Chapter 3
Collisions in Plasmas
3.1 Binary collisions between charged particles
Reducedmass for binary collisions:
Two particles interacting with each other have forces
F_{12} force on 1 from 2.
F_{21} force on 2 from 1.
By Newton's 3rd law, F_{12} = − F_{21}.
Equations of motion:
m_{1} 
⋅⋅
r

1

= F_{12} ; m_{2} 
⋅⋅
r

2

= F_{21} 
 (3.1) 
Combine to get

⋅⋅
r

1

− 
⋅⋅
r

2

= F_{12}  ⎛ ⎝

1
m_{1}

+ 
1
m_{2}
 ⎞ ⎠


 (3.2) 
which may be written

m_{1}m_{2}
m_{1} + m_{2}


d^{2}
dt^{2}

(r_{1} − r_{2} ) = F_{12} 
 (3.3) 
If F_{12} depends only on the difference vector r_{1} − r_{2},
then this equation is identical to the equation of a particle
of "Reduced Mass" m_{r} ≡ [(m_{1}m_{2})/(m_{1} + m_{2} )]
moving at position r ≡ r_{1} − r_{2} with respect
to a fixed center of force:
m_{r} 
⋅⋅
r

= F_{12} (r) . 
 (3.4) 
This is the equation we analyse, but actually particle
2 does move. And we need to recognize that when
interpreting mathematics.
If F_{21} and r_{1} − r_{2} are always parallel, then a
general form of the trajectory can be written as an integral.
To save time we specialize immediately to the Coulomb force
F_{12} = 
q_{1} q_{2}
4 πϵ_{0}


r
r^{3}


 (3.5) 
Figure 3.1: Geometry of the collision orbit
Solution of this standard (Newton's) problem:
Angular momentum is conserved:
m_{r} r^{2} 
⋅
θ

= const. = m_{r} bv_{1} (θ clockwise from symmetry) 
 (3.6) 
Substitute u ≡ ^{1}/_{r} then · θ = [(bv_{1})/(r^{2})] = u^{2} bv_{1}
Also
 


d
dt


1
u

= − 
1
u^{2}


du
dθ


⋅
θ

= − bv_{1} 
du
dθ


  (3.7) 
 

− bv_{1} 
d^{2}u
dθ^{2}


⋅
θ

= − ( bv_{1})^{2} u^{2} 
d^{2} u
dθ^{2}


  (3.8) 

Then radial acceleration is

⋅⋅
r

− r 
⋅
θ

2

= − (bv_{1})^{2} u^{2}  ⎛ ⎝

d^{2} u
dθ^{2}

+ u  ⎞ ⎠

= 
F_{12}
m_{r}


 (3.9) 
i.e.

d^{2} u
dθ^{2}

+ u = − 
q_{1} q_{2}
4 πϵ_{0}


1
m_{r} ( bv_{1} )^{2}


 (3.10) 
This orbit equation has the elementary solution
u ≡ 
1
r

= C cosθ − 
q_{1} q_{2}
4 πϵ_{0}


1
m_{r} ( bv_{1} )^{2}


 (3.11) 
The sinθ term is absent by symmetry.
The other constant of integration, C, must be determined
by initial condition. At initial (far distant) angle,
θ_{1}, u_{1} = [1/(∞)] = 0 .
So
0 = C cosθ_{1} − 
q_{1} q_{2}
4 πϵ_{0}


1
m_{r} ( bv_{1} )^{2}


 (3.12) 
There:

⋅
r

1

= − v_{1} = − bv_{1} 
du
dθ

_{1} = + bv_{1} C sinθ_{1} 
 (3.13) 
Hence
tanθ_{1} = 
sinθ_{1}
cosθ_{1}

= 
−1 / Cb
q_{1} q_{2}
4 πϵ_{0}


1
m_{r} ( bv_{1} )^{2}

/ C 

= − 
b
b_{90}


 (3.14) 
where
b_{90} ≡ 
q_{1} q_{2}
4 πϵ_{0}


1
m_{r} v_{1}^{2}

. 
 (3.15) 
Notice that tanθ_{1} = −1 when b = b_{90}. This
is when θ_{1} = −45^{°} and χ = 90^{°}. So
particle emerges at 90^{°} to initial direction when
b = b_{90} "90^{°} impact parameter" 
 (3.16) 
Finally:
C = − 
1
b

cosecθ_{1} = − 
1
b

 ⎛ ⎝

1 + 
b_{90}^{2}
b^{2}
 ⎞ ⎠

^{1}/_{2}


 (3.17) 
3.1.1 Frames of Reference
Key quantity we want is the scattering angle but we need
to be careful about reference frames.
Most "natural" frame of ref is "CenterofMass" frame,
in which C of M is stationary. C of M has position:
R= 
m_{1} r_{1} + m_{2} r_{2}
m_{1} + m_{2}


 (3.18) 
and velocity (in lab frame)
V = 
m_{1}v_{1} + m_{2}v_{2}
m_{1} + m_{2}


 (3.19) 
Now
So motion of either particle in C of M frame is a factor times
difference vector, r.
Velocity in lab frame is obtained by adding V to the C of M
velocity, e.g. [(m_{2} · r)/(m_{1} + m_{2})] + V.
Angles of position vectors and velocity differences
are same in all frames.
Angles (i.e. directions) of velocities are not same.
3.1.2 Scattering Angle
In C of M frame is just the final angle of r.
(θ_{1} is negative)
Figure 3.2: Relation between θ_{1} and χ.
χ = π+ 2 θ_{1} ; θ_{1} = 
χ− π
2

. 
 (3.23) 
tanθ_{1} = tan  ⎛ ⎝

χ
2

− 
π
2
 ⎞ ⎠

= − cot 
χ
2


 (3.24) 
So
But scattering angle (defined as exit velocity angle
relative to initial velocity) in lab frame is different.
Figure 3.3: Collisions viewed in Center of Mass
and Laboratory frame.
Final velocity in CM frame
 

v_{1CM} ( cosχ_{c}, sinχ_{c} ) = 
m_{2}
m_{1} + m_{2}

v_{1} ( cosχ_{c}, sinχ_{c} ) 
  (3.27) 

[ χ_{c} ≡ χ and v_{1} is initial relative velocity].
Final velocity in Lab frame
v′_{L} = v′_{CM} + V=  ⎛ ⎝

V + 
m_{2}v_{1}
m_{1} + m_{2}

cosχ_{c} , 
m_{2}v_{1}
m_{1} + m_{2}

sinχ_{c}  ⎞ ⎠


 (3.28) 
So angle is given by
cotχ_{L} = 
V + 
m_{2} v_{1}
m_{1} + m_{2}

cosχ_{c} 
m_{2}v_{1}
m_{1} + m_{2}

sinχ_{c} 

= 
V
v_{1}


m_{1} + m_{2}
m_{2}

cosecχ_{c} + cotχ_{c} 
 (3.29) 
For the specific case when m_{2} is initially a
stationary target in lab frame, then
 


m_{1}v_{1}
m_{1} + m_{2}

and hence 
  (3.30) 
 


m_{1}
m_{2}

cosecχ_{c} + cotχ_{c} 
  (3.31) 

