Instabilities

Instabilities
Susan Lea
January 2007
1
Two-stream instability
1.1
Dispersion relation
This instability occurs when the electrons and ions have a relative velocity.
Working in the frame moving with the ions, we have:
Ions: n = n0 + ni; ~v = ~v i; (~v0 ´ 0 for the ions.)
Electrons: n = n0 + ne ; ~v = ~v 0 + ~ve
~ 0 = 0, and we look for an electrostatic perturThere is no magnetic ¤eld, B
~
bation so B1 = 0 too. Then the equations of motion are:
ions:
and
electrons:
Poissonzs equation:
Continuity:
mn0
M n0
·
@~vi
~
= en0 E
@t
´ ¸
@~ve ³
~
~
+ ~v0 ¢ r ~v e = ¡en0E
@t
(1)
(2)
~ ¢E
~ = e (ni ¡ ne)
r
"0
(3)
@ni ~
+ r ¢ (n0~vi ) = 0
@t
(4)
for ions, and
@ne
~ ¢ ([n0 + ne ] [~v 0 + ~v e]) = 0
+r
(5)
@t
for electrons. Linearizing, and using the usual wave form for the perturbed
quantities, we have:
Eqn (1)
~
¡i!M n0~vi = en0 E
(6)
Eqn (2)
³
´
~
¡imn0 ~v e ! ¡ ~k ¢ ~v 0 = ¡en0 E
1
(7)
Eqn (3)
~ = e (ni ¡ ne )
i~k ¢ E
"0
(8)
¡i!ni + in0~k ¢ ~vi = 0
(9)
¡i!ne + i~k ¢ (n0~ve + ne~v0 ) = 0
(10)
Eqn (4)
and eqn (5)
Solving for the densities, we have from (9):
~k ¢ ~vi
!
(11)
~k ¢ ~ve
! ¡ ~k ¢ ~v0
(12)
ni = n0
while from (10)
ne = n0
The denominator here indicates the doppler-shifted frequency in the electron
rest frame.
Then from Poissonzs equation (8):
Ã
!
~k ¢ ~v e
en0 ~k ¢ ~v i
~
~
ik ¢ E =
¡
"0
!
! ¡ ~k ¢ ~v0
Finally we use the equations of motion (6) and (7):
0
1
~
~
~
~
k ¢ eE
k ¢ eE
C
~ = en0 B
i~k ¢ E
¡
@
³
´2 A
"0
¡i! 2 M
im ! ¡ ~k ¢ ~v0
~ to be non zero in an electrostatic wave, we divide it out
Since we expect ~k ¢ E
to get the dispersion relation:
2
3
6 m
1 = ! 2pe 4
+³
M !2
1.2
1
! ¡ ~k ¢ ~v0
7
´2 5
(13)
Growth rate
Letzs try to ¤nd the growth rate for the 2-stream instability. The dispersion
relation (13) is a fourth order equation for !:
³
´2
³
´2
! 2 ! ¡ ~k ¢ ~v0 = ! 2 !2pe + ! ¡ ~k ¢ ~v0 ! 2pi
If ~k ¢ ~v0 = 0; we get !2 = ! 2pe + !2pi with no imaginary part to !: These are
just plasma oscillations. Thus the interesting case is propagation parallel to
2
~ ¢ ~v 0 = kv 0: Calling the right hand side of equation (13) F (!); we
the stream: k
see that F ¡! 0 as ! ¡! §1; and F ¡! 1 as ! ¡! 0 or kv0 : Thus there
is a minimum at point A where 0 < ! < kv0 : If this minimum is less than 1,
then there will be 4 real roots, implying that the system is stable. But if the
minimum is greater than 1, there will be only 2 real roots, and two complex
roots (a complex conjugate pair). One of these complex roots corresponds to
growth, and one to damping. Thus the condition for instability is:
(14)
F (! A) > 1
8
6
y
4
2
-1
0
1
2
p
w/kv
Plot shows F (!) with kv 0 = ! p = 2 and m=M = 0:1
First we ¤nd the position of the minimum.
0
dF
B m=M
= !2p @¡2 3 ¡ ³
d!
!
Set this equal to zero:
¡2
m=M
¡³
!3
3
4
5
1
2
C
´3 A
~
! ¡ k ¢ ~v 0
³ m ´1=3
³ m ´1=3
!
´ = ¡
=¡
´3 = 0 =) ³
M
M
! ¡ ~k ¢ ~v 0
! ¡ ~k ¢ ~v0
2
where we have taken the real cube root. This gives:
! A = kv0
and
! A ¡ kv0 = kv 0
³ m ´1=3 µ
M
1+
Ã ¡ ¢1=3
m
M
1+
³ m ´1=3 ¶¡ 1
¡ m ¢1=3 ¡ 1
M
3
(15)
M
!
