Derivation of the Boltzmann Transport EquationAbhranil Das

Department of Physics, University of Texas at Austin

June 4, 2015

The following article aims to derive the Boltzmann transport equation as physically and clearly as pos-sible. Required concepts that are not covered are an understanding of basic scattering theory and what thedifferential scattering cross-section means in a scattering phenomenon.

Let f(r,p, t) be the number density of gas particles in phase space, i.e. f(r,p, t) dτ dπ (dτ = dxdydzand dπ = dpxdpydpz shall denote infinitesimal position and momentum volumes respectively in the article)is the number of particles in volume dτ at position r with momenta between p and p + dp. Now consider afinite volume ∆τ between r and r +4r in position-space of the gas, in which we consider only the moleculeswith momenta between p1 and p1 + dp1, i.e. velocities between v1 and v1 + dv1, where v1 = p1

m .

b b+db

g dt

Figure 1: The phase-space volume under consideration

The number of molecules in this phase-space volume is then n(r,p1, t) = f(r,p1, t)4τ dπ1 =: f14r dπ1.(dπ1 = dp1xdp1ydp1z etc.) The Boltzmann transport equation investigates the rate of change of this number.

There are two causes leading to the number of molecules in this phase space volume changing. The firstis free streaming of molecules into and out of the box in position space. Recall that we are only lookingat molecules moving with velocities between v1 and v1 + dv1. The flux of particles due to free streamingis J = v1c, where c is the spatial density of particles given by c = f1dπ1. Therefore, by the continuityequation, the rate of change of the number of molecules in the box due to free-streaming is given by:

∂n

∂t

∣∣∣∣fs

= ∂f1

∂t

∣∣∣∣fs4τ dπ1 = −∇.J4τ = −∇. (v1f1) dπ14τ.

(Here ∇ is the gradient operator in position-space.) Now we use a vector calculus identity to write:

∇. (v1f1) = f1∇.v1 + ∇f1.v1.

1

Now, since over the box we have assumed the velocities v1 of the particles to be constant (velocities don’tchange due to free-streaming, only due to collisions, which will be dealt with later), ∇.v1 = 0 and we have:

∂f1

∂t

∣∣∣∣fs4τ dπ1 = −∇f1.v1dπ14τ ⇒

∂f1

∂t

∣∣∣∣fs

= −∇f1.v1.

Note that ∇f1.v1 is different from the vector derivative v1.∇f1.The second reason for the change of f1 is intermolecular collisions which cause particles to drop into and

out of the momentum volume we are looking at. First we consider particles dropping out of this volume, andwe shall denote the corresponding rate of change by ∂f1

∂t

∣∣∣coll-

. Since the momentum volume we are lookingat is infinitesimal, we assume that particles in that volume undergoing collisions necessarily go out of thevolume (i.e. there are no infinitesimally glancing collisions). First consider the rate of particles scatteredout of the momentum volume by collisions. Consider one of the particles in the box we are looking at, withmomentum p1, suffering a collision p1p2 → p

′

1p′

2. There are two parameters of such a collision, the impactparameter and the relative speed before the collision |p1−p2|

m = g(p1,p2). (This is the standard notation inthis context; don’t shoot the messenger.) The average number of collisions undergone in time dt by particle1 with a particle with momentum p2 and with impact parameter between b and b + db is the number ofparticles with momentum p2 in the annular cylindrical volume depicted below (remember that for a generalinterparticle potential, ‘collision’ may or may not involve the particles coming in hard sphere contact):

b b+db

g dt

Figure 2: The annular cylinder of collisions

This number is given by 2πb db g dt f2 dπ2 (f2 := f(r,p2, t)). Thus, the rate of such collisions per unittime is 2πb db g f2 dπ2.

Here is a little catch. Particle 1 does not continue on undeterred through the cylinder after its collisions.It’s momentum changes after each collision, i.e. both the speed and direction. However, notice that wehave assumed that the distribution of momenta of the particle 2 is entirely statistically independent oruncorrelated with that of particle 1. This assumption is named ‘molecular chaos’, or Stosszahl-Ansatz asBoltzmann put it. With this assumption, statistically speaking we can use the above picture of a fixedcylinder.

To get the total rate of collisions suffered by particle 1, we should integrate over the momentum of thesecond particle and the impact parameter of the collision, to get:

ˆb

2πb dbˆ

p2

g(p1,p2) f2 dπ2.