This is exact.
Small angle approximation (cotχ→ [ 1/(χ)]) , cosecχ→ [ 1/(χ)] gives

1
χ_{L}

= 
m_{1}
m_{2}


1
χ_{c}

+ 
1
χ_{c}

⇔ χ_{L} = 
m_{2}
m_{1} + m_{2}

χ_{c} 
 (3.32) 
So small angles are proportional, with ratio set by the
massratio of particles.
3.2 Differential CrossSection for Scattering by
Angle
By definition the crosssection, σ, for any
specified collision process when a particle is passing
through a density n_{2} of targets is such that the
number of such collisions per unit path length is
n_{2}σ.
Sometimes a continuum of types of collision is considered,
e.g. we consider collisions at different angles (χ)
to be distinct. In that case we usually discuss
differential crosssections (e.g [(dσ)/(dχ)])
defined such that number of collisions in an (angle)
element dχ per unit path length is n_{2} [(dσ)/(dχ)] dχ.
[Note that [(dσ)/(dχ)] is just notation for a number.
Some authors just write σ(χ), but I find that less
clear.]
Normally, for scatteringangle discrimination we discuss the
differential crosssection per unit solid angle:
This is related to scattering angle integrated over all
azimuthal directions of scattering by:
Figure 3.4: Scattering angle and solid angle relationship.
So that since

dσ
dΩ_{s}

dΩ_{s} = 
dσ
dχ

dχ 
 (3.35) 
we have

dσ
dΩ_{s}

= 
1
2 πsinχ


dσ
dχ


 (3.36) 
Now, since χ is a function (only) of the impact
parameter, b, we just have to determine the number of
collisions per unit length at impact parameter b.
Figure 3.5: Annular volume corresponding to db.
Think of the projectile as dragging along an annulus
of radius b and thickness db for an elementary
distance along its path, dl. It thereby drags through
a volume:
Therefore in this distance it has encountered a total
number of targets
at impact parameter b(db). By definition this is equal to
dl[(dσ)/db] db n_{2}. Hence the
differential crosssection for scattering (encounter) at
impact parameter b is
Again by definition, since χ is a function of b

dσ
dχ

dx = 
dσ
db

db ⇒ 
dσ
dχ

= 
dσ
db

 ⎢ ⎢

db
dχ
 ⎢ ⎢

. 
 (3.40) 
[db/dχ is negative but differential crosssections are
positive.]
Substitute and we get

dσ
dΩ_{s}

= 
1
2 πsinχ


dσ
db

 ⎢ ⎢

db
dχ
 ⎢ ⎢

= 
b
sinχ

 ⎢ ⎢

db
dχ
 ⎢ ⎢

. 
 (3.41) 
[This is a general result for classical collisions.]
For Coulomb collisions, in C of M frame,
⇒ 
db
dχ

= b_{90} 
d
dχ

cot 
χ
2

= − 
b_{90}
2

cosec^{2} 
χ
2

. 
 (3.43) 
Hence
 


sinχ


b_{90}
2

cosec^{2} 
χ
2


  (3.44) 
 

  (3.45) 
 

  (3.46) 

This is the Rutherford CrossSection.
for scattering by Coulomb forces through an angle χ measured
in C of M frame.
Notice that [(dσ)/(dΩ_{s})] → ∞
as χ→ 0.
This is because of the longrange nature of the Coulomb force.
Distant collisions tend to dominate. (χ→ 0⇔b → ∞).
3.3 Relaxation Processes
There are 2 (main) different types of collisional relaxation
process we need to discuss for a test particle moving through
a background of scatterers:
 Energy Loss (or equilibrium)
 Momentum Loss (or angular scattering)
The distinction may be illustrated by a large angle (90^{°})
scatter
from a heavy (stationary) target.
If the target is fixed, no energy is transferred to it.
So the energy loss is zero (or small if scatterer is
just `heavy'). However, the momentum in the x direction
is completely `lost' in this 90^{°} scatter.
This shows that the timescales for Energy loss and momentum
loss may be very different.
3.3.1 Energy Loss
For an initially stationary target, the final velocity
in lab frame of the projectile is
v_{L}′ =  ⎛ ⎝

m_{1}v_{1}
m_{1} + m_{2}

+ 
m_{2}v_{1}
m_{1} + m_{2}

cosχ_{c} , 
m_{2}v_{1}
m_{1} + m_{2}

sinχ_{c}  ⎞ ⎠


 (3.48) 
So the final kinetic energy is
 


1
2

m_{1}v_{L}^{′2} = 
1
2

m_{1}v_{1}^{2}  ⎧ ⎨
⎩

 ⎛ ⎝

m_{1}
m_{1} + m_{2}
 ⎞ ⎠

2

+ 
2 m_{1}m_{2}
( m_{1}+m_{2} )^{2}

cosχ_{c} 
  (3.49) 
 

+ 
m_{2}^{2}
( m_{1} + m_{2} )^{2}

( cos^{2} χ_{c} + sin^{2} χ_{c} )  ⎫ ⎬
⎭


  (3.50) 
 


1
2

m_{1}v_{1}^{2}  ⎧ ⎨
⎩

1 + 
2m_{1}m_{2}
( m_{1}+m_{2} )^{2}

( cosχ_{c} − 1)  ⎫ ⎬
⎭


  (3.51) 
 


1
2

m_{1}v_{1}^{2}  ⎧ ⎨
⎩

1 − 
2m_{1}m_{2}
( m_{1}+m_{2} )^{2}

2 sin^{2} 
χ_{c}
2
 ⎫ ⎬
⎭


  (3.52) 

Hence the kinetic energy lost is ∆K = K − K′
 


1
2

m_{1}v_{2} 
4 m_{1}m_{2}
( m_{1}+m_{2} )^{2}

sin^{2} 
χ_{c}
2


  (3.53) 
 


1
2

m_{1}v_{1}^{2} 
4m_{1}m_{2}
( m_{1}+m_{2} )^{2}


1

[using cot 
χ_{c}
2

= 
b
b_{90}

] 
  (3.54) 

(exact).
For small angles χ << 1 i.e. b / b_{90} >> 1 this
energy lost in a single collision is approximately
 ⎛ ⎝