=
1+
kv0
¡ m ¢1=3
M
Then
F (! A)
0
B
= ! 2p @
=
=
³
¡ m ¢1=3 ´2
(m=M ) 1 + M
(kv0 )2 (m=M )2=3
+
³
1
¡ m ¢1=3´2
M
C
A
(kv0 )2
1+
³ m ´1=3 ¶2 µ
³ m ´1=3 ¶
1+
1+
2
M
M
(kv0 )
µ
¶
³ m ´1=3 3
! 2pe
1+
2
M
(kv0 )
!2p
µ
(16)
Thus instability should occur for:
³
´3=2
kv0 · ! pe 1 + (m=M )1=3
(17)
For a given v 0; this relation gives the maximum k (minimum ¸) for instability.
Small enough k (long enough ¸) are always unstable. Equivalently, for a given
v 0 and a given k; it gives a minimum density (in ! pe ):
Another way to look at this is that electron plasma oscillations, Doppler
shifted to the ion frame, resonate with the ion plasma oscillations.
Now rewrite the dispersion relation (13):
! 2pi
!2
Rearrange:
+³
! 2pe
! ¡ ~k ¢ ~v 0
³
´2
! ¡ ~k ¢ ~v 0 = ! 2pe
Ã
´2 = 1
! 2pi
1¡ 2
!
!¡ 1
We expect to ¤nd that the real part of the frequency is roughly comparable to
! pe À ! pi (we can check this later) so we approximate:
Ã
!
³
´
! 2pi
~
! ¡ k ¢ ~v0 = §! pe 1 +
(18)
2! 2
where the terms match up as follows:
~k ¢ ~v0 ' §! pe
and
! 30 ' §
so
!0 =
!pe !2pi
2
!pe ³ m ´2=3
! pe ³ m ´2=3 §i¼ =3
;
and
!
=
e
0
2 1=3 M
2 1=3 M
4
The last two roots form a complex conjugate pair:
p ´
! pe ³ m ´2=3 1 ³
! 0 = 1=3
1§ i 3
2
M
2
and one of these roots represents growth at rate
p
! pe ³ m ´2=3 3
° = 1=3
M
2
2
Check:
µ
¶1=3 µ ¶ 1=3
!0
! pe
M
'
=
' 12 À 1
!pi
!pi
m
so our approximation is valid.
2
Rayleigh Taylor Instability
2.1
Drift Analysis
~ The magConsider a plasma supported against gravity by a magnetic ¤eld B:
2
netic ¤eld acts like a }light ¢uid} of density » ¹0 B supporting the }heavy}
plasma. The system is unstable to overturning, putting the heavy ¢uid on the
bottom.
The gravitational drift (motion notes eqn 7) is
~vDg =
~
m ~g £ B
m (¡g^
y ) £ B z^
m g^
x
=
=¡
2
2
q B
q
B
q B
(19)
So electrons go in the plus x¡direction, and protons go in the ¡x¡direction.
Thus we get a charge separation as shown in the ¤gure, which, in turn, produces
an electric ¤eld in the x¡direction.
5
~ £B
~ drift
Now the E
~vE =
~ £B
~
E
B2
(20)
~ is in the x¡direction, ~v E is
is the same for both protons and electrons. Since E
in the direction of x^ £ z^ = ¡^y : Thus the perturbation continues to grow.
Let the perturbation be of the form
4y = a sin kx
(21)
where the amplitude a is intially small (a ¿ ¸). Since the gravitational drift
speed is proportional to m; we neglect the electrons. The equation of charge
continuity is:
@½q
~ ¢ ~j = 0
+r
(22)
@t
Thus at the boundary, the surface charge density ¾ grows because the current
density brings charge to the surface. Integrate over a pillbox crossing the
surface, as usual:
Z
I
@½q
dV = ¡ n
^ ¢ ~jdA
@t
Thus
@¾
= ¡^
n ¢ ~j
@t
(23)
where ^n is the inward (upward here) pointing normal. The surface has slope
@y
= ka cos kx = tan µ
@x
and thus a vector tangent to the surface has components:
0
1
1
ka cos kx
A
^t = (cos µ; sin µ) = @ q
;q
2
1 + (ka cos kx)
1 + (ka cos kx)2
and thus the normal we want has components:
0
1
¡ka cos kx
1
A
n
^ = @q
;q
2
2
1 + (ka cos kx)
1 + (ka cos kx)
For small amplitude perturbations, ka ¿ 1; we may approximate:
n
^ ¼ (¡ka cos kx; 1)
Then, with ~vDg in the x¡direction,
~j = n0 e~v i = ¡n0 M g x^
B
6
(24)
and equation (23) becomes
@¾
g
g
= ¡n0 M ka cos kx = ¡n0 M ka cos kx
@t
B
B
(25)
So we may write
¾ = ¾ 0 cos kx
where
d¾ 0
g
= ¡n0 M ka
(26)
dt
B
Now recall the dielectric constant for low frequency motions in a magnetized
plasma (Chen equation 3-28, ¢uids notes eqn 1)
µ
¶
¹ ½c2
¹ ½c2
" = 1+ 0 2
" 0 ¼ 0 2 "0 À "0
B
B
We use this with Poissonzs equation:
~ ¢ "E
~ =½
r
Integrating across the boundary, we have:
~ ¢n
~V ¢ n
~ ¢n
"E
^ ¡ "0E
^ ' "E
^=¾
~ V on the vacuum side is unimportant since " À "0 : Thus.
