From 6ce0bb9a8f9fd7d169cbb414a9537d68c5290aae Mon Sep 17 00:00:00 2001 From: Prefetch Date: Fri, 14 Oct 2022 23:25:28 +0200 Subject: Initial commit after migration from Hugo --- source/know/concept/boltzmann-equation/index.md | 357 ++++++++++++++++++++++++ 1 file changed, 357 insertions(+) create mode 100644 source/know/concept/boltzmann-equation/index.md (limited to 'source/know/concept/boltzmann-equation') diff --git a/source/know/concept/boltzmann-equation/index.md b/source/know/concept/boltzmann-equation/index.md new file mode 100644 index 0000000..6193fe6 --- /dev/null +++ b/source/know/concept/boltzmann-equation/index.md @@ -0,0 +1,357 @@ +--- +title: "Boltzmann equation" +date: 2022-10-02 +categories: +- Physics +- Thermodynamics +- Fluid mechanics +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: + +$$\begin{aligned} + N = \iint_{-\infty}^\infty f(\vb{r}, \vb{v}, t) \dd{\vb{r}} \dd{\vb{v}} +\end{aligned}$$ + +At equilibrium, all processes affecting the particles +no longer have a net effect, so $f$ is fixed: + +$$\begin{aligned} + \dv{f}{t} + = 0 +\end{aligned}$$ + +If each particle's momentum only changes due to collisions, +then a non-equilibrium state can be described as follows, very generally: + +$$\begin{aligned} + \dv{f}{t} + = \bigg(\! \pdv{f}{t} \!\bigg)_\mathrm{\!col} +\end{aligned}$$ + +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: + +$$\begin{aligned} + \bigg(\! \pdv{f}{t} \!\bigg)_\mathrm{\!col} + &= \pdv{f}{t} + \bigg( \pdv{f}{x} \dv{x}{t} \!+\! \pdv{f}{y} \dv{y}{t} \!+\! \pdv{f}{z} \dv{z}{t} \bigg) + + \bigg( \pdv{f}{v_x} \dv{v_x}{t} \!+\! \pdv{f}{v_y} \dv{v_y}{t} \!+\! \pdv{f}{v_z} \dv{v_z}{t} \bigg) + \\ + &= \pdv{f}{t} + \bigg( v_x \pdv{f}{x} \!+\! v_y \pdv{f}{y} \!+\! v_z \pdv{f}{z} \bigg) + + \bigg( a_x \pdv{f}{v_x} \!+\! a_y \pdv{f}{v_y} \!+\! a_z \pdv{f}{v_z} \bigg) + \\ + &= \pdv{f}{t} + \vb{v} \cdot \nabla f + \vb{a} \cdot \pdv{f}{\vb{v}} +\end{aligned}$$ + +Where we have introduced the shorthand $\ipdv{f}{\vb{v}}$. +Inserting Newton's second law $\vb{F} = m \vb{a}$ +leads us to the **Boltzmann equation** or +**Boltzmann transport equation** (BTE): + +$$\begin{aligned} + \boxed{ + \pdv{f}{t} + \vb{v} \cdot \nabla f + \frac{\vb{F}}{m} \cdot \pdv{f}{\vb{v}} + = \bigg(\! \pdv{f}{t} \!\bigg)_\mathrm{\!col} + } +\end{aligned}$$ + +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, +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 $f_0$: + +$$\begin{aligned} + \pdv{f}{t} + \vb{v} \cdot \nabla f + \frac{\vb{F}}{m} \cdot \pdv{f}{\vb{v}} + = \frac{f_0 - f}{\tau} +\end{aligned}$$ + +Where $\tau$ is the average collision period. +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 $n$: + +$$\begin{aligned} + n(\vb{r}, t) = \int_{-\infty}^\infty f(\vb{r}, \vb{v}, t) \dd{\vb{v}} +\end{aligned}$$ + +Consequently, a purely velocity-dependent quantity $Q(\vb{v})$ can be averaged like so: + +$$\begin{aligned} + \Expval{Q} + = \frac{1}{n} \int_{-\infty}^\infty Q(\vb{r}, \vb{v}, t) \: f(\vb{r}, \vb{v}, t) \dd{\vb{v}} +\end{aligned}$$ + +With that in mind, we multiply the collisionless BTE equation by $Q(\vb{v})$ and integrate, +assuming that $\vb{F}$ does not depend on $\vb{v}$: + +$$\begin{aligned} + 0 + &= \int_{-\infty}^\infty Q \bigg( \pdv{f}{t} + \vb{v} \cdot \nabla f + \frac{\vb{F}}{m} \cdot \pdv{f}{\vb{v}} \bigg) \dd{\vb{v}} + \\ + &= \int Q \pdv{f}{t} \dd{\vb{v}} + \int (\vb{v} \cdot \nabla f) \: Q \dd{\vb{v}} + \frac{\vb{F}}{m} \cdot \int Q \pdv{f}{\vb{v}} \dd{\vb{v}} + \\ + &= \pdv{}{t}\int Q f \dd{\vb{v}} + \int \Big( \nabla \cdot (\vb{v} f) - f (\nabla \cdot \vb{v}) \Big) Q \dd{\vb{v}} + + \frac{\vb{F}}{m} \cdot \int \bigg( \pdv{}{\vb{v}} (Q f) - f \pdv{Q}{\vb{v}} \bigg) \dd{\vb{v}} +\end{aligned}$$ + +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 $\vb{v} \to \pm\infty$, +so the first term in the third integral vanishes after it is integrated: + +$$\begin{aligned} + 0 + &= \pdv{}{t}\big(n \Expval{Q}\big) + \int \nabla \cdot (\vb{v} f) \: Q \dd{\vb{v}} + + \frac{\vb{F}}{m} \cdot \bigg( \Big[ Q f \Big]_{-\infty}^\infty - \int f \pdv{Q}{\vb{v}} \dd{\vb{v}} \bigg) + \\ + &= \pdv{}{t}\big(n \Expval{Q}\big) + \nabla \cdot \int Q \vb{v} f \dd{\vb{v}} + - \frac{\vb{F}}{m} \cdot \int f \pdv{Q}{\vb{v}} \dd{\vb{v}} +\end{aligned}$$ + +We thus arrive at the prototype of the BTE's so-called **moment equations**: + +$$\begin{aligned} + \boxed{ + 0 + = \pdv{}{t}\big(n \Expval{Q}\big) + \nabla \cdot \big(n \Expval{Q \vb{v}}\big) - \frac{\vb{F}}{m} \cdot \bigg( n \Expval{\pdv{Q}{\vb{v}}} \bigg) + } +\end{aligned}$$ + +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: + +$$\begin{aligned} + \boxed{ + 0 + = \pdv{\rho}{t} + \nabla \cdot \big(\rho \vb{V}\big) + } +\end{aligned}$$ + +
+ + + +
+ +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/): + +$$\begin{aligned} + \boxed{ + 0 + = \pdv{}{t}\big(\rho \vb{V}\big) + \rho \vb{V} (\nabla \cdot \vb{V}) + \nabla \cdot \hat{P} - n \vb{F} + } +\end{aligned}$$ + +
+ + + +
+ +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: + +$$\begin{aligned} + \boxed{ + 0 + = \pdv{}{t}\bigg(\frac{\rho}{2} |\vb{V}|^2 + U \bigg) + + \nabla \cdot \bigg(\frac{\rho}{2} |\vb{V}|^2 \vb{V} + \vb{V} \cdot \hat{P} + U \vb{V} + \vb{J} \bigg) + - \vb{F} \cdot \big( n \vb{V} \big) + } +\end{aligned}$$ + +
+ + + +
+ + + +## References +1. M. Salewski, A.H. Nielsen, + *Plasma physics: lecture notes*, + 2021, unpublished. -- cgit v1.2.3