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Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
Lecture 3 “Introduction to material processing” is covered in a separate document (power point
slides).
Lecture 4 Basic Plasma Equations
In our first lecture, we dealt with some simple properties of fluids. While plasmas have many of
the qualities of fluids, they also have charged particles which are subject to electromagnetic
forces.
In undergraduate electromagnetism, we learned that the E-M forces were governed by a set of
equations known as Maxwell’s Equations. (Here we also include the equation for charge
conservation.)
∇ ∧ E = −∂ t B
∇ ∧ H = J free + ∂ t D
Maxwell’s Equations Point Form:
∇ ∑D = ρ free
∇ ∑B = 0
∇ ∑J = −∂ t ρ
∫
E • dl = −
d
∫∫ B ∑dS
dt Surface
∫
H • dl =
∫∫ J
Enclo sin g
curve
Enclo sin g
curve
Maxwell’s Equations Integral Form:
∑dS +
Surface
∫∫ D ∑dS = ∫∫∫ ρ
Enclo sin g
surface
free
free
d
∫∫ D ∑dS
dt Surface
dτ
Volume
∫∫ B ∑dS = 0
Enclo sin g
surface
d
∫∫ J ∑dS = − dt ∫∫∫ ρdτ
Enclo sin g
surface
Volume
We can readily transfer from the point form of Maxwell’s equations to the integral from using
the Divergence and Stoke’s Theorems.
Divergence Theorem
∫∫∫ ∇ • Vdτ = ∫∫ V • ndσ
Volume
Stoke’s Theorem
enclo sin g
surface
∫∫ (∇ ∧ V) • ndσ = ∫ V • dr
surface
enclo sin g
curve
In addition, we also found Lorentz Force Equation:, F = q(E + v ∧ B) , and Ohm’s Law:,
J = σE .
Page 1
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
To keep things simple, we will at first just focus on plasma-induced fields. Of particular interest
is the first equation in our list of Maxwell’s equations, ∇ ∧ E = −∂ t B (Faraday’s Law). We
know that we can let B = ∇^ A , where A is know as the magnetic vector potential. Thus,
rewriting Faraday’s law, we find
∇ ∧ E = −∂ t ∇ ∧ A
⇓
∇ ∧ (E + ∂ t A) = 0
⇓
E = −∇Φ − ∂ t A
where Φ is the scalar potential. Often in laboratory plasmas there is no time varying magnetic
field and hence
∂t A = 0
E = ∇Φ
NOTE THAT THIS IS NOT TRUE IN FUSION PLAMSAS NOR IS IT TRUE IN SOME ARC
TYPE PLASMAS! IN BOTH CASES THERE EXISTS A STRONG SELF-INDUCED TIMEVARYING MAGNETIC FIELDS.
We can now plug our scalar potential back into Gauss’ Law to get
∇ ∑E = −∇ ∑∇Φ
= −∇ 2 Φ .
ρ
=
ε
This is known as Poisson’s equation. Assume that all of our charged particles are singly charged
positive ions, Ion+, and negatively charged electrons, e-. (Electron-positron, Ion+-Ion-, etc,
plasmas also exist. We are just looking at the most typical.) Then Poisson’s equation becomes
ρ
∇2Φ = −
ε
,
e
= (ne − ni )
ε
where ne is the electron density and ni is the ion density. This tells us that a plasma can induce
an electric field that depends on the local densities of the charge carrier. Because of this, we will
begin to look at the velocity distribution function f (r, v, t ) = f x, y, z, vx , vy , vz , t . The number of
particles that are inside a volume of dxdydzdvx dvy dvz is simply
(
(
)
)
f x, y, z, vx , vy , vz , t dxdydzdvx dvy dvz . Often the six coordinates are considered to be independent.
Then the rate of change of particles at point is
Page 2
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
df ∂f dr ∂f dv ∂f
=
+
+
dt ∂r dt ∂v dt ∂t
∂f
∂t
F
∂f
= v ∑(∇ r f ) + ∑(∇ v f ) +
m
∂t
This is the Boltzmann equation. If particles are not sourced/sunk at the point then the total
number does not change and hence
∂f
df
= 0 = (∇ r f ) ∑v + (∇ v f ) ∑a +
∂t
dt
This is known as the collisionless Boltzmann equation or the Vlasov equation.
= (∇ r f ) ∑v + (∇ v f ) ∑a +
The velocity distribution and macroscopic quantities.
In the very first class, we talked about the velocity distribution of a gas, and an assortment of
different velocity distributions. Further, we related the velocity distribution, f(v), to a probability
distribution, p(y). Now, we want to return to the concept.
Expectation value
Let us assume that we can randomly pick a y from a set, having a probability p(y0) of picking y0.
What value would we expect to get – on average – for y? If I were to draw 100 y’s from a
basket, we would expect p(y0)*100 to be y0. Assume that I can also pick y1, y2, and y3. Then the
average of my 100 picks would be
( y0 p( y0 ) + y1 p( y1 ) + y2 p( y2 ) + y3 p( y3 ))
y =
( p( y0 ) + p( y1 ) + p( y2 ) + p( y3 ))
∑ y p( y )
=
∑ p( y )
i
i
i
i
i
If I were to have a continuous set of y’s then the equation becomes
∫ yp( y)dy
y =
∫ p( y)dy
as we have already used. (Often we normalize the probability such that
∫ p( y)dy = 1.)
This
averaging is important as when we make a measurement, we are typically making a series of
measurements – even if we don’t know it – and our result is the average of our series of
measurements.
f(r,v,t) is our velocity and density distribution function and it is in a very real sense a probability
distribution – with 6 dimensions (r,v) and one free parameter (t)! (Because of this definition, r
and v are truly independent quantities – v does not depend on the location only f. Thus, if we
where to desire to know the average velocity then we look at
∫ vf (r, v, t )dv .
v =
∫ f (r, v, t )dv
Page 3
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
Likewise the average of the v2 is simply
v 2 f (r, v, t )dv
∫
2
v =
.
r
v
v
f
,
,
t
d
(
)
∫
We can do this for all sorts of averages
0
1
2
∫ r f (r, v, t )dr ; ∫ r f (r, v, t )dr ; ∫ r f (r, v, t )dr ;....
∫ f (r, v, t )dr ∫ f (r, v, t )dr ∫ f (r, v, t )dr
∫ v f (r, v, t )dv ; ∫ v f (r, v, t )dv ; ∫ v f (r, v, t )dv ;....
∫ f (r, v, t )dv ∫ f (r, v, t )dv ∫ f (r, v, t )dv
0
2
1
However, we are not interested in most of them. In fact we typically only look at
n(r, t ) = ∫ v 0 f (r, v, t )dv
density
Γ(r, t ) = n v = ∫ v1 f (r, v, t )dv
particle flux
Etotal = 12 mn v 2 = 12 m ∫ v 2 f (r, v, t )dv
Erandom = 23 P = 23 nkT = 12 mn (v − v
= m ∫ (v − v
1
2
)
) f (r, v, t )dv
2
Edirected = Etotal − Erandom = 12 mn v
2
total kinetic energy
2
random kinetic energy / pressure ( P)
directed kinetic energy
(These are known as the zeroth, first, and second moments of the distribution function.)
Why are we bringing this up again? Well we can do the same thing to Boltzmann’s Equation.
Zeroth moment of the Boltzmann Equation – The Equation of Continuity (Particle conservation)
 df 
∫  dt  dv = Gain − Loss = f
c
F
∂f 

= ∫  v ∑(∇ r f ) + ∑(∇ v f ) +  dv

m
∂t 
F

 ∂f  
=  ∫ (v ∑(∇ r f ))dv + ∫  ∑(∇ v f ) dv + ∫   dv
m

 ∂t  

F


 ∂
=  ∇ r ∑∫ (vf )dv + ∫  ∑(∇ v f ) dv +   ∫ fdv


 ∂t 


m
F
∂n 

=  ∇ r ∑n v + ∫  ∑(∇ v f ) dv + 
m


∂t 
Now what happens to the central term on the right hand side? First let us assume that force, F, is
independent of the velocity. (Strictly speaking, this is not true as F = e(v ∧ B) .) Then the
integral becomes
Page 4
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
F
F

∫  m ∑(∇ f ) dv = m ∑∫ ((∇ f ))dv
v
v
.
∞
F
F
= ( f ) −∞ = (0 − 0) = 0
m
m
The distribution goes to zero for physical reasons – we don’t want an infinite density. If the
force is dependent on velocity, it is typically just to first order as in the magnetic force. Then
q
F

∫  m ∑(∇v f ) dv = m ∫ ((v ∧ B) ∑∇v f )dv
To solve this, we must deal with some vector identities. First,
A ∧ (B ∧ C) = B(A ∑C) − C(A ∑B) and
.
A ∑(B ∧ C) = (A ∧ B) ∑C = −(B ∧ A) ∑C....
Thus,
(v ∧ B) ∑∇ v f = (v ∧ B) ∑∇ v [ f ]
= ∇ v [ f ] ∑(v ∧ B)
= (∇ v [ f ] ∧ v) ∑B
= (∇ v ∧ [ fv]) ∑B − ( f∇ v ∧ [v]) ∑B
= ∇ v ∑([ fv] ∧ B) − f∇ v ∑(v ∧ B)
First we will look at the second term on the right.
∇ v ∑(v ∧ B) = ∂Vx Vy Bz − Vz By + ∂Vy (Vz Bx − Vx Bz ) + ∂Vz Vx By − Vy Bx
(
)
(
)
=0
Now we can go back to the integral and we find
−q
F

∫  m ∑(∇v f ) dv = m ∫ ((v ∧ B) ∑∇v f )dv
−q
=
∇ v ∑([ fv] ∧ B) dv
m ∫
−q
=
∫ ([ fv] ∧ B) dv
m surface
(
)
(
)
at ∞
=0
Here fv goes to zero on the surface for physical reasons – we don’t want an infinite energy so f
must go to zero faster than v.
This leaves the continuity equation
∂n 

f c =  ∇ r ∑(n v ) + 

∂t 
First moment of the Boltzmann equation – Momentum Conservation
Now we multiple Boltzmann’s equation
F
∂f
df
= v ∑(∇ r f ) + ∑(∇ v f ) +
∂t
dt
m
Page 5
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
by mv and integrate over velocity to get
df
F
∂f 

m ∫ v  dv = ∆Momentum = m ∫ v v ∑(∇ r f ) + ∑(∇ v f ) +  dv
 dt 

m
∂t 
F

 ∂f  
= m ∫ v(v ∑(∇ r f ))dv + ∫ v ∑(∇ v f ) dv + ∫ v  dv
m

 ∂t  

First we will look at the third part of the right-hand side
∂
∂v
 ∂f 
∫ v ∂t  dv = ∂t ∫ v( f )dv − ∫ ∂t ( f )dv
.
[
]
∂
= [ ∫ v( f )dv] − ∫ a( f )dv
∂t
1
424
3
Which is zero
from before
[
]
∂
v( f )dv
∂t ∫
∂
= [n v ]
∂t
Now for the second term
q
F

∫ v m ∑(∇v f ) dv = m ∫ v (E + v ∧ B) ∑(∇v f ) dv u sin g the chain rule
q
= ∫ ∇v
∑
[ fv(E + v ∧ B3)] dv
{
m
This operates 144244
=
[
]
on the force Notice that this
notation is correct
− This is a tensor ( matrix )
q
q
fv∇ v ∑[(E + v ∧ B)]dv − ∫ [ f (E + v ∧ B)] ∑∇ v vdv
∫
m
m
= 0 from before
644474448
q
=
∫ fv(E + v ∧ B)dv
m surface
−
= 0 from before
I = identity tensor
644
7444
8
}
q
q
− ∫ fv∇ v ∑[(E + v ∧ B)]dv − ∫ [ f (E + v ∧ B)] ∑ ∇ v v dv
m
m
q
= − ∫ [ f (E + v ∧ B)] ∑Idv
m
q
= − ∫ [ f (E + v ∧ B)]dv
m
q
= − n(E + v ∧ B)
m
Finally, we can deal with the first term…
(HOLD ON THIS GETS EASIER!)
Page 6
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
∫ v(v ∑(∇ f ))dv = ∇ ∑∫ fvvdv
r
r
= ∇ r ∑(n vv )
What precisely is this last term? Well let us consider what v is…
v = v directed + v random
Thus
∇ r ∑[n vv ] = ∇ r ∑ n (v dir + v rnd )(v dir + v rnd )
= ∇r
[
∑[n (v
]
dir
v dir + v dir v rnd + v rnd v dir + v rnd v rnd )
]
but
v dir = v dir so




= ∇ r ∑(nv dir v dir ) + ∇ r ∑ 2 nv dir v rnd  + ∇ r ∑(n v rnd v rnd
123

= 0 − it is

random! 
)
The last term we have dealt with before. We know for a Maxwellian that
P
E = 12 m v 2 = 23 kT = 23
but for our calculations
n
v = v rnd !
Thus we find
P = mn v rnd v rnd
(Notice that P is a tensor – as might make sense – Pressure is direction dependent)
Now going back to our momentum conservation equation we find
F

 ∂f  
∆Momentum ≡ ∆M c = m ∫ v(v ∑(∇ r f ))dv + ∫ v ∑(∇ v f ) dv + ∫ v  dv
m

 ∂t  

P
q
∂


= m ∇ r ∑(n v v ) + ∇ r ∑ + − n(E + v ∧ B) + [n v ]


m
m
∂t
∂


=  m v ∇ r ∑(n v ) + mn v ∑∇ r ( v ) + m∇ r ∑P − qn(E + v ∧ B) + m [n v ]


