Landau (Fokker-Planck) kinetic equation

PHYS 7500 Plasma Transport Theory #2 c○Jeong-Young Ji 1
Kinetic equation for plasmas
The exact (microscopic) distribution function F(t,x,v) satisfies
∂F
∂t + v · ∂F
∂x + F micro
m · ∂F
∂v = 0 (1)
which is a manifestation of Liouville theorem. For N particles with initial
conditions x i (0) and v i (0), the exact solution is
F(t,x,v) =
N∑
δ[x − x i (t)]δ[v − v i (t)]. (2)
i=1
Finding F is equivalent to finding all trajectories of N particles, which is practically
impossible.
Problem 2-1 (10 pts). Prove that (2) is the solution of (4).
Therefore, we want to play with a statistical (macroscopic) distribution function
f which is smooth (mist) in contrast to a spiky (rain) microscopic distribution
function F. This smooth function can be obtained by taking an ensemble average:
f = 〈F〉 ens
, and the governing equation for f is obtained from (1):
∂f
∂t + v · ∂f 〈
∂x + Fmicro
m · ∂F 〉
= 0. (3)
∂v
ens
By writing F micro = ∑ j F a←j + F ext = 〈F〉 ens
+ ˜F = F + ˜F and F = f + ˜F,
〈
∂f
∂t + v · ∂f
∂x + F m · ∂f ˜F
∂v = − m · ∂ ˜F
〉
, (4)
∂v
ens
where the F term explains the response of a particle to a mean field and ˜F
explains the effect of occasional collision events due to close approach. Basically
the ˜F term is called a collision operator:
〈 ∑ ˜F a←j
C(f a , f b ) = −
· ∂ ˜F
〉
a
m a ∂v
j∈b
If one solves Eq. (1) for the fluctuation ˜F (the Fourier transform method),
one can derive Balescu-Lenard operator. However, we start from the intuitive
Boltzmann operator (binary collision is dominant)
∫ ∫
C(f a , f b )(v a ) = dv b dΣ|v a − v b | [f a (v a ′ )f b(v b ′ ) − f a(v a )f b (v b )] (5)
and derive its approximate version for a Coulomb interaction (small angle scattering
is dominant, plasma parameter (4πnλ 3 D )−1 n
=
1/2
≪ 1). The
T 3/2
ens
e3
4πɛ 3/2
0

PHYS 7500 Plasma Transport Theory #2 c○Jeong-Young Ji 2
Boltzmann operator satisfies: (i) f is always nonnegative (ii) the operator conserves
particle number, momentum, and energy (iii) ∫ dvf lnf is always decreasing
(non-Maxwellian) or constant (Maxwellian for equilibrium). To summarize,
the state of an ionized gas for plasma species a is described by a distribution
function f a and its evolution is governed by the (approximate) Boltzmann equation
or Landau-Fokker-Planck equation
∂f a
∂t + v · ∂f a
∂x + F a
m a
· ∂f a
∂v = ∑ b
C(f a , f b ),
where a and b stand for particle species, e.g., ions (i) and electrons (e).
Rutherford scattering
The scattering of a particle a by a target particle b with a Coulomb potential
α
r
is given by (this is a good exercise to refresh your memory of sophomore
physics)
sin θ 2 = 1
√
1 + (ρ/ρ0 ) , (6)
2
where
and
ρ 0 =
|α|
m ab u2, (7)
m ab =
m am b
,
m a + m b
(8)
u = v a − v b . (9)
Here ρ is the impact parameter, and θ is the scattering angle.
Problem 2-2 (10 pts). Show from energy and momentum conservation that
|u ′ | = |u| where u ′ = v a ′ − v b ′ is the relative velocity after the collision.
In the target particle’s rest frame, the scattered velocity is
u ′ = u(ˆxsin θ cosφ + ŷ sin θ sin φ + ẑ cosθ), (10)
or
u ′ = u + ∆u ‖ + ∆u ⊥ , (11)
∆u ‖ = u(cos θ − 1) = uẑ(cosθ − 1), (12)
∆u ⊥ = u(ˆxsin θ cosφ + ŷ sinθ sinφ) (13)
where ‖ and ⊥ are parallel and perpendicular to u (the relative velocity before
the collision). The differential cross section for a Coulomb interaction is
dΣ
dΩ = d(πρ2 )
2π sinθdθ = −
ρ
sinθ
dρ
dθ = − ρ2 0
4 sin 4 θ 2
ρ 2 0
= −
(1 − cosθ) 2 . (14)

