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+---
+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:
+
+$$\begin{aligned}
+ \boxed{
+ 0
+ = \pdv{\rho}{t} + \nabla \cdot \big(\rho \vb{V}\big)
+ }
+\end{aligned}$$
+
+
+
+
+
+
+We insert $Q = m$ into our prototype,
+and since $m$ is constant, the rest is trivial:
+
+$$\begin{aligned}
+ 0
+ &= \pdv{}{t}\big(n \Expval{m}\big) + \nabla \cdot \big(n \Expval{m \vb{v}}\big) - \frac{\vb{F}}{m} \cdot \bigg( n \Expval{\pdv{m}{\vb{v}}} \bigg)
+ \\
+ &= \pdv{\rho}{t} + \nabla \cdot \big(\rho \Expval{\vb{v}}\big) - 0
+\end{aligned}$$
+
+
+
+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/):
+
+$$\begin{aligned}
+ \boxed{
+ 0
+ = \pdv{}{t}\big(\rho \vb{V}\big) + \rho \vb{V} (\nabla \cdot \vb{V}) + \nabla \cdot \hat{P} - n \vb{F}
+ }
+\end{aligned}$$
+
+
+
+
+
+
+We insert $Q = m \vb{v}$ into our prototype and recognize $\rho$ wherever possible:
+
+$$\begin{aligned}
+ 0
+ &= \pdv{}{t}\big(n \Expval{m \vb{v}}\big) + \nabla \cdot \big(n \Expval{m \vb{v} \vb{v}}\big)
+ - \frac{\vb{F}}{m} \cdot \bigg( n \Expval{\pdv{(m \vb{v})}{\vb{v}}} \bigg)
+ \\
+ &= \pdv{}{t}\big(\rho \Expval{\vb{v}}\big) + \nabla \cdot \big(\rho \Expval{\vb{v} \vb{v}}\big)
+ - \vb{F} \cdot \bigg( n \Expval{\pdv{\vb{v}}{\vb{v}}} \bigg)
+\end{aligned}$$
+
+With $\vb{v} \vb{v}$ being a dyadic product.
+To give it a physical interpretation,
+we split $\vb{v} = \vb{V} \!+\! \vb{w}$,
+where $\vb{V}$ is the average velocity vector,
+and $\vb{w}$ is the local deviation from $\vb{V}$:
+
+$$\begin{aligned}
+ \Expval{\vb{v} \vb{v}}
+ &= \Expval{(\vb{V} \!+\! \vb{w}) (\vb{V} \!+\! \vb{w})}
+ = \Expval{\vb{V} \vb{V} + 2 \vb{V} \vb{w} + \vb{w} \vb{w}}
+ = \vb{V} \vb{V} + 2 \vb{V} \Expval{\vb{w}} + \Expval{\vb{w} \vb{w}}
+\end{aligned}$$
+
+Since $\vb{w}$ represents a deviation from the mean, $\Expval{\vb{w}} = 0$.
+We define the pressure tensor:
+
+$$\begin{aligned}
+ \hat{P}
+ \equiv \rho \Expval{\vb{w} \vb{w}}
+ = \rho \Expval{(\vb{v} \!-\! \vb{V}) (\vb{v} \!-\! \vb{V})}
+\end{aligned}$$
+
+This leads to the expected result,
+where $\nabla \cdot (\rho \vb{V}\vb{V})$ represents the fluid momentum,
+and $\nabla \cdot \hat{P}$ the viscous/pressure momentum:
+
+$$\begin{aligned}
+ 0
+ &= \pdv{}{t}\big(\rho \vb{V}\big) + \nabla \cdot \big(\rho \vb{V} \vb{V} + \hat{P}\big) - n \vb{F}
+\end{aligned}$$
+
+
+
+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:
+
+$$\begin{aligned}
+ \boxed{
+ 0
+ = \pdv{}{t}\bigg(\frac{\rho}{2} |\vb{V}|^2 + U \bigg)
+ + \nabla \cdot \bigg(\frac{\rho}{2} |\vb{V}|^2 \vb{V} + \vb{V} \cdot \hat{P} + U \vb{V} + \vb{J} \bigg)
+ - \vb{F} \cdot \big( n \vb{V} \big)
+ }
+\end{aligned}$$
+
+