where the ¤eld E
~ ¢ ^n = ¾ 0 cos kx = Ex nx + E y ny ¼ E y
E
"
(27)
~
With our usual assumed form for perturbations, ei(k¢~x¡ !t) ; and uniform density
away from the boundary, the electric potential satis¤es Laplacezs equation:
r2 © = 0 =) k 2 © = 0 =) kx2 + ky2 = 0 =) ky = §ikx
Thus
Ey =
and
¾ 0 cos kx ¡ ky
e
"
(28)
(29)
¾ 0 sin kx ¡ky
e
(30)
"
¡
¢
= ¾"0 k cos kxe ¡ky + (¡k) cos kxe¡ky = 0 as
Ex =
~ ¢E
~ = @E x + @E y
(Check : r
@x
@y
required)
~ £B
~ drift moves the boundary, so:
Now the y¡component of the E
@
Ex
¾ 0 sin kx
da
4y = vEy (y = 0) = ¡
=¡
=
sin kx
@t
B
"B
dt
7
(31)
where we used equation (21) in the last step. Thus, di¤erentiatinbg wrt t and
using equation 26, we get
d2 a
dt2
=
=
1 d¾ 0
1 ³
g ´
1
=¡
¡n0 M ka =
n0M gka
"B dt
"B
B
"B 2
1
1
n0 M gka =
n0 M gka = gka
¹ 0½c 2
2
"0 ¹0 ½c2
2 "0 B
¡
(32)
B
p
This is the exponential equation with solutions e §°t and ° = gk: The positive
root indicates growth of the perturbation.
Note that any drift that has a similar e¤ect to the gravitaional drift (e.g.
curvature drift) would cause a similar instability.
2.2
Normal mode analysis
We can get the same result by several di¤erent methods. Next we consider a
somewhat more powerful method that allows to consider an equilibrium state
that is non-uniform.
Consider a plasma at rest in the lab, supported by a magnetic ¤eld.
We
~ 0 = B0 (y)^z: First we investigate
choose coordinates such that ~g = ¡g^
y and B
the initial equilibrium. From the momentum equation:
³
´
³
´
~ P0 + ¹0 B02 = ½0~g + ¹0 B
~0 ¢ r
~ B
~ 0 = ½0~g
r
(33)
2
~ 0 is not a function of z in this case. We also have Maxwellzs equation
since B
~
~
~ 0 : Finally,
r¢B = 0; which is automatically satis¤ed by our assumed form for B
we choose ~v0 = 0: Since ~g is in the ¡y direction, equation 33 may be written
as:
´
@ ³
¹
P 0 + 0 B02 = ¡½0 g
(34)
@y
2
Next we perturb and linearize. :
Momentum equation:
³
´
³
´
³
´
@~v 1
~ 1 ¡ ¹0 r
~ B
~0 ¢ B
~ 1 + ¹0 B
~0 ¢ r
~ B
~ 1 +¹0 B
~1 ¢ r
~ B
~ 0 + ½1~g (35)
½0
= ¡rP
@t
Maxwell equation:
~ ¢B
~1 = 0
r
(36)
@½1
~ (½ ~v 1 ) = 0
+r
0
@t
(37)
Continuity equation:
Equation of state:
1 @
=) ° P1 + p 0
½0 @t
Ã
d
dt
µ
P
½°
°
¶
@½
¡ °+1 1
@t
½0
8
=0
!
³
´µ P ¶
~
+ ~v1 ¢ r
=0
½°
(38)
Magnetic ¤eld evolution:
´
³
´ ³
´
~1 ³
@B
~ B
~0 + B
~0 r
~ ¢ ~v 1 ¡ B
~0 ¢ r
~ ~v1 = 0
+ ~v1 ¢ r
@t
(39)
¡i!½0 vx = ¡ikx P1 ¡ ikx ¹0 B0 Bz
(40)
Now letzs look at these equations component by component.
Since the
equilibrium state has gradients in the y¡direction, then.we assume all perturbations are of the form s 1 (y)ei(kx¡!t) ; with no z¡dependence.
The momentum equation has three components:
¡i!½0 v y = ¡
@P 1
@
¡ ¹0
(B0 Bz ) ¡ ½1 g
@y
@y
¡i!½0 vz = ¹0 By
@B0
@y
(41)
(42)
~ 1:
Maxwell equation for B
ikxBx +
@By
=0
@y
(43)
Continuity equation:
~ ¢ ~v1 + v y @½0 = 0
¡i!½1 + ½0 r
@y
(44)
Equation of state:
i!
i!°
@
¡ ° P1 + °+1 P0 ½1 + v y
½0
@y
½0
µ
P0
°
½0
¶
=0
(45)
Expand out the last term, and multiply by ½°0 :
¡i!P1 + i!°
P0
@P 0
P0 @½0
½ + vy
¡ °v y
=0
½0 1
@y
½0 @y
(46)
Magnetic ¤eld evolution (3 components)
¡i!Bx : = 0
(47)
¡i!By = 0
´
@B0 ³ ~
¡i!Bz + vy
+ r ¢ ~v 1 B0 = 0
@y
(48)
So Bx = By = 0: Thus the ¤eld lines do not bend.
