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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 6 Basic plasma dynamics
Our fluid equations that we developed before are:
∂n 

f c =  ∇ r ∑(n v ) + 

∂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
We have looked collisions in the last section. Now we will look at the Lorentz force terms.
Often, it is best to look at how a single particle reacts to the fields, so we will start there.
Lorentz force
The equation of motion for a single particle of charge q and mass m is given by the Lorentz force
dv
law: m
= q(E + v ∧ B) .
dt
While this equation can be quite complex, for complex fields, it is often easiest to look at the
cases when the external fields are simple.
Case 1: E = Ez z√; B = 0
dv q
= (E + v ∧ B)
dt m
q
= Ez z√
m
Integrating gives
x = x0 + vx 0 t
y = y0 + vy 0 t
Letting E = E 0 ; B = 0



q

2
 Gives r = r0 + v 0 + 2 m E 0 t 
q
Ez t 2
2m
This is quite simple and not terribly informative.
z = z0 + vz 0 t +
Cases 2: E = 0; B = B0 z√
dv q
= (E + v ∧ B)
dt m
q
= v ∧ B0 z√
m
Thus we find
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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
dvx q
= vy B0
dt
m
dvy
q
= − vx B0
dt
m
dvz
=0
dt
To solve this set of equations, we must separate the components of the velocity. This is simple
to do by differentiating the equations again and substituting to give.
2
d 2 vx q dvy
q 

B0 = −
B v
=
 m 0 x
dt 2
m dt
2
d 2 vy
q dvx
q 

B0 = −
B v
=−
 m 0 y
dt 2
m dt
d 2 vz
=0
dt 2
These second order equations are of course are easily solved as
vx = vx 0 e ± iω c t
vy = vy 0 e ± iω c t where ω c =
q
B0
m
vz = vz 0
Now taking into account the original coupled first order equations we find
vx = v⊥ 0 cos(ω c t + ϕ )
vy = − v⊥ 0 sin(ω c t + ϕ )
vz = vz 0
Integrating a second time we find,
v
v
x = ⊥ 0 sin(ω c t + ϕ ) + x0 − ⊥ 0 sin(ϕ )
ωc
ωc
y=
v⊥ 0
v
cos(ω c t + ϕ ) + y0 − ⊥ 0 cos(ϕ )
ωc
ωc
z = vz 0 t + z0
rc =
v⊥ 0
ωc
v⊥ 0
is the radius of a circular orbit around the magnetic field line; it is
ωc
known as the Larmor radius or the cyclotron radius. Further we note that the positively charged
particles orbit in a left-hand orbit while the negatively charged particles orbit in a right-hand
orbit.
It is easy to see that rc =
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Class notes for EE6318/Phys 6383 – Spring 2001
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Example
Of particular interest is the magnetic field required to give an electron a gyro-frequency of 2.45
GHz. This is of interest because it is required to understand Electron Cyclotron Resonance,
ECR, plasma sources. (Given time at the end of the semester, we will discuss these sources in
detail.)
ωc
2π
q
=
B0
2 πm
= 2.8 E 6 B0 ( Hz / Gauss)
fc =
⇓
B0 ≈ 875G
Case 3: E = Ez z√; B = B0 z√
dv q
= (E + v ∧ B)
dt m

x√ y√ z√ 
q
Ez z√ + vx vy vz 
=

m 
0 0 B0 

or
dvx q
= vy B0
dt
m
dvy
q
= − vx B0
dt
m
dvz q
= Ez
dt
m
This is easy to solve as we have already done this as parts.
x = rc sin(ω c t + ϕ ) + x0 − rc sin(ϕ )
y = rc cos(ω c t + ϕ ) + y0 − rc cos(ϕ )
z = z0 + vz 0 t +
rc =
q
Ez t 2
2m
v⊥ 0
ωc
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Case 4: E = Ex x√; B = B0 z√: Note that E = Ey y√ would also work here
dv q
= (E + v ∧ B)
dt m

