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+---
+title: "Maxwell-Boltzmann distribution"
+date: 2021-05-08
+categories:
+- Physics
+- Statistics
+- Thermodynamics
+layout: "concept"
+---
+
+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](/know/concept/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:
+
+$$\begin{aligned}
+    f_K(\tilde{p})
+    \:\propto\: \exp\!\big(\!-\! \beta K(\tilde{p}) \big)
+    \qquad \qquad
+    f_U(\tilde{r})
+    \:\propto\: \exp\!\big(\!-\! \beta U(\tilde{r}) \big)
+\end{aligned}$$
+
+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.
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