P 1 ; and Bz : From (44)
µ
¶
1
@½0
~
½1 =
½0 r ¢ ~v 1 + vy
i!
@y
9
(49)
Now we solve for ½1 ;
(50)
From (46) and (50)
µ
¶
µ
¶
P0 1
@½0
1
@P0
P 0 @½0
~
P1 = °
½0 r ¢ ~v1 + v y
+
vy
¡ °vy
½0 i!
@y
i!
@y
½0 @y
µ
¶
1
@P
~ ¢ ~v1 + v y 0
=
°p0 r
i!
@y
From (49)
Bz =
·
´ ¸
1
@B0 ³ ~
vy
+ r ¢ ~v1 B0 = 0
i!
@y
(51)
(52)
Notice that these three expressions have very similar forms. Now we put these
expressions into the momentum equation components and solve for the components of ~v :
µ
´ ¶¶
@p 0
@B0 ³ ~
+ ¹0 B0 vy
+ r ¢ ~v 1 B0
@y
@y
µ
µ
¶¶
¡
¢
i
~ ¢ ~v 1 + vy @p 0 + ¹ B0 @B0
= ¡ 2 kx °p0 + ¹0 B02 r
(53)
0
! ½0
@y
@y
vx
=
1
!½0
µ
kx
i!
¶µ
~ ¢ ~v 1 + v y
°p 0 r
From the intial equilibrium equation (34), we may simplify the factor multiplying v y :
³¡
´
¢
i
~ ¢ ~v1 ¡ ½0 gv y
vx = ¡ 2 kx °p 0 + ¹0 B02 r
(54)
! ½0
y component:
i!½0 vy
vy
µ
¶
·
µ
´ ¶¸
@ 1
@p 0
@ B0
@B0 ³ ~
~
=
°p 0 r ¢ ~v1 + v y
+ ¹0
vy
+ r ¢ ~v1 B0
@y i!
@y
@y i!
@y
µ
¶
1
~ ¢ ~v1 + v y @½0 g
+
½0 r
i!
@y
·
µ
µ
¶¶¸
¢
¡1
@ ¡
@p 0
@B0
2 ~
=
°p 0 + ¹0 B0 r ¢ ~v 1 + vy
+ ¹0 B0
! 2 ½0 @y
@y
@y
µ
¶
1
~ ¢ ~v1 + v y @½0 g
¡ 2
½0 r
! ½0
@y
·
¶ ¸
³
´ µ
¡
¢
¡1
@
@½0
2 ~
~
=
°p 0 + ¹0 B0 r ¢ ~v 1 ¡ ½0 gvy + ½0 r ¢ ~v1 + vy
(55)
g
! 2 ½0 @y
@y
The z¡component is trivial:
(56)
¢
~
Now we use the equation (54) for v x to eliminate the quantity °p0 + ¹0 B20 r¢
vz = 0
~v 1
i
Then vy becomes:
¡
¢
! 2 ½0
~ ¢ ~v1
v x + ½0 gv y = °p 0 + ¹0 B02 r
kx
10
¡
(57)
vy
=
=
¶ µ
¶ ¸
! 2 ½0
@½0
~
i
v x + ½0 gv y ¡ ½0 gv y + ½0 r ¢ ~v 1 + vy
g
kx
@y
· 2
µ
¶ ¸
¡1
! @
~ ¢ ~v1 + v y @½0 g
i
(½0 vx ) + ½0 r
(58)
2
! ½0 kx @y
@y
¡1
! 2 ½0
·
@
@y
µ
At this point we can simplify by considering only incompressible modes:
~ ¢ ~v1 = 0: We couldnzt do this before, because incompressible is equivalent
r
~ ¢ ~v 1 ; which is therefore
to ° ¡! 1; and our equations contained a term ° r
indeterminate.
Now that we have eliminated this term, we are free to make
the incompressible assumption. Note this does not mean the entire plasma is
incompressible, but only that the modes we are investigating are incompressible.