x√ y√ z√ 
q
=
Ex x√ + vx vy vz  or


m
0 0 B0 

dvx q
?
= Ex + ω c v y ⇒
dt
m
dvy
?
= −ω c vx
⇒
dt
dvz
=0
⇒ vz = vz 0 ; z = vz t + z0
dt
The third equation is easy to solve while the first two are more difficult. Differentiating gives
d 2 vx
= −ω c2 vx
2
dt
⇓
vx = vx 0 e ± iω c t
d 2 vy
q
= −ω c  Ex + ω c vy 
2


dt
m
 qB E

= −ω c  0 x + ω c vy 
 m B0

d 2 vy d 2 
 Ex

E 
= −ω 
+ vy  but
= 2  vy + x  so
2
dt
dt 
B0 
 B0

2
c
vy +
Ex
= vy 0 e ± iω c t
B0
Ex
B0
Plugging these into our initial equations gives

dvx q
E 
= Ex + ω c  vy 0 e ± iω c t − x 
dt
m
B0 

vy = vy 0 e ± iω c t −
= ω c vy 0 e ± iω c t
= ±iω c vx 0 e ± iω c t
⇓
vy 0 = ±ivx 0
Let v⊥ = mivy 0 = vx 0 giving
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This document is for instructional use only and may not be copied or distributed outside of EE6318/Phys 6383
vx = v⊥ e ± iω c t ⇒ v⊥ cos(ω c t + φ )
vy = miv⊥ e ± iω c t −
Ex
E
⇒ v⊥ sin(ω c t + φ ) − x
B0
B0
vz = vz 0 ; z = vz t + z0
This means that the particle travels along the as it would with just the magnetic field but it also
have a drift in the E ∧ B direction. We can calculate this in general. First the average force is
F = q(E + v ∧ B) = 0
Therefore
F ∧ B = q(E ∧ B + ( v ∧ B) ∧ B) = 0
⇓
E ∧ B = −( v ∧ B) ∧ B
= B ∧ ( v ∧ B)
= v (B ∑B) − B( v ∑B)
Now if there is no drift along B then we get
E∧B
v =
B2
What we have described above is true in general. Assuming that we have any constant force that
is a right angle to the magnetic field.
Ftotal ∧ B = 0 = F⊥ ∧ B + q(v ∧ B) ∧ B
⇓
v drift =
F⊥ ∧ B
qB2
Case 5: E = 0; B = B0 + (r ∑∇)B + ... : Non - uniform magnetic field
Here we look at a magnetic field that is non-uniform in space. The Taylor series expansion of
such a field will be of the form
B = B0 + (r ∑∇)B + ...
This should be straight forward from
Bz ( y) = Bz 0 + y∂ y Bz + ...
Now from Lorentz’s Force Law we have
dv
m
= q(E + v ∧ B) = q(v ∧ B)
dt
or in the y-direction
dv
m y = − qvx Bz
dt
(
)
= − qvx Bz 0 + y∂ y Bz + ...
(We can do the same thing in the x-direction.)
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The force averaged over one gyration is
dvy
m
= − q vx Bz 0 + yvx ∂ y Bz + ...
dt
The first term is clearly zero. The second term is not as
yvx = (rc cos(ω c t + ϕ ) + y0 − rc cos(ϕ ))(v⊥ 0 cos(ω c t + ϕ ))
(
)
= rc v⊥ 0 cos2 (ω c t + ϕ ) - all of the other terms are zero
1
rc v⊥ 0
2
We can now plug this into our average force equation to get
dvy
1
m
= − q rc v⊥ 0 ∂ y Bz − or in general
2
dt
=
1
= − q rc v⊥ 0 ∇Bz
2
1
1
= mv⊥2 0 ∇Bz
2
B
In the x-direction this becomes
dvx
m
= − q vy Bz 0 + yvy ∂ y Bz + ...
dt
but
(
)
yvy = (rc cos(ω c t + ϕ ) + y0 − rc cos(ϕ ))(v⊥ 0 sin(ω c t + ϕ ))
= rc v⊥ 0 cos(ω c t + ϕ ) sin(ω c t + ϕ )
=0
Thus, we have from above a drift velocity
F ∧B
v drift = ⊥ 2
qB
B ∧ ∇Bz
1
rc v⊥ 0
B2
2
1
1 B ∧ ∇Bz
= mv⊥2 0
B B2
2
This leads into a topic known as magnetic mirrors. Magnetic mirrors are naturally occurring
phenomena that happen at the magnetic poles of planets and stars. In laboratory-based plasmas
magnetic mirror are used in some process systems to confine the plasma. (They were also used –
quite unsuccessfully - as a confinement mechanism for fusion plasmas.)
=
In a mirror the gradient of the magnetic field is parallel to the direction of the field lines. This
sort of arrangement is known as a cusp field and looks like the figure below. (This is the
geometry one finds with permanent magnets.)
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z
θ
From Maxwell’s Equations we have
∇ ∑B = 0
⇓ - in cylindrical coordinates
r ∂ r (rBr ) + ∂ z ( Bz ) = 0 - or
-1
r -1∂ r (rBr ) = −∂ z ( Bz )
Integrating over r and assuming ∂ z ( Bz ) is not a function r gives
∫ ∂ (rB )dr = − ∫ r∂ ( B )dr
r
r
z
rBr ≈ −
z
r2
∂ z ( Bz )
2
r=0
r
Br ≈ − ∂ z ( Bz )
2
r=0
Now from Lorentz,
F = q(E + v ∧ B)
⇓
=0
}


Fr = q vθ Bz − vz Bθ  = qvθ Bz


Fθ = q(vz Br − vr Bz )
=0
 }

r
Fz = q vr Bθ − vθ Br  = − qvθ Br = qvθ ∂ z ( Bz )
2


r=0
Now
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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
vθ = vx2 + vy2
has to do with the direction of θ.
= m v⊥
and
v
q
r = rc = ⊥ where ω c = B0
m
ωc
Thus
r
Fz = m qv⊥ c ∂ z ( Bz )
2
2
v
= m q ⊥ ∂ z ( Bz )
2ω c
= − 12 mv⊥2
1
∂ z ( Bz )
B
= −µ∂ z ( Bz ) where µ = 12 mv⊥2
1
B
µ is the magnetic moment of a particle gyrating around a point. This can be easily seen from
µ = IA - where I and A are the current and area
qω
=  c  ( πrc2 )
 2π 
qv 2⊥
=
2ω c
=
1
2
mv 2⊥
B
What does this mean?
As a particle moves into a region of increasing B, the Larmor radius shrinks but the magnetic
moment remains constant. (This is shown in a number of books, including Chen.) Since the B
field strength is increasing the particles tangential velocity must increase to keep µ constant.
The total energy of the particle must also remain constant and thus the particle velocity parallel
to the magnetic field must decrease. This causes the particle to bounce off of the ‘magnetic
mirror’. (There are still ways for some of the particles to ‘leak’ through the mirror. This is one
of the major reasons that magnetic mirrors did not work in the fusion field.)
Additional drift motions
There are a number of additional drift motions that occur for single particles. We unfortunately
do not have time to cover these drifts. The other drifts include:
Curved B: Curvature drift
Non-uniform E
Curved Vacuum fields
Polarization Drift.
Most of these are covered in the main text or in Chen.
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Bulk Motions
At this point, we need to deal with some of the bulk motions that occur in plasmas. These are
not single particle motions but rather collective motion of all/most of the charge species in the
plasma. The first, and most important is the electrostatic plasma oscillation, giving rise to the
plasma frequency. [This but just one of a very wide variety of waves in plasmas.] These
oscillations occur because one of the species becomes displaced from the other. When it
accelerates back toward the other species, in gains too much energy and over shoots. To derive
the plasma frequency, we will assume the simplest of geometries and plasmas.
1) No external fields. (This can be relaxed and the same result can be obtained.)
2) No random motion of the particles. (Hence, all particles of a species move at the same
velocity at the same point in space. This can be relaxed and one can get the same result – it is
just harder to do.)
3) Only the electrons move. (This is not a bad assumption for many aspects of plasmas.)
4) The plasma is one-dimensional and of such a length that the walls do not influence the result.
(This implies that we are considering just regions that are at least several λdebye from the walls.)
From Maxwell’s equations we have,
∇ ∑E = ρ ε
∇∧E= 0
(We will ignore the induced magnetic field.) Then our equation of motion (momentum
conservation) and continuity (particle conservation) become
Continuity Equation
∂n 