1
2

m_{1}v_{1}^{2}  ⎞ ⎠


4 m_{1}m_{2}
( m_{1}+m_{2} )^{2}

 ⎛ ⎝

b_{90}
b
 ⎞ ⎠

2


 (3.55) 
If what we are asking is: how fast does the projectile lose
energy? Then we need add up the effects of all
collisions in an elemental length dl at all relevant
impact parameters.
The contribution from impact parameter range db at b will equal
the number of targets encountered times ∆K:

n_{2}dl2 πb d b encounters



1
2

m_{1}v_{1}^{2} 
4m_{1}m_{2}
( m_{1}+m_{2} )^{2}

 ⎛ ⎝

b_{90}
b
 ⎞ ⎠

2

Loss per encounter (∆K)


 (3.56) 
This must be integrated over all b to get total energy loss.
dK = n_{2} dlK 
4 m_{1}m_{2}
( m_{1}+m_{2} )^{2}

 ⌠ ⌡

 ⎛ ⎝

b_{90}
b
 ⎞ ⎠

2

2 πb d b 
 (3.57) 
so

dK
dl

= K n_{2} 
m_{1} m_{2}
( m_{1} + m_{2} )^{2}

8 πb_{90}^{2} [ ln b  ]^{max}_{min} 
 (3.58) 
We see there is a problem both limits of the integral
(b → 0, b → ∞) diverge logarithmically.
That is because the formulas we are integrating are approximate.
 We are using smallangle approx for ∆K.
 We are assuming the Coulomb force applies but this is a
plasma so there is screening.
3.3.2 Cutoffs Estimates
1. Smallangle approx breaks down around b = b_{90}. Just truncate
the integral there; ignore contributions from b < b_{90}. Actually
this apparently arbitrary approximation is rigorously justified by an
integration of the exact (not small angle) loss expression.
 ⌠ ⌡


2bdb
(b/b_{90})^{2}+1

= b_{90}^{2}ln[(b/b_{90})^{2}+1] →0 as b→0. 

So if we had not made the smallangle approximation (which is here not
mathematically essential) we would not have had a smallb divergence.
2. Largeb cutoff arises because the Coulomb potential does not apply to
arbitrarily large distances. Debye Shielding says really the potential varies as
so approximate this by cutting off integral at b = λ_{D}
equivalent to
So the cutoffs can be taken as b_{min} = b_{90} and b_{max} = λ_{D} .
 

K n_{2} 
m_{1}m_{2}
(m_{1}+m_{2} )^{2}

8 πb_{90}^{2} ln Λ  
  (3.60) 
 


λ_{D}
b_{90}

=  ⎛ ⎝

ϵ_{0} T_{e}
n e^{2}
 ⎞ ⎠

^{1}/_{2}

 / 
 ⎛ ⎝

q_{1}q_{2}
4 πϵ_{0} m_{r}v_{1}^{2}
 ⎞ ⎠


  (3.61) 

So Coulomb Logarithm is `lnΛ'
Λ = 
λ_{D}
b_{90}

=  ⎛ ⎝

ϵ_{0} T_{e}
n e^{2}
 ⎞ ⎠

^{1}/_{2}

 / 
 ⎛ ⎝

q_{1}q_{2}
4 πϵ_{0} m_{r}v_{1}^{2}
 ⎞ ⎠


 (3.62) 
Because these cutoffs are in ln term result is not
sensitive to their exact values.
One commonly uses Collision Frequency. Energy Loss
Collision Frequency is
ν_{K} ≡ v_{1} 
1
K


dK
dl

= n_{2} v_{1} 
m_{1}m_{2}
( m_{1} + m_{2} )^{2}

8 πb_{90}^{2} ln Λ  
 (3.63) 
Substitute for b_{90} and m_{r} (in b_{90})
 

n_{2} v_{1} 
m_{1}m_{2}
( m_{1}+m_{2} )^{2}

8 π  ⎡ ⎢
⎣

q_{1}q_{2}
4 πϵ_{0} 
m_{1}m_{2}
m_{1}+m_{2}

v_{1}^{2} 
 ⎤ ⎥
⎦

2

lnΛ 
  (3.64) 
 

n_{2} 
q_{1}^{2}q_{2}^{2}
( 4 πϵ_{0} )^{2}


8 π
m_{1}m_{2}v_{1}^{3}

lnΛ 
  (3.65) 

Collision time τ_{K} ≡ 1/ν_{K}
Effective (Energy Loss) Crosssection
[ ^{1}/_{K} [dK/(d l)] = σ_{K}n_{2} ]
σ_{K} = ν_{K} / n_{2}v_{1} = 
q_{1}^{2}q_{2}^{2}
( 4 πϵ_{0} )^{2}


8 π
m_{1}m_{2}v_{1}^{4}

lnΛ. 
 (3.66) 
3.3.3 Momentum Loss
Loss of xmomentum in 1 collision is
 

  (3.67) 
 

m_{1}v_{1}  ⎧ ⎨
⎩

1 −  ⎛ ⎝

m_{1}
m_{1}+m_{2}

+ 
m_{2}
m_{1}+m_{2}

cosχ_{c}  ⎞ ⎠
 ⎫ ⎬
⎭


  (3.68) 
 

p_{x} 
m_{2}
m_{1}+m_{2}

( 1 − cosχ_{c} ) 
  (3.69) 
 

p_{x} 
m_{2}
m_{1}+m_{2}


χ_{c}^{2}
2

= p_{x} 
m_{2}
m_{1}+m_{2}


2 b_{90}^{2}
b^{2}


  (3.70) 

(small angle approx).
Hence rate of momentum loss can be obtained using an integral
identical to the energy loss but with the above parameters:
 

n_{2}p  ⌠ ⌡

b_{max}
b_{min}


m_{2}
m_{1} + m_{2}


2 b_{90}^{2}
b^{2}

2 πb d b 
  (3.71) 
 

n_{2} p 
m_{2}
m_{1} + m_{2}

4π b_{90}^{2} lnΛ 
  (3.72) 

Note for future reference:

dp
dt

= v_{1} 
dp
dl

= n_{2}v_{1}^{2} 
m_{1}m_{2}
m_{1} + m_{2}

4 πb_{90}^{2} lnΛ . 
 (3.73) 
Therefore Momentum Loss.
Collision Frequency
 

v_{1} 
1
p


dp
dl

= n_{2}v_{1} 
m_{2}
m_{1}+m_{2}

4 πb_{90}^{2} lnΛ 
  (3.74) 
 

n_{2}v_{1} 
m_{2}
m_{1}+m_{2}

4 π  ⎡ ⎢
⎣

q_{1}q_{2}
4 πϵ_{0} 
m_{1}m_{2}
m_{1}+m_{2}

v_{1}^{2} 
 ⎤ ⎥
⎦

2

lnΛ 
  (3.75) 
 

n_{2} 
q_{1}^{2}q_{2}^{2}
( 4 πϵ_{0} )^{2}


4 π(m_{1}+m_{2})
m_{2}m_{1}^{2}v_{1}^{3}

lnΛ 
  (3.76) 

