Page 317 - Intro to Tensor Calculus

311

~ F ~
and the velocity vector changes because of the acceleration .Here f(~r, V, t)dτdτ v represents the expected
m
number of particles in the phase space element dτdτ v at time t.
Assume there are no collisions, then each of the gas particles in a volume element of phase space centered
~
r
at position ~ and velocity V 1 move during a time interval dt to a phase space element centered at position
~
~
r ~ + V 1 dt and V 1 + ~ F dt. If there were no loss or gains of particles, then the number of particles must be
m
conserved and so these gas particles must move smoothly from one element of phase space to another without
any gains or losses of particles. Because of scattering collisions in dτ many of the gas particles move into or
~
~
~
out of the velocity range V 1 to V 1 + dV 1 . These collision scattering processes are denoted by the collision
~
operator D C f(~r, V, t) in the Boltzmann equation.
Consider two identical gas particles which experience a binary collision. Imagine that particle 1 with
~ ~ ~ ~ 0 ~ ~ 0
2
1
velocity V 1 collides with particle 2 having velocity V 2 .Denote by σ(V 1 → V , V 2 → V ) dτ V 1 dτ V 2 the
~
~ 0
~ 0
~ 0
conditional probability that particle 1 is scattered from velocity V 1 to between V and V + dV and the
1
1
1
~
~ 0
~ 0
struck particle 2 is scattered from velocity V 2 to between V and V +dV . We will be interested in collisions
~ 0
2
2
2
~ ~
~
~ 0 ~ 0
of thetype(V , V ) → (V 1 , V 2 ) for a fixed value of V 1 as this would represent the number of particles scattered
2
1
~ ~
~
. Also of interest are collisions of the type (V 1 , V 2 ) → (V , V ) for a fixed value V 1 as this represents
~ 0 ~ 0
into dτ V 1 1 2
~ 0
. Imagine a gas particle in dτ with velocity V subjected to a beam of particles
particles scattered out of dτ V 1 1
~ 0
~ 0
with velocities V . The incident flux on the element dτdτ V is |V − V |f(~r, V ,t)dτ V and hence
~ 0
~ 0
0
1
2
2
2
0
2
1
~ ~ 0 ~ ~ 0 ~ 0 ~ 0 ~ 0
1
2
2
2
1
σ(V 1 → V , V 2 → V ) dτ V 1 dτ V 2 dt |V − V |f(~r, V ,t) dτ V 2 0 (2.5.79)
~ 0 ~ ~ ~
represents the number of collisions, in the time interval dt, which scatter from V to between V 1 and V 1 +dV 1
1
~
~
~
as well as scattering V to between V 2 and V 2 + dV 2 . Multiply equation (2.5.79) by the density of particles
~ 0
2
~
~ 0 ~ 0
in the element dτdτ V and integrate over all possible initial velocities V ,V and final velocities V 2 not equal
1
0
2
1
~
to V 1 . This gives the number of particles in dτ which are scattered into dτ V 1 dt as
Z Z
~
~ 0
~ 0
~ 0
~ 0
~ 0
~ ~ 0
dt dτ V σ(V → V 1 , V → V 2 )|V − V |f(~r, V ,t)f(~r, V ,t). (2.5.80)
Ns in = dτdτ V 1 dτ V 2 dτ V 0 0 1 2 1 2 1 2
2 1
dt is
In a similar manner the number of particles in dτ which are scattered out of dτ V 1
Z Z Z
~
~
~
~
~
~ 0
~ ~ 0
dtf(~r, V 1 ,t) dτ V σ(V → V 1 , V → V 2 )|V 2 − V 1 |f(~r, V 2 ,t). (2.5.81)
Ns out = dτdτ V 1 dτ V 2 dτ V 0 0 1 2
2 1
Let
~
~
~
~
~ ~ 0
~ 0
~ 0
~ ~ 0
W(V → V 1 , V → V 2 )= |V 1 − V 2 | σ(V → V 1 , V → V 2 ) (2.5.82)
2
1
1
2
~
define a symmetric scattering kernel and use the relation D C f(~r, V, t)= Ns in − Ns out to represent the
Boltzmann equation for gas particles in the form

F
∂ ~
~
~
+ V ·∇ ~r + ·∇ ~ V f(~r, V 1 ,t)=
∂t m
(2.5.83)
Z Z Z

~ 0 ~
~
~
~
~ 0
~ 0
~ 0
dτ 0 dτ 0 dτ V 2 W (V 1 → V , V 2 → V ) f(~r, V ,t)f(~r, V ,t) − f(~r, V 1,t)f(~r, V 2,t) .
V
1 V 2 1 2 1 2
~
Take the moment of the Boltzmann equation (2.5.83) with respect to an arbitrary function φ(V 1 ). That
~
is, multiply equation (2.5.83) by φ(V 1 ) and then integrate over all elements of velocity space dτ V 1 . Define
the following averages and terminology: