Diffusion
S.M.Lea
January 07
Now we are going to take a look at some of the consequences of collisions between the
plasma particles. Wezll start by looking at collisions between the charged particles and
neutral particles (neutral atoms) in a partially-ionized plasma, because these collisions are
easier to analyze than those between charged particles.
1 Effect of collisions on the equation of motion
Each collision is described by the collision cross section ¾: Classically, ¾ is just the area
of the particle we are colliding with. Electric and quantum effects provide corrections to the
classical result.
We focus attention on a slab of area A and thickness dx: This slab contains
dN = n0 dV = n0 Adx
target particles, where n0 is the number density of these target particles (neutral atoms, for
now). The total area of the slab blocked to incoming charged particles is then
Abloc ked = ¾dN = ¾n0 Adx
Here we have assumed that dx is so small that target particles do not get in front of each
other, thus reducing the total blocked area. The -ux of outgoing particles is reduced because
of the ones that get blocked. Then if the -ux of incoming particles is F in particles/m 2 ; the
fraction that is blocked is just equal to Ablocked =A; and so
Fout = F in (1 ¡ ¾n0 dx)
and thus
dF = F out ¡ Fin = ¡F in ¾ n0dx
and hence we form the differential equation
dF
= ¡¾n0 F
dx
which is easily integrated if n0 and ¾ are independent of x :
F = F 0 exp (¡¾n0 x)
The quantity
¸=
is the mean free path between collisions.
1
¾n0
1
(1)
(2)
The mean time between collisions is
¿ =
¸
1
=
v
¾ n0v
and the collision frequency is
1
º coll = = n0 ¾v
(3)
¿
(Strictly, to obtain a more accurate result, we should average over the distribution function.)
Now we can amend the -uid momentum equation to account for the momentum removed by
the collisions.
¯
d~p ¯¯
~p
= ¡ = ¡mnº coll ~u
dt ¯coll
¿
Thus the momentum equation (plas-uid notes equation 11) becomes:
·
¸
³
´
d~u
@~u ³ ~ ´
~ + ~u £ B
~ ¡ rP
~ ¡ mnº coll ~u
mn
= mn
+ ~u ¢ r ~u = qn E
(4)
dt
@t
~ = 0 (unmagnetized plasma), and look for a steady
Next letzs specialize to the case B
u
state solution, as we did before when looking at particle drifts. Then @~
= 0; and if the drift
@t ³
´
~ ~u: We are
speed is small (very subsonic) we can also neglect the non-linear term ~u ¢ r
left with:
which we can solve to get
~u =
~ ¡ rP
~ ¡ mnº coll ~u = 0
qnE
~ ¡ rP
~
~
qnE
q ~
kB T rn
=
E¡
mnº coll
mº coll
mº c oll n
The quantity
q
´ §¹
mº coll
is called the mobility, (the § being the sign of the charge q) and
kB T
´D
mº coll
(5)
(6)
is the diffusion coef®cient. Notice that the diffusion coef®cient has dinmensions of L2 =T :
The -ux of particles is then:
~ ¡ D rn
~
F~ = n~u = §¹nE
(7)
When the electric ®eld is zero, the -ux is proportional to the density gradient:
~
F~ = ¡Drn
(8)
a result known as Fickzs Law.
In a plasma it is unusual for the electric ®eld to be exactly zero. Because the electrons
have a higher thermal speed than the ions (at the same temperature), the diffusion coef®cient
is larger for the electrons than for the ions. The electrons diffuse down the density gradient
faster, leading to an imbalance between the electron and ion densities, that in turn creates
electric ®eld. The electric ®eld acts to slow the electrons and speed up the ions until both
diffuse at the same rate. The resulting diffusion is called ambipolar diffusion, because both
charges (ambi means both) diffuse together.
