NRE6102 Plasma Physics
Spring 2006
By Andrew Seltzman
Note convention:
Within these notes T is taken to be kT where k is the Boltzmann constant and T is in
Kelvin.
Fusion
To fuse light nuclei, the coulomb repulsion must be overcome by initial kinetic energy.
The initial energy is stored in the form of thermal motion of the ions in a plasma. For
fusion the ions must collide at approximately T  107 K .
While fusion relies primarily on collisions, to successfully overcome the coulomb
barrier these collisions must be nearly head on. Since the majority of collisions occur at
small scattering angles, the confinement time must be large in order to provide a higher
probability for a head on collision to occur.
Plasma
A plasma is an ionized combination of electrons and ions that exhibits the following
properties:
 Quasineutrality: the average charge over a large volume is 0, therefore the
electron and ion densities are equal
 Dimensions exceed the debye length: plasma can provide electrostatic screening
of a test charge within it’ volume
 Minimal impurities
 Collective behavior: The plasma acts as a single fluid
Reaction rate
The reaction rate in a plasma is dependant on the ion collision energy and therefore the
temperature. The temperature of a plasma has a Maxwellian distribution.
Maxwellian distribution
The reaction rate is calculated from the cross section.
n1n2  v   dv1dv2 f 2 (v2 ) f1 (v1 ) ( v1  v2 ) v1  v2 where fi (vi ) is the velocity distribution
function of a given species.
Fusion Ignition
In a reactor system we want the energy balance to allow alpha heating to maintain
fusion temperatures within the plasma without external power input.
In a fusion reaction the neutron carries approximately 80% of the reaction energy and
escapes the plasma to be used in power production when it is absorbed externally.
The ignition criterion is given by
1 2
3nT
n  v E f  Prad 
4
E
where E f is the fusion energy (3.5MeV for deuterium)
Prad are the radiative power losses and
 E is the confinement time
Using a best case scenario with no radiative losses, the fusion parameter can be solved
for by simple algebra.
3k
is approximately 3E21 keV( s 1 ) m3
nT E E f 
v
T2
For a tokamak
n  1020 m 3
T  10 KeV
 E  1  10sec
Plasma
Since plasmas have a high particle density, it is impossible to numerically solve for the
interactions between any particle and every other one in the plasma.
Consider a plasma with a non-uniform charge distribution given by the MaxwellBoltzmann distribution.
e
ne  n0e T
for a small exponent the first order approximation is
 e 
ne  n0  1   and the local charge density is   e(n1  ne )  e(n0  ne )
T 

The generated potential is given by Poisson’s equation
e  ne  n0  1
 2 
 2
0
e
where the debye length is
1/ 2
 T 
e   0 e2 
 n0 e 
This is the farthest distance that a test charge within a plasma can interact electrostaticly
with other particle before its field is screened by the neighboring charges.
The electrostatic potential of a point charge is given by
1 e
 (r ) 
4 0 r
so we look for a solution in this form
Within a plasma the potential of a point charge is attenuated exponentially due to the
shielding effect from other charges.
 (r ) 
e
 r 
 
 e 
e
4 0 r
This approximates a pure coulomb potential for r
We now introduce the plasma approximation

Since the electric field farther then e from a particle is attenuated by a factor of e, we
can claim that outside of e , the particles electric field does not interact with other
particles. We further claim that within e , the particle interacts with all other particles
with a 1/ r potential function.
At any given point in the plasma, the equation of motion of a given particle is
determined by the coulomb scattering and the Lorenz force law.
mx  Fscat  e  E  x  B 
For the plasma approximation to be valid, it must satisfy the plasma condition. That is to
say that the number of particles within the debye sphere must be sufficiently dense to
screen out the particles charge, and allow the exponent to be approximates as 1.
4
n  n0  3 1
3
Collective plasma effects
Plasma frequency
Consider a section of plasma where you could move a section of electrons between x1
and x0 to the area located just beyond x1 and then release them.
The excess charge at x1 is given by   n0e( x0  x1 ) generating an electric field by
ne
gausses law for a sheet charge E ( x1 )    0 ( x0  x1 ) which will generate a force on
0
the charge distribution at x1 , attempting to restore equilibrium.
The equation of motion for this system is given by
n0e2
me x1  eE ( x1 )   
( x0  x1 )
0
The solution to this differential equation (second order homogeneous) is
  x0  x1
 n0 e 2 
  0
 me 0 
 