Collision Time τ_{p} = 1/ν_{p}
CrossSection (effective) σ = ν_{p}/n_{2}v_{1}
Notice ratio

Energy Loss ν_{K}
Momentum loss ν_{p}

= 
2
m_{1}m_{2}

 / 

m_{1}+m_{2}
m_{2}m_{1}^{2}

= 
2 m_{1}
m_{1} + m_{2}


 (3.77) 
This is
Third case, e.g. electrons → ions shows that mostly
the angle of velocity scatters. Therefore Momentum
`Scattering' time is often called `90^{°} scattering'
time to `diffuse' through 90^{°} in angle.
3.3.4 `Random Walk' in angle
When m_{1} << m_{2}, energy loss << momentum loss.
Hence v_{L}′ ≅ v_{1}. All that matters is
the scattering angle: χ_{L} ≅ χ_{c} ≅ 2 b_{90}/b.
Mean angle of total deviation ∆α in length L is zero because all
directions are equally likely.
But:
Mean square angle is
 

n_{2} L  ⌠ ⌡

b_{max}
b_{min}

χ^{2} 2 πb d b 
  (3.81) 
 

L n_{2} 8 π b_{90}^{2} lnΛ 
  (3.82) 

Spread is `all round' when ―∆α^{2} ≅ 1.
This is roughly when a particle has scattered 90^{°} on
average. It requires
L n_{2} 8 πb_{90}^{2} lnΛ = 1 . 
 (3.83) 
So can think of a kind of `crosssection' for `σ_{90}'
90^{°} scattering as such that
 

1 when L n_{2} 8 π b_{90}^{2} lnΛ = 1 
  (3.84) 
 

πb_{90}^{2} 8 lnΛ ( = 2 σ_{p}) 
  (3.85) 

This is 8 lnΛ larger than crosssection for 90^{°}
scattering in single collision.
Be Careful! `σ_{90}' is not a usual type of
crosssection because the whole process is really diffusive in
angle.
Actually all collision processes due to coulomb force are best
treated (in a Mathematical way) as a diffusion in velocity space
→ FokkerPlanck equation.
3.3.5 Summary of different types of collision
The Energy Loss collision frequency is to do with slowing
down to rest and exchanging energy. It is required for
calculating
Equilibration Times (of Temperatures)
Energy Transfer between species.
The Momentum Loss frequency is to do with loss of
directed velocity. It is required for calculating
Mobility: Conductivity/Resistivity
Viscosity
Particle Diffusion
Energy (Thermal) Diffusion
Usually we distinguish between electrons and ions because
of their very different mass:
Energy Loss [Stationary Targets]
Momentum Loss
 

n_{e} 
e^{4}
( 4 πϵ_{0} )^{2}


8 π
m_{e}^{2} v_{e}^{3}

lnΛ ^{p}ν_{ee} = ^{K}ν_{ee} ×  ⎡ ⎣

m_{e}+m_{e}
2m_{e}

= 1  ⎤ ⎦


 
 

n_{i} 
Z^{2}e^{4}
( 4 πϵ_{0} )^{2}


8 π
m_{e}m_{i}v_{e}^{3}

lnΛ ^{p}ν_{ei} = ^{K}ν_{ei} ×  ⎡ ⎣

m_{e}+m_{i}
2m_{e}

≅ 
m_{i}
2m_{e}
 ⎤ ⎦


 
 

n_{i} 
Z^{2}e^{4}
( 4 πe_{o} )^{2}


8 π
m_{i}^{2}v_{i}^{3}

lnΛ ^{p}ν_{ii} = ^{K}ν_{ii} ×  ⎡ ⎣

m_{i}+m_{i}
2m_{i}

= 1  ⎤ ⎦


  (3.86) 
 

n_{e} 
Z_{e}^{2}e^{4}
( 4 πϵ_{0} )^{2}


8 π
m_{i}m_{e}v_{i}^{3}

lnΛ ^{p}ν_{ie} = ^{K}ν_{ie} = ×  ⎡ ⎣

m_{e}+m_{i}
2mi_{i}

≅ 
1
2
 ⎤ ⎦


 

Sometimes one distinguishes between `transverse diffusion' of
velocity and `momentum loss'. The ratio of these two is


p^{2}∆l

 / 
 ⎢ ⎢

∆p_{}
p ∆l
 ⎢ ⎢





d χ_{L}^{2}
d l

 / 
 ⎢ ⎢

1
p


dp
dl
 ⎢ ⎢


  (3.87) 
 


 ⎛ ⎝

m_{2}
m_{1} + m_{2}

χ_{c}  ⎞ ⎠

2

m_{2}
m_{1}+m_{2}


χ_{c}^{2}
2


= 
2m_{2}
m_{1}+m_{2}

. 
  (3.88) 

So
 


2 m_{2}
m_{1}+m_{2}

= 1 like particles 
  (3.89) 
 

  (3.90) 
 

≅ 
2m_{2}
m_{1}

m_{2} << m_{1} . 
  (3.91) 

Hence
 

^{p}ν_{ee} = ^{K}ν_{ee} ( = `ν_{ee}′!!) 
  (3.92) 
 

2 ^{p}ν_{ei} = ^{K}ν_{ee} 
n_{i}
n_{e}

Z^{2} ( = Zν_{ee} ) ( = `ν_{ei}′) 
  (3.93) 
 

^{p}ν_{ii} = ^{K}ν_{ii} ( = ν_{ii}!! ) (Like Ions) 
  (3.94) 
 


2m_{e}
m_{i}

^{p}ν_{ie} = 
m_{e}
m_{i}

^{K}ν_{ie} = ^{K}ν_{ii} = ν_{ii} 
  (3.95) 

[But note: ions are slowed down by electrons long before being
angle scattered.]
3.4 Thermal Distribution Collisions
So far we have calculated collision frequencies with stationary
targets and singlevelocity projectiles but generally we shall
care about thermal (Maxwellian) distributions (or nearly
thermal) of both species. This is harder to calculate and
we shall resort to some heuristic calculations.
3.4.1 e → i
Very rare for thermal ion velocity to be ∼ electron.
So ignore ion motion.
Average over electron distribution.
Momentum loss to ions from (assumed) drifting
Maxwellian electron distribution:
f_{e} (v) = n_{e}  ⎛ ⎝

m_{e}
2 πT_{e}
 ⎞ ⎠

^{3}/_{2}

exp  ⎡ ⎣

− 
m( v− v_{d} )^{2}
2T
 ⎤ ⎦


 (3.96) 
Each electron in this distribution is losing momentum to the ions
at a rate given by the collision frequency
ν_{p} = n_{i} 
q_{e}^{2}q_{i}^{2}
( 4 πϵ_{0} )^{2}


4 π( m_{e}+m_{i} )
m_{i}m_{e}^{2}v^{3}

lnΛ 
 (3.97) 
so total rate of loss of momentum is given by (per unit volume)
− 
dp
dt

