Non-linear effects
S.M.Lea
January 2007
Now it is time to take a brief look at some phenomena that arise when instabilities lead
to large amplitude ®elds and density perturbations.
1 The ponderomotive force
The ponderomotive force is akin to radiation pressure. It is a force on the plasma that
arises from the presence of a large-amplitude wave. The electron equation of motion is
h
i
d~v (~r ; t)
~ (~r; t) + ~v (~r ; t) £ B
~ (~r; t)
m
= ¡e E
(1)
dt
where the position of the electron is ~r (t) : There are non-linear effects due to
~ term
1. the ~v £ B
~ (~r ; t) ; the ®eld at the perturbed position of the electron, not
2. the fact that we must use E
~
E (~r0 ) :
Now let us assume that the electric ®eld is due to a plasma wave and has the form
~ (~r; t) = E
~ s (~r) cos !t
E
~ s (~r ) contains all the information about the spatial dependence of the
where the function E
electric ®eld, and the ®eld is oscillating in time as expected for a wave. We cannot use the
~
exponential form ei(k¢~x¡!t) here because of the non-linearity. (At least, we cannot do it
easily, because of all the products that occur. Extreme care is needed with taking the real
part.) We use a "bootstrap" method to build the solution, as follows.
1st order
The ®rst order response is the response due to the electric ®eld at the unperturbed
position:
d~v1 (~r; t)
~ (~r0) = ¡eE
~ s (~r0 ) cos !t
m
= ¡e E
dt
(Note that the time average of this force is zero.) We can integrate this equation with respect
to time to get the velocity
Z t
e ~
e ~
~v1 (~r; t) ¡ v 1 (~r; 0) = ¡ E
(~
r
)
cos !tdt = ¡
E s (~r0 ) sin !t
(2)
s
0
m
m!
0
1
and integrate again to get the ®rst order displacement. Take ~v 1 (~r ; 0) ´ 0: Then
Z t
e ~
e ~
±~r1 (~r ;t) = ¡ E s (~r0 )
sin !tdt =
E (~r0) (cos !t ¡ 1)
2 s
m
m!
0
(3)
Now we can use the dieplacement to calculate the electric ®eld at the perturbed position
³
´ ¯
~ s (~r ) = E
~ s (~r0 ) + ±~r1 ¢ r
~ E
~ s ¯¯ + ¢ ¢ ¢
E
(4)
r0
~
~
Next we use Faradayzs Law to ®nd the corresponding B:
~
~ £E
~ = ¡ @B
r
@t
~ has the same form as E
~
Assume that B
~ (~r; t) = B
~ s (~r) sin !t
B
(The reason for the sine rather than the cosine is made obvious in the line below.) Then,
substituting into Faradayzs law, we have
~ £E
~ s (~r) cos !t = ¡! B
~ s (~r ) cos !t
r
and so the ®rst order term is:
~ 1 (~r;t) = ¡ 1 r
~ £E
~ s (~r0 ) sin !t
B
(5)
!
We now have the complete set of ®rst order quantities that we need, and we can use these to
®nd the next order.
2nd order
We look for corrections to the ®rst order terms. The second order terms in the equation
of motion (1) are:
³
´
d~v2 (~r; t)
~ 1 + ~v1 £ B1
m
= ¡e E
dt
·³
¶¸
´ ¯
³
´ µ
~ E
~ s¯¯ cos !t + ¡ e E
~ s (~r0 ) sin !t £ ¡ 1 r
~ £E
~ s (~r0 ) sin !t
= ¡e ±~r1 ¢ r
~r0
m!
!
i
³
´
2
e2 h ~
e
~ E
~ s (~r0 ) cos !t ¡
~ s (~r0 ) £ r
~ £E
~ s (~r0 ) sin2 !t
= ¡
E
(~
r
)
(cos
!t
¡
1)
¢
r
E
s
0
m! 2
m! 2
Now we time average. The term in cos !t averages away, and the terms in sin2 and cos2
average to 1/2. Thus
´
³
´i
d~v 2 (~r )
e 2 h³ ~
~ E
~ s (~r0 ) + E
~ s (~r0 ) £ r
~ £E
~ s (~r0 )
m<
> =¡
E
(~
r
)
¢
r
s
0
dt
2m! 2
Expand the second term:
³
´
~ £E
~ s (~r0 )
"ijk E j r
= "ijk E j " klm @l E m
k
=
(±il ±jm ¡ ±jl ± im ) E j @l E m
=
Em @iE m ¡ E j @j E i
³
´
1
~ ¢r
~ Ei
@ iE 2 ¡ E
2
=
2
Thus
m
<
·³
¸
´
d~v2 (~r )
e2
1~ 2 ³~ ~ ´ ~
~
~
~
>= ¡
E s (~r0 ) ¢ r E s (~r0 ) + rE s ¡ E ¢ r E s (~r0 )
dt
2m! 2
2
e2 ~ 2
rE s
4m! 2
Since the time average of the ®rst-order force is zero, the time averaged non-linear force on
the electron is
e2 ~ 2
F~ = ¡
rEs
(6)
4m! 2
The force on a unit volume of the plasma is then the force exerted on all the electrons in the
plasma:
= ¡
F~ NL
! 2p " 0 Es2
n0 e2 " 0 ~ 2
~
rE
=
¡
r
s
4m! 2 " 0
!2
4
! 2p "0 < E 2 >
! 2p
~
~ < uE >
= ¡ 2r
= ¡ 2r
!
