Download FB Plasma Notes - School of Physics

PLASMA PHYSICS
Dr Ferg Brand
School of Physics
University of Sydney NSW 2006, AUSTRALIA
Notes for a 20 lecture course in PLASMA PHYSICS for
Senior and Honours (3rd and 4th year) Physics given 1997 –
2000.
This course in Plasma Physics is not simply aimed at people
specializing in industrial plasmas or fusion plasmas. But it is one
which will provide a useful background for a broad range of
topics in physics. It will introduce a number of techniques that
will serve you well in many other Physics applications.
Illustrative examples will be drawn from astrophysics,
ionospheric and magnetospheric physics, solid state plasma
physics as well as the more traditional plasma areas.
January 2001
CONTENTS
I
Introduction
II
Motion of ions and electrons in E and B fields
III
Fluid description of a plasma
Boltzmann equation approach
IV
Diffusion
Key ideas – 1
V
Waves in plasmas – 1
VI
Waves in plasmas – 2
Waves I have known
Key ideas – 2
VII
Plasma diagnostics
VIII
Plasma processing
IX
Fusion
Solutions to exercises
Assignment questions, 2000
Examination paper, 2000
Note. Some of the material will not be examinable, it is included for completeness.
This material is indented from the margin and is in smaller type.
PLASMA PHYSICS
Physical Constants
magnitude of charge on electron e = 1.60 × 10−19 C
mass of electron me = 9.11 × 10−31 kg
mass of hydrogen ion mi = 1.67 × 10−27 kg
Boltzmann’s constant k = 1.38 × 10−23 J K−1
velocity of light in vacuum c = 3.00 × 108 m s−1
permittivity of vacuum ε0 = 8.85 × 10−12 F m−1
permeability of vacuum µ0 = 4π × 10−7 H m−1
1 eV = 1.60 × 10−19 J
I. INTRODUCTION
What is a plasma?
An ionized gas made up of electrons, ions and neutral particles, but electrically
neutral.
The word was first used by Irving Langmuir in 1928 to describe the ionized gas in an
electric discharge.
Fourth state of matter. Consider the series of phase transitions solid-liquid-gas. If we
continue to increase the temperature above, say, 20 000 K (lower if there is a
mechanism for ionizing the gas) we obtain a plasma.
(Note solid state physicists talk about electron-hole plasmas.)
A plasma has interesting properties because the electrostatic force is a long range
force and every charged particle interacts with many of its neighbours. We can get
collective behaviour. We can treat the plasma as an electrical fluid.
Examples
It has been said that 99% of matter in the universe is in the plasma state.
lightning
earth’s ionosphere, aurora, earth’s magnetosphere, radiation belts
interplanetary medium, solar wind
solar corona
stellar interiors
interstellar medium
laboratory plasmas such as glow discharges, arcs
fluorescent lamps, neon signs
electrical sparks
thermonuclear fusion experiments
a homely examples: flame
other examples: rocket exhaust
How to characterize a plasma
plasma density n (m−3) (often cm−3. 1 cm−3 = 106 m−3)
temperature T (K or more conveniently eV)
Temperature and energy
For an ideal gas in thermal equilibrium, the probability that velocity lies in the range
 1 mv 2 


dvxdvydvz around velocity (vx,vy,vz) is proportional to exp − 2
 dv x dv y dv z .
kT 