=  ⌠ ⌡

f_{e}(v) ν_{p}(v) m_{e}v d^{3}v

 (3.98) 
To evaluate this integral approximately we adopt the following
simplifications.
 Ignore variations of lnΛ with v and just
replace a typical thermal value in Λ = λ_{D}/b_{90}(v_{1}).
 Suppose that drift velocity v_{d} is small relative
to the typical thermal velocity, written
v_{e} ≡ √{T_{e}/m_{e}} and express f_{e} in terms of
u ≡ [(v)/(v_{e})] to first order in
u_{d} ≡ [(v_{d})/(v_{e})]:
 

n_{e} 
1
( 2 π)^{3/2} v_{e}^{3}

exp  ⎡ ⎣

−1
2

( u − u_{d} )^{2}  ⎤ ⎦


  (3.99) 
 


n_{e}
( 2π)^{3/2} v_{e}^{3}

( 1 + u . u_{d} ) exp  ⎡ ⎣

−u^{2}
2
 ⎤ ⎦

= ( 1 + u_{x} u_{d} ) f_{o} 
  (3.100) 

taking xaxis along u_{d} and denoting by f_{o} the
unshifted Maxwellian.
Then momentum loss rate per unit volume
 

 ⌠ ⌡

f_{e} ν_{p} m_{e}v_{x} d^{3} v 
 
 

ν_{p} (v_{e}) m_{e}  ⌠ ⌡

(1 + u_{x} u_{d}) f_{o} 
v_{e}^{3}
v^{3}

v_{x} d^{3} v 
  (3.101) 
 

ν_{p} (v_{e}) m_{e}v_{d}  ⌠ ⌡


u_{x}^{2}
u^{3}

f_{o} d^{3} v 
 

To evaluate this integral, use the spherical symmetry
of f_{o} to see that:

 ⌠ ⌡


u_{x}^{2}
u^{3}

f_{o} d^{3} v 



1
3

 ⌠ ⌡


u_{x}^{2} + u_{y}^{2} + u_{z}^{3}
u^{3}

f_{o} d^{3} v= 
1
3

 ⌠ ⌡


u^{2}
u^{3}

f_{o} d^{3} v 
 
 


1
3

 ⌠ ⌡

α
0


v_{e}
v

f_{o} 4 πv^{2} d v 
 
 


2 π
3

v_{e}  ⌠ ⌡

α
0

f_{o} 2 v d v 
 
 


2 π
3

v_{e} 
n_{e}
( 2 π)^{3/2} v_{e}^{3}

 ⌠ ⌡

α
0

exp  ⎛ ⎝

−v^{2}
2v_{e}^{2}
 ⎞ ⎠

dv^{2} 
 
 


2 π
3


n_{e}
( 2 π)^{3/2}

2 = 
2
3 ( 2 π)^{1/2}

n_{e} . 
  (3.102) 

Thus the Maxwellaveraged momentumloss frequency is
− 
1
p


dp
dt

≡ 
ν

ei

= 
2
3 ( 2 π)^{1/2}

ν_{p} (v_{e}) 
 (3.103) 
(where p = m_{e}v_{d}n_{e} is the momentum per unit volume
attributable to drift).
 


2
3 ( 2 π)^{1/2}

n_{i} 
q_{e}^{2}q_{i}^{2}
( 4 πϵ_{0} )^{2}


4 π( m_{e}+m_{i} )
m_{i} m_{e}^{2} v_{e}^{3}

lnΛ_{e} 
  (3.104) 
 


2
3 ( 2 π)^{1/2}

n_{i}  ⎛ ⎝

Ze^{2}
4 πϵ_{0}
 ⎞ ⎠

2


4 π
m_{e}^{1/2} T_{e}^{3/2}

lnΛ_{e} 
  (3.105) 

(substituting for thermal electron velocity, v_{e},
and dropping [(m_{e})/(m_{i})] order term), where
Ze = q_{i}.
This is the standard form of electron collision frequency.
3.4.2 i → e
Ion momentum loss to electrons can be treated by a
simple Galilean transformation of the e → i
case because it is still the electron thermal motions
that matter.
Figure 3.6: Ionelectron collisions are
equivalent to electronion collisions in a moving reference frame.
Rate of momentum transfer, [dp/dt], is same in both
cases:
Hence p_{e}ν_{ei} = p_{i} ν_{ie} or

ν

ie

= 
p_{e}
p_{i}


ν

ei

= 
n_{e}m_{e}
n_{i}m_{i}


ν

ei


 (3.107) 
(since drift velocities are the same).
Ion momentum loss to electrons has much lower collision
frequency than e → i because ions possess
so much more momentum for the same velocity.
3.4.3 i → i
Ionion collisions can be treated somewhat like e → i
collisions except that we have to account for moving targets
i.e. their thermal motion.
Consider two different ion species moving relative to each other
with drift velocity v_{d}; the targets' thermal motion
affects the momentum transfer crosssection.
Using our previous expression for momentum transfer, we can
write the average rate of transfer per unit volume as:
[see 3.73 "note for future reference"]
− 
dp
dt

=  ⌠ ⌡

 ⌠ ⌡

v_{r} 
m_{1}m_{2}
m_{1}+m_{2}

v_{r} 4 π b_{90}^{2} lnΛ f_{1}f_{2} d^{2} v_{1} d^{3} v_{2} 
 (3.108) 
where v_{r} is the relative velocity (v_{1}−v_{2}) and
b_{90} is expressed
b_{90} = 
q_{1}q_{2}
4 πϵ_{0}


1
m_{r}v_{r}^{2}


 (3.109) 
and m_{r} is the reduced mass
[(m_{1}m_{2})/(m_{1}+m_{2})].
Since everything in the integral apart from f_{1}f_{2} depends only on
the relative velocity, we proceed by transforming the velocity
coordinates from v_{1},v_{2} to being expressed in terms of relative
(v_{r}) and average (i.e. individual center of mass velocity, V)
v_{r} ≡ v_{1} − v_{2} ; V ≡ 
m_{1}v_{1} + m_{2}v_{2}
m_{1}+m_{2}

. 
 (3.110) 
Take f_{1} and f_{2} to be shifted Maxwellians in the overall
C of M frame:
f_{j} = n_{j}  ⎛ ⎝

m_{j}
2 πT
 ⎞ ⎠

^{3}/_{2}

exp  ⎡ ⎣

− 
m_{j} ( v_{j} − v_{dj} )^{2}
2 T
 ⎤ ⎦

(j = 1, 2) 
 (3.111) 
where m_{1}v_{d1} + m_{2}v_{d2} = 0. Then
 

n_{1}n_{2}  ⎛ ⎝

m_{1}
2πT
 ⎞ ⎠

^{3}/_{2}

 ⎛ ⎝

m_{2}
2 πT
 ⎞ ⎠

^{3}/_{2}

exp  ⎡ ⎣

− 
m_{1}v_{1}^{2}
2 T

− 
m_{2}v_{2}^{2}
2T
 ⎤ ⎦


 
 