f c =  ∇ r ∑(n v ) + 

∂t 
⇓
0 = ∇ r ∑(n v ) +
∂n
∂t
Energy Equation
∂ 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
⇓
∂ v

mn
+ v ∑∇ r v  = qn(E)
 ∂t

For this particular wave, we are considering deviations from charge neutrality. Thus we will
have an induced electric field given by
ε∇ ∑E = ρ = e(ni − ne )
We have three items that are changing with time, E, ne and <ve>. We will expand each of these
items to produce a time average term, denoted with a ‘0’ and an oscillating term denoted with a
‘1’. Thus
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E = E 0 + E1 - but E 0 ≈ 0!
ne = n0 + n1
v e = v 0 + v1 - but v 0 ≈ 0!
Then our conservation of momentum (energy) equation becomes
 ∂(v 0 + v1 )
 q
+ (v 0 + v1 ) ∑∇ r (v 0 + v1 ) = (E 0 + E1 )

∂t

 m
 ∂(v1 )
 q
+ (v1 ) ∑∇ r (v1 ) = (E1 )

 ∂t
 m
Now the second term on the left-hand side is small compared to the other two. (Two oscillating
terms as opposed to one.) Thus we are left with
∂(v1 ) q
 implying we have 
i ωt − βx )
- 
= (E1 ) - let v1 ∝ e (

 a travelling wave 
∂t
m
⇓
iωv1 = −
e
E1
m
Now we can do the same thing to the continuity equation
∂n
∇ r ∑(n v ) +
=0
∂t
∂(n0 + n1 )
∇ r ∑((n0 + n1 )(v1 )) +
=0
∂t
∂(n1 )
=0
∇ r ∑((n0 )(v1 )) +
∂t
Where again we have dropped the higher order terms. Thus
i ωt − βx )
n0 ∇ r ∑(v1 ) = −∂t (n1 ) - letting n1 ∝ e (
⇓
in0 βv1 = −iωn1
Finally, we solve Poisson’s Equation
ε∇ ∑E = e(ni − ne )
ε∇ ∑E1 = e(n0 − (n0 + n1 ))
−iβεE1 = −en1
This gives us three equations and three unknowns
e
in0 βv1 = −iωn1 , iβεE1 = en1 , iωv1 = − E1.
m
Combining the last two to eliminate E1 gives
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e 2 n1
.
ωmβε
Placing this into the first to eliminate n1 gives
n e2
ω 2pe = 0 ; ω pe = f pe / 2 π ,
mε
the angular electron frequency of the plasma. This is also known as the dispersion relation.
v1 = −
Typically for process plasmas the density is ~1010-12 cm-3. Thus, fpe ~ 1 to 10 GHz.
This is, in some sense, the simplest oscillation that can exist in a plasma. Note that this is not a
dω
wave in the typical sense! (Energy does not move in this oscillation, the group velocity,
, is
dβ
zero. Here ‘ β ’ is the wave vector.)
There are numerous other oscillations that are waves that that can transfer energy. They can be
divided into electrostatic waves and electromagnetic waves. We will deal first with the
electrostatic waves.
Electrostatic waves in plasmas
Let us go back to our fundamental fluid equations, the continuity equation, the energy equation,
and Poisson’s equation. As before we will ignore the collision terms and assume that the
magnetic field is zero. Here however, we will include the pressure variations of the species.
Poisson’s equation
ε∇ ∑E = ρ = e(ni − ne )
Continuity Equation
∂n 

f c =  ∇ r ∑(n v ) + 

∂t 
⇓
0 = ∇ r ∑(n v ) +
∂n
∂t
Energy Equation
∂ 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
⇓
∂ v