Then:
~ ¢ ~v1 = ikxv x + @vy = 0
r
(59)
@y
and the equation (58) for vy becomes:
vy =
µ
vy 1 +
g @½0
! 2 ½0 @y
¶
¡i @
v y @½0
(½0 vx ) ¡ 2
g
½0 kx @y
! ½0 @y
=
=
¡i
½0 kx
1
½0 k2x
(60)
µ
¶
µ
¶
@
½0 @v y
1 @
@vy
¡
=
½0
@y
ikx @y
½0 kx2 @y
@y
2
@½0 @v y
1 @ vy
+ 2
(61)
@y @y
kx @y 2
Some possible solutions of this equation are:
1. Exponential density variations:
1 @½0
= constant = ®
½0 @y
(62)
The equation for v y becomes:
³
@ 2v y
@vy
g ´
2
+
®
¡
k
v
1
+
® =0
y
x
@y 2
@y
!2
(63)
This di¤erential equation has a solution of the form v y = v1 epy where p is a
solution of the equation:
³
g ´
p2 + ®p ¡ k2x 1 + 2 ® = 0
(64)
!
r³
µ q
¶
´
k2
k2
k2
1
1
®
2
2
x
x
x g
The solution is p = § 2
® + 4kx + 4 ! 2 g® ¡ 2 ® = 2 § 1 + 4 ®2 + 4 ! 2 ® ¡ 1 ;
and we expect p · 0 (the velocity dies away as we go away from the boundary)
we need the minus sign if ® > 0. Then we see that jpj > ®: Thus
g®
p 2 + ®p ¡ k2x
g®k2x
2
=
)
!
=
!2
kx2
p (p + ®) ¡ kx2
11
We have ! 2 < 0; and thus instability, if kx2 > p (p + ®) ; and then with ! = i°;
the growth rate ° is
p
1
° = g® q
p(p+ ®)
1 ¡ k2
x
2
If ! + g® = 0; there is a solution in which vy does not vary with y (p = 0).
Then we have an instability (! = i°) with growth rate
°=
p
g®
: Note we have instability only if the density is increasing upward.
If the
density decreases upward (® is negative) then the plasma is stable.
0
2. No gradients in the initial state: @½
= 0: Instead we have a sharp
@y
boundary across which ½0 goes to zero, and the magnetic ¤eld suddenly increases.
(This is the case that we anlayzed using drifts above.) Then for
y > 0 we have from equation (61):
vy =
1 @ 2 vy
=) vy = v 1 e§k x y
k2x @y2
For y > 0; the minus sign in the exponent is the appropriate choice.
@½
boundary itself, @y0 ¡! 1; and equation (61) becomes:
vy
g @½0
1 @½0 @v y
=
! 2½0 @y
½0 k2x @y @y
or
(65)
At the
(66)
g
1
=
(¡kx vy ) =) ! 2 = ¡gkx
(67)
! 2 ½0
½0 k2x
p
Again we have an instability with growth rate ° = gkx ; the same result we
got from the drift analysis.
Note this is a pure instability (not overstability)
since the frequency has
p
no real part. The growth rate increases as k; so smaller wavelengths grow
faster. At some point this trend is limited by e¤ects we have ignored in this
analysis, such as viscosity, which is more important on small scales. Also, once
the perturbations become non-linear, this growth rate no longer applies.
vy
2.3
Instability of magnetically-con¤ned plasma
~ drift is:
The curvature plus grad-B
~vt otal
d rift
m
=
qB2
µ
2
v?
+ vk2
2
12
¶ ~
Rc
~
£B
R2c
Comparing with the gravitational drift, we have an e¤ective gravitational acceleration of
¶
~c µ
R
1 2
2
~ge ¤ = 2 vk + v?
Rc
2
~ c points out of the plasma, as shown below:
and thus we have an instability if R
Unstable situation
2.3.1
Energy analysis
The stable state is one of minimum energy. The energy of this system is the
magnetic ¤eld energy (u B = B2 =2¹0 ) and internal (thermal) energy. For an
adiabatic equation of state, P / ½° ; the internal energy density is:
uin t =
N
nkT
2
where the number of degrees of freedom N is related to ° by
°
=
N
=
Thus
uin t =
N +2
N
2
° ¡1
nkT
P
=
°¡1
° ¡1
and the internal energy in volume V is
Uint =
PV
° ¡1
In the Rayleigh-Taylor instability driven by gravity, ¢ux tubes are interchanged
without any change in volume. There is no change in magnetic or internal
energy. The gravitational energy is decreased as the plasma falls. The system
is unstable.
Now we look at the magnetically con¤ned plasma without gravity. Seen end
on, the system looks like this:
13
¢utes- end view
The system is low ¯, the plasma pressure is small compared with the magnetic pressure, so the magnetic energy density does not change much as the
¢ux-tube 1 moves to position 2. The magnetic energy in a ¢ux tube is:
Z
Z
B2
B2
EM =
dV =
Ad`
tu b e 2¹0
tu b e 2¹0
where A is the cross-sectional area of the tube and we integrate along its length.