×  ⎧ ⎨
⎩

1 + 
v_{1} . m_{1}v_{d1}
T

+ 
v_{2} . m_{2}v_{d2}
T
 ⎫ ⎬
⎭


  (3.112) 

to first order in v_{d}. Convert to local CM vcoordinates and
find (after algebra)
 

n_{1}n_{2}  ⎛ ⎝

M
2πT
 ⎞ ⎠

^{3}/_{2}

 ⎛ ⎝

m_{r}
2πT
 ⎞ ⎠

^{3}/_{2}

exp  ⎡ ⎣

− 
MV^{2}
2T

− 
m_{r}v_{r}^{2}
2T
 ⎤ ⎦


 
 

×  ⎧ ⎨
⎩

1 + 
m_{r}
T

v_{d} . v_{r}  ⎫ ⎬
⎭


  (3.113) 

where M = m_{1}+m_{2}. Note also that (it can be shown) that the
Jacobian of the transformation is unity
d^{3}v_{1} d^{3}v_{2} = d^{3}v_{r}d^{3}V.
Hence
 

 ⌠ ⌡

 ⌠ ⌡

v_{r} m_{r} v_{r} 4 π b_{90}^{2}lnΛ n_{1}n_{2}  ⎛ ⎝

M
2πT
 ⎞ ⎠

^{3}/_{2}

 ⎛ ⎝

m_{r}
2πT
 ⎞ ⎠

^{3}/_{2}


 
 

exp  ⎛ ⎝

− 
MV^{2}
2T
 ⎞ ⎠

exp  ⎛ ⎝

− 
m_{r}v_{r}^{2}
2T
 ⎞ ⎠

 ⎧ ⎨
⎩

1 + 
m_{r}
T

v_{d} . v_{r}  ⎫ ⎬
⎭

d^{3} v_{r} d^{3} V 
  (3.114) 

and since nothing except the exponential depends on V, that
integral can be done:
− 
dp
dt

=  ⌠ ⌡

v_{r} m_{r}v_{r}4πb_{90}^{2} lnΛ n_{1}n_{2}  ⎛ ⎝

m_{r}
2πT
 ⎞ ⎠

^{3}/_{2}

exp  ⎛ ⎝

−m_{r}v_{r}^{2}
2π
 ⎞ ⎠

 ⎧ ⎨
⎩

1 + 
m_{r}
T

v_{d} . v_{r}  ⎫ ⎬
⎭

d^{3} v_{r} 
 (3.115) 
This integral is of just the same type as for e−i collisions,
i.e.
 

v_{d} v_{rt} m_{r} 4 π b_{90}^{2} (v_{rt}) lnΛ_{t} n_{1}n_{2}  ⌠ ⌡


u_{x}^{2}
u^{3}


^
f

o

(v_{r}) d^{3} v_{r} 
 
 

v_{d}v_{rt} m_{r} 4 π b_{90}^{2} (v_{rt}) ln Λ_{t} n_{1}n_{2} 
2
3 ( 2π)^{1/2}


  (3.116) 

where v_{rt} ≡ √{[T/(m_{r})]}, b_{90}^{2} (v_{rt}) is
the ninety degree impact parameter evaluated at velocity v_{tr},
and ∧f_{o} is the
normalized Maxwellian.
− 
dp
dt

= 
2
3 ( 2π)^{1/2}

 ⎛ ⎝

q_{1}q_{2}
4 πϵ_{0}
 ⎞ ⎠

2


4 π
m_{r}^{2}v_{rt}^{3}

lnΛ_{t} n_{1}n_{2} m_{r} v_{d} 
 (3.117) 
This is the general result for momentum exchange rate
between two Maxwellians drifting at small relative
velocity v_{d}.
To get a collision frequency is a matter of deciding which
species is stationary and so what the momentum density of
the moving species is. Suppose we regard 2 as targets then
momentum density is n_{1}m_{1}v_{d} so

ν

12

= 
1
n_{1}m_{1}v_{d}


dp
dt

= 
2
3 ( 2 π)^{1/2}

n_{2}  ⎛ ⎝

q_{1}q_{2}
4 πϵ_{0}
 ⎞ ⎠

2


4 π
m_{r} v_{rt}^{3}


lnΛ_{t}
m_{1}

. 
 (3.118) 
This expression works immediately for electronion collisions
substituting m_{r} ≅ m_{e}, recovering previous.
For equalmass ions m_{r} = [(m_{i}^{2})/(m_{i}+m_{i})] = ^{1}/_{2} m_{i} and
v_{rt} = √{[T/(m_{r})]} = √{[2T/(m_{i})]}.
Substituting, we get

ν

ii

= 
1
3 π^{1/2}

n_{i}  ⎛ ⎝

q_{1}q_{2}
4 πϵ_{0}
 ⎞ ⎠

2


4 π
m_{i}^{1/2} T_{i}^{3/2}

lnΛ 
 (3.119) 
that is, [ 1/(√2 )] times the e−i
expression but with ion parameters substituted.
[Note, however, that we have considered the ion
species to be different.]
3.4.4 e → e
Electronelectron collisions are covered by the same formalism,
so

ν

ee

= 
1
3 π^{1/2}

n_{e}  ⎛ ⎝

e^{2}
4 πϵ_{0}
 ⎞ ⎠

2


4 π
m_{e}^{1/2} T_{e}^{3/2}

lnΛ . 
 (3.120) 
However, the physical case under discussion is not so
obvious; since electrons are indistiguishable how do we
define two different "drifting maxwellian"
electron populations? A more specific discussion
would be needed to make this rigorous.
Generally ν_{ee} ∼ ν_{ei}/ √2 :
electronelectron collision frequency ∼ electronion
(for momentum loss).
3.4.5 Summary of Thermal Collision Frequencies
For momentum loss:
 


√2

n_{i}  ⎛ ⎝

Ze^{2}
4 πϵ_{0}
 ⎞ ⎠

2


4 π
m_{e}^{1/2} T_{e}^{3/2}

lnΛ_{e} . 
  (3.121) 
 


1
√2


ν

ei

. (electron parameters) 
  (3.122) 
 


n_{e}m_{e}
n_{i}m_{i}


ν

ei

. 
  (3.123) 
 


√2

n_{i′}  ⎛ ⎝

q_{i}q_{i′}
4 πϵ_{0}
 ⎞ ⎠

2


4 π
m_{i}^{1/2} T_{i}^{3/2}

 ⎛ ⎝

m_{i′}
m_{i}+m_{i′}
 ⎞ ⎠

^{1}/_{2}

lnΛ_{i} 
  (3.124) 

