The Boltzmann Equation
June 2024
This article gives a brief introduction to the distribution function and a derivation of its governing equation of motion, the Boltzmann Equation.
The Distribution Function
The central object of interest in kinetic theory is the so-called (velocity) distribution function $f(t, \mathbf x, \mathbf q)$. Imagine that we have a gas in some container, where the gas is composed of a single type of molecule or atom. If we draw a grid over the container, then any one cell will contain many molecules with many different velocities. For each cell we can think about the distribution of velocities at a given moment in time - $f_i(t, \mathbf q)$. $f_i(t,\mathbf q)\delta \mathbf q$ gives us the total mass of molecules in cell $i$ with velocity $\mathbf q$. Working with the mass directly baked into the distribution function like this avoids needing to carry around factors of the molecular mass. This means that the total mass of molecules in the cell is given by
$$\int f_i(t, \mathbf q) d\mathbf q = \rho_i(t) \,\Delta V$$where $\Delta V$ is the cell volume and $\rho_i$ is the particle density. If we take the continuum limit, by making the cells ever smaller, we get our object of study; the distribution function $f(t, \mathbf x, \mathbf q)$.
Moments
By taking the moments of the distribution function at each point in space we can recover our macroscopic observables. The zeroth, first and second moments give us the mass density, velocity density and total energy density respectively
$$\begin{align} \int f(\mathbf x, \mathbf q)\, d\mathbf q &= \rho(\mathbf x) &\text{Mass density} \\ \int \mathbf q\,f(\mathbf x, \mathbf q) \,d\mathbf q &= \rho(\mathbf x)\mathbf u(\mathbf x) &\text{Momentum density} \\ \frac{1}{2} \int |\mathbf q|^2 \, f(\mathbf x, \mathbf q) \, d\mathbf q &= \rho(\mathbf x)E(\mathbf x) &\text{Total energy density} \end{align}$$We can split the total energy density into overall bulk kinetic energy, and internal energy caused by the thermal motion of the molecules. Define the relative velocity to be $\mathbf v = \mathbf q - \mathbf u$. The integral of the second moment of the relative velocity gives us the internal energy density.
$$\begin{align} \frac{1}{2} \int |{\mathbf v}|^{2}\, f(\mathbf x, \mathbf q)\, d\mathbf q &= \rho(\mathbf x) U(\mathbf x) &\text{Internal energy density} \end{align}$$Note this decomposition gives us $\rho E = \rho (U + |\mathbf u|^2)$, that is the total energy density is the sum of internal and kinetic energies:
$$\begin{align} \rho E &= \frac{1}{2} \int |\mathbf q|^2 \,f\,d\mathbf q = \frac{1}{2} \int |\mathbf v + \mathbf u|^2 \, f\,d\mathbf q = \frac{1}{2}\int \big(|\mathbf v|^2 + |\mathbf u|^2 + 2\mathbf u \cdot \mathbf v \big) \, f \,d\mathbf q \\ &= \rho (U + |\mathbf u|^2) + \mathbf u\cdot\!\! \cancelto{0}{\int \mathbf v\,f\,d\mathbf q} = \rho(U + |\mathbf u|^2) \end{align}$$The Boltzmann Equation
We have defined the distribution function, whose moments give us our macroscopic properties of interest. The macroscopic dynamics are then completely determined by how $f$ changes as a function of time. We need to write down an equation of motion for $f$; this is precisely what the Boltzmann equation is.
We can derive the Boltzmann equation in two exactly equivalent ways, taking either the Eulerian or the Lagrangian approach.
Eulerian
The Eulerian approach considers a fixed point in phase-space and asks how $f$ changes there, i.e. what is $\frac{\partial f}{\partial t}$. Ultimately that is what we are after in order to implement our simulation.
There are 2 ways in which $f$ can change;
- Advection - Particles arrive at $(\mathbf x, \mathbf q)$ from elsewhere due to their velocities & accelerations.
- Collision - Particles with differing velocities interact at $\mathbf x$.
The exact change that the second mechanism causes in $f$ is down to modelling choices. In this unspecified form it is denoted $\Omega$ in the literature, thus $\delta f_\text{col} = \Omega \delta t$ for some small time interval $\delta t$.