mn
+ v ∑∇ r v  = qn(E) − ∇ r ∑P
 ∂t

This would be identical to the plasma oscillations except for the pressure term. We will deal
with that term first.
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We know from the ideal gas law that p = nkT . Then, assuming an isotropic, or one dimensional,
plasma
∇ r ∑P = ∇ r p
= kT∇ r n
This is true provided that the compression is ‘isothermal’. In other words the temperature stays
the same during the compression. Often, this is not true. Rather, we have ‘adiabatic’
compression, where the temperature changes. In this case, it can be shown using
thermodynamics that
p = Cnγ .
Here C is a constant and γ = Cp CV is the ratio of the specific heats. We can see from the above
equation that
∇p = ∇(Cnγ )
= C∇(nγ )
= Cγnγ −1∇n
∇n
n
∇n
= γp
n
= γCnγ
N +2
where N is the number of degrees of freedom. Thus for
N
N=1, γ = 3 . (This is a crude approximation but it works for our needs.)
Further, it can be shown that γ =
Thus, our equation of motion becomes
∂ v

mn
+ v ∑∇ r v  = qn(E) − γkT∇ r n .
 ∂t

As before, we will assume that the density, velocity and electric field consists of a time-averaged
term and an oscillating term. Thus,
E = E 0 + E1 - but E 0 ≈ 0!
ne = n0 + n1
v e = v 0 + v1 - but v 0 ≈ 0!
letting
i ωt − βx )
v1 , E1 ,n1 ∝ e (
We can now follow the derivation that we made before but this time we will add our additional
term.
Assuming that we are examining electrons, our conservation of momentum (energy) equation
becomes
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=0

 }

∂
v
0
=0
=
=0
  0 + v1   }
}


}



+  v 0 + v1  ∑∇ r  v 0 + v1  = −e(n0 + n1 ) E 0 + E1  − γkTe ∇(n0 + n1 )
m(n0 + n1 )
∂t











mn0
∂(v1 )
= −en0 E1 − γkTe ∇n1
∂t
⇓
imn0ωv1 = −en0 E1 + iγkTe βn1
Now we can do the same thing to the continuity equation
∂n
∇ r ∑(nv) +
=0
∂t
∂(n0 + n1 )
∇ r ∑((n0 + n1 )(v1 )) +
=0
∂t
∂(n1 )
=0
∇ r ∑((n0 )(v1 )) +
∂t
n0 ∇ r ∑(v1 ) = −∂t (n1 )
⇓
in0 βv1 = iωn1
Finally, we solve Poisson’s Equation
ε∇ ∑E = e(ni − ne )
ε∇ ∑E1 = e(n0 − (n0 + n1 ))
−iβεE1 = −en1
This gives us three equations and three unknowns
in0 βv1 = iωn1 , −iβεE1 = −en1 , imn0ωv1 = −en0 E1 + iγkTe βn1 .
Combining the last two to eliminate E1 gives
 e2
γkTe β 
v1 = 
+
 n1
 mβεω mn0ω 
which is very similar to what we got before except we now have a second term.
Placing this into the first to eliminate n1 gives
e 2 n0 γkTe 2
ω2 =
+
β
mε
m
γkTe
= ω 2pe + cγ2 β 2 - where cγ2 =
is the electron sound speed
m
Note that here the group velocity is non-zero, meaning that energy can be carried by the wave.
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Now, let us assume that we are dealing with (positive) ions. Here, however, the electric field is
determined by the electrons, not the ions. Thus, we need to replace the electric field with the
gradient of the potential and use Boltzmann’s relation on the electron density. Thus, our
equations become
∂n
∇ r ∑(nv) +
=0
∂t
⇓
n0 βv1 = ωn1
(as before)
∂ v