To determine the electric ®eld and the diffusion rate, set the ion -ux equal to the electron
2
-ux:
~ ¡ Dirn
~ = F~e = ¡¹ nE
~ ¡ Dern
~
F~i = ¹i nE
e
~ 6= 0:) We can solve this
(Here we are using the plasma approximation: ni ' ne but E
~ :
relation for E
~
~ = (Di ¡ De) rn
E
(9)
¹i + ¹e n
and then the -ux (of either species) is:
~
(Di ¡ De ) rn
~
F~ = ¹i n
¡ Dirn
¹i + ¹e n
·
¸
¹i D i ¡ ¹i D e ¡ ¹i D i ¡ ¹e D i ~
=
rn
¹i + ¹e
µ
¶
¹i D e + ¹e D i ~
= ¡
rn
(10)
¹i + ¹e
which is Fickzs law with a new diffusion coef®cient:
kB T e
kB Ti
e
+ mºecoll M
¹ i D e + ¹ eD i
º coll
Da =
= M º coll mºecoll
e
¹i + ¹e
+ mº coll
M ºcoll
kB (T e + Ti )
(11)
(m + M ) º coll
The diffusion is dominated by the higher mass ions. Note the symmetry between properties
of electrons and ions in this expressin for Da :
=
2 Solutions to the diffusion equation
Inserting our expression for the particle -ux into the continuity equation, we have:
@n ~ ~
+r¢F = 0
@t
@n
¡ Dr2 n = 0
(12)
@t
We may amend this equation to include a source (or sink) of particles, S:
@n
¡ Dr2 n = S
@t
@
In a steady state, @t
´ 0; and so the equation simpli®es:
¡Dr2 n = S
(13)
(In the absence of a source, there is no steady state solution.)
Now letzs consider what the source might be.
2.1
Ionization
The rate of collisional ionization (production of new ions) is proportional to the density of
3
plasma particles (and also to the density of neutrals)
S = Zn
Then the diffusion equation (13) takes the form
¡Dr2 n
= Zn
Z
r n = ¡ n
D
In a one-dimensional problem, the solutions to this equation are sines and cosines:
Ãr
!
Ãr
!
Z
Z
n = A cos
x + B sin
x
D
D
p
The wavelength is ®xed (¸ = 2¼ D=Z) and so we must ®t the boundary conditions by
adjusting the constants A and B: If n = 0 at x = d then
Ãr
!
Ãr
!
Z
Z
A cos
d + B sin
d =0
D
D
2
determines
A = ¡B tan
and so
(
Ãr
Z
d
D
!
Ãr
!
Ãr
!
Ãr
!)
Z
Z
Z
n = N sin
x ¡ tan
d cos
x
D
D
D
In cylindrical coordinates with azimuthal symmetry, the diffusion equation takes the
form:
µ
¶
1 @
@n
Z
r
+ n=0
r
@r
@r
D
p
Changing variables to w = Z=Dr; we have:
1 0
n +n = 0
w
This is Besselzs equation of order zero (Lea eqn 8.69 and 8.70), with solution
Ãr
!
Z
n = J0
r
D
n00 +
and n ! 0 as r ! 1:
With a source term along the cylinder axis, we have:
S0
r2 n = ¡ ± (r)
D
which has the solution:
n = n0 ln a=r
This solution gives n = 0 at the edge of the cylinder, r = a; and since (see Lea eqn 6.27)
1
r2 ln = ¡2¼± (r)
r
4
then
S0
2¼D
The solution is divergent at r = 0 because we chose the source to be a delta-function.
n0 =
20
15
10
5
0
0.2
0.4
r
0.6
0.8
1
n versus r with a line source along the axis
2.2
Recombination
When an electron and an ion recombine to form a neutral atom, the plasma density
decreases. We have a sink. Since the two particles must collide in order to recombine, the
recombination rate is proportional to n2 :
dn
= ¡®n2
dt
and our diffusion equation takes the form
@n
¡ Dr2 n = ¡®n2
@t
¡
¢
D
In the absence of gradients, r2 n = 0 or when n is large, (n À ®L
2 ; where L is a relevant
length scale) the equation becomes
@n
= ¡®n2
@t
with solution
1
1
=
+ ®t
n
n0
n0
n =
(14)
1 + ®tn0
5
1
0.8
0.6
n
0.4
0.2
0
1
2
t
3
4
5
Recombination: n=n0 versus ®n0 t
This solution will hold until n drops to a low enough value that the diffusion term
becomes important. This happens when
n0
D
»
1 + ®tn0
®L2
or
µ 2
¶
1
®L n0
L2
1
t»
¡1 =
¡
®n0
D
D
®n0
2.3
Time-dependent solutions
To obtain the full solution, we try separation of variables. In one dimension, look for a
solution of the form
n (x; t) = X (x) T (t)
Then equation (12) takes the form
X
dT
d2 X
¡ DT
=0
dt
dx 2
Dividing through by XT ; we get:
1 dT
1 d2 X
¡D
=0
T dt
X dx 2
The ®rst terms depends only on t; and the second only on x; so we require that each term be
separately equal to a constant.
