Boltzmann Equation II: Binary Collisions

Physics 127b: Statistical Mechanics Boltzmann Equation II: Binary Collisions Binary collisions in a classical gas Scattering out Scattering in v'1 v'2 v 1 v 2 R v 2 v 1 v'2 v'1 Center of V' Mass Frame V θ θ b sc sc R V V' Figure 1: Binary collisions in a gas: top—lab frame; bottom—centre of mass frame Binary collisions in a gas are very similar, except that the scattering is off another molecule. An individual scattering process is, of course, simplest to describe in the center of mass frame in terms of the relative E velocity V =Ev1 −Ev2. However the center of mass frame is different for different collisions, so we must keep track of the results in the lab frame, and this makes the calculation rather intricate. I will indicate the main ideas here, and refer you to Reif or Landau and Lifshitz for precise discussions. Lets first set things up in the lab frame. Again we consider the pair of scattering in and scattering out processes that are space-time inverses, and so have identical cross sections. We can write abstractly for the scattering out from velocity vE1 due to collisions with molecules with all velocities vE2, which will clearly be proportional to the number f (vE1) of molecules at vE1 (which we write as f1—sorry, not the same notation as in the previous sections where f1 denoted the deviation of f from the equilibrium distribution!) and the 3 3 number f2d v2 ≡ f(vE2)d v2 in each velocity volume element, and then we must integrate over all possible vE0 vE0 outgoing velocities 1 and 2 ZZZ df (vE ) 1 =− w(vE0 , vE0 ;Ev , vE )f f d3v d3v0 d3v0 . (1) dt 1 2 1 2 1 2 2 1 2 out vE0 vE0 Similarly for the scattering in from collisions between molecules with velocities 1 and 2 which collide to 1 give outgoing velocities vE1 and vE2 ZZZ df (vE ) 1 = w(vE , vE ;Ev0 , vE0 )f0f 0d3v d3v0 d3v0 (2) dt 1 2 1 2 1 2 2 1 2 in f 0 = f(vE0 ) (with 1 1 etc.). The space-time inversion symmetry tells us that w(vE0 , vE0 ;Ev , vE ) = w(vE , vE ;Ev0 vE0 ), 1 2 1 2 1 2 1 2 (3) so that ZZZ df (vE ) 1 = w(vE0 , vE0 ;Ev , vE )(f0f 0 − f f )d3v d3v0 d3v0 . (4) dt 1 2 1 2 1 2 1 2 2 1 2 coll These expressions are good for setting up the problem and making general arguments, but there is a 9 dimensional integral to do, and the scattering function w is comprised of many delta functions that express vE0 vE0 the conservation of energy and momentum (only a very limited set of 1 and 2 are consistent with the incoming velocities) that are complicated to write down directly in terms of the lab-frame velocities. We consequently transform to the center of mass frame, defining center of mass and relative coordinates m rE + m rE 1 1 2 2 E rEc = , R =Er1 −Er2, (5) m1 + m2 the corresponding velocities E E cE = drEc/dt, V = dR/dt, (6) and similar primed expressions in terms of the outgoing velocities. The conservation laws are now easy to state cE =Ec0, VE = VE 0 , (7) and we can simply describe the scattering rate, which in the CM frame is entirely analogous to scattering off a fixed center, in terms of a differential (angular) cross section scattering rate into solid angle dsc = nscVσ(θsc, V )dsc, (8) with nsc the density of scatterers, here the particles of velocity v2. Thus we can write the collision term: Z Z df (vE ) 1 0 0 3 = f(vE )f (vE ) − f(vE )f (vE ) |vE −Ev | σ (|vE −Ev | ,θsc) dscd vE (9) dt 1 2 1 2 1 2 1 2 2 coll vE2 sc with the geometry as shown in Fig. 1 and σ the differential cross section in the center of mass frame. So now we are doing the integral in the centre of mass frame for each vE2—a much simpler integration—but must vE0 vE0 vE , vE θ evaluate 1 and 2 in terms of 1 2 and sc. Thus the collision term can be evaluated as a four dimensional integral For more details consult Reif, chapter 14 and particularly §14.8. Boltzmann’s H-theorem An interesting application of the Boltzmann equation is Boltzmann’s H-theorem. Boltzmann showed from the Boltzmann equation that the quantity H (not the Hamiltonian!) defined by Z H = d3vfln f (10) 2 never increases dH ≤ . dt 0 (11) Thus, within the limitations of the Boltzmann equation, −H plays the role of a generalization of the