Vlasov theory of Landau damping
S.M.Lea
January 2007
In Vlasov theory we use the full power of the Vlasov equation to discuss the evolution of
a plasma, including the effects of details of the particle velocity distribution on the evolution
of perturbations. We will study growth and damping of perturbations. The basic technique
is the standard "perturb and linearize" approach that we have been using. Our ®rst topic is
the Landau damping of Langmuir waves.
1 Landau damping
1.1
Qualitative discussion
First a qualitative analysis. The effect arises from an interaction of the individual particles
with the wave potential, which translates at the wave phase speed vÁ = !=k. Some particles
travel faster than the wave and move to the right in the diagram below. The total energy of
an electron is
1
1
E = mv 2 ¡ eÁ = mv 2 + UE
2
2
One such particle that starts with UE = U1 < eÁmax will ultimately be re-ected by the wave
potential: it will bounce back and forth in the potential well. But a particle with an initial
UE > eÁmax will go over the hill (with some loss of speed). Similarly, a particle going
slightly slower than the wave will move to the left in the wave frame and will be trapped. A
particle going a lot slower is not trapped but continues to move to the left in the wave frame.
A particle that is initially moving with v > vÁ and becomes trapped ends up with an average
1
speed equal to vÁ : Since its energy is decreased, the wave gains energy. But a particle
moving with v < vÁ that gets trapped gains energy (in the lab frame) and so the wave loses
energy. If v Á is in the tail of the Maxwellian distribution, there are more particles that gain
energy than particles that lose energy. and so the wave loses energyw it is damped. This is
non-linear Landau damping.
To understand linear Landau damping we have to look at the initiation of this process.
Letzs see what happens to the particle velocities: (diagram from Chen p 255). The bottom
graph shows the particle potential energy ¡eÁ: The upper panel shows the particle velocities
in the wave frame. A particle initially at A gains energy during the ®rst quarter cycle of the
wave, while a particle intially at C loses energy. Similarly a particle initially at B loses
energy during the ®rst quarter cycle, while a particle at D gains energy. Since there are more
particles at A than at C and more at D than at B; (see distribution to the left) the particles
as a whole lose energy and so the wave damps. Thus linear Landau damping is a start-up
effect. This is a cluew initial conditions are going to be important in our analysis.
2
1.2
Mathematical analysis.
Langmuir waves are high-frequency waves so the ions are unperturbed. The Vlasov equation
(plas-uid notes eqn 9) with electrostatic ®elds only is:
@f
~ + qE
~ ¢ @f = 0
+ ~v ¢ rf
@t
m
@~v
Now we perturb and linearize. A new feature we have not used before is the perturbation to
the distribution function:
f (~v ) = f0 (~v ) + f1 (~v )
~
Then, with E0 = 0; and keeping only ®rst order terms,
@f 1
~ 1+ q E
~ ¢ @f 0 = 0
+ ~v ¢ rf
(1)
@t
m
@~v
~ Remember (plas-uid notes
Poissonzs equation will allow us to relate the integral of f to E:
eqn 6):
Z
n=
f (~v )d3 ~v
Z
e
e
~
r ¢ E = ¡ n1 = ¡
f1 (~v ) d3 ~v
(2)
"0
"0
Next Fourier transform
³ these two´equations (or, equivalently, assume that the perturbation is
proportional to exp i ~k ¢ ~x ¡ !t ):
Thus:
e ~ @f0
¡i!f1 + i~k ¢ ~v f1 ¡ E
¢
=0
m
@~v
and
Z
~ =¡e
i~k ¢ E
f1 (~v)d3~v
"0
~ ; we have:
Choosing coordinates with the x¡axis along E
Z
e
Ex = i
f 1 (~v )d3~v
"0 kx
and we ®nd f 1 from (3).