We can construct the Maxwellian velocity distribution function f(v)
 1 mv 2 
3

2
m
 exp − 2
f ( v) = n


,
 2πkT 
kT 



and use it to calculate average values.
e.g.,
Calculate the particle density
n=
∫ f ( v) dv dv dv
x
y
z
calculate the mean velocity
u=
1
vf ( v) dv x dv y dv z = 0
n∫
calculate the mean speed
v=
1
vf ( v) dv x dv y dv z =
n∫
8kT
πm
calculate the rms speed
2
v rms
=
1 2
3kT
v f ( v) dv x dv y dv z =
so vrms =
∫
n
m
You do. Show the average kinetic energy per particle is
E av =
3kT
m
3
kT
2
Temperature T in K can be expressed in eV simply by calculating the energy kT in J
and converting to eV.
You do. Show 1 eV corresponds to 11 600 K.
(So, by a 2 eV plasma we mean T = 23 200 K. Note that Eav = 3 eV.)
Formation of a plasma
Ionization at high temperatures
We have said that a sufficiently hot gas becomes a plasma.
Atoms in a gas have a spread in thermal energy and they collide with each other.
Sometimes there is a collision with high enough energy to knock an electron out of
the atom and ionize it. Energy must exceed ionization potential, 13.6 eV for hydrogen.
In a cold gas such collisions are very infrequent, in a hot gas more likely.
From the Maxwellian velocity distribution function we can derive the Saha equation
which gives the fraction of ionization we can expect in a gas in thermal equilibrium
at temperature T,
3
ni
T2
 U 
≅ 3 × 10 27
exp − i 
nn
ni
 T 
where ni is the ion density, nn is the neutral particle density and Ui is the ionization
energy. (In this expression Ui and T are both in eV.)
Note: There is significant ionization below 13.6 eV.
Ionization in an electric field, gas discharges
Another way of achieving ionization.
The term discharge was first used when a capacitor was discharged across the gap
between two electrodes placed closed to each other. If the voltage is sufficiently high,
electric breakdown of the air occurs. The air is ionized and the conducting path closes
the circuit and a current can flow. Later the term was applied to any situation where a
gas was ionized by an electric field and a current flowed.
Discharge may give off light.
The simplest gas discharge is a glass tube with a metal electrode sealed into each end.
The tube is evacuated and filled with various gases at different pressures. The
electrodes are connected to a dc supply.
Raise the voltage.
(i) Low voltage (10s of volts)
no visible effect.
very small currents ( ≈ 10−15 A), ionization by cosmic rays and natural
radioactivity. The discharge is called non-self-sustaining as an external ionizing agent
is required.
increase voltage and a saturation current is reached when all the charges are
collected. Townsend discharge.
(ii) Increase voltage
breakdown occurs. e.g., if the gap is 10 mm and the pressure is 1 torr then this
happens at 400 V, if 1 atmosphere then 30 kV.
current increases by several orders of magnitude, but voltage does not change.
discharge becomes independent of an external ionizing source; it is selfsustaining.
ionization is caused by electrons colliding with atoms.
This is one of the most important mechanism in an electric discharge. So we
will examine it in some detail.
Electrons are accelerated by the electric field and gain energy.
They collide with atoms. If their energy is small, the collision will be elastic
and they will lose only a small fraction ≈ m/M of their energy in the encounter.
After the collision they will gain more energy from the field.
Their energy increases until it is large enough that the collision is inelastic; the
atom is excited or ionized. For ionization, the electron energy must exceed the
ionization potential of the atom.
(Note that positive ions lose a large fraction of their energy in each elastic
collision and it is much more unlikely that their energy will increase sufficiently to
ionize.)
You do. Estimate fractions of energy lost by collisions between (a) electrons with
atoms and (b) ions with atoms.
After an ionizing collision the second electron is then available to ionize.
There is an avalanche effect.
the gas is appreciably ionized in µs to ms.
We can plot Townsend’s ionization coefficient, η (in V−1), the number of ionizing
collisions caused by an electron as it falls through a p.d. of 1 V
Note the units. 1 V/cm. mmHg or 1 V/cm. torr = 0.75 V m−1 Pa−1.
η=a