dT
= ¡kT
dt
We chose a negative separation constant because we expect n to decrease with time in the
absence of sources. Thus
T = e¡ kt
6
Then the equation for X is
¡k ¡ D
1 d2 X
X dx 2
d2 X
dx 2
and the solutions are sines and cosines
X = A cos
r
= 0
= ¡
k
x + B sin
D
k
X
D
r
k
x
D
the complete solution is of the form
µ
¶·
¸
t
x
x
n = n0 exp ¡
cos p
+ B sin p
¿
D¿
D¿
where we wrote k = 1=¿ :
p
Writing this in terms of the wavelength ¸; with ¸ = 2¼ D¿ ; we get:
Ã
!·
¸
(2¼)2 Dt
2¼x
2¼x
n = n0 exp ¡
cos
+ B sin
¸
¸
¸2
Here we can see that the shortest wavelengths decay the fastest. Thus the density distribution
will be smoothed in time.
4
3
2
1
-4
-2
0
2 x
4
2
distribution at t=t0 = 0; .25, .5 and 1, where t0 = ¸ 20 = (2¼) D
In a cylindrical system, the equation (12) is:
µ
¶
@n
1 @
@n
¡D
r
=0
@t
r @r
@r
and again we separate, with n = R (r) T (t):
dT
1 d
R
¡ DT
(rR0 ) = 0
dt
r dr
7
and dividing by RT we have
1 dT
D
¡
T dt
R
µ
R00 ¡
R0
r
¶
=0
5
x
The solution for T is the same as before, and then the R equation is
µ
¶
R0
1
D R00 ¡
+ R=0
r
¿
which is again the Bessel equation of order zero. The solution is:
µ
¶
r
R = J0 p
D¿
and
µ
¶
r
n (r; t) = n0 J0 p
e¡ t=¿
D¿
The function J 0 is like a sine function with decreasing amplitude and variable wavelength.
1
0.8
J0(x)
0.4
0.2
-15
-10
-5
0
10
15
-0.2
-0.4
The ®rst zero occurs at argument 2.4048. Thus n will become zero at
r
p
= 2:4048
D¿
This value of r takes the place of the wavelength in the previous solution, and again we see
that with ¿ / r2 ; the shortest length scales decay the fastest.
2.4
General solution
When a plasma is set up, it ®rst decays by recombination, and later diffusion becomes
8
n0
dominant. When recombination dominates, n decreases relatively slowly, n = 1+®tn
0
(equation 14), but once diffusion sets in, the decay is exponential, n » exp (¡t=¿ ) with
¿ » L2 =D:
2
In the plot below, t is in terms of the time constant L2 =D; and we chose ® LD n0 = 2:(Blue
recombination, red diffusion, solid combination.)
1
0.8
0.6
n
0.4
0.2
0
2.5
0.2
0.4
0.6
0.8
1t
1.2
1.4
1.6
1.8
The whole enchilada
Including everything, (ionization, recombination, diffusion, and external sources/sinks) the
equation for evolution of the plasma density would be:
@n
¡ Dr2 n = Q0 n ¡ ®n2 + S
@t
This is a big mess! We almost always approximate by including only the dominant effects.
For example, we can ®nd a uniform equilibrium solution (@=@t ´ 0; r2 ´ 0; S = 0) by
balancing ionization and recombination:
Q0
n=
®
9
2
This is often a good solution in astronomical situations, but is not so good in laboratory
plasmas.