PHYS 7500 Plasma Transport Theory #2 c○Jeong-Young Ji 3
It follows from the momentum conservation
0 = m a ∆v a + m b ∆v b (15)
that
and
∆v b = − m a
m b
∆v a (16)
∆u = ∆v a − ∆v b = (1 + m a
m b
)∆v a = m a
m ab
∆v a = − m b
m ab
∆v b (17)
Collision operator for Coulomb interaction
The Boltzmann operator is difficult to calculate for the Coulomb interaction.
Fortunately, the scattering cross section (14) for a large angle is extremely small
compared to a small angle. Therefore, Taylor-expanding distribution functions
after the collision in Eq. (5) in power of ∆v and retaining only terms through
the second order, we obtain a practically tractable collision operator:
C(f a , f b ) =
∫
dv b
∫
(
dΣ|v a − v b |
∆v a · ∂f a
∂v a
f b + f a ∆v b · ∂f b
∂v b
+ 1 2 ∆v ∂ ∂
a∆v a : f a + ∆v a ∆v b : ∂f a ∂f b
∂v a ∂v a ∂v a ∂v b
+ 1 )
2 f ∂ ∂
a∆v b ∆v b : f b .
∂v b ∂v b
Using Eq. (17), we can write
∫ [
mab ∂f a
C(f a , f b ) = dv b u ̂∆u · f b − m ab ∂f b
f â∆u · )
m a ∂v a m b ∂v b
+ 1 2 (m ab
m a
) 2 ̂∆u∆u :
+( m ab
m b
) 2 f a ̂∆u∆u :
∂
∂v a
∂
∂v b
∂
∂v a
f a − m ab
m a
m ab
̂∆u∆u : ∂f a ∂f b
m b ∂v a ∂v
]
b
∂
f b , (18)
∂v b
where we have defined ̂∆u = ∫ dΣ∆u and ̂∆u∆u = ∫ dΣ∆u∆u.
Calculation of velocity change
Here we calculate
Using Eq. (14),
̂∆u =
∫ ρmax
0
∫ ρmax
dΣ∆u = u dΣ(cosθ − 1).
0
∫ θmin
2πρ
̂∆u 2 0
= u d cosθ (cosθ − 1)
π (1 − cosθ)
2 θ min
= 2πρ 2 0u ln(1 − cosθ)
∣ ,
π

PHYS 7500 Plasma Transport Theory #2 c○Jeong-Young Ji 4
where θ min is the angle for ρ = ρ max = λ D . From (6),
where
Then
1 − cosθ min
2
= sin 2 θ min
2
=
1
1 + Λ2, (19)
Λ ab = λ D
= λ Dm ab u 2
≈ 3λ D m b T a + m a T b
. (20)
ρ 0 |α| |α| m a + m b
̂∆u = −2πρ 2 0u ln(1 + Λ 2 ) ≈ −4πρ 2 0u ln Λ (Λ ≫1), (21)
where ln Λ is called the Coulomb logarithm.
Also, we calculate
∫
̂∆u∆u = dΣ∆u∆u
∫
= dΣ(∆u ‖ ∆u ‖ + ∆u ‖ ∆u ⊥ + ∆u ⊥ ∆u ‖ + ∆u ⊥ ∆u ⊥ ).
From Eqs. (12) and (13), ∫ dΣ∆u ‖ ∆u ⊥ = 0, ∫ dΣ∆u ⊥ ∆u ‖ = 0,
∫
∫
dΣ∆u ‖ ∆u ‖ = uu dΣ(cosθ − 1) 2
and
∫
dΣ∆u ⊥ ∆u ⊥ =
where we have used
and
Finally, we have
∫ θmin
= uu
π
2πρ 2 0
d cosθ (cos θ − 1)2
(1 − cosθ)
2
= 2πρ 2 0 uu(1 + cosθ min) = 4πρ 2 Λ2
0uu 1 + Λ 2 ,
∫
−ρ 2 0
dΩ
(1 − cosθ) 2 u2 [ˆxˆx sin 2 θ cos 2 φ
+(ˆxŷ + ŷˆx)sin 2 θ cosφsin φ + ŷŷ sin 2 θ sin 2 φ]
= ρ 2 0 u2 ∫ θmin
π
d cosθ
(1 − cosθ) 2 π(ˆxˆx + ŷŷ)sin2 θ
= πρ 2 0 u2 (ˆxˆx + ŷŷ)[− 2Λ2
1 + Λ 2 − 2 ln 1
1 + Λ 2 ],
1 + cosθ min = 2(1 − sin 2 θ 2 ) = 2(1 − 1
1 + Λ 2 ) = 2Λ2
1 + Λ 2 , (22)
cosθ min = − 1 − Λ2
1 + Λ 2 . (23)
̂∆u∆u ≈ 4πρ 2 0 (u2 I − uu)ln Λ (Λ ≫ 1). (24)

