Boltzmann distribution

**Maxwell-Boltzmann**
Probability density function ?
Cumulative distribution function ?
Parameters	$a>0\,$
Support	$x\in [0;\infty)$
pdf	$\sqrt{\frac{2}{\pi}} \frac{x^2 e^{-x^2/(2a^2)}}{a^3}$
cdf	$\textrm{erf}\left(\frac{x}{\sqrt{2} a}\right) -\sqrt{\frac{2}{\pi}} \frac{x e^{-x^2/(2a^2)}}{a}$
Mean	$\mu=2a \sqrt{\frac{2}{\pi}}$
Median
Mode	$\sqrt{2} a$
Variance	$\sigma^2=\frac{a^2(3 \pi - 8)}{\pi}$
Skewness	$\gamma_1=\frac{2 \sqrt{2} (5 \pi - 16)}{(3 \pi - 8)^{3/2}}$
Kurtosis	$\gamma_2=-4\frac{96-40\pi+3\pi^2}{(3 \pi - 8)^2}$
Entropy
mgf
Char. func.

The Maxwell-Boltzmann distribution is a probability distribution with applications in physics and chemistry. The distribution can be thought of as the magnitude of a 3-dimensional vector if its components are distributed as a normal distribution with standard deviation $a$ . If $X i$ are distributed as $X \sim N(0, a^2)$ , then

$Z = \sqrt{X_1^2+X_2^2+X_3^2}$

is distributed as a Maxwell-Boltzmann distribution with parameter $a$ .

1 Properties
2 Physical applications of the Maxwell-Boltzmann distribution
3 See also
4 External links

Properties

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The Maxwell-Boltzmann distribution with $a = 1$ is equivalent to the chi distribution with three degrees of freedom. Additionally, if Z is distributed as a Maxwell-Boltzmann distribution with parameter a, then

$W = \frac{Z}{a}$

will be distributed as a chi distribution with three degrees of freedom.

The root-mean-square of a Maxwell-Boltzmann distribution is $\sqrt{3}a$ . Since $\sqrt{2} < 2\sqrt{2/\pi} < \sqrt{3}$ , it follows that the mode is less than the mean, which is always less than the root-mean-square.

Physical applications of the Maxwell-Boltzmann distribution

It has been suggested that this article or section be merged into Maxwell-Boltzmann statistics. (Discuss)

The Maxwell-Boltzmann distribution forms the basis of the kinetic theory of gases, which explains many fundamental gas properties, including pressure and diffusion. The Maxwell-Boltzmann distribution is usually thought of as the distribution of molecular speeds in a gas, but it can also refer to the distribution of velocities, momenta, and magnitude of the momenta of the molecules, each of which will have a different probability distribution function, all of which are related.

The Maxwell-Boltzmann distribution can be derived using statistical mechanics (see the derivation of the partition function). It corresponds to the most probable speed distribution in a collisionally-dominated system consisting of a large number of non-interacting particles in which quantum effects are negligible. Since interactions between the molecules in a gas are generally quite small, the Maxwell-Boltzmann distribution provides a very good approximation of the conditions in a gas.

In many other cases, however, the condition of elastic collisions dominating all other processes is not even approximately fulfilled. That is true, for instance, for the physics of the ionosphere and space plasmas where recombination and collisional excitation (i.e. radiative processes) are of far greater importance: in particular for the electrons. Not only would the assumption of a Maxwell distribution yield quantitatively wrong results, but even prevent a correct qualitative understanding of the physics involved. Also, in cases where the quantum thermal wavelength of the gas is not small compared to the distance between particles, there will be deviations from the Maxwell distribution due to quantum effects.

Originally suggested by Maxwell using an invariance argument, and later derived by Boltzmann using kinetic theory, the Maxwell-Boltzmann distribution can now most readily be derived from the Boltzmann distribution for energies:

$\frac{N_i}{N} = \frac{g_i \exp\left(-E_i/kT \right) } { \sum_{j}^{} g_j \,{\exp\left(-E_j/kT\right)} } \qquad\qquad (1)$

where N_i is the number of molecules at equilibrium temperature T, in a state i which has energy E_i and degeneracy g_i, N is the total number of molecules in the system and k is the Boltzmann constant. (Note that sometimes the above equation is written without the degeneracy factor g_i. In this case the index i will specify an individual state, rather than a set of g_i states having the same energy E_i.) Because velocity and speed are related to energy, Equation 1 can be used to derive relationships between temperature and the speeds of molecules in a gas. The denominator in this equation is known as the canonical partition function.