2
!
= ¡
(7)
where
"0 E 2
2
is the electric energy density and <> denotes the time average.
We may now apply this result to some of the waves we have studied.
Electrostatic waves
The electric ®eld accelerates the electron, which moves farther in the half cycle in which
the ®eld is decreasing than it does in the half cycle with increasing ®eld. There is a net drift.
~ is along ~k; so is the force and the drift.
Since E
Electromagnetic waves
~ and
The wave magnetic ®eld pushes electrons in the direction of ~k (~v is parallel to E
~£B
~ is parallel to ~k): So there is a drift along ~k:
E
The ponderomotive force acts primarily on electrons (notice the mass in the denominator
of equation (6). The resulting charge separation generates additional electric ®elds that
transmit the force to the ions. This ®eld also acts on the electrons, of course. The net result
is to slow the electrons and speed the ions. (We saw a similar effect in ambipolar diffusion.)
The net force is given by equation (7).
uE =
2 Parametric instabilities
These instabilities are also known as wave-wave interactions. The ponderomtive force
plays an important role in these instabilities.
From a QM point of view, we can regard the plasma waves as particlesw plasmonsw with
energy }! and momentum }~k: When one wave (called the pump wave) interacts with a
plasma, it can generate two |daughter} waves, provided that energy and momentum are
conserved. That is, we need
!0 = !1 + !2
3
and
~k0 = ~k1 + ~k2
We can see how this works for Langmuir waves and EM waves by looking at the ! versus k
plot. Remember
! 20 = ! 2p + k 2 c2
for EM waves (blue curve) and
2
2
! 21;2 = ! 2p + k1;
2 vth
for Langmuir waves (black curve). The plot shows an EM wave (~k0 ; red arrow) with two
Langmuir wave daughters (green arrows).
Mechanism of the instability:
~ 0 and a perturbation with
Suppose we have a pump wave with electric ®eld amplitude E
~
wave number k1 forms in the plasma, giving rise to E1 : Then there will be a force on the
plasma
´2
!2p "0 ³
~
~0 + E
~1
F~ NL = ¡ 2 r
E
!
4
If the pump wave has wavelength long compared with the perturbation (k0 ¿ k1 ), and
E 1 ¿ E 0 initially, then
´
! 2p "0 ³
~
~0 ¢ E
~1
F~N L ' ¡ 2 r
2E
!
4
h
³
´
³
´i
!2p
~ E
~ 0a ¢ E
~ 1a cos ~k0 ¢ ~x ¡ ! 0 t cos ~k1 ¢ ~x ¡ ! 1 t
= ¡ 2 "0r
2!
h
n
h³
´
i
h³
´
ioi
!2p
~ E
~ 0a ¢ E
~ 1a cos ~k0 + ~k1 ¢ ~x ¡ (!0 + ! 1 ) t + cos ~k0 ¡ ~k1 ¢ x ¡ (!0 ¡ ! 1 ) t
= ¡ 2 "0r
4!
Thus this force drives disturbances at the two frequencies
! = !0 § ! 1
4
and also
~k = ~k0 § ~k1
This is the frequency matching condition described above. We also need
~ 0a ¢ E
~ 1a 6= 0
E
So if the pump wave is an EM wave and the daughters are Langmuir waves, the Langmuir
waves cannot propagate in the same direction as the pump.
If there is no damping, a pump wave of any amplitude can drive a parametric instability.
But usually there is some form of damping, and then there is a threshold amplitude below
which the instability does not go. If ¡1 and ¡2 are the damping rates for the two daughters,
then we ®nd
E 20 > C! 1 ! 2 ¡1 ¡2
where C is a constant. (See Chen sec 8.5.3 for the details.)
3 Shock waves and solitons
These are intrinsically non-linear phenomena. When a wave grows to large amplitude,
different parts of the wave travel with different speeds. This leads to steepening of the
wave form. In a soliton, the wave has a pro®le that moves without change of shape as the
non-linear effects of steepening and dispersion balance each other. To see how this can
work, look at an ion sound wave with dispersion relation (waves notes page 10)
!2
v 2s
=
k2
1 + k2 ¸ 2D
Since
vs2
1
/
- 2p
n
the phase speed vÁ = !=k increases with n: Thus a wave pro®le that is initially sinusoidal
will steepen.
¸2D =
5
Clearly the effect is negligible unless k¸ D is not much less than one. Thus shock waves and
solitons have structure on the order of ¸ D :
Now for the details. The ion-sound wave is a longitudinal wave, so everything is
one-dimensional. We choose x as a coordinate measured along the direction of wave
propagation. From energy conservation, if the ion speed is u 0 at a point where the potential
is zero, then
1
1
M u20 = M u 2 + eÁ
2
2
and thus
r
eÁ
u = u 20 ¡ 2
M
From the continuity equation for the ions,
@n ~
+ r ¢ (n~u ) = 0
@t
in a steady state (@=@t ´ 0); nu is constant:
µ
¶ ¡1=2
u0
eÁ
ni = n0
= n0 1 ¡ 2
u
M u 20
The electron density is given by the Boltzmann relation
µ
¶
eÁ
ne = n0 exp
kT e
Then Poissonzs equation gives
"
µ
¶ µ
¶¡ 1=2 #
d2 Á
e
en0
eÁ
eÁ
=
(ne ¡ ni ) =
exp
¡ 1¡2
(8)
dx2
"0
"0
kT e
M u 20
To simplify this, letzs introduce the dimensionless variables
eÁ
e
Â=
=
Á
kT e
M vs2
6
(9)
and, since we expect structure on a scale ¸ D ;
x
-p
x
»=
=x
=
¸D
vs
vs
s
ne2
"0M
(10)
Finally, we introduce the Mach number
p
u0
u0 M
M=
= p
vs
kTe
(11)
Then equation (8) becomes:
"
µ
¶ ¡1=2 #
en0
ÂkT e
=
exp  ¡ 1 ¡ 2
"0
M u 20
³
d2 Â
 ´¡ 1=2
Â
=
e
¡
1
¡
2
(12)
M2
d» 2
Now if we de®ne the pseudo-potential function
µ
r
¶
Â
V (Â) = 1 ¡ e + M2 1 ¡ 1 ¡ 2 2
M
which has the property that
V (0) = 0
then
µ
µ
¶³
¶
³
dV
1
2
 ´¡ 1=2
 ´¡ 1=2
= ¡e + M2 ¡
¡ 2
1¡ 2 2
= ¡e + 1 ¡ 2 2
dÂ
2
M
M
M
and thus Possionzs equation (12) takes the form:
M v 2s d2 Â -2p
e d» 2 v 2s
d2 Â
dV (Â)
=¡
(13)
2
dÂ
d»
We can solve this by analogy to the problem of a particle moving in a potential well. The
potential  is like a coordinate x and the spatial coordinate » is like t: Let y = dÂ=d»:
(This is the electric ®eld and acts like the velocity in our comparison problem.) The diagram
below shows what the potential well looks like for three values of the Mach number. The
vertical axis is V and the horizontal axis is Â:
7
1
2
chi
3
4
5
0
0.2
-20
0.15
-40
V
V
0.1
0.05
-60
-80
0
0.1
0.2 chi 0.3
0.4
0.5
M = :9
M=3
0.2
V
0.1
0
0.2
0.4
chi
0.6
0.8
1
-0.1
p
M= 2
Looking at the bottom plot for a moment, a particle entering at  = 0 will ®nd its
velocity increasing for a while, and then decreasing again as it reaches the wall at the far
side of the well. Its "velocity" will become zero, so it will re-ect and return to the origin,
repeating its motion (but in reverse). The other plot shows that if the Mach number is too
large or too small, the "potential" does not have a well, and we do not get this behavior.
