PHYSICS 140B : STATISTICAL PHYSICS
HW ASSIGNMENT #8 SOLUTIONS
(1) Consider a monatomic ideal gas in the presence of a temperature gradient ∇T . Answer
the following questions within the framework of the relaxation time approximation to the
Boltzmann equation.
(a) Compute the particle current j and show that it vanishes.
(b) Compute the ‘energy squared’ current,
Z
jε2 = d3p ε2 v f (r, p, t) .
(c) Suppose the gas is diatomic, so cp = 27 kB . Show explicitly that the particle current j
is zero. Hint: To do this, you will have to understand the derivation of eqn. 8.85 in
the Lecture Notes and how this changes when the gas is diatomic. You may assume
Qαβ = F = 0.
Solution :
(a) Under steady state conditions, the solution to the Boltzmann equation is f = f 0 + δf ,
where f 0 is the equilibrium distribution and
δf = −
τ f 0 ε − cp T
·
v · ∇T .
kB T
T
For the monatomic ideal gas, cp = 52 kB . The particle current is
Z
α
j = d3p v α δf
Z
∂T
τ
=−
d3p f 0 (p) v α v β ε − 52 kB T
2
kB T
∂xβ
2nτ
∂T 5
=−
k
T
,
ε
ε
−
B
2
3mkB T 2 ∂xα
where the average over momentum/velocity is converted into an average over the energy
distribution,
dv
P (v) = √2π (kB T )−3/2 ε1/2 φ(ε) e−ε/kB T .
P̃ (ε) = 4πv 2
dε M
As discussed in the Lecture Notes, the average of a homogeneous function of ε under this
distribution is given by
α
ε = √2π Γ α + 32 (kB T )α .
Thus,
ε ε − 52 kB T =
√2
π
n
(kB T )2 Γ
1
7
2
− 25 Γ
5
2
o
=0.
(b) Now we must compute
jεα2
=
Z
d3p v α ε2 δf
=−
∂T 3
2nτ
ε ε − 52 kB T .
2
α
3mkB T ∂x
We then have
3
ε ε − 25 kB T =
√2
π
n
(kB T )4 Γ
and so
jε 2 = −
11
2
−
5
2
Γ
9
2
o
=
105
2
(kB T )4
35 nτ kB
(kB T )2 ∇T .
m
(c) For diatomic gases in the presence of a temperature gradient, the solution to the linearized Boltzmann equation in the relaxation time approximation is
δf = −
τ f 0 ε(Γ ) − cp T
·
v · ∇T ,
kB T
T
where
L21 + L22
,
2I
where L1,2 are components of the angular momentum about the instantaneous body-fixed
axes, with I ≡ I1 = I2 ≫ I3 . We assume the rotations about axes 1 and 2 are effectively
classical, so equipartition gives hεrot i = 2 × 21 kB = kB . We still have hεtr i = 32 kB . Now in
the derivation of the factor ε(ε − cp T ) above, the first factor of ε came from the v α v β term,
so this is translational kinetic energy. Therefore, with cp = 72 kB now, we must compute
ε(Γ ) = εtr + εrot = 12 mv 2 +
εtr εtr + εrot − 27 kB T
So again the particle current vanishes.
= εtr εtr − 25 kB T = 0 .
Note added :
It is interesting to note that there is no particle current flowing in response to a temperature
gradient when τ is energy-independent. This is a consequence of the fact that the pressure
gradient ∇p vanishes. Newton’s Second Law for the fluid says that nmV̇ + ∇p = 0, to
lowest relevant order. With ∇p 6= 0, the fluid will accelerate. In a pipe, for example, eventually a steady state is reached where the flow is determined by the fluid viscosity, which
is one of the terms we just dropped. (This is called Poiseuille flow.) When p is constant, the
local equilibrium distribution is
f 0 (r, p) =
2
p/kB T
e−p /2mkB T ,
(2πmkB T )3/2
2
where T = T (r). We then have
f (r, p) = f 0 (r − vτ, p) ,
which says that no new collisions happen for a time τ after a given particle thermalizes.
I.e. we evolve the streaming terms for a time τ . Expanding, we have
τ p ∂f 0
f = f0 −
·
+ ...
m ∂r
p
τ
5
·∇T + . . . f 0 (r, p) ,
ε(p) − 2 kB T
= 1−
2kB T 2
m
which leads to j = 0, assuming the relaxation time τ is energy-independent.