PHYS 7500 Plasma Transport Theory #2 c○Jeong-Young Ji 5
Landau collision operator
Introducing the Landau tensor
and its derivative
U = u2 I − uu
u 3 ,
∂
∂v a
· U = − 2u
u 3 ,
we can write Eqs. (21) and (24), respectively, as
and
̂∆u ≈ −4π α2 ln Λ
m 2 u = γ abm a 1 ∂
ab u4 2m 2 · U, (25)
ab
u ∂v a
̂∆u∆u ≈
4π α2 ln Λ U
m 2 ab
u = γ abm a U
m 2 ab
u , (26)
where
γ ab = 4πα2 ln Λ ab
m a
= q2 a q2 b ln Λ ab
4πɛ 2 0 m .
a
Let’s write Eq. (18) term by term (t i denotes the ith term and t ij denotes the
jth term in t i )
t 1 = γ ∫ (
ab ∂ 1 ∂f a
dv b · U · f b + 1 )
∂f a
f b = t 11 + t 12 ,
2 ∂v a m a ∂v a m b ∂v a
t 2 = γ ∫ (
ab ∂
dv b · U · − 1 ∂f b
f a − m )
a ∂f b
2 ∂v a m b ∂v b m 2 f a = t 21 + t 22 ,
b
∂v b
t 3 = γ ∫
ab 1 ∂ ∂f a
dv b U : f b ,
2 m a ∂v a ∂v a
t 4 = − γ ∫
ab 2
2 m b
= γ [ ∫
ab 1
−
2 m b
= t 41 + t 42 ,
dv b U : ∂f a
∂v a
∂f b
dv b U : ∂f a
∂v a
∂f b
∂v b
−
( 1 IBP →)
∂v b 2
∫
dv b
( ∂
∂v a
· U ) · ∂f a
∂v a
f b
]
and
t 5 = γ ab
2
= γ ab
2
m a
m 2 b
f a
∫
∫
m a
m 2 f a
b
dv b U :
∂
∂v b
∂f b
∂v b
(IBP →)
dv b
( ∂
∂v a
· U ) · ∂f b
∂v b
.

PHYS 7500 Plasma Transport Theory #2 c○Jeong-Young Ji 6
From
and
t 11 + t 3 = γ ab
2
t 21 + t 41 = − γ ab
2
∫
∂
·
∂v a
dv b U ·
t 12 + t 42 = 0,
∫
∂
·
∂v a
t 22 + t 5 = 0,
1
m a
∂f a
∂v a
f b ,
dv b U · f a
1
m b
∂f b
∂v b
,
we have the Landau collision operator
C(f a , f b ) = γ ∫ ( )
ab ∂
1 ∂f a 1 ∂f b
· dv b U · f b − f a . (27)
2 ∂v a m a ∂v a m b ∂v b
Problem 2-3 (15 pts). Prove the conservation laws for the particle number,
momentum, and energy, i.e.,
∫
dv a C ab (v a ) = 0,
∫
∫
dv a m a v a C ab (v a ) +
dv b m b v b C ba (v b ) = 0,
and
∫
∫
1
dv a
2 m ava 2 C ab(v a ) +
dv b
1
2 m bv 2 b C ba(v b ) = 0.
Hint: Use Gauss’ theorem and the integration by parts.
Problem 2-4 (20 pts). The entropy for species a is defined as
∫
S(t,x) = − dvf a (v)ln f a (v).
(i) Assume F a (t,x,v) = F a (t,x) + q a v × B. Show that the entropy production
rate for a due to collisions is
∫
Θ a = − dv lnf a (v)C a (v),
where [use Eq. (28)]
C a (v) = ∑ b
C(f a , f b ).
(ii) Show that the total entropy production rate, Θ = ∑ a Θ a, is nonnegative,
i.e.,
Θ ≥ 0.
(iii) Show that the general solution for Θ = 0 is Maxwellian distributions with
the equal temperatures and flow velocities.

