From 1d700ab734aa9b6711eb31796beb25cb7659d8e0 Mon Sep 17 00:00:00 2001 From: Prefetch Date: Tue, 20 Dec 2022 20:11:25 +0100 Subject: More improvements to knowledge base --- .../maxwell-boltzmann-distribution/index.md | 85 +++++++++++----------- 1 file changed, 42 insertions(+), 43 deletions(-) (limited to 'source/know/concept/maxwell-boltzmann-distribution') diff --git a/source/know/concept/maxwell-boltzmann-distribution/index.md b/source/know/concept/maxwell-boltzmann-distribution/index.md index 318e659..946525c 100644 --- a/source/know/concept/maxwell-boltzmann-distribution/index.md +++ b/source/know/concept/maxwell-boltzmann-distribution/index.md @@ -13,6 +13,7 @@ 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/) @@ -24,55 +25,51 @@ $$\begin{aligned} \:\propto\: \exp\!\big(\!-\! \beta E\big) \end{aligned}$$ -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 $$\tilde{r}$$, +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 -$$(\tilde{r}, \tilde{p})$$ is as follows: +$$(\va{r}, \va{p})$$ is as follows: $$\begin{aligned} - f(\tilde{r}, \tilde{p}) - \:\propto\: \exp\!\Big(\!-\! \beta \big( K(\tilde{p}) + U(\tilde{r}) \big) \Big) + f(\va{r}, \va{p}) + \:\propto\: \exp\!\Big(\!-\! \beta \big( K(\va{p}) + U(\va{r}) \big) \Big) \end{aligned}$$ -Since this is classical physics, -we can split the exponential. -In quantum mechanics, -the canonical commutation relation would prevent that. -Anyway, splitting yields: +Since this is classical physics, we can split the exponential +(in quantum mechanics, the canonical commutation relation would prevent that): $$\begin{aligned} - f(\tilde{r}, \tilde{p}) - \:\propto\: \exp\!\big(\!-\! \beta K(\tilde{p}) \big) \exp\!\big(\!-\! \beta U(\tilde{r}) \big) + f(\va{r}, \va{p}) + \:\propto\: \exp\!\big(\!-\! \beta K(\va{p}) \big) \exp\!\big(\!-\! \beta U(\va{r}) \big) \end{aligned}$$ Classically, the probability distributions of the momenta and positions are independent: $$\begin{aligned} - f_K(\tilde{p}) - \:\propto\: \exp\!\big(\!-\! \beta K(\tilde{p}) \big) + f_K(\va{p}) + \:\propto\: \exp\!\big(\!-\! \beta K(\va{p}) \big) \qquad \qquad - f_U(\tilde{r}) - \:\propto\: \exp\!\big(\!-\! \beta U(\tilde{r}) \big) + f_U(\va{r}) + \:\propto\: \exp\!\big(\!-\! \beta U(\va{r}) \big) \end{aligned}$$ -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: +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: $$\begin{aligned} - K(\tilde{p}) - = \sum_{n = 1}^N K_n(\vec{p}_n) + K(\va{p}) + = \sum_{n = 1}^N K_n(\vb{p}_n) \qquad \mathrm{where} \qquad - K_n(\vec{p}_n) - = \frac{|\vec{p}_n|^2}{2 m} + K_n(\vb{p}_n) + = \frac{|\vb{p}_n|^2}{2 m} \end{aligned}$$ Consequently, the probability distribution $$f(p_x, p_y, p_z)$$ for the @@ -100,10 +97,10 @@ so the velocity in each direction is independent of the others: $$\begin{aligned} f(v_x) - = \sqrt{\frac{m}{2 \pi k_B T}} \exp\!\Big( \!-\!\frac{m v_x^2}{2 k_B T} \Big) + = \sqrt{\frac{m}{2 \pi k_B T}} \exp\!\bigg( \!-\!\frac{m v_x^2}{2 k_B T} \bigg) \end{aligned}$$ -The distribution is thus an isotropic gaussian with standard deviations given by: +The distribution is thus an isotropic Gaussian with standard deviations given by: $$\begin{aligned} \sigma_x = \sigma_y = \sigma_z @@ -111,17 +108,18 @@ $$\begin{aligned} \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}$$, only its magnitude, +That was the distribution of the velocities along each axis, +but what about the speed $$v = |\vb{v}|$$? +Because we do not care about the direction of $$\vb{v}$$, only its magnitude, the [density of states](/know/concept/density-of-states/) $$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) + = \dv{}{v} \bigg( \frac{4 \pi}{3} v^3 \bigg) = 4 \pi v^2 \end{aligned}$$ @@ -132,7 +130,7 @@ then gives us the **Maxwell-Boltzmann speed distribution**: $$\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) + = 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}$$ @@ -144,10 +142,10 @@ and the root-mean-square speed $$v_{\mathrm{rms}}$$: $$\begin{aligned} f'(v_\mathrm{mode}) = 0 - \qquad + \qquad \quad v_\mathrm{mean} = \int_0^\infty v \: f(v) \dd{v} - \qquad + \qquad \quad v_\mathrm{rms} = \bigg( \int_0^\infty v^2 \: f(v) \dd{v} \bigg)^{1/2} \end{aligned}$$ @@ -159,12 +157,12 @@ $$\begin{aligned} v_{\mathrm{mode}} = \sqrt{\frac{2 k_B T}{m}} } - \qquad + \qquad \quad \boxed{ v_{\mathrm{mean}} = \sqrt{\frac{8 k_B T}{\pi m}} } - \qquad + \qquad \quad \boxed{ v_{\mathrm{rms}} = \sqrt{\frac{3 k_B T}{m}} @@ -172,6 +170,7 @@ $$\begin{aligned} \end{aligned}$$ + ## Kinetic energy distribution Using the speed distribution, @@ -194,7 +193,7 @@ so the energy distribution $$f(K)$$ is: $$\begin{aligned} f(K) = \frac{f(v)}{m v} - = \sqrt{\frac{2 m}{\pi}} \: \bigg( \frac{1}{k_B T} \bigg)^{3/2} v \exp\!\Big( \!-\!\frac{m v^2}{2 k_B T} \Big) + = \sqrt{\frac{2 m}{\pi}} \Big( \frac{1}{k_B T} \Big)^{3/2} v \exp\!\bigg( \!-\!\frac{m v^2}{2 k_B T} \bigg) \end{aligned}$$ Substituting $$v = \sqrt{2 K/m}$$ leads to @@ -203,7 +202,7 @@ the **Maxwell-Boltzmann kinetic energy distribution**: $$\begin{aligned} \boxed{ f(K) - = 2 \sqrt{\frac{K}{\pi}} \: \bigg( \frac{1}{k_B T} \bigg)^{3/2} \exp\!\Big( \!-\!\frac{K}{k_B T} \Big) + = 2 \sqrt{\frac{K}{\pi}} \Big( \frac{1}{k_B T} \Big)^{3/2} \exp\!\bigg( \!-\!\frac{K}{k_B T} \bigg) } \end{aligned}$$ -- cgit v1.2.3