Boltzmann equation

Set

context $ {\bf K}:\mathbb R^3\times\mathbb R^3\times\mathbb t\to\mathbb R $
range $ :: {\bf K}({\bf x},{\bf v},t)$
context $ S\in\mathbb N $
$i,j\in\text{range}(S)$
context $ m_i\in \mathbb R^* $
context $ I_{ij}: \mathbb R^3\times[-\tfrac{\pi}{2},\tfrac{\pi}{2}]\times[0,2\pi]\to\mathbb R $
range $ :: I_{ij}({\bf v},\vartheta,\varphi) $

The integer $S$ denote the species, $m_i$ their respective masses and $I_{ij}$ the differential cross sections for two particle collisions.

context $ {\bf v'},\ {\bf v'}_1 : \mathbb R^{2\times 3}\to \mathbb R^3 $
range $ :: {\bf v'}({\bf v'},{\bf v}_1),\ {\bf V}_1({\bf v},{\bf v}_1) $
definiendum $ {\bf f} \in \mathrm{it} $
postulate $ f_i:\mathbb R^3\times\mathbb R^3\times\mathbb R\to\mathbb R_+ $
range $ :: f_i({\bf x},{\bf v},t)$
range $ J[f_i|f_j]({\bf x},{\bf v},t) \equiv \int\int g\ I_{ij}(g,\vartheta,\varphi)\ \left(f_i({\bf x},{\bf v'}({\bf v},{\bf v}_1),t)\cdot f_j({\bf x},{\bf v'}_1({\bf v},{\bf v}_1),t)-f_i({\bf x},{\bf v},t)\cdot f_j({\bf x},{\bf v}_1,t)\right)\ \mathrm d\Omega(\vartheta,\varphi)\ \mathrm d^3v_1 $
postulate $ \left(\frac{\mathrm \partial}{\partial t}+{\bf v}\cdot\nabla_{\bf x}+\frac{1}{m_i}{\bf K}\cdot\nabla_{\bf v}\right)f_i = \sum_{j=1}^S J[f_i|f_j]$
I'm not sure about the summation “$\cdots = \sum_{j=1}^S J[f_i|f_j]$” here — check that.

Discussion

There are different routes to obtain the Boltzmann equation as approximation to the one-particle distribution function equation in the BBGKY hierarchy. In that case one also has

$ \int f_i({\bf x},{\bf v},t)\ \mathrm d^3x \mathrm d^3v \overset{!}{=} 1 $

Neglecting some arguments, we can read the “collision integral functional” $J$ as:

$ J[f_i|f_j] \equiv \int\int g\ I_{ij}(g,\vartheta,\varphi)\ \left(f_i({\bf v'})\cdot f_j({\bf v'}_1)-f_i({\bf v})\cdot f_j({\bf v}_1)\right)\ \mathrm d\Omega(\vartheta,\varphi)\ \mathrm d^3v_1. $

The velocity-after-collision-${\bf V}$-independent term contains a factor $f$ which you can pull out, leaving an integral which is a velocity average over a collision rate

$ -\ \left\langle \int g\ I_{ij}\ \mathrm d\Omega \right\rangle_{f_j({\bf v}_1)}\cdot f_i({\bf v}). $

The function $f_j$ arises as one-particle $({\bf x}_1,{\bf v}_1)$-density and so if $g\int I\ \mathrm d\Omega\equiv g\ \sigma_\text{tot}$ is volume transfer per time, then integrating out ${\bf v}$ makes the terms in bracket into a process characteristic time $\frac{1}{\tau}$. Hence the term is of the form of a decay rate

$r:=-\frac{1}{\tau} f.$

at the fixed point velocity ${\bf v}$. The second term is more involved, since it doesn't represent the loss at ${\bf v}$, but the gain: It's the sum of processes of pairs particles with velocities ${\bf v'},{\bf v'}_1$, which end up with particles having the velocity ${\bf v}$ and any other velocity ${\bf v}_1$.

Note

$Assumptions = {kT > 0, n > -1, \[CapitalTheta] > 0};

f[EE_] = E^(-(EE/kT))*
   E^(-(EE^2/\[CapitalTheta]^2))/(E^(\[CapitalTheta]^2/(2 kT)^2)
       Sqrt[\[Pi]] /2 \[CapitalTheta] Erfc[\[CapitalTheta]/(2 kT)]);
       
Integrate[f[EE] EE^n, {EE, 0, \[Infinity]}]

Reference

Wikipedia: Boltzmann equation

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