∂t
using the continuity equation
∂n 

f c =  ∇ r ∑(n v ) + 

∂t 
and rearranging…
Page 7
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
∂


∆M c =  m v ∇ r ∑(n v ) + mn v ∑∇ r ( v ) + ∇ r ∑P − qn(E + v ∧ B) + m [n v ]


∂t
∂n 



m v  f c −  + mn v ∑∇ r ( v ) + ∇ r ∑P



∂t 
=

 − qn E + v ∧ B + mn ∂ v + m v ∂n 
(
)
[ ]

∂t
∂t 
⇓
∂ v

+ v ∑∇ r v  = ∆M c −
mn
 ∂t
 123
momentum
lost via
collisions
mv fc
12
4 4
3
− ∇ r ∑P + qn(E + v ∧ B)
momentum change
via particle gain / loss
This is slightly different then Lieberman but it is correct.
This is also known as the fluid equation of motion
Energy Conservation (Heat flow equation)
Now we multiply Boltzmann’s equation
df
F
∂f
= v ∑(∇ r f ) + ∑(∇ v f ) +
dt
m
∂t
by 12 mv 2 and integrate over velocity to get
F

 ∂f  
2  df 
1
dv = ∆Energy = 12 m ∫ v 2 (v ∑(∇ r f ))dv + ∫ v 2  ∑(∇ v f ) dv + ∫ v 2   dv .
2 m∫ v
m

 dt 
 ∂t  

This is trivial and is left to the reader (yeah right!)
Rather than go through this derivation, we will accept Lieberman’s formula. (We don’t have
time to do this in class now anyway and the first two formula are more important.)
SIDE NOTE ON VECTORS
We know that in matrix notation the inner product is
 Ax   Bx 
A ∑B =  Ay  ∑ By 
   
 Az   Bz 
⇒
6447
44
8  Bx  

= Ax Ay Az  By  
 
 Bz  
= Ax Bx + Ay By + Az Bz
(
⇓
)
What we have above is the form
Page 8
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
⇒
}
⇓
multiply squential
 Ax   Bx  





AB = Ay By  row elements of A with
  
 Az   Bz   column elements of B
 Ax Bx
=  Ax By

 Ax Bz
Ay Bx
Ay By
Ay Bz
Az Bx 
Az By 

Az Bz 
Equilibrium Properties
Boltzmann’s Relation
Boltzmann’s relation makes use of the momentum conservation equation
∂ v

+ v ∑∇ r v  = ∆M c − m v f c − ∇ r ∑P + qn(E + v ∧ B) .
mn
 ∂t

Under many conditions, most of these terms are small, particularly for the electron. This is
particularly true for the electrons. If we assume that the electron is massless, me ≈ 0 , then our
equation reduces to
0 = −∇ r ∑P + qn(E + v ∧ B).
If we further assume that the electron is either traveling along the B field or that B=0 then
∇ r ∑P = qnE
but
Page 9
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
 Px
∇ r ∑P = ∇ ∑ 0

0
0
0

Pz 
0
Py
0
∂ x Px + ∂ y Py + ∂ z Pz = ∂ r ∑P = ∂ r (nkTe )
= kTe ∂ r (n)
= qnE = qn( −∇Φ)
⇓
∂rΦ =
kTe 1
∂ r (n)
e n
⇓
Φ=
kTe
ln(n) + const
123
e
kT
≡
e
ln ( n 0 )
⇓
 eΦ 
n = n0 exp

 kTe 
Boltzmann Relation
 eΦ 
n = n0 exp

 kTe 
Debye Length
We can now calculate the Debye length – an effective length over which a plasma will shield an
electric field. (The length is the 1/e distance for reducing a potential.)
First, we have Poisson’s equation
ρ
∇2Φ = −
ε
e
= (ne − ni )
ε
We make the further assumption that the density of the electrons in absence of the potential is the
same as the ion density. (We make this assumption because if either the electrons or the ions
were to leave an area, a significant electric field would be setup to try to pull them back
together.) Thus,
ni = n0
Plugging this and the Boltzmann relation into Poisson’s equation gives
en 
eΦ

∇ 2 Φ = 0  exp  − 1




kT
ε
Now, using the Taylor series expansion of the exponential, which we assume is approximately 1,
gives
Page 10
Class notes for EE6318/Phys 6383 – Spring 2001
This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
∇2Φ =
  eΦ  2  
en0  eΦ
1
+
+
O
    − 1
kT
ε 
 kT  
e 2 n0 Φ
εkT
Solving the differential equation leaves
 −r 
Φ ≈ Φ 0 exp
 where
 λ Debye 
≈
e n 
λ Debye =  0 
 εkT 
2
−
1
2
Page 11