e
@f0
1
f1 = i Ex
m
@v x ! ¡ ~k ¢ ~v
Substitute into the previous equation:
Z
Z
e
e
@f0
1
n0 e2
@ f^0
1
3
³
´ d ~v = ¡
Ex = i
i Ex
Ex
d3~v
"0 kx
m
@vx ! ¡ ~k ¢ ~v
"0 mk
@vx (! ¡ kvx )
(3)
(4)
(5)
~ parallel to ~k: Then for E x not
where in the last line we assumed longitudinal waves with E
zero, we have the dispersion relation:
Z
! 2p
@f^0
1
1= 2
dv y dvz dvx
(6)
k
@v x (vx ¡ !=k)
Now in principle the problem is solved. But how do we do that integral? There is no
3
problem with the integral over vy and vz : But since ! is, in general, a complex number, we
are actually doing an integral along the real axis in the complex vx plane, and such integrals
are path dependent unless the integrand is analytic everywhere. Our integrand has a pole at
vx = !=k: As the imaginary part of ! gets smaller, the pole approaches the real axis, and we
shall have to deform our path to go around the pole. So how do we know what path to use?
Well, remember that Landau damping is a result of initial conditions. So we should
really be solving an initial value problem, for which the Laplace transform usually works
better than the Fourier transform. So letzs go back to 1 and Laplace transform it in time
while retaining the Fourier transform in space:
e
@f 0
sF 1 ¡ f 1 (0) + i~k ¢ ~vF 1 ¡ L (Ex )
=0
m
@vx
where F1 is the Laplace transform of f1 and f 1 (0) is the intitial value of f1 : Now we solve
for F1 :
e
f 1 (0) + m
L (E x ) @@vfx0
F1 =
s + ikvx
From Poissonzs equation (2), we get:
Z
Z
e
f 1 (0) + m
L (E x ) @@vfx0 3
e
e
ikL (E x ) = ¡
F 1 (~v )d3~v = ¡
d ~v
"0
"0
s + ikvx
Factor out the density and write f 0 = n0 f^0 to get:
Ã
!
Z
Z
n0 e 2
@ f^0
1
e
f 1 (0) 3
3
L (E x ) ik +
d ~v = ¡
d ~v
"0 m
@v x s + ikvx
"0
s + ikvx
or
L (E x ) =
=
´
1+
R
f 1(0)
3
v
s+ ikv x d ~
R
@ f^0
n0e 2
1
d3~v
i"0mk
@vx s+ ikv x
i "e0k
i "0ek
R
f1(0) 3
d ~v
s+ikv x
^
@ f0
1
3v
@vx vx ¡is=k d ~
!2 R
1 ¡ : k2p
R f 1(0) 3
i "e0k s+ikv
d ~v
x
R f1(0) 3
i "0ek s+ikv
d ~v
x
=
R @ f^0
!2
p
1
1 + : k 2 @v x is=k¡vx d3~v
(7)
" (is; k)
where I have used the de®nition
Z
!2p
@ f^0
1
d3 ~v
2
k
@v x vx ¡ is=k
and " is the dielectric constant (see below). Here we see that our previous dispersion relation
6 is just " (!; k) = 0; where " appears in the denominator of L (E x ) ; with is replaced with
!:
To solve for Ex ; we use the Mellin inversion integral.
Z ° +i1
1
E x (t) =
L (E x ) es tds
2¼i ° ¡i1
" (is; k) ´ 1 ¡ :
4
The prescription says that we must choose a path of integration that passes to the right of all
the poles of L (E x) : Note that these poles are just the zeroes of the denominator, i.e. the
roots of the dispersion relation.
Thus on the integration path we have Re(s) = °: Noting that this s is related to the
previous ! by is = !; we have Re(s) = Re (¡i!) = Im (!) : Thus on the integration path
in the !¡plane, Im (!) = °:
Now we can take this information and use it in our previous problem. The poles of
L (E x ) determine the behavior of the integral that gives E x ; that is they determine the
behavior of the plasma. In fact, if we denote the roots of " = 0 as !n ; (and assume these are
simple poles) then we can evaluate the integral to get:
Z °+i1
Z °+i1 e R f1(0) 3
i "0k s+ikv x d ~v st
1
1
st
Ex (t) =
L (E x ) e ds =
e ds
2¼ i °¡ i1
2¼i °¡i1
" (is; k)
Z
X
s + i!n ie
f 1 (0)
=
lim
d3~v e¡i! m t
s!¡ i! n " (!; k) "0 k
s
+
ikv
x
n
where s is greater than Re (¡i!n ) until the limit is taken.
