Categories: Fluid mechanics, Physics, Thermodynamics.

# Boltzmann equation

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}

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:

\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 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}

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}

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}

## References

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 M. Salewski, A.H. Nielsen, Plasma physics: lecture notes, 2021, unpublished.

© Marcus R.A. Newman, CC BY-SA 4.0.