path: root/source/know/concept/boltzmann-equation/index.md

---
title: "Boltzmann equation"
sort_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 particle count 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:

$$\begin{aligned}
    \boxed{
        0
        = \pdv{\rho}{t} + \nabla \cdot \big(\rho \vb{V}\big)
    }
\end{aligned}$$


{% include proof/start.html id="proof-moment0" -%}
We insert $$Q = m$$ into our prototype,
and since $$m$$ is constant, the rest is trivial:

$$\begin{aligned}
    0
    &= \pdv{}{t}\big(n \Expval{m}\big) + \nabla \cdot \big(n \Expval{m \vb{v}}\big) - \frac{\vb{F}}{m} \cdot \bigg( n \Expval{\pdv{m}{\vb{v}}} \bigg)
    \\
    &= \pdv{\rho}{t} + \nabla \cdot \big(\rho \Expval{\vb{v}}\big) - 0
\end{aligned}$$
{% include proof/end.html id="proof-moment0" %}


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/):

$$\begin{aligned}
    \boxed{
        0
        = \pdv{}{t}\big(\rho \vb{V}\big) + \rho \vb{V} (\nabla \cdot \vb{V}) + \nabla \cdot \hat{P} - n \vb{F}
    }
\end{aligned}$$


{% include proof/start.html id="proof-moment1" -%}
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 desired result,
where $$\nabla \cdot (\rho \vb{V}\vb{V})$$ is the fluid momentum,
and $$\nabla \cdot \hat{P}$$ is the viscous/pressure momentum:

$$\begin{aligned}
    0
    &= \pdv{}{t}\big(\rho \vb{V}\big) + \nabla \cdot \big(\rho \vb{V} \vb{V} + \hat{P}\big) - n \vb{F}
\end{aligned}$$
{% include proof/end.html id="proof-moment1" %}


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:

$$\begin{aligned}
    \boxed{
        0
        = \pdv{}{t}\bigg(\frac{\rho}{2} |\vb{V}|^2 + U \bigg)
        + \nabla \cdot \bigg(\frac{\rho}{2} |\vb{V}|^2 \vb{V} + \vb{V} \cdot \hat{P} + U \vb{V} + \vb{J} \bigg)
        - \vb{F} \cdot \big( n \vb{V} \big)
    }
\end{aligned}$$


{% include proof/start.html id="proof-moment2" -%}
We insert $$Q = m |\vb{v}|^2 / 2$$ into our prototype and recognize $$\rho$$ wherever possible:

$$\begin{aligned}
    0
    &= \pdv{}{t}\bigg(n \Expval{\frac{m |\vb{v}|^2}{2}}\bigg)
    + \nabla \cdot \bigg(n \Expval{\frac{m |\vb{v}|^2}{2} \vb{v}}\bigg)
    - \frac{\vb{F}}{m} \cdot \bigg( n \Expval{\pdv{}{\vb{v}} \frac{m |\vb{v}|^2}{2}} \bigg)
    \\
    &= \pdv{}{t}\bigg(\frac{\rho}{2} \Expval{|\vb{v}|^2}\bigg)
    + \nabla \cdot \bigg(\frac{\rho}{2} \Expval{|\vb{v}|^2 \vb{v}}\bigg)
    - \frac{\vb{F}}{2} \cdot \bigg( n \Expval{\pdv{|\vb{v}|^2}{\vb{v}}} \bigg)
\end{aligned}$$

We handle these terms one by one. Substituting $$\vb{v} = \vb{V} + \vb{w}$$ in the first gives:

$$\begin{aligned}
    \Expval{|\vb{v}|^2}
    &= \Expval{(\vb{V} \!+\! \vb{w}) \cdot (\vb{V} \!+\! \vb{w})}
    = \Expval{|\vb{V}|^2 + 2 \vb{V} \cdot \vb{w} + |\vb{w}|^2}
    \\
    &= |\vb{V}|^2 + 2 \vb{V} \cdot \Expval{\vb{w}} + \Expval{|\vb{w}|^2}
    = |\vb{V}|^2 + \Expval{|\vb{w}|^2}
\end{aligned}$$