Now letzs look at the !¡plane. Remember that ! = is means the !¡plane is rotated
90 o relative to the s¡plane: s = +1 corresponds to ! = +i1; s = +i1 corresponds to
! = ¡1; and s = ° corresponds to ! = i°:
5
In this plane all the poles (normal mode frequencies) are below the path of integration
and ! > !n :
Now letzs go to the v plane. We integrate along (or close to) the real axis, but the
imaginary part of ! is positive and equals i° (as can be seen from the graph above). Thus
the path of integration passes below the poles of this integrand at v = !=k = i°=k: This is
the Landau prescription.
Now of course we (almost) never actually do the Laplace transform, but in order to get
correct results we must always have the integration path in the v¡plane pass beneath the
poles at v = !=k:
Now letzs evaluate the frequencies using equation 6. We can split the integral over vx up
into 2 pieces
1. The principal value:
P (I) = lim
"!0
µZ
! r¡ "
¡1
+
Z
+1
! r +"
¶
@f^0
1
dvx
@vx v x ¡ !=k
2. The contribution from the pole.
To get the principal value, ®rst integrate by parts:
¯ +1 Z
¯
f^0
1
¯
P (I) =
¯
+ f^0
2 dvx
¯
vx ¡ !=k
(v x ¡ !=k)
¡1
The integrated term is zero, since f0 ! 0 as vx ! §1; and the denominator helps. Now
6
letzs assume that the wave phase speed is large compared with the electron thermal speed.
Both f 0 and @f 0 =@v get small as v gets large, if f 0 is a Maxwellian. Thus the integral is
2
dominated by the range where v ¿ !=k: Thus we may expand the 1= (vx ¡ !=k) factor:
Ã
!
µ
¶2
µ
¶2
1
k2
1
k2
kv x
kvx
= 2
= 2 1+2
+3
+¢¢¢
v x ¡ !=k
! (1 ¡ kvx =!)2
!
!
!
so we have:
Z
P (I) dvy dvz
=
=
=
!
µ
¶2
kv x
kvx
1+ 2
+3
+ ::: dv xdv y dvz
!
!
µ
¶
Z
k2
3k2
2
^
1+0+ 2
f0 v xdv xdv y dvz + :::
!2
!
µ
¶
k2
3k 2 kB T e
1+ 2
+ :::
!2
!
m
Z
f^0
Ã
where the average of one component of ~v , squared, over the distribution function, is
kB T e=m; and the average of one component by itself is zero because of the symmetry of the
Maxwellian.
For the contribution from the pole, we have to deform the contour so that it passes under
the pole. It is a simple pole, and if the imaginary part of ! is small, we are getting 12 of a
circle around the pole. We are going around the pole counter-clockwise, and so we get
¯
@ f^ ¯¯
+¼i (residue at the pole) = ¼i
¯
@v x ¯
v x =!=k
Putting it all together, we have:
2
¯
µ
¶
!2p k2
3k 2 kB T e
@ f^ ¯¯
4
1 =
1+ 2
+ ::: + ¼i
¯
k2 ! 2
!
m
@vx ¯
=
!2p
!2
µ
vx = !=k
¯
¶
! 2p @ f^ (v x ) ¯¯
3k kB T e
1+ 2
+ ::: + i¼ 2
¯
!