Now the ¢ux © = BA is constant along the length of the tube, so
Z
©2
d`
EM =
(68)
2¹0 tu b e A
Thus the change in energy due to the interchange of the ¢ux tubes is:
Z
Z
Z
Z
©2
d`
©2
d`
©2
d`
©2
d`
¢EM = 1
+ 2
¡ 1
¡ 2
2¹0 2 A
2¹0 1 A
2¹0 1 A
2¹0 2 A
and this change is zero if the two ¢uxes are equal: © 1 = ©2 :
Now we look at the internal energy. As the ¢ux tubes are interchanged, the
pressure will change if the volume changes. The new pressure in region 2 is
µ ¶°
V1
0
P 2 = P1
V2
14
Thus the change in energy is
· µ ¶°
µ ¶°
¸
1
V1
V2
¢E in t =
P1
V2 ¡ P1 V 1 + P 2
V 1 ¡ P 2 V2
° ¡1
V2
V1
i
1 h
=
P1 V 1° V 21¡° ¡ P 1 V1 + P2 V 2° V11¡° ¡ P2 V 2
° ¡1
Now letting P2 = P 1 + ±P; V 2 = V 1 + ±V; and expanding to second order, we
have:
(
)
¡ ¢
°) ±V 2
1h + (1 ¡ °) ±V
+ (1¡ °)(¡
¡ 1+
P 1 V1
V
2
i
V
¡
¢
¡ ¢2
¡
¢¡
¢
¢E in t =
° ¡1
1 + ±P
1 + ° ±V
+ °(°2¡1) ±V
¡ 1 + ±P
1 + ±V
P
V
V
P
V
"
#
¡ ±V ¢2
P 1 V1
(1 ¡ °) ±V
¡ °( 1¡°)
+1
V
2
V
=
¡ ¢
° ¡ 1 +° ± V + °(°¡1) ±V 2 + ± P + ° ±P ±V ¡ 1 ¡ ±P ¡ ±V ¡ ±P ±V
V
2
V
P
PV
P
V
P V
"
µ
¶2 #
P 1 V1
±P ±V
±V
=
(° ¡ 1)
+ ° (° ¡ 1)
° ¡1
PV
V
·
µ
¶¸
±V ±P
±V
= P1 V 1
+°
V
P
V
µ
¶
P
= ±V ±P + ° ±V
V
The ¤rst order terms all disappear, as expected if the initial state is an equilibrium. Any equilibrium is an extremum of the energy, but a stable equilibrium
is a minimum. Thus the energy change must be positive for the system to be
stable. As P ! 0; the ¤rst term dominates. Thus for stability we need
±P > 0 and ±V > 0
or
±P < 0 and ±V < 0
Now with ©1 = © 2 ;
±V
=
=
Z
Z
d`
d`
Ad` = ± BA = ©±
B
B
·Z
Z
¸
d`
d`
©
¡
2 B
1 B
±
Z
Thus if 1 represents the region with the plasma, so that ±P = P 2 ¡ P 1 < 0; we
need ±V < 0 for stability. That is
Z
Z
d`
d`
<
for stability.
B
2
1 B
But since B2 > B1 to con¤ne the plasma, we need
l2 <
B2
l1
B1
15
stable system - side view
Thus the ¤eld lines curve as shown, with the plasma on the "outside". This is
the geometry of a neutron star magnetosphere, but is the opposite of what we
want for a CNF machine, where the plasma is on the "inside". This is a very
dangerous instability, as the growth rate is high and it moves a lot of plasma!
3
Resistive Drift Waves
3.1
Two-¢uid analysis
This is an example of a universal instability. We start o¤ by analyzing drift
waves with no resistivity, then add the resisitivity and look at its e¤ects.
Drift waves are driven by gradients in the intial state. Con¤ned plasmas
always have gradients, so these kinds of instabilities occur in many situations.
As an example, we consider electrostatic drift waves, which have a phase velocity
equal to the diamagnetic drift velocity:
~v D = ¡
~ £B
~
~ £B
~
kB T rn
kB T rn
=
q
nB2
m!c nB
We use the 2-¢uid equations and study small perturbations.
·
¸
³
´
@~v ³ ~ ´
~ + ~v £ B
~ ¡ rp
~
nm
+ ~v ¢ r ~v = nq E
@t
(69)
(70)
@n ~
+ r ¢ (n~v ) = 0
(71)
@t
Now we perturb and linearize. In the perturbed state there is an electric ¤eld:
~ = ¡r©
~
E
Electrons can move freely along the ¤eld lines, and so establish the Boltzmann
distribution:
e©
n = no exp(e©=kB T ) ) n1 = no
(72)
kB T
16
~ = Bo z^;
(We could also get this from the equation of motion.) Now let B
~ be along x
and let rn
^: Also take ~k in the y ¡ z plane, Ti = 0; ~vi;o = 0 and
kB T 1 dn o
~v e;o = ~vD = V D ^
y; with V D = ¡ m!
(eqn.69).
c n dx
Continuity equation (71)
dno
¡i!n1 + i~k ¢ ~v ino + vix
=0
dx
Solve for the ion density n1 :
~ ¢ ~v i
n1
vix dno
k
=
+
no
i!no dx
!
(73)
Ion equation of motion (eqn 70 for the ions)
³
´
~o
¡i!nM ~v i = en ¡i~k© + en~v i £ B
Solve the equation of motion for the components of ~v : (Remember that ~k is in
the y ¡ z plane.)