When the flow takes place in a restricted geometry, a dimensionless figure of merit known
as the Knudsen number, Kn = ℓ/L, where ℓ is the mean free path and L is the characteristic linear dimension associated with the geometry. For Kn ≪ 1, our Boltzmann transport
calculations of quantities like κ, η, and ζ hold, and we may apply the Navier-Stokes equations1 . In the opposite limit Kn ≫ 1, the boundary conditions on the distribution are
crucial. Consider, for example, the case ℓ = ∞. Suppose we have ideal gas flow in a cylinder whose symmetry axis is x̂. Particles with vx > 0 enter from the left, and particles with
vx < 0 enter from the right. Their respective velocity distributions are
Pj (v) = nj
m
2πkB Tj
3/2
e−mv
2 /2k
B Tj
,
where j = L or R. The average current is then
Z
n
o
jx = d3v nL vx PL (v) Θ(vx ) + nR vx PR (v) Θ(−vx )
r
r
2kB TL
2kB TR
− nR
.
= nL
m
m
(2) Suppose the relaxation time is energy-dependent, with τ (ε) = τ0 e−ε/ε0 . Compute the
particle current j and energy current jε flowing in response to a temperature gradient ∇T .
Solution :
Now we must compute
α α Z
v
j
3
δf
=
d
p
ε vα
jεα
nεo
∂T 2n
τ
(ε)
ε − 52 kB T ,
=−
2
α
2
3mkB T ∂x
ε
1
These equations may need to be supplemented by certain conditions which apply in the vicinity of solid
boundaries.
3
where τ (ε) = τ0 e−ε/ε0 . We find
−ε/ε0 α
e
ε
=
√2
π
(kB T )
−3/2
Z∞
1
dε εα+ 2 e−ε/kB T e−ε/ε0
0
=
√2
π
Γ α+
3
2
α
(kB T )
ε0
ε0 + kB T
α+ 3
2
.
Therefore,
5/2
ε0
kB T
e
ε0 + kB T
7/2
−ε/ε 2 15
ε0
0 ε
= 4 (kB T )2
e
ε0 + kB T
9/2
−ε/ε 3 105
ε0
3
0
ε = 8 (kB T )
e
ε0 + kB T
−ε/ε0
ε =
3
2
and
5/2
ε0
5nτ0 kB2 T
∇T
2m
(ε0 + kB T )7/2
7/2 ε0
2ε0 − 5kB T
5nτ0 kB2 T
∇T .
jε = −
4m
ε0 + kB T
ε0 + kB T
√
The previous results are obtained by setting ε0 = ∞ and τ0 = 1/ 2 nv̄σ. Note the strange
result that κ becomes negative for kB T > 25 ε0 .
j=
(3) Use the linearized Boltzmann equation to compute the bulk viscosity ζ of an ideal gas.
(a) Consider first the case of a monatomic ideal gas. Show that ζ = 0 within this approximation. Will your result change if the scattering time is energy-dependent?
(b) Compute ζ for a diatomic ideal gas.
Solution :
According to eqn. 8.111 in the Lecture Notes, the solution to the linearized Boltzmann
equation in the relaxation time approximation is
kB
τf0
α β ∂Vα
δf = −
− εtr + εrot
∇·V .
mv v
kB T
∂xβ
cV
We also have
Tr Π = nm hv 2 i = 2n hεtr i = 3p − 3ζ ∇·V .
4
We then compute Tr Π:
Tr Π = 2n hεtr i = 3p − 3ζ ∇·V
Z
= 2n dΓ (f 0 + δf ) εtr
The f 0 term yields a contribution 3nkB T = 3p in all cases, which agrees with the first term
on the RHS of the equation for Tr Π. Therefore
Z
2
ζ ∇·V = − 3 n dΓ δf εtr .
(a) For the monatomic gas, Γ = {px , py , pz }. We then have
Z
2nτ
ε
3
0
α β ∂Vα
ζ∇·V =
−
∇·V
d p f (p) ε mv v
3kB T
∂xβ
cV /kB
2nτ D 2 kB E
=
3 − cV ε ∇·V = 0 .
3kB T
Here we have replaced mv α v β → 31 mv 2 = 23 εtr under the integral. If the scattering time
is energy dependent, then we put τ (ε) inside the energy integral when computing the
average, but this does not affect the final result: ζ = 0.
(b) Now we must include the rotational kinetic energy in the expression for δf , and we
have cV = 52 kB . Thus,
Z
kB
2nτ
0
α β ∂Vα
ζ∇·V =
− εtr + εrot
∇·V
dΓ f (Γ ) εtr mv v
3kB T
∂xβ
cV
E
2nτ D 2 2
kB
ε
+
ε
ε
−
=
rot εtr ∇·V ,
3kB T 3 tr cV tr
and therefore
ζ=
2nτ 4 2
εtr − 52 kB T εtr =
15
3kB T
5
4
15 nτ kB T
.