8
0.6
0.58
0.56
0.54
dchi/dxi
0.52
0.5
0.48
0.46
0
0.2
0.4
xi
0.6
0.8
1
"velocity" versus »
In our case this |velocity} is dÂ=d»; so we may |integrate} to get the resulting Â: dÂ=d»
is zero (zero slope) at » = 0; then positive, then zero, then negative as the particle returns to
the origin.. We get a graph like this:
If the particle loses speed for some reason, it can get trapped in the well. This leads to an
oscillating velocity, and thus an oscillating Â:
Multiplying both sides of equation (13) by y; we get:
dy
dV (Â) dÂ
y
= ¡
d»
d d»
µ
¶
d 1 2
dV
y
= ¡
d» 2
d»
which we can integrate immediately to get
s
·
µ
r
¶¸
p
dÂ
Â
Â
2
y=
= § ¡2V = § ¡2 1 ¡ e + M 1 ¡ 1 ¡ 2 2
(14)
d»
M
This is not easily integrable, so now we approximate.
Let  be small (but not very small), so we expand the functions in equation (12) to
9
second order:
d2 Â
d»2
Ã
!
¡ 1¢¡ 3¢ µ
¶2
¡2 ¡2
Â2
Â
2Â
= 1+Â +
+ ¢ ¢¢ ¡ 1 +
+
+ ¢¢ ¢
2
M2
2
M2
µ
¶
µ
¶
1
Â2
3
= Â 1¡
+
1
¡
+¢¢¢
M2
2
M4
(15)
Now with 20-20 hindsight, we let
·
 = Âm cosh
Then
Â0 =
and
µ
»
¢
¶¸¡l
(16)
·
µ ¶¸¡(l+1)
¡l
»
»
Âm sinh
cosh
¢
¢
¢
·
µ ¶¸¡ l
·
µ ¶¸¡ (l+2) )
1
»
(l + 1)
»
2 »
Â
=
cosh
¡
sinh
cosh
¢
¢
¢
¢
¢
·
µ ¶¸¡l ½
µ
¶¾
¡l
»
1
=
Â
cosh
1 ¡ (l + 1) 1 ¡
¢2 m
¢
cosh2 » =¢
µ
¶
µ
¶ 2=l
¡l
l+1
l2
l (l + 1)
Â
=
 ¡l +
= 2Â ¡
Â
(17)
¢2
¢
¢2
Âm
cosh 2 »=¢
Now we compare this result (17) with equation (15) and ask what it will take to make them
the same.
00
¡l
Â
¢ m
(
1. To get the coef®cient of  right we must have
l2
1
=1¡
¢2
M2
Since the left hand side is a square, and thus positive, it is necessary that M > 1:
(18)
2. To make the second term be a square of  we need
l=2
3. To get the right coef®cient of Â2 we need
µ
¶
l (l + 1)
1
3
¡ 2
=
1¡
¢ Âm
2
M4
Soving for Âm ; using (18) and (19), we get
µ
¶
¡2l (l + 1)
1
3
¡
¢
¡
¢
Âm =
=
¡
1
¡
M2
¢2 1 ¡ M3 4
1 ¡ M3 4
¡ 2
¢
M ¡ 1 M2
= ¡3
(M4 ¡ 3)
(19)
(20)
(21)
For Âm to be positive, we must have1 < M < 3 1=4 = 1: 3: (A more exact treatment
10
without the approximations we have made gives Mmax = 1:6).
Finally, from (18) and (19) we obtain
¢=p
2
1 ¡ 1=M2
= p
2M
(22)
M2 ¡ 1
Thus the solution we have found is (16, 21, 22 and 19)
¡ 2
¢
M ¡ 1 M2
h
i
Â=3
p
»
(3 ¡ M4 ) cosh2 2M
M2 ¡ 1
(23)
The amplitude is < 1; justifying our approximations, if
3M4 ¡ 3M2
Ã
M2 ¡
3+
which means
p
8
57
!Ã
p
3 ¡ 57
8
¡: 56873
4M4 ¡ 3M2 ¡ 3
p !
3 ¡ 57
2
M ¡
8
<
<
<
<
<
3 ¡ M4
0
0
p
3 + 57
8
M2 < 1: 3187
M2 <
and thus
p
M < 1: 3187 = 1: 1483
With M = 1:125; the solution looks like:
 =
3³
(1: 3187 ¡ 1) 1: 3187
´
h
i
p
2
3 ¡ (1: 3187) cosh 2 2p 1:»3187 1: 3187 ¡ 1
: 99982
cosh 2 : 2458»
Or, In the original variables,
eÁ
: 99982
=
2
kT e
cosh : 2458x=¸ D
=
11
0.8
0.6
chi
0.4
0.2
-10
-8
-6
-4
-2
0
2
4 xi 6
8
10
This is a soliton pro®le. The width is of order 10¸D : The pulse gets narrower as M
increases.
