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| author | Prefetch | 2026-09-05 21:55:33 +0200 |
|---|---|---|
| committer | Prefetch | 2026-09-05 21:55:33 +0200 |
| commit | 5cacf4ffaf3a9621ab536195f6469f98a420f054 (patch) | |
| tree | 317b734468a287c50d2403fdfafa45434f538150 /source/know/concept/boltzmann-equation | |
| parent | 29b49508a751649310173e592b63415dbf563a2a (diff) | |
Diffstat (limited to 'source/know/concept/boltzmann-equation')
| -rw-r--r-- | source/know/concept/boltzmann-equation/index.md | 20 |
1 files changed, 12 insertions, 8 deletions
diff --git a/source/know/concept/boltzmann-equation/index.md b/source/know/concept/boltzmann-equation/index.md index 5f4add0..3821512 100644 --- a/source/know/concept/boltzmann-equation/index.md +++ b/source/know/concept/boltzmann-equation/index.md @@ -67,7 +67,7 @@ but unfortunately also quite difficult to work with. 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**: +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 $$f_0$$: @@ -90,14 +90,15 @@ $$\begin{aligned} n(\vb{r}, t) = \int_{-\infty}^\infty f(\vb{r}, \vb{v}, t) \dd{\vb{v}} \end{aligned}$$ -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: $$\begin{aligned} - \Expval{Q} - = \frac{1}{n} \int_{-\infty}^\infty Q(\vb{r}, \vb{v}, t) \: f(\vb{r}, \vb{v}, t) \dd{\vb{v}} + \Expval{Q}\!(\vb{r}, t) + = \frac{1}{n} \int_{-\infty}^\infty Q(\vb{v}) \: f(\vb{r}, \vb{v}, t) \dd{\vb{v}} \end{aligned}$$ -With that in mind, we multiply the collisionless BTE equation by $$Q(\vb{v})$$ and integrate, +With that in mind, we multiply the collisionless BTE by $$Q(\vb{v})$$ and integrate, assuming that $$\vb{F}$$ does not depend on $$\vb{v}$$: $$\begin{aligned} @@ -136,7 +137,8 @@ $$\begin{aligned} 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 $$\vb{V} \equiv \Expval{\vb{v}} = n^{-1} \int \vb{v} f \dd{\vb{v}}$$ +is the fluid velocity: $$\begin{aligned} \boxed{ @@ -231,7 +233,8 @@ $$\begin{aligned} {% include proof/start.html id="proof-moment2" -%} -We insert $$Q = m |\vb{v}|^2 / 2$$ into our prototype and recognize $$\rho$$ wherever possible: +We insert $$Q = m |\vb{v}|^2 / 2$$ into our prototype +and recognize $$\rho$$ wherever possible: $$\begin{aligned} 0 @@ -244,7 +247,8 @@ $$\begin{aligned} - \frac{\vb{F}}{2} \cdot \bigg( n \Expval{\pdv{|\vb{v}|^2}{\vb{v}}} \bigg) \end{aligned}$$ -We handle these terms one by one. Substituting $$\vb{v} = \vb{V} + \vb{w}$$ in the first gives: +We handle these terms one by one. +Substituting $$\vb{v} = \vb{V} + \vb{w}$$ in the first gives: $$\begin{aligned} \Expval{|\vb{v}|^2} |