 (t )  Ae
i pe t
 Be
 i pe t
In other words the electrons will accelerate towards the positively charged area, over
shoot and accelerate back causing an oscillation to occur at  pe for electrons and  pi
1/ 2
 e 2 n0 
 pe  

 me 0 
 pi  z
me
1
 pe   pe
mi
50
For fusion plasma  pe is about 1E11 rad/s
Coulomb scattering
Solving the two body motion problem with coulomb forces, we find that as one particle
approaches another it experiences a force due to coulomb repulsion. This will cause the
particles to deflect and scatter off one another at an angle related to the impact
parameter X which specifies the hypothetical distance of closest approach if the particle
had not been deflected. Since debye screening is present, we only solve for interactions
with particles closer then e .
Often we like to work this problem in the center of mass reference frame (right) instead
of the lab reference frame (left).
Converting between lab and COM reference frame.
m1r1  m2 r2  (m1  m2 )rcm
F12 
e1e2 (r1  r2 )
4 0 r1  r2
3
mr  m1r1  m2 r2  r  r1  r2
mrcm 
m1m2
m1m2
r m
m1  m2
m1  m2
(m1  m2 )rcom  m1r1  m2 r2
r  r1  r2
r  r1  r2
F12 
m1
e1e2 (r1  r2 )
4 0 r1  r2
3

e1e2 rˆ
4 0 r1  r2
2
 2 r1
ee
  1 2 2 rˆ   m2 r2
2
t2
4 0 r
r1 
1 e1e2
1 e1e2
rˆ, r2 
rˆ
2
m1 4 0 r
m2 4 0 r 2
 2 r  m1  m2  e1e2 r


t 2  m1m2  4 0 r 3
This equation can be solved as
e12e22
ee
1
 1  2 4 cos(c   )  1 22 2
r
xv
mr v x
x= impact parameter
c = angle in COM reference frame
 =phase
V r
The scattering angle can then be derived from
e1e2
 
tan  c  
2
 2  4 0 mr xv
Solving for the lab frame with
m 
cot( L )   1  cscc  cot c
 m2 
Using trigonometric identities, important approximations of this are
m1 m2   L   c
1
m1  m2   L   c
2
m
m1 m2   L  1 sin  c
m2
Cross sectional scattering area
For the previous scattering angle formula we find that particles in an annular ring
between x and x + dx will scatter into an angle between  L and  L  d L .
For a given particle flux, the amount scattered into the solid angle d   2 sin c dc is
defined as the cross sectional area of the annular ring between x and x + dx.
Scattering cross section is given by d   (c )d    (c )2 sin c dc  2 xdx
Rutherford scattering formula
 e1e2 
d
2 xdx
 ( c ) 


d  2 sin  c d c 
2
2  c
 8 0 mr v sin  2


2



2
Large scattering angles
The probability for a 90 degree or greater scattering event is the probability that a particle
approaches a second within a disk defined by an impact parameter Xmin that is the
largest X that will produce a 90 degree scattering angle.
  c  90    x902
The critical impact parameter can be derived from
e1e2
 
tan  c  
2
 2  4 0 mr xv
such that
 e1e2 
 c  90  
2
  4 0 mr v 2 
2
Small scattering angles
By observing the Rutherford cross section we see that is more probable that a particle
will scatter 90 degrees through a series of small angle collisions.
In the limit of small angle collisions ( c  0 ), x and  c are related by
 c 
2 e1e2
    0 
tan
since
in
the
small
angle
limit
 
2
4 0 mr v 2 x
2
We can use this formula to define the mean square deflection test
 c 
2
 n2 L
 max
        2 sin    d 


 c   n2 L
2
2
c
c
c
c
min
xmax
   
c
xmin
2
2 xdx using the substitution  (c )2 sin c dc  2 xdx
 c 
2
n L ee 
 2  1 22
2   0 mr v 
2 x
max

xmin
dx
plugging in  c
x
2
 c 
2

n2 L  e1e2   xmax 

 ln 
 integrating over X
2   0 mr v 2   xmin 
We now define the bounds of integration within the coulomb logarithm:
The maximum impact parameter that an interaction can take place over is the debye
length since all other charges further away then this are electrostaticly screened out.
1/ 2
 T 
xmax  2   0 e2 
 n2 e2 
The minimum impact parameter is that which will cause a 90 degree scattering to occur.
We define this as the upper limit on small scattering angles.
e1e2
xmin 
4 0 mr v 2
Plugging these values into the solution we say.
2
 c 
2
 ee 
 n2 L  1 2 2  ln( )
  0 mr v 
We now define the coulomb logarithm.
1/ 2

 xmax 
 ( 0T )3  
ln     ln 
  ln 12  4 2   which is about 15-20 for fusion plasmas
 xmin 
 n2e2 e1  