Remembering that the |velocity} in our analogy is essentially the electric ®eld, we can
determine it from Poissonzs equation (8 or 12). Integrating once,
Z
dÂ
¸2D »
=
(ne ¡ ni ) d»
d»
n0 0
Thus the |velocity} decreases if the density difference decreases. This may occur due to
re-ection of ions from the front of the shock, leading to an increased ion density upstream.
If the ions are not completely cold, some of them will not have enough kinetic energy to get
over the initial potential hill. Then the picture looks like this:
The thickness of the transition region is again a few ¸ D : This an ion shock wave.
12
4 Plasma sheaths
A sheath is basically a boundary layer between the plasma proper and, for example, a
container wall. With no magnetic ®eld, the electrons diffuse to the wall faster than the ions,
leaving a net positive charge in the plasma. An electric ®eld is set up by the resulting charge
distribution, and this ®eld slows the electron motion. Choosing the potential in the bulk
plasma to be zero, the potential at the wall will become negative, forming a potential barrier
that repels electrons.
To analyze the sheath, we assume that the ions are cold (T i ´ 0). Ions drift toward the
wall with speed u0 at x = 0: Neglect collisions. (This is OK. The electrons still reach
the wall faster because their thermal speed is greater.) Then from energy conservation, the
Boltzmann relation, the continuity equation and Poissonzs equation, we retrieve the same set
of equations that we had for solitons. In particular, from (14), we have
·
µ
r
¶¸
1 0 2
Â
Â
2
(Â ) = ¡V = ¡ 1 ¡ e + M 1 ¡ 1 ¡ 2 2
2
M
Again we chose the constant to make V = 0 and thus E = 0 in the bulk of the plasma, (at
» = 0).
The right hand side must be positive since the left is a square, and, as we found before,
this means that M > 1: This is the Bohm sheath criterion. (For small Â; the RHS is
Ã
!
¡ 1¢ ³
µ
¶
µ
¶
1
¡2
Â2
Â
 ´2
Â2
1
2
2
¡1 + 1 + Â +
¡M 1¡ 1+
¡
2
=
1
¡
2
M2
2
M2
2
M2
) The ions must be accelerated by the electric ®eld before entering the sheath region. This
means that E cannot be zero at x = 0; as we have assumed, but it can be very small. The
picture now looks like this:
The potential slope is negative and the curvature is also negative. This means that ni > ne
13
throughout this region.
Since  is negative, it is convenient to introduce a new variable ª = ¡Â: (My
ª =Chenzs  in section 8.2) Then Poissonzs equation (12) becomes:
µ
¶ ¡1=2
d2 ª
ª
¡ª
= ¡e
+ 1+2 2
M
d»2
When ª À 1 (close to the wall), this equation for ª becomes approximately
µ
¶ ¡1=2
ª
M
ª00 ' 1 + 2 2
' p
M
2ª
0
Multiply both sides by ª and integrate to get
p
1
2
(ª 0) = M 2ª + constant
2
Now it is convenient to rede®ne the zero level for potential so that ª = 0 at some value »s
of » where ª0 is very small (essentially zero). Then we may take the constant to be zero.
Taking the square root
ª0 = 2 3=4 M1=2 ª 1=4
and integrating again, we get
4 3=4
ª
= 2 3=4 M1=2 » + constant
3
Putting in the boundary conditions, ª = 0 at »s
4 3=4
ª
= 23=4 M1=2 (» ¡ »s )
3
We may now rewrite all this in terms of the physical variables Á; x; vs etc.
µ
¶3=4
µ
¶
4 ¡eÁ
x ¡ xs
= 2 3=4 M1=2
(24)
3 kT e
¸D
We are interested in the current toward the wall in terms of the potential at the wall, where
the ion current is
j = n0 eu 0 = n0 eMvs
Note that x wall ' x s + d where d is the thickness of the sheath. We ®nd M from (24),
µ
¶3=4
22
¡eÁ
¸D
1=2
M
=
3 £ 2 3=4 kTe
d
M
=
=
=
2 5=2
v 2s
3=2
(¡eÁw )
3=2
9
- 2p (kTe ) d2
p
4 2
"0 M
(¡eÁw ) 3=2
1=2
9
n0 e2 (kT e) M d2
p
4 2
"0
(¡eÁw ) 3=2
p
2
9
n0 e v s M d2
14
and thus
p
4 2
"0
j = n0 e
(¡eÁw )3=2
p
vs
2
9
n0 e vs M d2
r
4 2e (¡Áw )3=2
=
"0
(25)
9 M
d2
This is the Child-Langmuir Law. Notice that the current is independent of the plasma
density n0: The sheath thickness d is determined by equation (25) once the current and the
potential are measured.
15