3 Diffusion as a random walk
The diffusion coef®cient has dimensions
L2
T
In fact, for the unmagnetized plasma that we have discussed so far, we found (eg 11)
[D] =
2
kT
kT ¿ 2
(vth ¿ )
¸2
¿ =
=
=
m
m¿
¿
¿
where ¸ is the mean free path and ¿ is the mean time between collisions. Thus
D=
D»
(step) 2
time per step
(15)
Herezs another way to look at this:
A particle has a displacement ~si from collision i ¡ 1 to collision i: Each displacement ~si
is independent of all the others, in magnitude and direction, but the average (rms) length is
< j~si j > rms = ¸
the mean free path. i.e.
N
1 X
~si ¢ ~si
N i=1
Now after N collisions, the total displacement is:
¸2 =
N
X
~s =
(16)
~s i
i=1
and its average value < ~s > = 0: This happens because the direction of ~s is random. The
length of the total displacement (i.e. the total distance travelled), s; is given by:
1
ÃN ! 0N
X
X
s2 = ~s ¢ ~s =
~si ¢ @
~sj A
i=1
=
N X
N
X
i=1 j=1
~si ¢ ~sj
and the mean value is
< s2 > = <
N X
N
X
i= 1 j= 1
Now because the steps are independent,
j= 1
~s i ¢ ~sj >
< ~si ¢ ~s j > = 0 for i 6= j
10
Thus
< s2 > = <
N
X
i= 1
where we used equation (16). Thus
~si ¢ ~s i > = N ¸ 2
p
p
dr ms = < s2 > = N ¸
p
The mean distance travelled is N times the mean length of each step. Of course the
direction travelled is random, so the average displacement < ~s> = 0 as noted above.
We can now compute a diffusion speed of sorts:
p
distance travelled
N¸
¸
¸
=
=p
= p
time
N¿
N¿
¿t
Thus the longer the diffusion proceeds, the slower the speed.
The diffusion coef®cient (15) is
2
D=
(distance travelled)
time
4 Diffusion in magnetized plasmas
When the plasma is magnetized, the -ux of particles along B is the same as in an
unmagnetized plasma. Putting the z¡axis along the magnetic ®eld, we have:
@n
Fz = §¹nE z ¡ D
@z
The particles gyrate around the ®eld lines, and collisions abruptly change the velocity, and
cause abrupt jumps in the position of the guiding center:
This picture shows that the step size in the random walk is now roughly equal to the Larmor
radius, and thus we should expect the diffusion coef®cient for motion across B to be about
D=
11
r2L
¿
Here the ions diffuse faster, because the Larmor radius rL = mv=eB »
greater for the ions than for the electrons.
Now for the equations. The equation of motion is:
³
´
d~v
~ + ~v £ B
~ ¡ kB T rn
~ ¡ mnº~v
mn
= §en E
dt
Once again we look for a steady state solution:
³
´
~ + ~v £ B
~ ¡ kB T rn
~ ¡ mnº~v
0 = §en E
p
kT m=eB is
Now ~v appears in two terms. Letzs do it one component at a time:
@n
mnºv x = §en (E x +v y B) ¡ kB T
@x
D @n
v x = §¹ (E x + vy B) ¡
n @x
and similarly
D @n
v y = §¹ (Ey ¡ vx B) ¡
n @y
Stuff this back into the equation for vx :
µ
·
¸ ¶
D @n
D @n
v x = §¹ Ex + §¹ (E y ¡ v x B) ¡
B0 ¡
n @y
n @x
µ
¶
¡
¢
D @n
@n
v x 1 + ¹2 B2
= §¹Ex + ¹2 E y B ¡
§ ¹B
n @x
@y
But
e
¹B =
B = ! c¿
mº
Also recall the expression for diamagnetic drift (plas-uid eqn 15):
µ
¶
~ £ rn
~
kB T B
kB T 1
@n
@n
~vD =
=
¡ ^
x+ ^
y
qB
Bn
qB n
@y
@x
Thus
D @n
kB T 1 @n
eB kB T 1 @n
¹B
= ! c¿
= !c ¿
n @y
mº n @y
mº eB n @y
(17)
2
= ¡ (! c ¿ ) vDx
Thus equation (17) gives:
=
1
½
Ey
D @n
2
+ (!c ¿ ) v Dx ¡
B
n @x
1 + (! c ¿ )2
2
h³
´
i
(! c ¿ )
D? @n
~ £B
~
= §¹? E x +
E
+ v Dx ¡
2
x
n @x
1 + (! c ¿ )
with a similar expression for vy : Here we wrote
D
D? =
1 + (!c ¿ )2
vx
§¹E x + (! c ¿ )
12
2
¾
(18)
Thus
2
D? ~
(! c ¿ )
r? n +
vE + ~v D ]
(19)
2 [~
n
1 + (!c ¿ )
The usual perpendicular drifts, ~vE and ~vD ; are slowed by collisions if !c ¿ < 1. The
drifts parallel to the electric ®eld and the density gradient are reduced by the magnetic ®eld.
When ! c ¿ ¿ 1 the ®eld has little effect on the plasma con®nement, (diffusion is almost
~ but when ! c ¿ À 1 the magnetic ®eld signi®cantly retards
the same as in the absence of B)
diffusion and aids plasma con®nement.
Now
eB ¸ mfp
¸ mfp
! c¿ =
=
m v
rL
and if !c ¿ À 1; (¸ mfp À rL ) the perpendicular diffusion coef®cient (18) is approximately:
~? ¡
~v ? = §¹? E
D
kB T
kB T
º
2 =
2 =
m! 2c
(! c ¿ )
mº (! c ¿ )
In this form we can see that D? / º : this happens because collisions are necessary to move
~ = 0; D / 1=º : collisions slow the drift. Finally
the guiding center. In contrast, with B
we express D in terms of the Larmor radius, in the strong ®eld limit (! c ¿ À 1).
µ ¶2
rL
1
r2
2
D? ' v th
= L
v th
¿
¿
In this form we can see that the step size in the random walk is indeed the Larmor radius.
D? '
4.1
~
Ambipolar diffusion across B
With a magnetic ®eld, there are many different ways for the plasma to maintain quasi~ can be balanced by currents along B:
~ Here wezll consider one
neutrality. Currents across B
simple example to show some of the features that can arise. A second example is in the
problems.
~ and let the thickness perpendicular
Let the plasma be in®nite in the z¡direction (along B)
~
to B be much greater than the Debye length. Because the plasma is in®nite in z; motions
parallel to the ®eld cannot help to maintain the quasi-neutrality, and we can set the ion and
~ equal. Then we have:
electron -uxes perpendicular to B
~ ? ¡ D? ;e r
~ ? n = +n¹ E
~ ? ¡ D? ;ir
~ ?n
¡n¹ E
~ :
Solve for E
?; e
?;i
~
~ = D?;i ¡ D?; e r ?n
E
¹?;e + ¹? ;i n
13
and then the -ux is
F~
=
~ ?n
D?;i ¡ D?; e r
~ ?n
¡ D? ;ir
¹?;e + ¹? ;i n
¡
¢
(D?;i ¡ D?;e ) ¹? ;i ¡ D? ;i ¹?;e + ¹?;i
~ ?n
r
¹?; e + ¹? ;i
n~v = n¹? ;i
=
=
¡
¹?;i D?;e + D?; i¹?;e
~ ? n = ¡D?;a r
~ ?n
r
¹? ;e + ¹?;i
which is Fickzs law again. The ambipolar diffusion coef®cient for this particular situation is
¹? ;iD?; e + D?;i¹? ;e
D?;a =
¹?;e + ¹?; i
and in the strong ®eld case (! c ¿ À 1);
D?;i ¹?;e
=
=
and
kB T i
e
kB Ti M 2
em2
ºi
ºe =
ºi
ºe
2
2
2
2
M -c m! c
Me B
me2 B 2
¢
kB T i M m
Ti ¡
º iº e =
D?;e ¹?;i
3
4
e B
Te
em2
m
ºe =
ºe
me2 B 2
eB 2
But
r
kB T e
º e = n¾v ' n¾
m
So
r
r
m
kB T e
n¾ p
Te m
¹?;e '
n¾
=
kB T e m =
¹
eB2
m
eB 2
T i M ? ;i
Then the ambipolar diffusion coef®cient becomes
µ
¶
D?;e ¹? ;i
Ti
q
D?; a =
³
´ 1+
m
Te
¹?;i 1 + TTei M
µ
¶
D?;e
Ti
=
q
1+
' 2D?;e
m
Te
1 + TTei M
¹?;e =
where in the last step we assumed T e ' T i: Again the diffusion is dominated by the slower
speciesw this time the electrons.
