Boltzmann equation – Nonequilibrium States
Consider a system of gas in a domain R.
Starting from the Liouville equation, molecular chaos assumption together with a coarsegraining procedure leads to the Boltzmann equation.
Assumption:
There is no interaction outside a "hard" core, i.e.,
Fj = 0 if |xi − xj | > σ, ∀i, j, i 6= j
where Fj is the total force exerted on particle j. Note that this "hard core" does not have
to be infinite hard.
Define Region UN : the region in the phase space whose boundaries are ∂R and the
union of sets of the surfaces σ ij defined by
|xi − xj | = σ.
In UN ,
X
∂
∂
vj ·
ρN = 0
ρN +
∂t
∂x
j
j=1
N
Integrating
bution
(1)
ρN
:
R
d3 v · · · d3 vN d3 x2 · · · d3 xN yields an equation for one-particle density distri| 2 {z
}|
{z
}
R3N −3
UN
∂ (1)
∂ (1)
ρN + v1 ·
ρ
∂t
∂x1 N
N Z
X
(2)
−
ρN (x1 , xj ; v1 , vj )v1 · nj dSj d3 vj
+
N Z
X
j=2
N.B. vj is a 3-vector
→ volume element
j=2
(2)
ρN (x1 , xj ; v1 , vj )vj · nj dSj d3 vj = 0
dSj – the surface area element of the sphere |xi − xj | = σ
nP
j is chosen to point inwards to the center of the ball.
N.B. the second -term comes
P from the integration over N − 1 terms (j 6= 1) via Gauss
divergence theorem. The first -term comes from the interchanging the integration and
differentiation for the term j = 1, which introduces a surface integral over |x1 − xj | = σ.
1
Except on the surfaces |x1 − xj | = σ, the integral over other boundaries of UN vanishes
since there are no particles leaving or entering the region and the in-flux cancels the out-flux.
Note that xj, vj are dummy variables, all terms in the sums are equal, respectively. Therefore
Z
∂ (1)
∂ (1)
(2)
ρN + v1 ·
ρN + (N − 1) ρN (x1 , x2 ; v1 , v2 )(v2 − v1 ) · n2 dS2 d3 v2 = 0
(1)
∂t
∂x1
Define:
Positive hemisphere: (v2 − v1 ) · n2 > 0 :
Particle 2 (bullet) entering the interaction region of particle 1 (target)
Negative hemisphere: (v2 − v1 ) · n2 < 0 :
Particle 2 (bullet) leaving the interaction region of particle 1.
B: the bullet; T: the target.
Q : How to handle the integrals over Positive- and Negative-hemispheres?
therefore
Z
(2)
ρN (x1 , x2 ; v1 , v2 )(v2 − v1 ) · n2 dS2 d3 v2
Z h
i
(2)
(2)
−
= (N − 1)
ρN (x1 , x+
;
v
,
v
)
−
ρ
(x
,
x
;
v
,
v
)
|v2 − v1 | rdrdηd3 v2
1 2
1
1 2
2
2
N
(N − 1)
2
(2)
where x±
2 denotes the points located on the ±-hemisphere with the same (r, η).
(i)
N.B. Eq. (1) is not closed – infinite hierarchy ρN , i = 1, 2, · · · (BBGKY-hierarchy).
Q: How to close the equation?
Closure problem – involving molecular chaos assumption and coarse-graining. N.B.
Closure problems are a fundamental problem involving nonequilibrium phenomena.
Assumption: σ is very small such that there is no third particle entering the
interaction region while two particles collide, i.e., the intersection of the sets |xi − xj | > σ
and |xk − xj | > σ (i 6= j, k 6= i, k 6= j) is neglected.
Within the interaction sphere |x1 − x2 | ≤ σ, we have the Liouville equation:
X
∂
∂
∂
∂
∂
∂
ρN + v2 ·
ρN +
vj ·
ρN + X12 ·
ρN + X21 ·
ρN = 0
ρN + v1 ·
∂t
∂x1
∂x2
∂x
∂v
∂v
j
1
2
j=3
N
(3)
where X12 is the the force/mass of particle 2 exerted on particle 1. X21 is the the force/mass
of particle 1 exerted on particle 2. From Newton’s third law, we have
X21 = −X12 .
Integrating Eq. (3) over d3 v3 d3 v4 · · · d3 vN d3 x3 · · · d3 xN and the fourth term has a zero
contribution because the surface integral is neglected on account of the negligent 3-body
interaction.
