Processing math: 100%

Boltzmann equation

Set

context K:R3×R3×tR
range ::K(x,v,t)
context SN
i,jrange(S)
context miR
context Iij:R3×[π2,π2]×[0,2π]R
range ::Iij(v,ϑ,φ)

The integer S denote the species, mi their respective masses and Iij the differential cross sections for two particle collisions.

context v, v1:R2×3R3
range ::v(v,v1), V1(v,v1)
definiendum fit
postulate fi:R3×R3×RR+
range ::fi(x,v,t)
range J[fi|fj](x,v,t)g Iij(g,ϑ,φ) (fi(x,v(v,v1),t)fj(x,v1(v,v1),t)fi(x,v,t)fj(x,v1,t)) dΩ(ϑ,φ) d3v1
postulate (t+vx+1miKv)fi=Sj=1J[fi|fj]
I'm not sure about the summation “=Sj=1J[fi|fj]” 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

fi(x,v,t) d3xd3v!=1

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

J[fi|fj]g Iij(g,ϑ,φ) (fi(v)fj(v1)fi(v)fj(v1)) dΩ(ϑ,φ) d3v1.

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

 g Iij dΩfj(v1)fi(v).

The function fj arises as one-particle (x1,v1)-density and so if gI dΩg σtot is volume transfer per time, then integrating out v makes the terms in bracket into a process characteristic time 1τ. Hence the term is of the form of a decay rate

r:=1τf.

at the fixed point velocity v. The second term is more involved, since it doesn't represent the loss at v, but the gain: It's the sum of processes of pairs particles with velocities v,v1, which end up with particles having the velocity v and any other velocity v1.

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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