Assuming that the particle velocity is given by mr v 2  3T
By setting  c   1 the 90 degree scattering cross section through small angles can
be defined as.
2
 90 
1
n290
e e 
 1 2
2
ln   
2   0 mr v 2 
2
By comparing the cross sectional probabilities of small and large angle scattering we find
that a particle is about 100 times more likely to scatter through small angle collisions,
therefore we ignore large scattering events in our calculations.
 90
 8ln 
 ( c  90)
Characteristic times
The time required to scatter 90 degrees through small angle collisions can be found by
dividing the characteristic length by the velocity. The characteristic time is a good
indication of how fast particles will loose organized momentum and reach thermal
equilibrium with other particles.
We can use this to determine that an electron beam in a plasma will diverge very quickly,
however an ion beam will retain its focus longer, scattering electrons as it passes.
In COM frame
2 mr  02 (3T )3/ 2
L90
2 02 mr2v3
 90 


v
n2 (e1e2 )2 ln()
n2 (e1e2 )2 ln()
6 02 3 me (T )3/ 2
electron->electron
 
ion->ion
 
electron->ion
 90ei 
ion->electron
 90ie  
ee
90
ii
90
ne 4 ln( )
6 02 3 mi (T )3/ 2
ne4 ln( )
6 3 02 me1/ 2 (T )3/ 2
ni  ze  ei2 ln( )
2
 mi  ei
 90
 me 
about 1E-4s
about 1E-2s
about 1E-4s
about 1s
1/ 2
 
ee
90
ei
90
m 
 m  ie
ii
  e   90
  e  90
 mi 
 mi 
Energy transfer in collisions
The energy transferred between two particles in a collision can be determined from
collision kinematics.
Consider two particles m1 moving at v0 and m2 which is stationary, also M is reduced
mass and particles scatter at CM (particle 1) and CM (particle 2) in the CM frame and 
in the lab frame with final velocities v1 f and v2 f .
 m1  m2  vCM  m1v0
m1
m1m2
m1
M
 v0
 v0
m2
 m1  m2 
 m1  m2  m1m2
Assuming an elastic collision, both KE and P are conserved.
1
1
1
m1v02  mv12f  m2v22 f conservation of KE
2
2
2
vCM  v0
m1v0  m1v1 f cos CM  m2v2 f cos CM conservation of P in the x axis
0  m1v1 f sin CM  m2 v2 f sin CM conservation of P in the y axis
Final velocities of these particles are


 4M 2

4m1m2 2
v12f  v02 1 
cos 2 CM   v02 1 
cos 2 CM 
2
 m  m 

 m1m2

1
2


m1
M
v2 f  2v0
cos CM  2v0
cos CM
m2
 m1  m2 
The energy transfer between the two particles can now be determines as
4m1m2
1
E2  E  m2v22 f 
E0 cos 2 CM
2
2
 m1  m2 
4m1m2

E

sin 2 LAB
2
E0  m1  m2 
2
4m1m2
E
 

sin 2  c 
2
E0  m1  m2 
2
For multiple small angle collisions  c  90 and the equation simplifies to
2m1m2
E

E0  m1  m2 2
The times of energy equalization are given by
 Eee   90ee   90ei
1/ 2
m 
     i   90ei
 me 
m 
 Eei   i  Eei
 me 
ii
E
ii
90
 mi  ei
 90
 me 
 Eei  
Resistively
Previously we have determined that particle collisions scatter organized momentum.
The equation of motion with scattering included is
dve
me ve
e4 ne ln 
me
 eE  ei  eE 
dt
 90
2 02 me ve2
The electric field accelerates electrons and ions, while the scattering converts the
organized drift into thermal energy among the particles. Since the second term decreases
with increasing velocity, electrons which exceed the Dreicer velocity will not be slowed
down and will continue to accelerate (runaway electrons).
1/ 2
veD
 e3ne ln  


2
 2 0 me E 
The current density is now defined as
j  eneve
For the limit where drift velocity is greater then thermal velocity the equation of motion
is solved for resistivity.
me dj
 E  j
e 2 ne dt

me ze2 ln()
12 3 02 (T )3/ 2
resistivity