Therefore,
∂ (2)
∂ (2)
∂ (2)
∂ (2)
∂ (2)
ρN + v1 ·
ρN + v2 ·
ρN + X12 ·
ρN + X21 ·
ρ =0
∂t
∂x1
∂x2
∂v1
∂v2 N
(4)
for |x1 − x2 | ≤ σ.
Denote the time duration of the interaction by τ .
x01 , x02 , v10 , v20 – before collision
x1 , x−
2 , v1 , v2 – after collision
dx1
dx2
dv1
dv2
= v1 ,
= v2 ,
= X12 ,
= X21
dt
dt
dt
dt
which is no more than the characteristics for the PDE in Eq. (4).
(5)
Therefore,
(2)
(2)
(2)
0
0
0
0
ρN (x1 , x−
2 , v1 , v2 ; t) = ρN (x1 , x2 , v1 , v2 ; t − τ )
i.e., ρN is constant along the characteristic curve.
3
(6)
MOLECULAR CHAOS (Nσ 2 =finite, N → ∞, σ → 0 (Boltzmann-Grad limit))
– Statistical independence
Using Eq. (6), we have
Z
(2)
(N − 1) ρN (x1 , x2 ; v1 , v2 )(v2 − v1 ) · n2 dS2 d3 v2
Eq.(2)
=⇒
Z h
i
(2)
(2) 0
0
0
0
N
ρN (x1 , x+
;
v
,
v
;
t)
−
ρ
(x
,
x
;
v
,
v
;
t
−
τ
)
|v2 − v1 | rdrdηd3 v2
1
2
2
1
2 1 2
N
Z
£ (1)
¤
(1) 0
0
(1) 0
0
3
= N
ρ∞ (x1 , v1 , t)ρ(1)
∞ (x2 , v2 , t) − ρ∞ (x1 , v1 , t − τ )ρ∞ (x2 , v2 , t − τ ) |v2 − v1 | rdrdηd v2
N−1≈N
N.B.
1) there is no need to distinguish the superscript + and −. All the quantities on the last
line in the above equation are measured just right before collision. Therefore, we are dealing
with the positive hemisphere only.
2) N − 1 → N (N À 1) . We used the notation ρ∞ = limN→∞ ρN .
COARSE-GRAINING (ρ becomes an averaged distribution over coarse-grained scale):
N.B. the difference among x01 , x02 , x1 , x2 is O(σ)
(1)
In ρ∞ , let
x1 = x2 = x01 = x02
(7)
this identification is the spatial coarse-graining, which introduces an error of O(σ).
Note that
³σ ´
τ ∼O
,
v – typical velocity
v
Temporal coarse-graining is achieved by equating t with t − τ , which introduces an error
O(σ). Therefore the density distribution
¡ ¢ ρ now is an averaged distribution over spatial scales
of O(σ) and temporal scales of O σv . The information below these spatiotemporal scales is
averaged out and any details are lost.
In the Boltzmann-Grad limit (Nσ 2 =finite, N → ∞, σ → 0 ), we have the Boltzmann
equation:
∂ρ
∂ρ
+v·
= Nσ 2
∂t
∂x
zZ
Collision Integral
}|
{
dr}0 dηd3 v1
[ρ(x, v 0 )ρ(x, v10 ) − ρ(x, v)ρ(x, v1 )] |v1 − v| |r0{z
σ−scaled
(1)
where index 1 is removed and 2 is replaced by 1 and ρ ≡ ρ∞ (x, v, t). Note that all ρ’s in the
collision integral have the same x-coordinate because of the spatial coarse-graining (7) . and
have the same time t because of the temporal coarse-graining.
Note that
4
2
Nσ ×typical velocity
1. the collision integral is of order the
volume of the gas
coarse-graining are of order Nσ 3 .
and the errors we introduced in the
e.g., Nσ 2 = 104 cm2 (if N ∼ 1020 , σ ∼ 10−8 cm), then Nσ 3 ∼ 10−4 cm3 . Therefore,
our approximation is good.
2. The essence of molecular chaos approximation is the independence of the two-particle
distribution, i.e.,
(2)
(1)
ρN→∞ (x1, x2 , v1, v2 ) ≈ ρ(1)
∞ (x1 , v1 )ρ∞ (x2 , v2 ).
which we know holds for ideal gas in the limit of N → ∞.
Why cannot we use this in Eq. (2) directly?
N.B. It is fine to use this approximation for the term
(2)
ρN (x1 , x+
2 ; v1 , v2 )
since the two particles before collision can be viewed as statistically independent.