m
k
@v x ¯
2
3
5
(8)
vx =!=k
where f^ (vx ) denotes the distribution function integrated over vy and v z
The real part of (8) gives the usual dispersion relation for Langmuir waves:
kB T e
! 2 = ! 2p + 3k 2
m
where we took ! ' ! p in the second term. Now letzs assume that the imaginary part is
small, ! = ! r + i°; ° ¿ ! r ; and ! 2 ' !2r + 2i! r °; and letzs neglect the small correction
2 =v 2 to the real part. Then the imaginary part of (8) gives:
term 3vth
Á
¯
¯
2
¯
^
¼ !3p @ f^ (vx ) ¯¯
2 ! p @ f (vx ) ¯
2i! r ° = i¼!r 2
)° =
¯
¯
k
@vx ¯
2 k 2 @v x ¯
v x =!=k
vx =!=k
If vÁ = !=k is in the tail of the distribution, the derivative @ f^ (v x) =@v x is small and
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negative. Since ° is negative, the wave damps. Chen evaluates ° and gets:
p
µ
¶
0:22 ¼
¡1
°=¡
exp
3
2k 2 ¸2D
(k¸D )
for a Maxwellian. Thus the damping is very small for wavelengths that are much greater
than the Debye length (k¸ D ¿ 1): Damping is greatest when ¸ ¸ D : This theoretical
result is amply con®rmed by experiment.
2 The plasma dielectric constant
This theory gives us a lot more than Landau damping. Why did I call the denominator
of equation 7 "? Letzs imagine imposing an external charge density ½ext into the plasma.
Then the Fourier-transformed Poissonzs equation (4), expressed in terms of the potential,
becomes:
Ã
!
1 X
2
k ©=
qs n1s + ½ext
"0
s
As before we use the distribution function to get n1; and use the Vlasov equation to relate f 1
to © (eqn 5):
qs
@f 0s
1
f 1s = ¡
k©
ms
@vx ! ¡ kv x
Thus:
Ã
!
Z
1 X
qs
@f0s (vx )
1
2
k ©=
qs ¡
k©
dv x + ½ext
"0
ms
@vx ! ¡ kvx
s
So
Ã
!
X Z @ f^0s (v x )
1
½
© k2 +
! 2s k
dvx = ext
(9)
@v
!
¡
kv
"0
x
x
s
where ! s is the plasma frequency for species s; or, integrating by parts:
Ã
!
XZ
! 2s
2
^
k ©"0 1 ¡
f 0s (v x)
= ½ext
2 dv x
(! ¡ kvx )
s
Thus we can identify
Ã
"0 1 ¡
XZ
f^0s (vx )
s
!2s
2 dv x
(! ¡ kv x )
!
="
as the dielectric constant for the plasma, and our previous dispersion relation, or any
dispersion relation for electrostatic waves, is found by setting " = 0:
We can also write " = "0 (1 + Â) ; where  is the susceptibility, and
XZ
! 2s
Â=¡
f^0s (vx )
dvx
(10)
2
(! ¡ kv x)
s
The sum is over all the particle species in the plasma.
8
It is also possible to write the susceptibility as:
Z
X
@
1
Â=
! 2s
f^0s (v x)
dvx
@!r
(!r + i° ¡ kv x)
(11)
s
Then  will have a real and an imaginary part,  =  r + i i: When ° ¿ ! r
Z
X
1
2 @
Âr =
!s
P
f^0s (vx )
dvx
@!
(!
¡
kv x)
r
r
s
and if we do a Taylor series expansion, we get:
@Â
 (! r + i°) =  (! r ) + i°
+ ::: =  r + i i
@! r
In most cases the imaginary part of  is due to the contribution of the pole:
¼ X 2 @ ^ ³!´
Âi = ¡i
!
f0s
k s s @! r
k
And thus from equation 12 we have:
¡ ¼k
P
2 @ ^
s ! s @! r f0s
¡! ¢
P
2
s !s
@ ^
f
@v 0s
(12)
¯
¯
(v)¯
¼
v= !
k
2
@Â=@!r
k
@Â=@! r
However, we must be alert for situations where  has an additional imaginary part.
° =
k
=¡
3 The two-stream distribution revisited
If the electrons are moving at a non-zero velocity with respect to the ions, then a
disturbance with !=k < v0 in the lab (ion) frame falls on the electron distribution function
where the slope @f 0=@v is positive. These waves will be unstable. We previously found
kv0 =! » (M=m)2=3 > 1; which satis®es this condition.
9