And likewise for the second term,
where we recognize the stress tensor $$\Expval{\vb{w} \vb{w}}$$:

$$\begin{aligned}
    \Expval{|\vb{v}|^2 \vb{v}}
    &= \Expval{(\vb{V} \!+\! \vb{w}) \cdot (\vb{V} \!+\! \vb{w}) (\vb{V} \!+\! \vb{w})}
    = \Expval{(|\vb{V}|^2 + 2 \vb{V} \cdot \vb{w} + |\vb{w}|^2) (\vb{V} \!+\! \vb{w})}
    \\
    &= \Expval{|\vb{V}|^2 \vb{V} + |\vb{V}|^2 \vb{w}
    + 2 (\vb{V} \cdot \vb{w}) \vb{V} + 2 (\vb{V} \cdot \vb{w}) \vb{w}
    + |\vb{w}|^2 \vb{V} + |\vb{w}|^2 \vb{w}}
    \\
    &= |\vb{V}|^2 \vb{V} + |\vb{V}|^2 \Expval{\vb{w}}
    + 2 (\vb{V} \cdot \Expval{\vb{w}}) \vb{V} + 2 \Expval{(\vb{V} \cdot \vb{w}) \vb{w}}
    + \Expval{|\vb{w}|^2} \vb{V} + \Expval{|\vb{w}|^2 \vb{w}}
    \\
    &= |\vb{V}|^2 \vb{V} + 0 + 0 + 2 \vb{V} \cdot \Expval{\vb{w} \vb{w}}
    + \Expval{|\vb{w}|^2} \vb{V} + \Expval{|\vb{w}|^2 \vb{w}}
\end{aligned}$$

The third term is fairly obvious, but we calculate it rigorously just to be safe:

$$\begin{aligned}
    \pdv{|\vb{v}|^2}{\vb{v}}
    &= \pdv{}{\vb{v}} \big( v_x^2 + v_y^2 + v_z^2 \big)
    = \vu{e}_x \pdv{v_x^2}{v_x} + \vu{e}_y \pdv{v_y^2}{v_y} + \vu{e}_z \pdv{v_z^2}{v_z}
    = 2 \vb{v}
\end{aligned}$$

To clarify the physical interpretation,
we define $$U$$, $$\vb{J}$$ and $$\hat{P}$$ as follows:

$$\begin{aligned}
    U
    &\equiv \frac{\rho}{2} \Expval{|\vb{w}|^2}
    = \frac{\rho}{2} \Expval{(\vb{v} \!-\! \vb{V}) \cdot (\vb{v} \!-\! \vb{V})}
    \\
    \vb{J}
    &\equiv \frac{\rho}{2} \Expval{|\vb{w}|^2 \vb{w}}
    = \frac{\rho}{2} \Expval{(\vb{v} \!-\! \vb{V}) \cdot (\vb{v} \!-\! \vb{V})(\vb{v} \!-\! \vb{V})}
    \\
    \hat{P}
    &\equiv \rho \Expval{\vb{w} \vb{w}}
    = \rho \Expval{(\vb{v} \!-\! \vb{V}) (\vb{v} \!-\! \vb{V})}
\end{aligned}$$

Putting it all together, we arrive at the expected result, namely:

$$\begin{aligned}
    0
    &= \pdv{}{t}\bigg(\frac{\rho}{2} |\vb{V}|^2 + U \bigg)
    + \nabla \cdot \bigg(\frac{\rho}{2} |\vb{V}|^2 \vb{V} + \vb{V} \cdot \hat{P} + U \vb{V} + \vb{J} \bigg)
    - \vb{F} \cdot \big( n \vb{V} \big)
\end{aligned}$$

For the sake of clarity, we write out the pressure term, including the outer divergence:

$$\begin{aligned}
    \nabla \cdot (\vb{V} \cdot \hat{P})
    &= (\nabla \cdot \hat{P}{}^{\mathrm{T}}) \cdot \vb{V}
    = \nabla \cdot
    \begin{bmatrix}
        P_{xx} & P_{xy} & P_{xz} \\
        P_{yx} & P_{yy} & P_{yz} \\
        P_{zx} & P_{zy} & P_{zz}
    \end{bmatrix}
    \cdot
    \begin{bmatrix}
        V_x \\ V_y \\ V_z
    \end{bmatrix}
    \\
    &=
    \begin{bmatrix}
        \displaystyle \pdv{P_{xx}}{x} + \pdv{P_{xy}}{y} + \pdv{P_{xz}}{z} \\
        \displaystyle \pdv{P_{yx}}{x} + \pdv{P_{yy}}{y} + \pdv{P_{yz}}{z} \\
        \displaystyle \pdv{P_{zx}}{x} + \pdv{P_{zy}}{y} + \pdv{P_{zz}}{z}
    \end{bmatrix}^{\top}
    \cdot
    \begin{bmatrix}
        V_x \\ V_y \\ V_z
    \end{bmatrix}
    = \sum_{i=1}^{3} \sum_{j=1}^{3} \pdv{P_{ij}}{x_j} V_i
\end{aligned}$$
{% include proof/end.html id="proof-moment2" %}



## References
1.  M. Salewski, A.H. Nielsen,
    *Plasma physics: lecture notes*,
    2021, unpublished.