Let’s consider now the first mechanism, advection. Advection is the simple effect that the distribution function gets carried along for the ride with particles that are moving. In the absence of any collisions, $f$ follows trajectories in phase-space. Thus
$$\delta f_\text{adv} = f(t,\, \mathbf x - \delta \mathbf x, \mathbf q - \delta \mathbf q) - f(t, \mathbf x, \mathbf q)$$The Taylor expansion to first order of the first term is
$$f(t + \delta t, \mathbf x - \delta \mathbf x, \mathbf q - \delta \mathbf q) \approx f(t, \mathbf x, \mathbf q) - \delta \mathbf x\cdot \frac{\partial f}{\partial \mathbf x} - \delta \mathbf q\cdot\frac{\partial f}{\partial \mathbf q}$$So we get
$$\delta f_\text{adv} = - \delta \mathbf x \cdot \frac{\partial f}{\partial \mathbf x} - \delta \mathbf q\cdot\frac{\partial f}{\partial \mathbf q}$$We also know that (to first order in $\delta t$) we have $\delta \mathbf x = \delta t \mathbf q$ and $\delta \mathbf q = \delta t(\mathbf F / \rho)$. $\mathbf F$ is a body force density, which may depend on any or all of time, space, and velocity. Combining these with the above gives
$$\delta f_\text{adv} = -\delta t\left(\mathbf q \cdot \frac{\partial f}{\partial \mathbf x} + \frac{\mathbf F}{\rho} \cdot \frac{\partial f}{\partial \mathbf q}\right)$$Now the total change in $f(t, \mathbf x, \mathbf q)$ is then $\delta f = \delta f_\text{adv} + \delta f_{col}$ and if we write $\delta f / \delta t$ and let $\delta t \rightarrow 0$ we get, finally, the Boltzmann equation
$$\frac{\partial f}{\partial t} = \Omega - \mathbf q \cdot \frac{\partial f}{\partial \mathbf x} - \frac{\mathbf F}{\rho} \cdot \frac{\partial f}{\partial \mathbf q}$$Lagrangian
The Eulerian perspective asks what happens to $f$ at a fixed point in phase-space. In contrast the Lagrangian perspective asks what happens to $f$ as we follow trajectories through phase-space. For the Eulerian derivation, we looked at trajectories which arrived at a specific phase-space point. The Lagrangian derivation asks about trajectories which start from a specific phase-space point. It hinges on the total derivative of $f$.
If there were no collisions we know that $f$ must just be advected along phase-space trajectories, and so the total derivative $\frac{df}{dt}$ would be zero. Thus any non-zero $\frac{\mathrm df}{\mathrm dt}$ must be due to collisions and we can write
$$\begin{align} \Omega = \frac{\mathrm df}{\mathrm dt} &= \frac{\partial f}{\partial t} + \frac{\partial f}{\partial \mathbf x}\cdot \frac{d\mathbf x}{dt} + \frac{\partial f}{\partial \mathbf q}\cdot\frac{d\mathbf q}{dt} \\ &= \frac{\partial f}{\partial t} + \frac{\partial f}{\partial \mathbf x}\cdot\mathbf q + \frac{\partial f}{\partial \mathbf q}\cdot\frac{\mathbf F}{\rho} \end{align}$$which rearranges to give us exactly the same equation as above
$$\frac{\partial f}{\partial t} = \Omega - \frac{\partial f}{\partial \mathbf x}\cdot\mathbf q - \frac{\partial f}{\partial \mathbf q}\cdot\frac{\mathbf F}{\rho}$$Sometimes this is written using the ’nabla’/grad notation
$$\frac{\partial f}{\partial t} = \Omega - \mathbf q\cdot \nabla_\mathbf x\,f - \frac{\mathbf F}{\rho}\cdot \nabla_\mathbf q \,f$$or in component notation with Einstein summation notation
$$\frac{\partial f}{\partial t} = \Omega - q_\alpha\frac{\partial f}{\partial x_\alpha} - \frac{F_\alpha}{\rho}\frac{\partial f}{\partial q_\alpha}.$$Conservation Laws
Although the particles are interacting at a given point in space (not phase-space!) according the collision operator $\Omega$, they must still obey all the expected conservation laws. This gives us some constraints $\Omega$ must obey;
$$\begin{align} \int \Omega \,d\mathbf q &= 0 &\qquad \text{Mass conservation}\\ \int \mathbf q \,\Omega \,d\mathbf q &= \mathbf 0 &\qquad\text{Momentum conservation}\\ \int \!|\mathbf q|^2 \,\Omega \,d\mathbf q &= 0 &\qquad\text{Total energy conservation}\\ \int |\mathbf v|^2 \,\Omega \,d\mathbf q &= 0 &\qquad\text{Internal energy conservation}\\ \end{align}$$