The Maxwell-Boltzmann distribution is an important relationship that finds many applications in physics and chemistry. It forms the basis of the kinetic theory of gases, which accurately explains many fundamental gas properties, including pressure and diffusion. The Maxwell-Boltzmann distribution also finds important applications in electron transport and other phenomena.

The Maxwell-Boltzmann distribution can be derived using Statistical Mechanics (see Derivation of the partition function). It corresponds to the most probable energy distribution in a system consisting of a large number of non-interacting particles. 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 (except at relatively high pressures and low temperatures, where intermolecular interactions become important).

The Maxwell-Boltzmann distribution can be expressed as:

<math>

\frac{N_i}{N} = \frac{\exp(\frac{-E_i}{kT} ) } { \sum_j \exp(\frac{E_j}{kT}) } </math> (1)

where N_i is the number of molecules at equilibrium temperature T, having energy level E_i, N is the total number of molecules in the system and k is Boltzmanns constant. Essentially Equation 1 provides a means for calculating the fraction of molecules (N_i/N) that have energy E_i at a given temperature, T. 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.

Maxwell-Boltzmann Velocity Distribution

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. From the Particle in a box problem in Quantum mechanics we know that the energy levels for a gas in a rectangular box with sides of lengths a_x, a_y, a_z are given by:

<math>

E_i =

\left( \frac{n_x^2}{a_x^2} + \frac{n_y^2}{a_y^2} + \frac{n_z^2}{a_z^2} \right)

\left( \frac{h^2}{8m} \right) </math> (2)

where, n_x, n_y, and n_z are the quantum numbers for x,y, and z motion, respectively. However, for a macroscopic sized box, the energy levels are very closely spaced, so the energy levels can be considered continuous and we can replace the sum with an integral. Furthermore, we can recognize that (h²n_i²/4a_i²) corresponds to the square of the ith component of momentum, p_i² giving:

<math>

\frac{N_i}{N} = \frac{1}{q} \exp \left[ \frac{-(p_x^2 + p_y^2 + p_z^2)}{2mkT} \right] </math> (3)

where q corresponds to the denominator in Equation 1. This distribution of N_i/N is proportional to the probability distribution function f_p for finding a molecule with these values of of momentum components, so:

<math>

f_p (p_x, p_y, p_x) = \frac{c}{q} \exp \left[ \frac{-(p_x^2 + p_y^2 + p_z^2)}{2mkT} \right] </math> (4)

The constant of proportionality, 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:

<math>

\int \int \int dp_x dp_y dp_z \; \frac{1}{q} \exp \left[ \frac{-(p_x^2 + p_y^2 + p_z^2)}{2mkT} \right]

\frac{1}{q} \left( \frac{2m \pi}{kT} \right)^{3/2} </math> (5) so in order for the integral of equation 4 to be 1,

<math>

c

\frac{q}{(\sqrt{2 \pi mkT}) ^ 3} </math> (6)

Substituting Equation 6 into Equation 4 and using p_i=mv_i for each component of momentum gives:

<math>

f_p (p_x, p_y, p_x) = \sqrt{\left( \frac{1}{2 \pi mkT} \right)^3} \exp \left[ \frac{-m(v_x^2 + v_y^2 + v_z^2)}{2kT} \right] </math> (7)

Finally recognizing that the velocity probability distribution, f_v is proportional to the momentum probability distribution function as

<math>f_v = m^3 f_p</math>

we get:

<math>

f_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] </math> (8)

Which is the Maxwell-Boltzmann velocity distribution.

Velocity Distribution in One Direction

For the case of a single direction Equation 8 can be reduced to:

<math>

f_v (v_x) = \sqrt{\frac{m}{2 \pi kT}} \exp \left[ \frac{-mv_x^2}{2kT} \right] </math> (9)

This distribution has the form of a Gaussian error curve. As expected for a gas at rest, the average velocity in any particular direction is zero.

Distribution of Speeds

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

<math>

v^2 = v_x^2 + v_y^2 + v_z^2 </math> (10)

The corresponding speed distribution is:

<math>

F(v) = 4 \pi v^2 \left( \frac{m}{2 \pi kT} \right)^{3/2} \exp \left( \frac{-mv^2}{2kT} \right) </math> (11)

Average Speed

Although Equation 11 gives the distribution of speeds or in other words the fraction of molecules having a particular speed, we are often more interested in quantities such as the average speed of the particles rather than the actual distribution. In the following subsections we will define and derive the most probable speed, the mean speed and the root-mean-square speed.

Most Probable Speed

The most probable speed, v_p, is the speed most likely to be possessed by any molecule in the system and corresponds to the maximum value of F(v). To find it, we calculate dF/dv, set it to zero and solve for v:

<math>\frac{dF(v)}{dv} = \left( \frac{m}{2 \pi kT} \right)^{3/2} \exp \left( -mv^2/2kT \right) \left[ 8\pi v + 4 \pi v^2 (-mv/kT) \right] = 0</math> (12)

<math>v_p = (2kT/m)^{1/2} = (2RT/M)^{1/2}</math> (13)

Mean Speed

The mean speed, <v>, or average speed can be calculated using the expression:

<v> = &int[0,∞]; vF(v)dv (14)

Substituting in Equation 11 and performing the integration gives:

<v> = (8kT/πm)^1/2 = (8RT/πM)^1/2 (15)

Note that <v> and v_p differ by a constant factor (4/π)^1/2.

Root-mean-square Speed
The root mean square speed, v_rms is given by

v_rms = √∫v²F(v)dv. (16)

Substituting for F(v) and performing the integration, we get

v_rms = (3kT/m)^1/2 = (3RT/M)^1/2 (17)

Thus, v_p < <v> < v_rms.