From 4780106a4f191c41d3b82ca9d1327a1c95a72055 Mon Sep 17 00:00:00 2001 From: Prefetch Date: Thu, 27 May 2021 20:46:01 +0200 Subject: Expand knowledge base --- .../maxwell-boltzmann-distribution/index.pdc | 219 +++++++++++++++++++++ 1 file changed, 219 insertions(+) create mode 100644 content/know/concept/maxwell-boltzmann-distribution/index.pdc (limited to 'content/know/concept/maxwell-boltzmann-distribution/index.pdc') diff --git a/content/know/concept/maxwell-boltzmann-distribution/index.pdc b/content/know/concept/maxwell-boltzmann-distribution/index.pdc new file mode 100644 index 0000000..082d2db --- /dev/null +++ b/content/know/concept/maxwell-boltzmann-distribution/index.pdc @@ -0,0 +1,219 @@ +--- +title: "Maxwell-Boltzmann distribution" +firstLetter: "M" +publishDate: 2021-05-08 +categories: +- Physics +- Statistics + +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. -- cgit v1.2.3