Another example of ambipolar diffusion in a magnetized plasma is explored in the
problem set.
5 Diffusion in fully ionized plasmas
Collisions between like particles ( electron-electron or ion-ion collisons) give rise to little
diffusion, since the center of mass of the system remains ®xed. The two guiding centers
14
have equal and opposite displacements. (These collisions are important in establishing a
Maxwellian distribution, however. See Spitzerzs book }Physics of Fully ionized gases} for
more details.)
When particles of opposite charge collide, things get more interesting. Both guiding
centers move in the same direction, leading to diffusion. In the diagram below, both particles
reverse directions, and both guiding centers move downward. In a more realistic situation,
the ion guiding center moves little, while the electron guiding center moves a lot. The big
idea is the same.
5.1
Coulomb collisions and resistivity
We are interested in the electron-ion collisions, and in particular the term P ie = ¡Pei , the
momentum transfer term that appears in the equation of motion. We may write this term
(momentum transferred to electrons by ions) P ei as
P ei = mne (~vi ¡ ~ve ) º ei
(20)
We expect the collision frequency º ei to be proprtional to ni and also to the square of the
charge, because the Couolmb force is involved. So we may write
P~ei = e 2 n2 (~vi ¡ ~ve ) ´
(21)
where ´ is the speci®c resistivity. By analyzing the collisons, we hope to obtain an
15
expression for ´ : Comparing equations (20) and (21), we see how º ei and ´ are related:
º ei =
5.2
e2 n
´ = " 0 !2p ´
m
(22)
Coulomb collisions
If an electron collides with an ion, M À m; we may assume that the ion remains ®xed
during the collision. The ion exerts an impulsive force on the electron.
The force acting on the electron is
Zi e2
4¼"0 r 2
and it acts for a time approximately r0 =v where r0 is the impact parameter (see ®gure).
(This approximation gets better the faster the electron is travelling.) Then the impulse
delivered is
Zi e 2 r0
e2
I = ¢ (mv) =
=
Z
i
4¼"0 r20 v
4¼ "0 r0 v
For a 90± collision, ¢ (mv) » mv; and so the impact parameter for a 90± collision is
approximately:
e2
r0 = Zi
4¼"0 mv 2
(Note: you can also get this result by setting the initial KE approximately equal to the
electric PE at closest approach.) Then the collision cross section is
µ
¶2
e2
2
¾ = ¼r0 = ¼ Zi
4¼"0 mv 2
and the collison frequency is
µ
¶2
e2
ni Zi2 e4
º ei = ni ¾v = ni v¼ Zi
= 3
(23)
2
4¼" 0 mv
v 16¼"20 m 2
F =
and with v 2 ' kT =m;
º ei =
ni
3=2
(kT e )
16
Zi2 e4
p
16¼"20 m
(24)
and (equation 22)
ºe i
m
ni
Zi2 e4
p
=
2
2
3=2
"0 ! p
ne e (kT e )
16¼"20 m
p
m
=
Z e2
(25)
3=2 i
2
16¼" 0 (kT e )
where we used the fact that ne = Zi ni:
Although this calculation is very rough, it captures the major features of the exact result.