+ v ∑∇ r v  = qn(E) − γkT∇ r n
mi n
 ∂t

= −en(∇φ ) − γkTi ∇ r n
=0

 }

∂
+
v
v
0
1


=0
=0
=0
 
}


}

 }
mi (n0 + n1 )
+  v 0 + v1  ∑∇ r  v 0 + v1   = −e(n0 + n1 )∇ φ0 + φ1  − γkTi ∇(n0 + n1 )
∂t











mi n0
∂(v1 )
= −en0 ∇φ1 − γkTi ∇n1
∂t
⇓
mi n0ωv1 = en0 βφ1 + γkTi βn1
Now, we can’t use Poisson’s equation but rather we assume that the change in local density can
be modeled with Boltzmann’s equation. E.g. that the local ion density is the same as the local
electron density and that the electron density is given by
ni ≈ ne = n0 e eφ / kTe (This approximation causes some small error)
= n0 e
=0

}
e  φ 0 + φ1  / kTe


= n0 e eφ1 / kTe
 eφ 
≈ n0 1 + 1 
kTe 

⇓
eφ1
kTe
This gives us our three equations and three unknowns
eφ
n1 = n0 1 , n0 βv1 = ωn1 , mi n0ωv1 = en0 βφ1 + γkTi βn1
kTe
Putting together the first two gives
ω eφ1
v1 =
.
β kTe
Plugging this and the first into the third gives
n1 = n0
Page 14
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
ω 2 kTe + γkTi
=
β2
mi
This is the ion-acoustic or ion-sound waves.
All of the above are just a few examples of electrostatic waves. There are many more
electrostatic waves.
We will now deal briefly with electromagnetic waves. It is important to note that we will
only look at the very simplest cases. There are a wide variety of electromagnetic waves that
are sustained in plasmas.
First, standard light waves exist. This comes directly from Maxwell’s equation.
 E
 E
 E
∇ 2   = σµ∂ t   + εµ∂ t2  
 H
 H
 H
E =±
iµω
iµω 
H
; η=
=
=
η
γ
σ + iεω 

µ
if σ = 0
ε

sign det er min ed by growth / decay
( growth => −, decay => + )
2π c
1
β=
= ;c =
λ ω
εµ
In plasmas, this is not quite correct. What happens if the waves are interacting with the plasma?
From Maxwell’s equations we have
∇ ∧ E = −∂ t B
∇ ∧ H = J free + ∂ t D
∇ ∑D = ρ free = e(ni − ne ) ≈ 0
∇ ∑B = 0 ds
However the current is not zero. This changes the fields. To solve the problem, we will consider
only the time varying components. Taking the curl of the first equation and the time derivative
of the second equation gives
∇ ∧ (∇ ∧ E) = ∇(∇ ∑E) − ∇ 2 E = −∂ t ∇ ∧ B
µ −1∂ t ∇ ∧ B = ∂ t J free + ∂ t2εE
We can combine these to give
∇(∇ ∑E) − ∇ 2 E = − µ∂ t J free − εµ∂ t2 E
∝ρ = 0
}

β  β ∑E + β 2 E = −iωµJ free + εµω 2 E - c-2 = εµ


⇓
(β
2
− c −2ω 2 )E = −iωµJ free
Page 15
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
If the light is at a high frequency, then the ions are effectively fixed. Thus the current is almost
entirely due to the motion of the electrons. Then the current can be given as
J free = −en0 v e1
from the equation of motion
dv
F = m e1 = −eE
dt
so that
in e 2 E
J free = − 0
ωm
plugging this in to Maxwell’s equation gives
2
(β 2 − c −2ω 2 )E = − n0 µme E
⇓


n µe 2
− c2β 2  E
0 = ω 2 − c2 0


m
 2 n0 e 2

= ω −
− c2β 2  E


εm
(
)
= ω 2 − ω 2pe − c 2 β 2 E
⇓
ω 2 = ω 2pe + c 2 β 2
This is the dispersion relation of electromagnetic waves in a plasma.
There is a very useful application of this dispersion relation.
Then is a very interesting phenomenon that occurs, known as cutoff.
Page 16
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 7 Diffusion
Our fluid equations that we developed before are:
∂n 

f c =  ∇ r ∑(n v ) + 

∂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
We have looked collisions in the last section. Now we will look at the Lorentz force terms.
Often, it is best to look at how a single particle reacts to the fields, so we will start there.
Page 17