p
p
exp − b

E
E

η depends on E/p ( a common parameter in gas discharge work – it allows scaling),
the gas.
η is low at low pressures because an electron encounters hardly any atoms.
η is low at high pressures because elastic collisions are more frequent and it is more
difficult for the electron to gain sufficient energy to ionize.
The figure shows the Penning effect in a gas mixture where argon is ionized by
metastable excited neon atoms.
Photoionization is not important in this case but might provide the initial
ionization to start things off.
The second important process is positive ions bombarding the cathode and
knocking off electrons.
This is described by γ, the secondary ionization coefficient, defined as the number of
electrons knocked off the cathode by a single positive ion.
γ depends on E/p, the gas and the cathode material (and the state of the cathode
surface, whether it is a pure metal or has an oxide layer).
Consideration of these two processes allow us to estimate the p.d. for breakdown.
Suppose the p.d. between the electrodes is V.
ηV
at the
One electron leaving the cathode becomes e electrons arriving
anode.
ηV
And e −1 positive ions heading back towards the cathode.
One ion produces γ electrons.
So one electron gives rise to
For breakdown, this must be
γ ( e ηV − 1) electrons.
≥1.
(iii) Now increase current
if the tube is long can get a beautiful radiant column, glow discharge.
voltage 100s of volts, current milliamperes.
discharge is maintained by positive ions bombarding the cathode and knocking
off electrons.
the ion and electron densities are equal only in the positive column. This plasma is
weakly-ionized, fraction ionized is 10−8 to 10−6, Te is 104 K but Ti and Tn are 300 K.
Increase pressure to 100 torr, positive column becomes longer and thinner.
Increase electrode distance, higher voltage required, positive column longer to
occupy the extra length.
Increase current, cathode glow covers more of the cathode surface so the
current density and the voltage remain fairly constant.
Different gases yield different colours. “neon” signs
(iv) Suppose the pressure is high and any series resistance is low
arc discharge
voltage can be low 10 V, current > 1 A
fraction ionized is 10−3 to 10−1, Te and Ti are 104 K
types of arcs
thermionic arc - emission of electrons is due to cathode being heated by the large
current of ions bombarding it. Cathode must withstand very high temperature, e.g.,
carbon, tungsten. This arc is self-sustaining.
(Emission of electrons from a hot surface is described by
eφ
j = aT 2 exp − 
 kT 
where
φ is the work function.)
e.g., carbon and tungsten arc lamps
thermionic arc with cathode heated by external source - non-self-sustaining.
field emission arc - emission of electrons is due to very high E at cathode.
e.g., mercury arc lamp, mercury arc rectifier.
metal arc - heating the cathode vapourizes the metal.
high-pressure arc p > 1 atm; low pressure arc < 1 atm.
(v) Processes of deionization
Dissociative recombination
A2+ + e → A ∗ + A
is the fastest recombination process in a weakly-ionized gas like a glow discharge
Radiative recombination
A + + e → A ∗ + hν
is not important for electron removal but may be important for light emission.
Diffusion to wall is slower in a well-developed discharge.
Three-body electron-ion recombination
A+ + e + e → A + e
is main process in high density, low temperature laboratory plasmas.
Debye length and plasma frequency
A plasma has a characteristic length and a characteristic time.
Screening of electrostatic fields
This leads to the Debye length λD.
First, consider a positive charge q all by itself. The potential at a distance
charge is
φ=
q
4πε 0 r
r from the
.
Now, consider a positive charge q in the middle of a plasma. It attracts electrons into
its vicinity and repels positive ions. We will calculate φ for this case.
If we allow the particle to have both kinetic and potential energy, the probability
 1 mv 2 + qφ 


factor becomes exp − 2
 dv x dv y dv z . φ depends on position so the
kT




probability depends on position.
The particle density is given by
for electrons
n=
∫ f ( v)dv dv dv
x
y
−eφ 
ne = n0 exp −

 kT 
∇. E =
E = −∇φ so
−∇ 2 φ =
so
qφ
n ∝ exp − 
 kT 
eφ
ni = n0 exp − 
 kT 
for ions (we will suppose they are singly-ionized)
Gauss’ Law can be written as
z
σ
ε0
σ
ε0
.
This is Poisson’s equation.
eφ
− eφ 
+ exp
.

kT
kT 
Assume that this potential term is very small, eφ << kT
The charge density is
σ = − ene + eni = en0  − exp
σ ≅ − en0 1 +

2n e 2φ
eφ 
eφ 

.
 + en0 1 −  = − 0

kT 
kT 
kT
I am going to use spherical coordinates (and assume spherical symmetry)
∇ 2φ ≡
1 d  2 dφ 
r
.
r 2 dr  dr 
Poisson’s equation becomes
2 n 0 e 2φ
1 d  2 dφ 
− 2
r
=−
ε 0 kT
r dr  dr 
with solution


q
r

φ=
exp −
4πε 0 r
ε 0 kT

2 n0 e 2

The potential falls away exponentially.
ε kT
Call λ D = 0 2 the Debye length then
n0e
φ=



.



