Categories: Physics, Statistics, Thermodynamics.

The **Maxwell-Boltzmann distributions** are a set of closely related probability distributions with applications in classical statistical physics.

In the canonical ensemble (where a fixed-size system can exchange energy with its environment), the probability of a microstate with energy \(E\) is given by the Boltzmann distribution:

\[\begin{aligned} f(E) \:\propto\: \exp\!\big(\!-\! \beta E\big) \end{aligned}\]

Where \(\beta = 1 / k_B T\). We split \(E = K + U\), with \(K\) and \(U\) the total kinetic and potential energy contributions. If there are \(N\) particles in the system, with positions \(\tilde{r} = (\vec{r}_1, ..., \vec{r}_N)\) and momenta \(\tilde{p} = (\vec{p}_1, ..., \vec{p}_N)\), then \(K\) only depends on \(\tilde{p}\), and \(U\) only depends on \(\tilde{r}\), so the probability of a specific microstate \((\tilde{r}, \tilde{p})\) is as follows:

\[\begin{aligned} f(\tilde{r}, \tilde{p}) \:\propto\: \exp\!\Big(\!-\! \beta \big( K(\tilde{p}) + U(\tilde{r}) \big) \Big) \end{aligned}\]

Since this is classical physics, we can split the exponential. In quantum mechanics, the canonical commutation relation would prevent that. Anyway, splitting yields:

\[\begin{aligned} f(\tilde{r}, \tilde{p}) \:\propto\: \exp\!\big(\!-\! \beta K(\tilde{p}) \big) \exp\!\big(\!-\! \beta U(\tilde{r}) \big) \end{aligned}\]

Classically, the probability distributions of the momenta and positions are independent:

\[\begin{aligned} f_K(\tilde{p}) \:\propto\: \exp\!\big(\!-\! \beta K(\tilde{p}) \big) \qquad f_U(\tilde{r}) \:\propto\: \exp\!\big(\!-\! \beta U(\tilde{r}) \big) \end{aligned}\]

We cannot evaluate \(f_U(\tilde{r})\) further without knowing \(U(\tilde{r})\) for a system. We thus turn to \(f_K(\tilde{p})\), and see that the total kinetic energy \(K(\tilde{p})\) is simply the sum of the particlesâ€™ individual kinetic energies \(K_n(\vec{p}_n)\), which are well-known:

\[\begin{aligned} K(\tilde{p}) = \sum_{n = 1}^N K_n(\vec{p}_n) \qquad \mathrm{where} \qquad K_n(\vec{p}_n) = \frac{|\vec{p}_n|^2}{2 m} \end{aligned}\]

Consequently, the probability distribution \(f(p_x, p_y, p_z)\) for the momentum vector of a single particle is as follows, after normalization:

\[\begin{aligned} f(p_x, p_y, p_z) = \Big( \frac{1}{2 \pi m k_B T} \Big)^{3/2} \exp\!\Big( \!-\!\frac{(p_x^2 + p_y^2 + p_z^2)}{2 m k_B T} \Big) \end{aligned}\]

We now rewrite this using the velocities \(v_x = p_x / m\), and update the normalization, giving:

\[\begin{aligned} \boxed{ f(v_x, v_y, v_z) = \Big( \frac{m}{2 \pi k_B T} \Big)^{3/2} \exp\!\Big( \!-\!\frac{m (v_x^2 + v_y^2 + v_z^2)}{2 k_B T} \Big) } \end{aligned}\]

This is the **Maxwell-Boltzmann velocity vector distribution**. Clearly, this is a product of three exponentials, so the velocity in each direction is independent of the others:

\[\begin{aligned} f(v_x) = \sqrt{\frac{m}{2 \pi k_B T}} \exp\!\Big( \!-\!\frac{m v_x^2}{2 k_B T} \Big) \end{aligned}\]

The distribution is thus an isotropic gaussian with standard deviations given by:

\[\begin{aligned} \sigma_x = \sigma_y = \sigma_z = \sqrt{\frac{k_B T}{m}} \end{aligned}\]

We know the distribution of the velocities along each axis, but what about the speed \(v = |\vec{v}|\)? Because we do not care about the direction of \(\vec{v}\), only its magnitude, the density of states \(g(v)\) is not constant: it is the rate-of-change of the volume of a sphere of radius \(v\):

\[\begin{aligned} g(v) = \dv{v} \Big( \frac{4 \pi}{3} v^3 \Big) = 4 \pi v^2 \end{aligned}\]

Multiplying the velocity vector distribution by \(g(v)\) and substituting \(v^2 = v_x^2 + v_y^2 + v_z^2\) then gives us the **Maxwell-Boltzmann speed distribution**:

\[\begin{aligned} \boxed{ f(v) = 4 \pi \Big( \frac{m}{2 \pi k_B T} \Big)^{3/2} v^2 \exp\!\Big( \!-\!\frac{m v^2}{2 k_B T} \Big) } \end{aligned}\]

Some notable points on this distribution are the most probable speed \(v_{\mathrm{mode}}\), the mean average speed \(v_{\mathrm{mean}}\), and the root-mean-square speed \(v_{\mathrm{rms}}\):

\[\begin{aligned} f'(v_\mathrm{mode}) = 0 \qquad v_\mathrm{mean} = \int_0^\infty v \: f(v) \dd{v} \qquad v_\mathrm{rms} = \bigg( \int_0^\infty v^2 \: f(v) \dd{v} \bigg)^{1/2} \end{aligned}\]

Which can be calculated to have the following exact expressions:

\[\begin{aligned} \boxed{ v_{\mathrm{mode}} = \sqrt{\frac{2 k_B T}{m}} } \qquad \boxed{ v_{\mathrm{mean}} = \sqrt{\frac{8 k_B T}{\pi m}} } \qquad \boxed{ v_{\mathrm{rms}} = \sqrt{\frac{3 k_B T}{m}} } \end{aligned}\]

Using the speed distribution, we can work out the kinetic energy distribution. Because \(K\) is not proportional to \(v\), we must do this by demanding that:

\[\begin{aligned} f(K) \dd{K} = f(v) \dd{v} \quad \implies \quad f(K) = f(v) \dv{v}{K} \end{aligned}\]

We know that \(K = m v^2 / 2\), meaning \(\dd{K} = m v \dd{v}\) so the energy distribution \(f(K)\) is:

\[\begin{aligned} f(K) = \frac{f(v)}{m v} = \sqrt{\frac{2 m}{\pi}} \: \bigg( \frac{1}{k_B T} \bigg)^{3/2} v \exp\!\Big( \!-\!\frac{m v^2}{2 k_B T} \Big) \end{aligned}\]

Substituting \(v = \sqrt{2 K/m}\) leads to the **Maxwell-Boltzmann kinetic energy distribution**:

\[\begin{aligned} \boxed{ f(K) = 2 \sqrt{\frac{K}{\pi}} \: \bigg( \frac{1}{k_B T} \bigg)^{3/2} \exp\!\Big( \!-\!\frac{K}{k_B T} \Big) } \end{aligned}\]

- H. Gould, J. Tobochnik,
*Statistical and thermal physics*, 2nd edition, Princeton.

© Marcus R.A. Newman, a.k.a. "Prefetch".
Available under CC BY-SA 4.0.