PHYS 7500 Plasma Transport Theory #2 c○Jeong-Young Ji 7
Hint: (i) Multiply (1 + lnf) to the kinetic equation and integrate over the
velocity variable. (ii) Θ can be written in the form of
Θ = ∑ ∫ ∫
m a γ ab
dv dv ′ f a (v)f b (v ′ )A · U · A,
4
a,b
where A should be found. Then show that A · U · A ≥ 0. (iii) Note that
A · U · A = 0 if and only if A is parallel to u.
Rosenbluth potentials
The Landau collision operator can be rewritten as
C(f a , f b ) = γ ∫ ( )
ab ∂
2m a ∂v · dv ′ ∂fa
U ·
∂v f m a ∂f b
b − f a
m b ∂v ′ , (28)
where f a = f a (v) and f b = f b (v ′ ) and u = v − v ′ . Defining the Rosenbluth
potentials:
∫
G b (v) = dv ′ f b (v ′ )|v − v ′ |, (29)
∫
H b (v) = dv ′ f b (v ′ )|v − v ′ | −1 , (30)
and using
∫
− m a
m b
∫
dv ′ U · ∂f a
∂v f b =
(
dv ′ ∂f
)
b
U · f a
∂v ′
U = ∂ ∂
∂v ∂v u,
∂ v ∂ v : U = −8πδ(v − v ′ ),
∫
= ∂
∂v ·
dv ′ ∂ ∂
∂v ∂v u · ∂f a
∂v f b
(
∂ ∂G b
f a
∂v ∂v
= − m a
m b
∫
= −2 m a
m b
f a
∂H b
∂v ,
)
− 2f a
∂H b
∂v ,
dv ′ ∂ ∂
(
∂v ∂v u · ∂f
)
b
f a
∂v ′
we can write the RMJ (Rosenbluth-MacDonald-Judd) form of the collision operator,
[
]
C(f a , f b ) = γ ab ∂
2m a ∂v · ∂
(
∂v · ∂ ∂G
) (
b
f a − 2 1 + m )
a ∂H b
f a .
∂v ∂v m b ∂v
Furthermore, we can also write the operator in the Fokker-Planck form,
C(f a , f b ) = − ∂ (
∂v · 〈∆v〉
)
f a + 1 ∂ ∂
(
∆t 2 ∂v ∂v : 〈∆v∆v〉
)
f a
∆t

PHYS 7500 Plasma Transport Theory #2 c○Jeong-Young Ji 8
where, from (17), (25), (26), (29), and (30),
∫ ∫
〈∆v〉
= dv b dΣu∆v a f b = γ ab ∂H b
, (31)
∆t
m ab ∂v a
and
〈∆v∆v〉
∆t
∫ ∫
= dv b
dΣu∆v a ∆v a f b = γ ab ∂ ∂G b
. (32)
m a ∂v a ∂v a
In the moment approach, the following form
[
]
C(f a , f b ) = γ ab 1 ∂ ∂f a
m a 2 ∂v ∂v : ∂ ∂G
(
b
∂v ∂v + 1 − m )
a ∂fa
m b ∂v · ∂H b
∂v + 4πm a
f a f b .
m b
is used to calculate collision terms [Ji and Held, Phys. Plasmas 13, 102103
(2006)].
Fokker-Planck operator
Consider a random-walk like process in the velocity space where velocity can
change from v to v + ∆v with a probability P(v, ∆v) independent of time:
∫
f(t,x,v) = d(∆v)f(t − ∆t,x,v − ∆v)P(v − ∆v, ∆v). (33)
From the Taylor-expansion,
∫ [
f(t,x,v) = d(∆v) fP − ∆t ∂fP − ∆v · ∂fP
∂t ∂v + 1 ]
2 ∆v∆v : ∂ ∂
∂v ∂v fP
= f − ∆t ∂f
∂t − ∂
∂v · (f 〈∆v〉) + 1 ∂ ∂
: (f 〈∆v∆v〉)
2 ∂v ∂v
where we have used ∫ d(∆v)P(v, ∆v) = 1, and defined
∫
〈∆v〉 = d(∆v)P(v, ∆v)∆v (34)
and
∫
〈∆v∆v〉 =
d(∆v)P(v, ∆v)∆v∆v. (35)
Then the change rate of f due to collisions is
∂f
= − ∂ (
∂t ∣ ∂v · f 〈∆v〉 )
+ 1 (
∂ ∂
∆t 2 ∂v ∂v : f 〈∆v∆v〉 )
.
∆t
col
The operator of this form is called the Fokker-Planck operator. Here the first
and second terms are the dynamical friction and the velocity diffusion, respectively.
It is interesting to compare (34) to (31), and (35) to (32).
Homework due: 9/30/2009