But it is not a reasonable approximation for the term
(2)
ρN (x1 , x−
2 ; v1 , v2 )
because collisions introduce a correlation between the particles after collision. We need
to pull the event back via characteristics to their conditions before collision, which
then is statistically independent again! Had we used the approximation without the
characteristics pull-back, we would have
(1) +
(1) −
ρ(1)
∞ (x1 , v1 , t)(ρ∞ (x2 , v2 , t) − ρ∞ (x2 , v2 , t)) ≈ 0
after identifying x+ and x− (since |x+ − x− | ≤ 2σ). That is, no collision will ever
matter – not physical results.
3. Common notations:
f ≡ Nmρ,
m – the mass of the molecules
f is the mass density of a single particle in the phase space (x, v). f /m is the number
density.
Normalization:
Z
Z
ρd3 xd3 v = 1
fd3 xd3 v = Nm = M
– the total mass of the gas
then the Boltzmann equation becomes
Z
∂f
∂f
1
+v·
=
(f 0 f10 − ff1 ) |V | rdrdηdv1 ≡ Q(f, f )
∂t
∂x
m
where |V | = |v1 − v| . The Boltzmann equation is a nonlinear integral, partial-differential
functional equation – “Functional” because f depends not only f (v) but also f (v 0 )
before collision. v0 , v10 and v, v1 are related.
5
4. Generalization:
Under an external force field (e.g. gravity):
Z
∂f
∂f
F ∂f
1
+v·
+ ·
=
(f 0 f10 − f f1 ) |V | rdrdηdv1
∂t
∂x m ∂v
m
Also it can easily be generalized into a situation with multiple species of particles:
fα , α = 1, · · · , nα . nα is the number of species. The interaction amongst different
species is modeled by the collision integral.
5. Collision invariants.
For any function φ(v),
φ = φ(v) – collision invariant if it satisfies
φ(v) + φ(v1 ) = φ(v 0 ) + φ(v10 )
There is a theorem stating that
φ(v) = a + b · v + c |v|2 ,
a, c ∈ R, b ∈ R3
For example, the elementary collision invariants:
φ0 ≡ 1
(φ1 , φ2 , φ3 ) ≡ v
φ4 ≡ |v|2
– mass conservation
– momentum conservation
– kinetic energy conservation
That is
v + v1 = v0 + v10
v2 + v12 = v02 + v102
The theorem says any other collision invariants are merely linear combinations of these
elementary collision invariants.
Theorem: if φ = φ(v) is a collision invariant, then
Z
d3 vφ(v)Q(f, f ) = 0
for any distribution f.
6
6. Maxwell distribution
Theorem: There exists a distribution f ≥ 0 that makes the collision integral Q(f, f )
vanish. Furthermore, it has the form:
2
f (v) = ea+b·v+c|v|
Due to normalizability on v ∈ (−∞, ∞)⊗3 , we have
c<0
Let c = −α, b = 2αV0 , where V0 ∈ R3 is a constant vector, then
2
f0 (v) = Ae−α(v−V0 )
or using the momentum notation:
2
f0 (p) = Ce−α(p−P0 )
which is the Maxwell distribution. Using normalization and expressed in terms of the
average kinetic energy α,C are found to be
µ 2¶
Z
p
3
3
, ε≡ d p
f0 (p) – average kinetic energy of a particle
α =
4εm
2m
³ α ´3/2
n,
n – the number density
C =
π
¶3/2
µ
3
= n
4πεm
Since ε = 32 kB T, we have
f0 (p) =
n
(2πmkB T )3/2
−
e
(p−P0 )2
2mkB T
which is the distribution for a rare gas in equilibrium (i.e., ∂f0 /∂t = 0)
Under the external field
F = −∇Φ(x)
the equilibrium distribution is
− kΦ(x)
T
f (x, p) = f0 (p)e
e.g., Φ(x) = mgh, then
− kmgh
T
f = f0 (p)e
7
B
B
(8)
Check: whether the distribution (8) satisfies the Boltzmann equation. First note that,
obviously,
∂f
=0
∂t
since Φ(x) is independent of p, the collision integral Q(f, f ) = 0, because Q(f0 , f0 ) = 0.
Finally we need to verify
´
³p
· ∇x + F · ∇p f (x, p) = 0
m
− Φ(x)
which is obviously true. Absorbing e kB T into n, the local density can be defined by
Z
− Φ(x)
n(x) ≡ d3 pf (x, p) = n0 e kB T
then
f (x, p) =
n(x)
−
e
(2πmkB T )3/2
(p−P0 )2
2mkB T
7. H-Theorem: For any solution of the Boltzmann equation, define
Z
H ≡ f log f d3 vd3 x
H-Theorem: For an isolated system:
dH
≤0
dt
Note that the quantity H is the negative entropy!
