From 16555851b6514a736c5c9d8e73de7da7fc9b6288 Mon Sep 17 00:00:00 2001 From: Prefetch Date: Thu, 20 Oct 2022 18:25:31 +0200 Subject: Migrate from 'jekyll-katex' to 'kramdown-math-sskatex' --- source/know/concept/boltzmann-equation/index.md | 85 +++++++++++++------------ 1 file changed, 44 insertions(+), 41 deletions(-) (limited to 'source/know/concept/boltzmann-equation/index.md') diff --git a/source/know/concept/boltzmann-equation/index.md b/source/know/concept/boltzmann-equation/index.md index 20399df..9ed2fd2 100644 --- a/source/know/concept/boltzmann-equation/index.md +++ b/source/know/concept/boltzmann-equation/index.md @@ -10,17 +10,17 @@ layout: "concept" --- Consider a collection of particles, -each with its own position $\vb{r}$ and velocity $\vb{v}$. -We can thus define a probability density function $f(\vb{r}, \vb{v}, t)$ -describing the expected number of particles at $(\vb{r}, \vb{v})$ at time $t$. -Let the total number of particles $N$ be conserved, then clearly: +each with its own position $$\vb{r}$$ and velocity $$\vb{v}$$. +We can thus define a probability density function $$f(\vb{r}, \vb{v}, t)$$ +describing the expected number of particles at $$(\vb{r}, \vb{v})$$ at time $$t$$. +Let the total number of particles $$N$$ be conserved, then clearly: $$\begin{aligned} N = \iint_{-\infty}^\infty f(\vb{r}, \vb{v}, t) \dd{\vb{r}} \dd{\vb{v}} \end{aligned}$$ At equilibrium, all processes affecting the particles -no longer have a net effect, so $f$ is fixed: +no longer have a net effect, so $$f$$ is fixed: $$\begin{aligned} \dv{f}{t} @@ -35,7 +35,7 @@ $$\begin{aligned} = \bigg(\! \pdv{f}{t} \!\bigg)_\mathrm{\!col} \end{aligned}$$ -Where the right-hand side simply means "all changes in $f$ due to collisions". +Where the right-hand side simply means "all changes in $$f$$ due to collisions". Applying the chain rule to the left-hand side then yields: $$\begin{aligned} @@ -49,8 +49,8 @@ $$\begin{aligned} &= \pdv{f}{t} + \vb{v} \cdot \nabla f + \vb{a} \cdot \pdv{f}{\vb{v}} \end{aligned}$$ -Where we have introduced the shorthand $\ipdv{f}{\vb{v}}$. -Inserting Newton's second law $\vb{F} = m \vb{a}$ +Where we have introduced the shorthand $$\ipdv{f}{\vb{v}}$$. +Inserting Newton's second law $$\vb{F} = m \vb{a}$$ leads us to the **Boltzmann equation** or **Boltzmann transport equation** (BTE): @@ -64,41 +64,41 @@ $$\begin{aligned} But what about the collision term? Expressions for it exist, which are almost exact in many cases, but unfortunately also quite difficult to work with. -In addition, $f$ is a 7-dimensional function, +In addition, $$f$$ is a 7-dimensional function, so the BTE is already hard to solve without collisions. We only present the simplest case, known as the **Bhatnagar-Gross-Krook approximation**: -if the equilibrium state $f_0(\vb{r}, \vb{v})$ is known, -then each collision brings the system closer to $f_0$: +if the equilibrium state $$f_0(\vb{r}, \vb{v})$$ is known, +then each collision brings the system closer to $$f_0$$: $$\begin{aligned} \pdv{f}{t} + \vb{v} \cdot \nabla f + \frac{\vb{F}}{m} \cdot \pdv{f}{\vb{v}} = \frac{f_0 - f}{\tau} \end{aligned}$$ -Where $\tau$ is the average collision period. +Where $$\tau$$ is the average collision period. The right-hand side is called the **Krook term**. ## Moment equations -From the definition of $f$, -we see that integrating over all $\vb{v}$ yields the particle density $n$: +From the definition of $$f$$, +we see that integrating over all $$\vb{v}$$ yields the particle density $$n$$: $$\begin{aligned} n(\vb{r}, t) = \int_{-\infty}^\infty f(\vb{r}, \vb{v}, t) \dd{\vb{v}} \end{aligned}$$ -Consequently, a purely velocity-dependent quantity $Q(\vb{v})$ can be averaged like so: +Consequently, a purely velocity-dependent quantity $$Q(\vb{v})$$ can be averaged like so: $$\begin{aligned} \Expval{Q} = \frac{1}{n} \int_{-\infty}^\infty Q(\vb{r}, \vb{v}, t) \: f(\vb{r}, \vb{v}, t) \dd{\vb{v}} \end{aligned}$$ -With that in mind, we multiply the collisionless BTE equation by $Q(\vb{v})$ and integrate, -assuming that $\vb{F}$ does not depend on $\vb{v}$: +With that in mind, we multiply the collisionless BTE equation by $$Q(\vb{v})$$ and integrate, +assuming that $$\vb{F}$$ does not depend on $$\vb{v}$$: $$\begin{aligned} 0 @@ -110,10 +110,10 @@ $$\begin{aligned} + \frac{\vb{F}}{m} \cdot \int \bigg( \pdv{}{\vb{v}} (Q f) - f \pdv{Q}{\vb{v}} \bigg) \dd{\vb{v}} \end{aligned}$$ -The first integral is simply $n \Expval{Q}$. -In the second integral, note that $\vb{v}$ is a coordinate -and hence not dependent on $\vb{r}$, so $\nabla \cdot \vb{v} = 0$. -Since $f$ is a probability density, $f \to 0$ for $\vb{v} \to \pm\infty$, +The first integral is simply $$n \Expval{Q}$$. +In the second integral, note that $$\vb{v}$$ is a coordinate +and hence not dependent on $$\vb{r}$$, so $$\nabla \cdot \vb{v} = 0$$. +Since $$f$$ is a probability density, $$f \to 0$$ for $$\vb{v} \to \pm\infty$$, so the first term in the third integral vanishes after it is integrated: $$\begin{aligned} @@ -134,9 +134,9 @@ $$\begin{aligned} } \end{aligned}$$ -If we set $Q = m$, then the mass density $\rho = n \Expval{Q}$, +If we set $$Q = m$$, then the mass density $$\rho = n \Expval{Q}$$, and we find that the **zeroth moment** of the BTE describes conservation of mass, -where $\vb{V} \equiv \Expval{\vb{v}} = \int \vb{v} f \dd{\vb{v}}$ is the fluid velocity: +where $$\vb{V} \equiv \Expval{\vb{v}} = \int \vb{v} f \dd{\vb{v}}$$ is the fluid velocity: $$\begin{aligned} \boxed{ @@ -150,8 +150,8 @@ $$\begin{aligned} -If we instead choose the momentum $Q = m \vb{v}$, +If we instead choose the momentum $$Q = m \vb{v}$$, we find that the **first moment** of the BTE describes conservation of momentum, -where $\hat{P}$ is the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/): +where $$\hat{P}$$ is the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/): $$\begin{aligned} \boxed{ @@ -178,7 +179,7 @@ $$\begin{aligned} -Finally, if we choose the kinetic energy $Q = m |\vb{v}|^2 / 2$, +Finally, if we choose the kinetic energy $$Q = m |\vb{v}|^2 / 2$$, we find that the **second moment** gives conservation of energy, -where $U$ is the thermal energy density and $\vb{J}$ is the heat flux: +where $$U$$ is the thermal energy density and $$\vb{J}$$ is the heat flux: $$\begin{aligned} \boxed{ @@ -240,7 +242,7 @@ $$\begin{aligned} -- cgit v1.2.3