate system, or as d${\displaystyle ^{3}v}$ = ${\displaystyle v^{2}}$d${\displaystyle v}$d${\displaystyle \Omega }$ in a standard spherical coordinate system, where d${\displaystyle \Omega }$ is an element of solid angle. Here ${\displaystyle f(v)}$ is given as a probability distribution function, properly normalized so that ${\displaystyle \int f(v)}$d${\displaystyle ^{3}v}$ over all velocities equals one. In plasma physics, the probability distribution is often multipled by the particle density, so that the integral of the resulting distribution function equals the density.

Undergraduate students are likely to work with the Maxwellian distribution function for particles moving in only one direction. If this direction is ${\displaystyle x}$, then one has

${\displaystyle f(v_{x})~\mathrm {d} v_{x}=\left({\frac {m}{2\pi kT}}\right)^{1/2}\,e^{-{\frac {mv_{x}^{2}}{2kT}}}~\mathrm {d} v_{x},}$

which can be obtained by integrating the three-dimensional form given above over ${\displaystyle v_{y}}$ and ${\displaystyle v_{z}}$.

Recognizing the symmetry of ${\displaystyle f(v)}$, one can integrate over solid angle and write a probability distribution of speeds as the function^[6]

${\displaystyle f(v)~\mathrm {d} v=\left({\frac {m}{2\pi kT}}\right)^{3/2}\,4\pi v^{2}e^{-{\frac {mv^{2}}{2kT}}}~\mathrm {d} v,}$

This probability density function gives the probability, per unit speed, of finding the particle with a speed near ${\displaystyle v}$. This equation is simply the Maxwell-Boltzmann distribution (given in the infobox) with distribution parameter ${\displaystyle a={\sqrt {kT/m}}}$. The Maxwell–Boltzmann distribution is equivalent to the chi distribution with three degrees of freedom and scale parameter ${\displaystyle a={\sqrt {kT/m}}}$.

The simplest ordinary differential equation satisfied by the distribution is:

${\displaystyle kTvf'(v)+f(v)(mv^{2}-2kT)=0,}$

${\displaystyle f(1)={\sqrt {\frac {2}{\pi }}}e^{-{\frac {m}{2kT}}}\left({\frac {m}{kT}}\right)^{3/2}}$

or in unitless presentation:

${\displaystyle a^{2}xf'(x)+\left(x^{2}-2a^{2}\right)f(x)=0,}$

${\displaystyle f(1)={\frac {{\sqrt {\frac {2}{\pi }}}e^{-{\frac {1}{2a^{2}}}}}{a^{3}}}.}$

With the Darwin–Fowler method of mean values the Maxwell–Boltzmann distribution is obtained as an exact result.^{[clarification needed]}

Typical speeds[edit]

The mean speed ${\displaystyle \langle v\rangle }$, most probable speed (mode) v_p, and root-mean-square speed ${\displaystyle {\sqrt {\langle v^{2}\rangle }}}$ can be obtained from properties of the Maxwell distribution.

This works well for nearly ideal, monatomic gases like helium, but also for molecular gases like diatomic oxygen. This is because despite the larger heat capacity (larger internal energy at the same temperature) due to their larger number of degrees of freedom, their translational kinetic energy (and thus their speed) is unchanged.^[7]

In summary, the typical speeds are related as follows:

${\displaystyle v_{p}\approx 88.6\%\ \langle v\rangle <\langle v\rangle <108.5\%\ \langle v\rangle \approx v_{\mathrm {rms} }.}$

The root mean square speed is directly related to the speed of sound c in the gas, by

${\displaystyle c={\sqrt {\frac {\gamma }{3}}}\ v_{\mathrm {rms} }={\sqrt {\frac {f+2}{3f}}}\ v_{\mathrm {rms} }={\sqrt {\frac {f+2}{2f}}}\ v_{p},}$

where ${\displaystyle \gamma =1+{\frac {2}{f}}}$ is the adiabatic index, f is the number of degrees of freedom of the individual gas molecule. For the example above, diatomic nitrogen (approximating air) at 7002300000000000000♠300 K, ${\displaystyle f=5}$^[9] and

${\displaystyle c={\sqrt {\frac {7}{15}}}v_{\mathrm {rms} }\approx 68\%\ v_{\mathrm {rms} }\approx 84\%\ v_{p}\approx 353\ \mathrm {m/s} ,}$

the true value for air can be approximated by using the average molar weight of air (6998290000000000000♠29 g/mol), yielding 7002347000000000000♠347 m/s at 7002300000000000000♠300 K (corrections for variable humidity are of the order of 0.1% to 0.6%).

