Categories: Physics, Statistics, Thermodynamics.

# Maxwell-Boltzmann distribution

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

## Velocity vector distribution

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 $$E = K + U$$, where $$K$$ and $$U$$ are 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 particles’ 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}

## Speed distribution

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}

## Kinetic energy distribution

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}

## References

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