-c
¡i!M vix = ev iy Bo ) vix = i viy
(74)
!
¡i!M viy = ¡iky e© ¡ ev ixBo
¡i!M v iz = ¡ikz e© ) viz =
e©kz
!M
Substitute vix into equation for v iy:
¡i!M viy = ¡iky e© ¡ ei
-c
ky e©
viy Bo ) viy =
!
!M
µ
¶ ¡1
-2
1 ¡ c2
!
(75)
Then we can ¤nd v ix from eqn (74).
vix = i
-c ky e©
! !M
µ
¶ ¡1
-2
1 ¡ 2c
!
Now we specialize to low frequency waves with ! ¿ - c; so that 1 ¡
Then:
- c ky e© ! 2
e©
vix = ¡i
= ¡iky
! !M - 2c
- cM
and
v iy = ¡
-2
c
!2
-2
¼ ¡ ! c2 :
ky e© ! 2
ky e© !
=¡
2
!M -c
M -c - c
and jv iy j ¿ jvix j : Put all this into eqn. 73 for n1 ; ignoring the term in viy
compared with v ix:
µ
¶
n1
1
e©
dno
kz e©kz
=
¡iky
+
no
i!no
- cM
dx
! !M
µ
¶
2 2
e©
ky V D
v k
=
+ s 2z
kB T
!
!
17
BT
where v 2s = kM
: Now we use the plasma approximation. Set this n1 equal to
the value in eqn. 72:
µ
¶
e©
e©
ky VD
v2k 2
=
+ s 2z ) !2 ¡ !ky VD ¡ v 2s k2z = 0
(76)
kB T
kB T
!
!
which is the dispersion relation for the drift waves. Notice that if ky ! 0 we
get ion acoustic waves, and if kz ! 0 we get ! = ky VD ; waves with phase speed
equal to the diamagnetic drift. These are the drift waves.
3.2
MHD analysis
We can analyze the same system with single ¢uid MHD. We do not have to use
the plasma approximation, as it is "built-in", but now we include a non-zero
current, carried by the electrons, in the initial state.
The single-¢uid MHD equations are:
Momentum equation:
·
¸
@~v ³ ~ ´
~ ¡ rP
~
½
+ ~v ¢ r ~v = ~j £ B
(77)
@t
Ohmzs law:
³
´
~ + ~v £ B
~ = ´~j + 1 ~j £ B
~ ¡ rP
~ e
E
en
Continuity equation:
@½ ~
+ r ¢ (½~v ) = 0
@t
Note that in the initial equilibrium we have:
(78)
(79)
~
~
~j o £ B
~ o = rp
~ ) ~jo = ¡ rp £ Bo = ¡noe~vD
B2o
i.e. the current is due to the electronzs diamagnetic drift.
Now perturb and linearize. Because n0 is not uniform, we assume n1 _ n0;
all other quantities are independent of x and proportional to exp [i (kz z + ky y)] :
For now, we neglect resistivity. Ohmzs law gives the velocity:
~ = ¡ r©;
~
E
~ = ¡i~k©
so E
y¡component of Ohmzs law:
¡iky © ¡ v xB o ¼ 0 ) v x = ¡iky
©
Bo
(80)
where we neglected the higher order terms.
x¡component of Ohmzs law:
vy Bo ¼ 0
18
(81)
and z¡component
¡ikz © =
1
e©
(¡ikz n1 kB Te ) ) n1 = no
eno
kB T e
(82)
which is just the Boltzmann relation. To get v z we use the equation of motion.
The x¡ and y¡components give the current.
x¡component
µ
¶
1 dno
1
1 dno
¡i!½vx = j y Bo ¡n1 kB T e
) jy =
¡i!½vx + n1 kB T e
(83)
no dx
Bo
no dx
y¡component
¡i!½vy = ¡jx Bo ¡¡iky n1 kB T e ) jx =
1
n1
(i!½v y ¡ iky n1 kB T e ) = ¡iky
kB T e
Bo
Bo
(84)
z¡component
¡i!½v z = ¡ikz n1 kB Te ) vz =
kz n1 kB T e
! no M
(85)
The continuity equation gives:
¡i!½1 +
d
(½ vx ) + iky ½o v y + ikz ½o vz = 0
dx o
Substituting for the velocity components (80), (81) and (85), we have:
µ
¶
n1
1 dno
©
kz n1 kB T e
¡i!
+
¡iky
+ ikz
=0
no
no dx
Bo
! no M
(86)
Now substitute in from eqn.82:
¡i!
e©
1 dno e©
¡ iky
kB T e
no dx kB T e
or
µ
kB T e
eBo
¶
+i
k2z e© kB T e
=0
! kB T e M
! 2 ¡ !ky V D ¡ v2s kz2 = 0
as before. (Remember that V D is negative.)