2r 
exp −
.
4πε 0 r
 λD 
q
Beyond a few Debye lengths, shielding by the plasma is quite effective and the
potential due to our charge is negligible.
This provides condition to determine if we have a plasma or not.
(i) the system must be large enough L >> λ D , and
(ii) there must be enough electrons to produce shielding N D >>> 1 , where
N D is the number of electrons in a Debye sphere.
Suppose there is a local concentration of charge. If plasma dimensions are much
greater than λD, then on the whole plasma is still neutral (we can describe the plasma
as quasineutral) and we can take ne ≅ ni ≅ n0 .
If we put an electrode into the plasma, it becomes shielded by a sheath ot thickness
≈ λD.
λ D = 69.0
T
m (T in K, ne in m -3 )
ne
Plasma oscillations
This leads to the plasma frequency ωpe.
What would happen if electrons were displaced from their equilibrium positions? The
electrostatic force due to the ions would pull them back, but the electrons would
overshoot and oscillations would ensue. These are known as plasma oscillations.
This is a very fast oscillation, so fast that the massive ions do not have time to
respond.
A simple calculation of the frequency follows.
Consider an infinite plasma. We will ignore thermal motions. We will treat the
massive ions as not moving.
Suppose a slab of electrons is displaced a small distance x (so we are dealing with a
1-dimensional problem). The slab has thickness L. Consider an area A.
F=m
Equation of motion for the slab of electrons is
mass of electrons in slab m
d2x
dt 2
= meneLA
What is the force on the electrons? We have two oppositely charged sheets facing
each other. The electric field between them is
E=
σs
ε0
where
σs is the surface
charge density or charge /area.
σs = enexA/A so the restoring force is equal to charge of electrons in slab × electric
field, so,
force on electrons = −ne LAe
ene x
ε0
.
Equation of motion becomes
ne LAme
en x
d2x
= − ne LAe e or
2
dt
ε0
ne e 2
d2x
=−
x
dt 2
ε 0 me
with solution
(
)
x = A cos ω pe t ,
where the (electron) plasma frequency is
ω pe =
ne e 2
.
ε 0 me
This is a ‘natural’ frequency for the plasma.
f pe = 8.98 ne Hz (ne in m −3 ).
We encounter ωpe when discussing wave propagation.
You do. Show ω pe λ D =
kT
.
me
We can now use λD and ωpe to classify plasmas.
Exercises
How to characterize a plasma
1.
Calculate the number density of an ideal gas at (a) 0°C and 760 torr, and (b)
20°C and 1 micron.
Note: Units of pressure
1 atm = 1.013 × 105 Pa = 760 mmHg = 760 torr
1 micron = 1 mtorr
Temperature and energy
2.
Calculate vrms for protons and electrons at 106 K.
3.
Are you surprised to learn that a plasma of 1 million K can be contained in a
steel vessel without melting it? You should not be if you understand the difference
between heat energy and temperature.
Consider a plasma in the Plasma Department’s TORTUS tokamak where ne =