Now, the total number of particles with momentum p1 that this applies to is the number of particles inthe phase-space volume we were looking at, f1 4τ dπ1. So we can say that the total rate of change of thenumber of particles in the phase-space volume due to collisions removing them from dπ1 is:

∂n

∂t

∣∣∣∣coll-

= ∂f1

∂t

∣∣∣∣coll-4τ dπ1 = −

ˆb

2πb dbˆ

p2

g f1 f2 dπ2 4τ dπ1 (1)

⇒ ∂f1

∂t

∣∣∣∣coll-

= −ˆ

b

2πb dbˆ

p2

g f1 f2 dπ2

2

Here the integration over p2 includes automatically the integration over g.Now, in analogy with 1, the rate of collisions of the form p

′

1p′

2 → p1p2 introducing particles from outsideinto dπ1 is given by:

∂n

∂t

∣∣∣∣coll+

= 2πˆ

b

b db

ˆp

′2

g f′

1 f′

2 dπ′

24r dπ′

1

But note that for a collision, given p1 and p2 we know p′

1 and p′

2 and vice versa uniquely throughbijective functions. So f ′

1 = f(r,p′

1, t) and f ′

2 = f(r,p′

2, t) can also be rewritten in terms of p1 and p2 astheir arguments. Assuming such a rewritten function, we can write:

∂n

∂t

∣∣∣∣coll+

= ∂f1

∂t

∣∣∣∣coll+

4τ dπ1 = 2πˆ

b

b db

ˆp2

g f′

1 f′

2 dπ24τ dπ′

1

In the above, dπ1 and dπ′

1 represent the same infinitesimal momentum volume of the first collidingparticle under consideration, just from two opposite perspectives. These volumes are therefore same and canbe cancelled. So we get:

∂f1

∂t

∣∣∣∣coll+

= 2πˆ

b

b db

ˆp2

g f′

1 f′

2 dπ2

Now, in ∂f1∂t

∣∣∣coll-

and ∂f1∂t

∣∣∣coll+

, instead of integrating over the impact parameter, we can equivalentlyintegrate over the scattered solid angle once we relate the two using the definition of the differential scatteringcross-section. In the center-of-mass frame, this relation is:

2πb db = σcm(Θ)dΩ,

where Θ and σcm(Θ) are respectively the scattering angle and the differential scattering cross-section inthe center-of-mass frame, and dΩ is the solid angle 2π sin ΘdΘ. (The scattering phenomenon is azimuthallysymmetric.)

Then we can then write:

∂f1

∂t

∣∣∣∣coll-

= −ˆ

ΩσcmdΩ

ˆp2

g f1 f2 dπ2,∂f1

∂t

∣∣∣∣coll+

=ˆ

ΩσcmdΩ

ˆp2

g f′

1 f′

2 dπ2

Putting all the terms together then, we get:

∂f1

∂t= ∂f1

∂t

∣∣∣∣fs

+ ∂f1

∂t

∣∣∣∣coll

= ∂f1

∂t

∣∣∣∣fs

+ ∂f1

∂t

∣∣∣∣coll+

+ ∂f1

∂t

∣∣∣∣coll-

= −∇f1.v1 +ˆ

ΩσcmdΩ

ˆp2

g f′

1 f′

2 dπ2 −ˆ

ΩσcmdΩ

ˆp2

g f1 f2 dπ2.

⇒ ∂f1

∂t+ ∇f1.v1 =

ˆΩσcmdΩ

ˆp2

g(f

′

1 f′

2 − f1f2

)dπ2.

This is the Boltzmann transport equation.In the case of elastic collisions, the after-collision momenta p

′

1 and p′

2 are found uniquely from thebefore-collision momenta p1 and p2 using conservation of momentum and energy: p1 + p2 = p

′

1 + p′

2 andm1v

21 +m2v

22 = m1v

′21 +m2v

′22 . So we can express f ′

1 and f ′

2 in terms of their natural arguments p′

1 and p′

2 andwrite, integrating over all primed momenta and including the conservation constraints using delta-functions:

∂f1

∂t+∇f1.v1 =

ˆΩσcmdΩ

ˆp2

ˆp

′1

ˆp

′2

g(f

′

1 f′

2 − f1f2

)δ3(p1+p2−p

′

1−p′

2)δ(m1v21+m2v

22−m1v

′21 +m2v

′22 )dπ

′

2dπ′

1dπ2.

3