The distribution of the momentum vector

What follows is a derivation wildly different from the derivation described by James Clerk Maxwell and later described with fewer assumptions by Ludwig Boltzmann. Instead it is close to Boltzmann's later approach of 1877.

For the case of an "ideal gas" consisting of non-interacting atoms in the ground state, all energy is in the form of kinetic energy. The relationship between kinetic energy and momentum for massive particles is

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

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

$\frac{N_i}{N} = \frac{1}{Z} \exp \left[ \frac{-(p_x^2 + p_y^2 + p_z^2)}{2mkT} \right] \qquad\qquad (3)$

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:

$f_\mathbf{p} (p_x, p_y, p_z) = \frac{c}{Z} \exp \left[ \frac{-(p_x^2 + p_y^2 + p_z^2)}{2mkT} \right]. \qquad\qquad (4)$

The normalizing constant c, can be determined by recognizing that the probability of a molecule having any 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:

$c = \frac{Z}{(\sqrt{2 \pi mkT}) ^ 3}. \qquad\qquad (5)$

Substituting Equation 5 into Equation 4 and using p_i = mv_i for each component of momentum gives:

$f_\mathbf{p} (p_x, p_y, p_z) = \sqrt{\left( \frac{1}{2 \pi mkT} \right)^3} \exp \left[ \frac{-(p_x^2 + p_y^2 + p_z^2)}{2mkT} \right]. \qquad\qquad (6)$

The distribution is seen to be the product of three independent normally distributed variables $p x$ $p y$ , and $p z$ , with variance $m k T$ . Additionally, it can be seen that the magnitude of momentum will be distributed as a Maxwell-Boltzmann distribution, with $a=\sqrt{mkT}$ .

Distribution of the energy

Using p² = 2mE we get the energy distribution:

$f_E\,dE=f_p\left(\frac{dp}{dE}\right)\,dE =2\sqrt{\frac{E}{\pi(kT)^3}}~\exp\left[\frac{-E}{kT}\right]\,dE.$

Since the energy is proportional to the sum of the squares of the three normally distributed momentum components, this distribution is a chi-square distribution with three degrees of freedom:

$f_E(E)\,dE=\chi^2(x;3)\,dx$

where

$x=\frac{2E}{kT}.\,$

The Maxwell-Boltzmann distribution can also be obtained by considering the gas to be a quantum gas.

Distribution of the velocity vector

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

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

and using p = mv we get

$f_\mathbf{v} (v_x, v_y, v_z) = \sqrt{ \left(\frac{m}{2 \pi kT} \right)^3} \exp \left[ \frac{-m(v_x^2 + v_y^2 + v_z^2)}{2kT} \right], \qquad\qquad (8)$

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 $k T / m$ . 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

$f_v (v_i) = \sqrt{\frac{m}{2 \pi kT}} \exp \left[ \frac{-mv_i^2}{2kT} \right]. \qquad\qquad (9)$

This distribution has the form of a normal distribution, with variance $\frac{m}{kT}$ . As expected for a gas at rest, the average velocity in any particular direction is zero.

Distribution of speeds

A chart displaying the speed probability density functions for a few noble gases at a temperature of 298.15K (25 C).

Usually, we are more interested in the speed of molecules rather than the component velocities, where the speed, v is defined as

$v = \sqrt{v_x^2 + v_y^2 + v_z^2}. (10)$

Since the speed is the square root of the sum of squares of the three independent, normally distributed velocity components, this distribution is a Maxwell-Boltzmann distribution, with $a=\sqrt{m/kT}$ .

We are often more interested in quantities such as the average speed of the particles rather than the actual distribution. The mean speed, most probable speed (mode), and root-mean-square can be obtained from properties of the Maxwell-Boltzmann distribution.