ni = 1019 m−3, T = 100 eV and volume of plasma = 1 m−3. How much would the
energy in this plasma raise the temperature of 200 ml of water?
Ionization at high temperatures
4.
Use the Saha equation to calculate the percentage of ionization in nitrogen
over a range of temperatures from say 300 K to 100 000 K. Plot as a function of log
T.
Use ntotal = ni + nn = 3 × 1025 m−3 (about what it is at room temperature, 1
atmosphere) and ionization energy for nitrogen 14.5 eV.
Ionization in an electric field, gas discharges
5.
Choose one of the following plasmas.
lightning
aurora
interstellar medium
ionosphere
solar wind
van Allen belts
solar corona
Do some searching and find out
typical values of n, Te, Ti, fractional ionization
what is the magnetic field environment?
what are the principal ionization and deionization processes?
Give references.
Debye length and plasma frequency
6.
For the radio-frequency discharge in the Senior Physics Lab, T = 3 eV, ne =
17
−3
10 m and diameter about 100 mm,
(i) calculate λD,
(ii) calculate ND the number of electrons in a Debye sphere.
7.
Add some curves of constant Debye length and constant plasma frequency to
the figure on p.2.
8.
Where would a solid state plasma fit on the figure on p.2?
Take T = 300 K, and estimate ne by assuming the solid is sodium and that each
atom contributes one electron.
Summary of chapter
You should be able to
Describe ionization by electrons in a gas discharge, the role of positive ions,
deionization.
Do calculations of
ε kT
Debye length λ D = 0 2 ,
n0e
plasma frequency ω pe =
ne e 2
.
ε 0 me
The derivations will not be examinable.
In subsequent sections you will be expected to use Maxwell’s equations in differential
form.
∇⋅E =
σ
ε0
∇×E= −
∂B
∂t
∇⋅B = 0
∇ × B = µ0 j+
where, in cartesian coordinates
∂E x ∂E y ∂Ez
∇.E =
+
+
.
∂y
∂z
∂x
(∇ × E) x
1 ∂E
c 2 ∂t
=
∂Ez ∂Ey
, etc.
−
∂z
∂y
and in cylindrical coordinates,
1 ∂ (rE r ) 1 ∂Eφ ∂E z
∇⋅E =
+
+
.
r ∂r
r ∂φ
∂z
 1 ∂E z ∂Eφ   ∂E r ∂E z 
 1 ∂ (rEφ ) ∂E r 
rˆ + 
zˆ
∇ × E = 
−
−
−
φˆ + 
∂z   ∂z
∂r 
∂φ 
 r ∂φ
 r ∂r
Note that we use σ for (volume) charge density to distinguish it from ρ for mass
density.
PLASMA PHYSICS
II. MOTION OF IONS AND ELECTRONS IN E AND B
FIELDS
We consider the paths of ions and electrons in E and B fields for some simple cases.
The Lorentz force on a point charge is F = q( E + v × B) .
E is measured in V m−1. B in T (often in gauss. 10000 gauss = 1 T)
1. E = constant, uniform
Suppose E = Ex$ . The Lorentz force equation becomes
dv x
q
= E
dt
m
dv y
=0
dt
dv z
= 0.
dt
This describes a constant acceleration along x.
2. B = constant, uniform
Suppose B = B$z .
Here is how we might produce a uniform magnetic field.
dv x
q
= vy B
(1)
dt
m
dv y
q
= − vx B
(2)
dt
m
dv z
= 0.
(3)
dt
d
of (1), substitute using (2)
dt
d 2 vx
q dvy
q q
B =  vx B B .
=
2