Energy loss ^{K}ν related to the above (^{p}ν) by
^{K}ν = 
2m_{1}
m_{1}+m_{2}

^{p}ν . 
 (3.125) 
Transverse `diffusion' of momentum ^{⊥} ν,
related to the above by:
^{⊥} ν = 
2m_{2}
m_{1}+m_{2}

^{p}ν . 
 (3.126) 
3.5 Applications of Collision Analysis
3.5.1 Energetic (`Runaway') Electrons
Consider an energetic (^{1}/_{2} m_{e} v_{1}^{2} >> T)
electron travelling through a plasma. It is slowed down
(loses momentum) by collisions with electrons and
ions (Z), with collision frequency:
 

ν_{ee} = n_{e} 
e^{4}
( 4 πϵ_{0})^{2}


8 π
m_{e}^{2} v_{1}^{3}

lnΛ 
  (3.127) 
 


n_{i}
2n_{e}

Z^{2} ν_{ee} = 
Z
2

ν_{ee} 
  (3.128) 

Hence (in the absence of other forces)
 

− ( ^{p}ν_{ee} + ^{p}ν_{ei} ) m_{p} v 
  (3.129) 
 

−  ⎛ ⎝

1 + 
Z
2
 ⎞ ⎠

ν_{ee} m_{e}v 
  (3.130) 

This is equivalent to saying that the electron experiences an
effective `Frictional' force
 


d
dt

(m_{e}v) = −  ⎛ ⎝

1 + 
Z
2
 ⎞ ⎠

ν_{ee} m_{e} v 
  (3.131) 
 

−  ⎛ ⎝

1 + 
Z
2
 ⎞ ⎠

n_{e} 
e^{4}
( 4 πϵ_{0} )^{2}


8 π lnΛ
m_{e}v^{2}


  (3.132) 

Notice
 for Z=1 slowing down is ^{2}/_{3} on
electrons ^{1}/_{3} ions
 F_{f} decreases with v increasing.
Suppose now there is an electric field, E. The electron
experiences an accelerating Force.
Total force
F = 
d
dt

(mv) = −eE + F_{f} = −eE −  ⎛ ⎝

1 + 
Z
2
 ⎞ ⎠

n_{e} 
e^{4}
( 4 πϵ_{0} )^{2}


8 πlnΛ
m_{e}v^{2}


 (3.133) 
Two Cases (When E is accelerating)
  eE  <  F_{f} : Electron Slows Down
  eE  >  F_{f} : Electron Speeds Up!
Once the electron energy exceeds a certain value its velocity
increases continuously and the friction force becomes less
and less effective. The electron is then said to have
become a `runaway'.
Condition:

1
2

m_{e}v^{2} >  ⎛ ⎝

1 + 
Z
2
 ⎞ ⎠

n_{e} 
e^{4}
( 4 πϵ_{0} )^{2}


8 πlnΛ
2 e E


 (3.134) 
3.5.2 Plasma Resistivity (DC)
Consider a bulk distribution of electrons in an electric field.
They tend to be accelerated by E and decelerated by
collisions.
In this case, considering the electrons as a whole, no loss of
total electron momentum by e−e collisions. Hence the friction
force we need is just that due to ―ν_{ei}.
If the electrons have a mean drift velocity v_{d} ( << v_{the})
then

d
dt

(m_{e}v_{d}) = − e E − 
ν

ei

m_{e} v_{d} 
 (3.135) 
Hence in steady state
The current is then
j = − n_{e} e v_{d} = 
n_{e} e^{2} E


 (3.137) 
Now generally, for a conducting medium we define the
conductivity, σ, or resistivity, η, by
j = σE ; ηj = E  ⎛ ⎝

σ = 
1
η
 ⎞ ⎠


 (3.138) 
Therefore, for a plasma,
Substitute the value of ―ν_{ei} and we get
 


n_{i} Z^{2}
n_{e}

· 
e^{2} m_{e}^{1/2} 8 π lnΛ
( 4 πϵ_{0} )^{2} 3  √

2 π

T_{e}^{3/2} 


  (3.140) 
 


Z e^{2} m_{e}^{1/2} 8 π lnΛ
( 4 πϵ_{0} )^{2} 3  √

2 π

T_{e}^{3/2} 

(for a single ion species). 
  (3.141) 

Notice
 Density cancels out because more electrons means (a) more
carriers but (b) more collisions.
 Main dependence is η ∝ T_{e}^{−3/2}.
High electron temperature implies low resistivity
(high conductivity).
 This expression is only approximate because the current
tends to be carried by the more energetic electrons, which have
smaller ν_{ei}; thus if we had done a proper average over
f(v_{e}) we expect a lower numerical value. Detailed
calculations give
η = 5.2 ×10^{−5} 
lnΛ
( T_{e}/eV )^{3/2}

Ωm 
 (3.142) 
for Z = 1 (vs. our expression ≅ 10^{−4}).
This is `Spitzer' resistivity. The detailed calculation value is
roughly a factor of two smaller than our calculation, which is not a
negligible correction!
3.5.3 Diffusion
For motion parallel to a magnetic field if we take a
typical electron, with velocity v_{} ≅ v_{te}
it will travel a distance approximately
before being pitchangle scattered enough to have its velocity
randomised. [This is an orderofmagnitude calculation so we
ignore ―ν_{ee}.] l is the mean free path.
Roughly speaking, any electron does a random walk along the
field with step size l and step frequency ―ν_{ei}.
Thus the diffusion coefficient of this process is
D_{e} ≅ l_{e}^{2} 
ν

ei

≅ 
v_{te}^{2}

. 
 (3.144) 
Similarly for ions
D_{i} ≅ l_{i}^{2} 
ν

ii

≅ 
v_{ti}^{2}


 (3.145) 
Notice

ν

ii

/ 
ν

ei

≅  ⎛ ⎝

m_{e}
m_{i}
 ⎞ ⎠

^{1}/_{2}

≅ 
v_{ti}
v_{te}

(if T_{e} ≅ T_{i}) 
 (3.146) 
Hence l_{e} ≅ l_{i}
Mean free paths for electrons and ions are ∼ same.
The diffusion coefficients are in the ratio