Derivation and related distributions[edit]

Maxwell–Boltzmann statistics[edit]

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.^[5][10] After Maxwell, Ludwig Boltzmann in 1872^[11] 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)^[12] 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, under certain assumptions, the logarithm of the fraction of particles in a given microstate is proportional to the ratio of the energy of that state to the temperature of the system:

${\displaystyle -\log({\frac {N_{i}}{N}})\propto {\frac {E_{i}}{N}}.}$

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.^[1][13]

This relation can be written as an equation by introducing a normalizing factor:

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

(1)

where:

• N_i is the expected number of particles in the single-particle microstate i,
• N is the total number of particles in the system,
• E_i is the energy of microstate i,
• the sum over index j takes into account all microstates,
• T is the equilibrium temperature of the system,
• k is the Boltzmann constant.

The denominator in Equation (1) is simply a normalizing factor so that the ratios ${\displaystyle N_{i}:N}$ add up to unity — in other words it is a kind of 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 (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[edit]

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

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

(2)

where p² is the square of the momentum vector p = [p_x, p_y, p_z]. We may therefore rewrite Equation (1) as:

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

(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 ${\displaystyle N_{i}:N}$ is proportional to the probability density function f_p for finding a molecule with these values of momentum components, so:

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

(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 (4) over all p_x, p_y, and p_z yields a factor of

${\displaystyle \iiint _{-\infty }^{+\infty }\exp \left[-{\frac {p_{x}^{2}+p_{y}^{2}+p_{z}^{2}}{2mkT}}\right]dp_{x}\ dp_{y}\ dp_{z}={({\sqrt {\pi }}{\sqrt {2mkT}})^{3}}}$

So that the normalized distribution function is:

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

The distribution is seen to be the product of three independent normally distributed variables ${\displaystyle p_{x}}$, ${\displaystyle p_{y}}$, and ${\displaystyle p_{z}}$, with variance ${\displaystyle mkT}$. Additionally, it can be seen that the magnitude of momentum will be distributed as a Maxwell–Boltzmann distribution, with ${\displaystyle a={\sqrt {mkT}}}$. 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[edit]

The energy distribution is found imposing

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

(7)

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

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

(8)

Using then (8) in (7), and expressing everything in terms of the energy ${\displaystyle E}$, we get

${\displaystyle f_{E}(E)dE={\frac {1}{(2\pi mkT)^{3/2}}}e^{-E/kT}4\pi m{\sqrt {2mE}}dE=2{\sqrt {\frac {E}{\pi }}}\left({\frac {1}{kT}}\right)^{3/2}\exp \left({\frac {-E}{kT}}\right)dE}$

and finally

${\displaystyle f_{E}(E)=2{\sqrt {\frac {E}{\pi }}}\left({\frac {1}{kT}}\right)^{3/2}\exp \left({\frac {-E}{kT}}\right)}$   (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 chi-squared 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 chi-squared distribution with one degree of freedom:^[14]

${\displaystyle f_{\epsilon }(\epsilon )\,d\epsilon ={\sqrt {\frac {1}{\pi \epsilon kT}}}~\exp \left[{\frac {-\epsilon }{kT}}\right]\,d\epsilon }$

where ${\displaystyle \epsilon }$ 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 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.

