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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 $\beta \equiv 1 / k_B T$. We split $E = K + U$, where $K$ and $U$ are the total contributions from the kinetic and potential energies of the system. For $N$ particles with positions $\va{r} \equiv (\vb{r}_1, ..., \vb{r}_N)$ and momenta $\va{p} = (\vb{p}_1, ..., \vb{p}_N)$, then $K$ only depends on $\va{p}$ and $U$ only on $\va{r}$, so the probability of a specific microstate $(\va{r}, \va{p})$ is as follows:

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

Since this is classical physics, we can split the exponential (in quantum mechanics, the canonical commutation relation would prevent that):

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

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

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

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

$$\begin{aligned} K(\va{p}) = \sum_{n = 1}^N K_n(\vb{p}_n) \qquad \mathrm{where} \qquad K_n(\vb{p}_n) = \frac{|\vb{p}_n|^2}{2 m} \end{aligned}$$

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

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

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

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

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:

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

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

That was the distribution of the velocities along each axis, but what about the speed $v = |\vb{v}|$? Because we do not care about the direction of $\vb{v}$, 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 $v$:

$$\begin{aligned} g(v) = \dv{}{v} \bigg( \frac{4 \pi}{3} v^3 \bigg) = 4 \pi v^2 \end{aligned}$$

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:

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

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 $v_{\mathrm{rms}}$:

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

Which can be calculated to have the following exact expressions:

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

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:

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

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

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

References

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

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