3.3
The instability
Inclusion of resistivity causes the electron density to deviate slightly from the
Boltzmann relation. The z¡component of Ohmzs law becomes:
µ
¶
1
n1 kB T e
¡ikz © = ´jz +
(¡ikz n1 kB T e) ) ¡ikz © 1 ¡
= ´j z
eno
no e©
19
Resistivity also produces a second-order correction to jy and jx ; which we could
get from the equation of motion, but we donzt need those here. . Instead, we
use the equation of charge continuity plus the plasma approximation to get:
~ ¢ ~j = 0 = @j x + iky jy + ikz jz
r
@x
Then using equations 83 and 84
µ
¶
µ
¶
·
µ
¶¸
@
n1
1
1 dno
©
n1 kB Te
¡iky
kB T e +iky
¡i!½v x + n1 kB T e
+ikz ¡ikz
1¡
=0
@x
Bo
Bo
no dx
´
no e©
µ
¶
n1 dno kB Te
M
kB T e 1 dno
n1 kB T e
2©
¡iky
+ !no
ky v x + iky
+ kz
1¡
=0
no dx Bo
Bo
Bo no dx
´
no e©
The terms in ky cancel. Then using equation 80 we have:
µ
¶
µ
¶
M
©
©
n1 kB T e
!no
ky ¡iky
+ kz2
1¡
=0
Bo
Bo
´
no e©
which gives the density:
"
#
µ ¶2
n1
e©
ky
M no
=
1 ¡ i´!
no
kB T e
kz
Bo2
(87)
wthe Boltzmann relation with a correction proportional to ´: As expected. ´
causes the electron density to lag (notice the i indicating a phase shift of ¼ =2):
Now we use the continuity equation (equation 86) again, eliminating n1:
"
#µ
µ ¶2
¶
µ
¶
e©
ky
M no
k2z kB T e
e© 1 dno kB T e
1 ¡ i´ !
¡i!
+
i
¡ik
=0
y
kB T e
kz
Bo2
! M
kB T e no dx
eBo
"
#
µ ¶2
¢
ky
M no ¡ 2
1 ¡ i´!
! ¡ kz2 v2s ¡ ky !V D = 0
(88)
kz
Bo2
which is the new dispersion relation. Check that we get back eqn (13) as ´ ! 0:
We look for a solution of the form ! = !r + i° and expect to ¤nd ° ¿ ! r if
´ is small. Then
! 2 ¼ !2r + 2i!r °
and the dispersion relation becomes:
"
#
µ ¶2
¡ 2
¢
ky
M no
2 2
!r + 2i! r ° ¡ kz vs 1 ¡ i´ (! r + i°)
¡ ky (! r + i°) VD = 0
kz
B 2o
Expanding, and keeping only ¤rst order terms, we have:
"
#
µ ¶2
¡ 2
¢
ky
M no
2 2
! r + 2i! r ° ¡ kz vs 1 ¡ i´! r
¡ ky (! r + i°) VD
kz
B2o
"
#
µ ¶2
¡ 2
¢
ky
M no
2 2
! r ¡ kz vs 1 ¡ i´! r
+ +2i! r ° ¡ ky (! r + i°) VD
kz
B2o
20
=
0
=
0
Real part:
¡
¢
! 2r ¡ k2z v 2s ¡ ky !r V D = 0
which is the same dispersion relation (76) that we obtained with ´ = 0:
Imaginary part:
Solve for ° :
¡
¢
2!r ° ¡ ! 2r ¡ k2z v 2s ´!r
°=
¡
µ
ky
kz
¶2
³ ´2
¢
! 2r ¡ k2z v 2s ´! r kkyz MBn2o
o
2!r ¡ ky VD
=
M no
¡ ky °V D = 0
Bo2
ky V D ´!2r
³
ky
kz
´2
M no
B2
o
2! r ¡ ky V D
where we used the dispersion relation for ! r to simplify the expression in the
numerator. Then if kz ¿ ky ; ! r ¼ ky V D ; the denominator 2! r ¡ ky V D ¼ ky VD ;
and we ¤nd:
µ ¶2
µ ¶2
ky
M no
ky
M no
2
2
° ' ´ !r
= ´ (ky VD )
kz
Bo2
kz
Bo2
We can see that this equals Chenzs expression in equation 6-66 by noting that:
´=
Then:
2
° = (ky VD )
µ
ky
kz
¶2
mº ei
e2 no
mº ei M no
2
= (ky V D )
e 2 no B2o
as given by Chen.
21
µ
ky
kz
¶2
º ei
- c! c
The diagram shows what is going on in this wave. Here ~k is in the y¡direction.
~ = ¡i~k© is ¼=2 out of phase.
Without resistivity, © is in phase with n1 and E
~
E is maximum where n1 is zero, and has the direction, shown. As the wave
~ £B
~ drift serves to move the particles so as to maintain the
propagates, the E
wave. When there is a non-zero resistivity, the potential, and hence the ¤eld,
gets further out of phase and the drift moves particles that are already displaced
left further left, and the amplitude grows.
Here we have seen several di¤erent techniques for analyzing stabilty. Wezll
see some more when we tackle Vlasov theory.
22

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