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[edit]

Recognizing that the velocity probability density f_v is proportional to the momentum probability density function by

${\displaystyle f_{\mathbf {v} }d^{3}v=f_{\mathbf {p} }\left({\frac {dp}{dv}}\right)^{3}d^{3}v}$

and using p = mv we get

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

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

${\displaystyle f_{\mathbf {v} }\left(v_{x},v_{y},v_{z}\right)\,dv_{x}\,dv_{y}\,dv_{z}.}$

Like the momentum, this distribution is seen to be the product of three independent normally distributed variables ${\displaystyle v_{x}}$, ${\displaystyle v_{y}}$, and ${\displaystyle v_{z}}$, but with variance ${\displaystyle {\frac {kT}{m}}}$. 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:

${\displaystyle f_{\mathbf {v} }\left(v_{x},v_{y},v_{z}\right)=f_{v}(v_{x})f_{v}(v_{y})f_{v}(v_{z})}$

where the distribution for a single direction is

${\displaystyle f_{v}(v_{i})={\sqrt {\frac {m}{2\pi kT}}}\exp \left[{\frac {-mv_{i}^{2}}{2kT}}\right].}$

Each component of the velocity vector has a normal distribution with mean ${\displaystyle \mu _{v_{x}}=\mu _{v_{y}}=\mu _{v_{z}}=0}$ and standard deviation ${\displaystyle \sigma _{v_{x}}=\sigma _{v_{y}}=\sigma _{v_{z}}={\sqrt {\frac {kT}{m}}}}$, so the vector has a 3-dimensional normal distribution, a particular kind of multivariate normal distribution, with mean ${\displaystyle \mu _{\mathbf {v} }={\mathbf {0} }}$ and standard deviation ${\displaystyle \sigma _{\mathbf {v} }={\sqrt {\frac {3kT}{m}}}}$.

Distribution for the speed[edit]

The Maxwell–Boltzmann distribution for the speed follows immediately from the distribution of the velocity vector, above. Note that the speed is

${\displaystyle v={\sqrt {v_{x}^{2}+v_{y}^{2}+v_{z}^{2}}}}$

and the volume element in spherical coordinates

${\displaystyle dv_{x}\,dv_{y}\,dv_{z}=v^{2}\sin \theta \,dv\,d\theta \,d\phi =v^{2}dv\,d\Omega }$

where ${\displaystyle \phi }$ and ${\displaystyle \theta }$ are the spherical coordinate angles of the velocity vector. Integration of the probability density function of the velocity over the solid angles ${\displaystyle d\Omega }$ yields an additional factor of ${\displaystyle 4\pi }$. The speed distribution with substitution of the speed for the sum of the squares of the vector components:

${\displaystyle f(v)=\left({\frac {2}{\pi }}\right)^{1/2}\left({\frac {m}{kT}}\right)^{3/2}v^{2}\exp \left[-{\frac {mv^{2}}{2kT}}\right].}$

See also[edit]

• Quantum Boltzmann equation
• Maxwell–Boltzmann statistics
• Maxwell–Jüttner distribution
• Boltzmann distribution
• Boltzmann factor
• Rayleigh distribution
• Kinetic theory of gases

References[edit]

Further reading[edit]

• Physics for Scientists and Engineers – with Modern Physics (6th Edition), P. A. Tipler, G. Mosca, Freeman, 2008, ISBN 0-7167-8964-7
• Thermodynamics, From Concepts to Applications (2nd Edition), A. Shavit, C. Gutfinger, CRC Press (Taylor and Francis Group, USA), 2009, ISBN 978-1-4200-7368-3
• Chemical Thermodynamics, D.J.G. Ives, University Chemistry, Macdonald Technical and Scientific, 1971, ISBN 0-356-03736-3
• Elements of Statistical Thermodynamics (2nd Edition), L.K. Nash, Principles of Chemistry, Addison-Wesley, 1974, ISBN 0-201-05229-6
• Ward, CA & Fang, G 1999, 'Expression for predicting liquid evaporation flux: Statistical rate theory approach', Physical Review E, vol. 59, no. 1, pp. 429–40.
• Rahimi, P & Ward, CA 2005, 'Kinetics of Evaporation: Statistical Rate Theory Approach', International Journal of Thermodynamics, vol. 8, no. 9, pp. 1–14.

External links[edit]

• "The Maxwell Speed Distribution" from The Wolfram Demonstrations Project at Mathworld