D_{i}
D_{e}

≅  ⎛ ⎝

m_{e}
m_{i}
 ⎞ ⎠

^{1}/_{2}

: Ions diffuse slower in parallel direction. 
 (3.147) 
Diffusion Perpendicular to Mag. Field is different
Figure 3.7: Crossfield diffusion by collisions
causing a jump in the gyrocenter (GC) position.
Roughly speaking, if electron direction is changed by
∼ 90^{°} the Guiding Centre moves by a distance
∼ r_{L}. Hence we may think of this as a random walk
with step size ∼ r_{L} and frequency ―ν_{ei}.
Hence
D_{e⊥} ≅ r_{Le}^{2} 
ν

ei

≅ 
v_{te}^{2}
Ω_{e}^{2}


ν

ei


 (3.148) 
Ion transport is similar but requires a discussion of the effects
of like and unlike collisions.
Particle transport occurs only via unlike collisions.
To show this we consider in more detail the change in guiding
center position at a collision.
Recall
m· v = q v∧B
which leads to
v_{⊥} = 
q
m

r_{L} ∧B (perp. velocity only). 
 (3.149) 
This gives
At a collision the particle position does not change
(instantaneously) but the guiding center position
(r_{0}) does.
r_{0}′+ r_{L}′ = r_{0} + r_{L} ⇒ ∆r_{0} ≡ r_{0}′− r_{0} = − (r_{L}′− r_{L}) 
 (3.151) 
Change in r_{L} is due to the momentum change caused
by the collision:
r_{L}′− r_{L} = 
B
q B^{2}

∧m(v′− v) ≡ 
B
q B^{2}

∧∆(m v) 
 (3.152) 
So
∆ r_{0} = − 
B
q B^{2}

∧∆(m v). 
 (3.153) 
The total momentum conservation means that ∆(mv)
for the two particles colliding is equal and opposite. Hence, from
our equation, for like particles, ∆r_{0} is equal and
opposite. The mean position of guiding centers of two colliding like
particles (r_{01} + r_{02})/2 does not change.
No net cross field particle (guiding center) shift.
Unlike collisions (between particles of different charge q)
do produce net transport of particles of either type. And
indeed may move r_{01} and r_{02} in same direction if they have
opposite charge.
D_{i⊥} ≅ r_{Li}^{2} 
^{p}ν

ie

≅ 
v_{ti}^{2}
Ω_{i}^{2}


^{p}ν

ie


 (3.154) 
Notice that
r_{Li}^{2} / r_{Le}^{2} ≅ m_{i}/m_{e} ; ―^{p}ν_{ie} /―ν_{ei} ≅ [(m_{e})/(m_{i})]
So D_{i⊥}/D_{e⊥} ≅ 1 (for equal temperatures).
Collisional diffusion rates of particles are same for
ions and electrons.
However energy transport is different because it
can occur by likelike collisions.
Thermal Diffusivity:
 

r_{Le}^{2}  ⎛ ⎝

ν

ei

+ 
ν

ee
 ⎞ ⎠

∼ r_{Le}^{2} 
ν

ei

 ⎛ ⎝

ν

ei

∼ 
ν

ee
 ⎞ ⎠


  (3.155) 
 

r_{Li}^{2}  ⎛ ⎝

^{p}ν

ie

+ 
ν

ii
 ⎞ ⎠

≅ r_{Li}^{2} 
ν

ii

 ⎛ ⎝

ν

ii

>> 
ν

ie
 ⎞ ⎠


  (3.156) 
 


r_{Li}^{2}
r_{Le}^{2}



≅ 
m_{i}
m_{e}


m_{e}^{1/2}
m_{i}^{1/2}

=  ⎛ ⎝

m_{i}
m_{e}
 ⎞ ⎠

^{1}/_{2}

(equal T) 
  (3.157) 

Collisional Thermal transport by Ions is
greater than by electrons
[factor ∼ (m_{i}/m_{e})^{1/2} ].
3.5.4 Energy Equilibration
If T_{e} ≠ T_{i} then there is an exchange of
energy between electrons and ions tending to make
T_{e} = T_{i}. As we saw earlier
^{K}ν_{ei} = 
2m_{e}
m_{i}

^{p}ν_{ei} = 
m_{e}
m_{i}

^{⊥} ν_{ei} 
 (3.158) 
So applying this to averages.

^{K}ν

ei

≅ 
2m_{e}
m_{i}


ν

ei

( ≅ 
ν

ie

) 
 (3.159) 
Thermal energy exchange occurs ∼ m_{e}/m_{i}
slower than momentum exchange.
(Allows T_{e} ≠ T_{i}).
So

dT_{e}
dt

= − 
dT_{i}
dt

= − 
^{K}ν

ei

( T_{e} − T_{i} ) 
 (3.160) 
From this one can obtain the heat exchange rate (per unit
volume), H_{ei}, say:
 

− 
d
dt

 ⎛ ⎝

3
2

n_{e}T_{e}  ⎞ ⎠

= 
d
dt

 ⎛ ⎝

3
2

n_{i}T_{i}  ⎞ ⎠


  (3.161) 
 

− 
3
4

n 
d
dt

( T_{e} − T_{i} ) = 
3
2

n 
^{K}ν

ei

( T_{e} − T_{i} ) 
  (3.162) 

Important point:

^{K}ν

ei

≅ 
m_{e}
m_{i}

Z ν_{ee} ≅ 
1
Z^{2}

 ⎛ ⎝

M_{e}
m_{i}
 ⎞ ⎠

^{1}/_{2}


ν

ii

. 
 (3.163) 
`Electrons and Ions equilibrate among themselves much faster
than with each other'.
3.6 Some Orders of Magnitude
 lnΛ is very slowly varying.
Typically has value ∼ 12 to 16 for laboratory plasmas.
 ―ν_{ei} ≈ 6 ×10^{−11} ( n_{i}/m^{3} )/ ( T_{e}/eV )^{3/2} (lnΛ = 15, Z = 1).
e.g. = 2 ×10^{5} s^{−1} (when n = 10^{20}m^{−3} and T_{e} = 1 keV.)
For phenomena which happen much faster than this, i.e.
τ << 1 / ν_{ei} ∼ 5 μs, collisions
can be ignored.
Examples: Electromagnetic Waves with high frequency.
 Resistivity. Because most of the energy of a
current carrying plasma is in the B field not the K.E. of
electrons. Resistive decay of current can be much slower than
―ν_{ei}. E.g. Coaxial Plasma: (Unit length)
Inductance L = μ_{o} ln^{b}/_{a}
Resistance R = η 1 / πa^{2}
L/R decay time
 


μ_{o} πa^{2}
η

ln 
b
a

≅ 
n_{e}e^{2}

μ_{o} πa^{2} ln 
b
a


 
 


n_{e}e^{2}
m_{e}ϵ_{0}


a^{2}
c^{2}


1

= 
ω_{p}^{2} a^{2}
c^{2}

· 
1

>> 
1

. 
  (3.164) 

Comparison
1 keV temperature plasma has ∼ same (conductivity/)
resistivity as a slab of copper ( ∼ 2 ×10^{−8} Ωm).
Ohmic Heating
Because η ∝ T_{e}^{−3 / 2}, if we try
to heat a plasma Ohmically, i.e. by simply passing a
current through it, this works well at low temperatures but
its effectiveness falls off rapidly at high temperature.
Result for most Fusion schemes it looks as if Ohmic
heating does not quite yet get us to the required ignition
temperature. We need auxilliary heating, e.g. Neutral Beams.
(These slow down by collisions.)