path: root/content/know/concept/maxwell-boltzmann-distribution/index.pdc

---
title: "Maxwell-Boltzmann distribution"
firstLetter: "M"
publishDate: 2021-05-08
categories:
- Physics
- Statistics
- Thermodynamics

date: 2021-05-08T18:35:37+02:00
draft: false
markup: pandoc
---

# 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](/know/concept/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.