dt
m dt
mm
Take
Write
qB
m
the (angular) cyclotron frequency or gyrofrequency. (Note the symbol Ω is often
used.)
ωc =
d 2 vx
+ ω c 2 vx = 0 .
2
dt
Similarly,
d 2 vy
dt
The solutions can be written as
2
+ ω c 2 vy = 0 .
vx = − v⊥sinω c t
vy = m v⊥ cosω c t
(The signs and phase angles have been chosen to match the sketches below. The upper
sign is for a positive charge, the lower for a negative.)
Integrate again
x=
v⊥
ωc
y=m
rL =
v⊥
ωc
cosω c t = rL cosω c t
v⊥
ωc
sinω c t = m rL sinω c t
is called the Larmor radius, radius of gyration, or gyroradius.
So a charge in a constant, uniform B moves in a circle with constant speed. Note that
the cyclotron frequency does not depend on how fast the charge is moving.
You do. Check that the directions of motion are correct and that the equations above
match the sketches.
You do. Calculate the cyclotron frequency (in Hz) for (a) hydrogen ions and (b)
electrons in a magnetic field of 1 T.
If the charge has a vz, this z-component of the motion is unchanged. The charge moves
in a helical path.
3. E constant, uniform. B constant, uniform.
Suppose B = B$z .
(
)
(1)
(
)
(2)
dv x
q
Ex + vy B
=
dt
m
dv y
q
Ey − v x B
=
dt
m
dv z
q
= Ez .
(3)
dt
m
(3) gives constant acceleration along z.
You do. Suppose E has a z-component only. Describe the motion and sketch the path
for this case.
(1) and (2) are manipulated as before, they give
Ey
d 2 vx
+ ω c 2 vx = ω c 2
and
2
dt
B
d 2 vy
E
+ ω c 2 vy = −ω c 2 x ,
2
dt
B
with solutions
Ey
v x = − v ⊥ sinω c t +
B
Ex
v y = m v ⊥ cosω c t −
.
B
The path of an electron is a combination of uniform circular motion plus a drift, called
an E × B drift.
1
E×B
v E×B =
E y x$ − E x y$ =
.
B
B2
Note that the drift term is independent of the charge and its sign, so all the charges
will drift together. The paths are cycloids.
(
)
If E⊥B then v E × B =
E
.
B
Here is an example where the E × B drift can cause a plasma to rotate.
4. B = constant, non-uniform.
We will consider two distinct cases.
Case (a):
Bz = B0 (1 + αz )
Bx = − B0
By = − B0
α
2
α
2
x
y
where α is small.
You do. Show Maxwell’s equations ∇. B = 0 and ∇ × B = 0 are satisfied as long as we
include these small Bx and By terms.
The Lorentz force equation becomes
dv x
q
=
v y Bz − v z By =
dt
m
dv y
q
= (v z Bx − v x Bz ) =
dt
m
dv z
q
=
v x By − v y Bx =
dt
m
(
(
)
α
q
 v y B0 + v y B0αz + v z B0 y
m
2 
)
q
α
 − v z B0 x − v x B0 − v x B0αz