(See Jackson Chapter 13). In fact, many small angle collisions prove to be more important
than the large angle collisions we considered. To account for these, we multiply by a factor
ln ¤; where
s
r
¸D
"0 kT 8¼"0 kT
("0 kT )3 8¼
¤=
'
=
r0
e2 ne Z ie2
ne
Zi e 3
is the ratio of the Debye length to the impact parameter for a 90 ± collision. (The reason
for this is that since the electric ®elds are shielded at distances greater than ¸ D ; that is an
effective upper limit to the impact parameter. ) Then the resistivity is:
´
=
´=
Zi e2 m1=2
ln ¤
(26)
3=2
16¼"20 (kTe )
Ignoring the weak dependence in ln ¤; the resistivity is independent of the plasma density.
The exact expression for ´ is
p
16 ¼Zi e2 m 1=2
´=
ln ¤
(27)
2
3=2
3 (4¼" 0 ) (2kT e )
which differs from our approximate value by a factor
16
p
= 2:13
3 2¼
At a temperature kT ' 100 eV, equation (27) gives
¢2 ¡
¢1=2 ¡
¢2
p ¡
16 ¼ 1:6 £ 10¡ 19 C
9 £ 10 ¡31 kg
9 £ 10 9 N ¢ m 2 =C2
´ =
ln ¤
3 (2 £ 100 £ 1:6 £ 10 ¡19 J)3=2
p N 2 m4
= 1: £ 10 ¡7 kg 2 3=2 ln ¤
C J
Now since 1 J = 1 N¢m = 1 kg¢m2 =s2 ; the units are:
kg 1=2 ¢ kg1=2 ¢ m3 =s
C2
We expect resistivity to be -¢m.
=
kg ¢ m 3
C2 ¢ s
V
J s
kg ¢ m3
¢m =
¢ m: = 2
A
CC
C ¢s
¡7
and so it checks. The value 10 -¢m is comparable to that of some metals, e.g. steel (see
LB pg 851).
-¢m =
17
5.3
Ohmzs law
The equation of motion for an electron in an unmagnetized plasma, taking collisions into
account, becomes:
@~v e
~ + P~ ei
mn
= ¡enE
@t
Assuming the ions remain immobile, the current density is
~j = ¡ne~v e
In a steady state,
@
@t
´ 0; using equation (21) with ~vi = 0,
~
enE
=
=
P~ei = e2 n2 (¡~v e) ´
´~jen
and thus
~ = ´~j
E
(28)
which is Ohmzs law with the usual de®nition of resistivity.
Since ´ is independent of the plasma density, so is the current. This might seem like
an odd result. The current should increase as ne increases, but the drag increases as ni
increases. Thus higher ni decreases v e: And because of quasi-neutrality, ni = ne :
Compare the result for a weakly ionized plasma, where (equation 7)
~
~j = ¡e¹nE
and the current density is proportional to ne but the drag is proportional to n0 :
Since the resistivity is strongly temperature dependent, ´ / T ¡3=2 ; a cold plasma is
very resistive. As the plasma is heated, its resistance decreases. This is the opposite of the
effect we expect in most metals. This means that we cannot effectively use ohmic heating to
raise the plasma temperature. As the temperature rises, the heating becomes less effective.
Practically, the temperature limit we can achieve by this process is about 1 keV.
Since the collision frequency is proportional to 1=v 3 ; (equation 23), fast electrons make
fewer collisions. Thus the fastest electrons effectively carry the current. This dependence
also leads to an effect called electron runaway. When the electric ®eld is turned on, some
~:
electrons in the tail of the Maxwellian happen to be moving fast in the direction opposite E
These electrons are now accelerated to even higher speeds, and consequently make very few
~ is large enough, there
collisions, and thus continue to accelerate to even higher speeds. If E
are electrons that never make another collision. They form a beam of runaway electrons.
p How large an electric ®eld do we need? We want to accelerate to a speed of order
kT =m in a time less than one collision time. Thus
r
e
kT
E¿ c >
m
m
p
kT m
E >
º ei
pe
kT m
ni
Z i2e 4
p
=
3=2 16¼"2 m
e
(kT e)
0
18
ne Zie3
kT e 16¼"20
Since E crit / 1=T ; and current heats the plasma, (up to a point), electron runaway occurs
relatively easily as E crit decreases to meet the applied E:
E cr it =
19