entropy (actually S/kT )toarbitrary distributions, not restricted to ones characterized by macroscopic thermodynamic variables. A sketch of the derivation is as follows. The time derivative of H is Z dH df = d3v 1 ( + f ) dt 1 dt 1 ln 1 (12) ZZZZ coll = w(vE0 , vE0 ;Ev , vE )(f0f 0 − f f )( + f )d3v d3v d3v0 d3v0 1 2 1 2 1 2 1 2 1 ln 1 1 2 1 2 (13) Using the symmetry of the scattering process under spatial inversions, time reversal, and interchange of the two incoming (and outgoing) molecules this can be written in four different ways, with all terms the same ( + f ) ( + f ), −( + f 0), −( + f 0) except 1 ln 1 is replaced by 1 ln 2 1 ln 1 1 ln 2 respectively. Summing these four expressions then gives ZZZZ dH 1 =− w(vE0 , vE0 ;Ev , vE ) ln f 0f 0 − ln f f f 0f 0 − f f d3v d3v d3v0 d3v0 (14) dt 4 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 But for any positive x and y it can be seen that (ln y − ln x)(y − x) ≥ 0 (15) y = x dH/dt ≤ f 0f 0 = f f with the equal sign only valid for . Hence 0, and the equality only holds if 1 2 1 2. f 0f 0 = f f The equilibrium condition 1 2 1 2 is equivalent to f 0 + f 0 = f + f ln 1 ln 2 ln 1 ln 2 (16) i.e. the sum of quantities before the collision equals the sum of corresponding quantities after the collision, for all collisions. Since the equilibrium distribution must also be isotropic, so that the momentum vector cannot occur in the distribution, this shows that the equilibrium distribution must be the exponential of a term proportional to the kinetic energy of the particle, i.e. the distribution must be the Maxwell distribution, and then the proportionality constant is β. Conservation laws and hydrodynamic equations For quantities that are conserved in the collisions, such as mass, momentum and energy, if we sum these quantities over all the molecules, we can derive macroscopic conservation laws which are the equations of hydrodynamics and heat transport. The conservation laws take the form of the time derivative of a density given by the divergence of the corresponding current or flux. These currents can then be evaluated first using the equilibrium distribution function f0 and then suing the correction f1 coming from deviations from equilibrium driven by gradients of the temperature, pressure or chemical potential, and velocity. Conservation laws First consider the mass. Then Z d3vmf(x,E v,t)E = n(x,t)mE = ρ(x,t)E (17) 3 with ρ the mass density. We can use the Boltzmann equation ∂f ∂f FE ∂f df +Ev · + · = ∂t ∂xE m ∂vE dt (18) coll to derive the equation of motion for the density Z Z Z ∂ρ ∂f ∂f df =− d3vmvE · − d3vFE · + d3vm . ∂t ∂xE ∂vE dt (19) coll The lastR term on the fight hand side is zero because mass is conserved in the collisions, the second term E 3 E E −F · d v ∇vf is also zero, and the first term is −∇·Eg with gE = ρuE the momentum density Z g(E x,t)E = d3vmvf(E x,E v,t)E (20) and uE the average velocity R d3v vf(E x,E v,t)E u(E x,t)E = R (21) d3vf(x,E v,t)E Thus we have derived the macroscopic conservation law for mass ∂ρ + ∇·E (ρu)E = . ∂t 0 (22) Now we investigate the consequence of the conservation of momentum mvE in the collision. Integrating over the velocity distribution gives Z Z Z ∂(ρu)E ∂f (x,E v,t)E ∂f ∂f = d3vmvE =− d3vmvE vE · − d3v vE FE · ∂t ∂t ∂xE ∂vE (23) where again the collision term drops out. This simplifies to Z ∂(ρu)E =−∇·E m d3v vE vfE + 8E ∂t (24) (you might want to write this in component notation if the double vector is confusing) where integration by parts is used in the last term, and 8(E x,t)E = F(E x,E t)n(x,t)E is the force per unit volume. Now write the molecular velocity in terms of the mean velocity uE and a correction VE with zero mean vE =Eu + V,E (25) R 3 E with d vVf= 0. Then Z Z d3v vE vfE = nuEuE + d3v VE Vf,E (26) since the cross term gives zero. Thus Eq. (24) reduces to ∂(ρu)E ←→ + ∇·E (ρuEu)E = 8E − ∇·E 6 ∂t (27) or in component form ∂(ρui) ∂ ∂ + (ρuiuj ) = 8i − 6ij (28) ∂t ∂xj ∂xj 4 ←→ (repeated index j summed over) with 6 the momentum flux or stress tensor Z R d3VVV f(x,E uE + V,t)E 3 E i j 6ij = m d VViVj f(x,E uE + V,t)= mn R = ρ ViVj .

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