Discussion of validity of the Boltzmann equation and consequence of molecular chaos
/ coarse-graining.
8
Homework:
HW1) Properties of Collision Operator:
1. Starting from the Newton second law for a central force F = −∇U (|x2 − x1 |) between
two bodies:
dx
= v,
dt
dx1
= v1 ,
dt
where ρ = x2 − x1 , show that
and
m
dv
∂U
=−
ρ̂,
dt
∂ρ
m
dv1
∂U
=
ρ̂
dt
∂ρ
½
v 0 = v + α̂ (α̂ · V )
v10 = v1 − α̂ (α̂ · V )
(9)
½
v = v0 + α̂ (α̂ · V 0 )
v1 = v10 − α̂ (α̂ · V 0 )
(10)
Hence
V 0 = V − 2α̂ (α̂ · V )
where the prime indicates that the variables denote the quantities before the collision,
V ≡ v1 − v, V 0 ≡ v10 − v0 are the relative velocities, and α̂ is the unit vector which
bisects the angle between the vectors V, and −V 0 . (see figure)
(a)
2. Assuming that the two-body potential is completely repulsive, then the angle θ = θ(r)
is an increasing function of r, the impact parameter.
Define
B(θ, V ) = |V | r
∂r
,
∂θ
show that the Boltzmann equation can be written in the form
Z
Z 2π Z π/2
∂
∂
1
3
d v1
dη
dθ (f 0 f10 − f f1 ) B(θ, V )
f +v·
f=
∂t
∂x
m v∈R3
0
0
9
(11)
For example, if the interparticle potential is hard-core:
½
0, r ≥ σ
U=
∞ r≤σ
then,
r = σ sin θ
B(θ, V ) = |V | σ 2 cos θ sin θ.
3. Define bilinear integral operator:
Z
1
(f 0 g10 + f10 g 0 − f g1 − f1 g) B(θ, V )d3 v1 dηdθ
Q(f, g) =
2m
for any functions f and g. Obviously, Q(f, g) = Q(g, f ), Q (f, g) = Q (f, f ), i.e., the
collision integral when f = g. Consider the eight-fold integral:
Z
Z
1
3
d vQ(f, g)φ(v) =
φ(v) (f 0 g10 + f10 g 0 − fg1 − f1 g) B(θ, V )d3 v1 d3 vdηdθ
2m
3
v∈R
for any function φ(v). Using Eqs (9) and (10) , show that
Z
d3 vQ(f, g)φ(v) =
3
v∈R
Z
1
(φ + φ1 − φ0 − φ01 ) (f 0 g10 + f10 g0 − f g1 − f1 g) B(θ, V )d3 v1 d3 vdηdθ
8m
where φ = φ(v), φ1 = φ(v1 ), φ0 = φ(v 0 ), φ01 = φ(v10 ). If f = g, then
Z
Z
1
3
d vQ(f, f )φ(v) =
(φ + φ1 − φ0 − φ01 ) (f 0 f10 − ff1 ) B(θ, V )d3 v1 d3 vdηdθ.
4m
v∈R3
Therefore if φ + φ1 = φ0 + φ01 holds almost everywhere, then
Z
d3 vQ(f, g)φ(v) = 0
v∈R3
independent of the functions f, g. Thus, φ is referred to as the collision invariant if
φ + φ1 = φ0 + φ01 .
4. Show that for any f ≥ 0,
Z
(log f ) Q(f, f )d3 v ≤ 0
2
the equality holds iff f (v) = ea+b·v+c|v| . (Hint: (x − y) log xy ≥ 0, and note that
B(θ, V ) ≥ 0 ).
10
HW2) H-Theorem:
1. Show that H-theorem holds for the BGK equation:
¶
µ
f − f (0)
∂
+ v · ∇x f (x, v, t) = −
∂t
τ
where
f
(0)
µ
m
(x, v, t) = ρ (x, t)
2πkB T (x, t)
¶3/2
− 2km T (v−u(x,t))2
e
B
with the assumption that the local equilibrium hydrodynamic fields are identical to
the true fields, i.e.,
R 3
R 3
d pAf
d pAf (0)
hAi ≡ R 3
(12)
is replaced with hAi0 ≡ R 3 (0)
d pf
d pf
In the proof, you may need
Z
3
d vf log f
(0)
=
Z
d3 vf (0) log f (0)
which is a special case of the assumption (12) .
2. Show that the H-theorem holds for the Boltzmann equation (11) for an isolated system.
11