m
2
α
α
q
 − v x B0 y + v y B0 x

m
2
2 
We will write v = v 0 + v 1 where 0 indicates the uniform constant or zero-order part
and 1 a small first-order correction, of the same order as α, and substitute in the
equations. This is a standard approach and we will use it frequently.
The zero-order equations. If we write down the zero-order terms, i.e., the terms in
v 0 and those that do not contain α, the equations that remain describe motion in a
constant, uniform B. This was discussed earlier.
You do. Show this.
The first-order equations. We first solve the zero-order equations to obtain
v x0 , v y0 , v z0 , x 0 , y 0 , z 0 ; then we write down the first order terms, i.e., the terms in v 1 and
those containing α; then substitute for v x0 , v y0 , v z0 , x 0 , y 0 , z 0 .
This gives
α
dv1x
q
=  v1y B0 + v⊥0 cosω c t B0αvz0t + vz0 B0 rL sinω c t 

dt
m
2
dv1y
dt
=
q 0 α

 − vz B0 rL cosω c t − v1x B0 − v⊥0 sinω c t B0αvz0 t 


m
2
dv1z
q
α
α
=  v⊥0 sinω c t B0 rL sinω c t + v⊥0 cosω c t B0 rL cosω c t 

dt
m
2
2
which can be written as
dv 1x
α
= ±ω c v 1y − αv ⊥0 v z0ω c t cosω c t − v z0 vz0 sinω c t
dt
2
1
dv y
α
= m v ⊥0 vz0 cosω c t m ω c v 1x ± αv ⊥0 v z0ω c t sinω c t
dt
2
1
dv z
α 2
= − v ⊥0
dt
2
upper sign ions, lower sign electrons, with solutions
v 1x = −
v 1y = m
v 1z = −
α
2
α
2
α
2
v ⊥0 v z0 ω c t 2 cosω c t −
v ⊥0 v z0 t cosω c t ±
α
2
α
2
v ⊥0 v z0 t sinω c t
v ⊥0 v z0ω c t 2 sinω c t
2
v ⊥0 t.
Adiabatic invariant
1 2
mv ⊥
2
is a constant of the motion or adiabatic invariant. Adiabatic carries the idea of
B
slowly-changing.
(
You do. Show this is true to first order in α. Start with v ⊥2 = ( v x0 + v 1x ) + v y0 + v 1y
2
)
2
and
substitute using the solutions above. (Note. Chen p 31 gives an alternative derivation.)
You do. Use the definition of rL and this result to show that the magnetic flux
encircled by orbit Φ M = BA is constant.
It follows that the magnetic moment of the gyrating charge is constant. The magnetic
q
moment is defined as µ = iA where i = , where q is the charge and t is the time for
t
one gyration and A is the area encircled by the orbit.
1 2
mv⊥
e
2
2
.
πr =
µ=
2π L
B
So µ is constant.
ωc
Magnetic mirror
As an electron spirals into a higher B region, v ⊥ increases and rL decreases. Since the
1
total energy mv 2 is a constant, vz must decrease. Eventually vz = 0 and the electron
2
reverses direction. It has been reflected by a magnetic mirror.
e.g., magnetic mirror used to trap plasma in an experimental device.
1 2
mv ⊥
2
To do a magnetic mirror calculation use
= constant and conservation of
B
1
1
energy mv ⊥2 + mv z2 = constant.
2
2
Case (b):
Bz = B0 (1 + αx )
Bx = B0αz
You do. Show Maxwell’s equations are satisfied.
B0αx describes the gradient.
B0αz describes the curvature. In the derivations below, the curvature terms are
underlined.
dv x q
= (v y B0 + v y B0αx )
dt
m
dv y
q
=
v z B0αz − v x B0 − v x B0αx
dt
m
dv z
q
= (− v y B0αz )
dt
m
(
)
Proceed as before.
dv 1x
2
= ±ω c v 1y − αv ⊥0 cos 2ω c t
dt
dv 1y
2
2
= ± αω c v z0 t m ω c v1x ± αv ⊥0 sinω c t cosω c t
dt
dv 1z
= αω c v ⊥0 v z0 t cosω c t
dt
upper sign ions, lower sign electrons, with solutions
αv 0
2
v = − ⊥ sin 2ω c t + αv z0 t
2ω c
2
1
x
αv ⊥0
αv ⊥0 αv z0
v =m
cos2ω c t ±
±
ωc
2ω c
2ω c
2
2
2
1
y
αv ⊥0 v z0
(cosω c t − 1 − ω c t sinω c t )
v =
ωc
1
z
Consider v 1y . There are constant drift terms.
αv⊥0
±
due to the gradient
2ω c
2
αv 0
± z due to the curvature.
ωc
2
They combine to give
vd =
 v ⊥0 2
2
±α 
+ v z0 
 2

ωc
.
This is the expression we will use.
It is perhaps unfortunate that these drifts are in the same direction. We cannot devise a
B such that they cancel.
We can express α in terms of the gradient of the magnetic field or the radius of
curvature Rc of the field lines.
(i) From the equations for B above, α =
(ii) From the sketch above, α =
You do. Show this.gradient drift
1
Rc
∇B z
.
B0
It is easy to see why a gradient gives rise to a drift. Consider the path of a charge
where the B field is large above the line and small below it. Above, the Larmor radius
is small and below, it is large.
We can sketch the drift.
The drift, for positive ions, is in the direction of − ∇B × B or R c × B .
e.g. in a toroidal magnetic field
B=
µ 0 Ni
2πr
1
1
and ∇B ≈ − 2 . i.e., ∇B increases as you go radially in towards the axis.
r
r
∇B
1
In this case α =
=− .
B
r
so B ≈
e.g., radiation belts in the earth’s magnetic field. This illustrates the magnetic mirror
as well.
5. Magnetic field with time variation
Drift and mirroring equations do not allow the long range prediction of trajectories,
particularly if there is no symmetry. It is nice to have constants of the motion or
invariants.
Again it can be shown that even when the magnetic field varies in time, the magnetic
moment µ is constant. This is the first adiabatic invariant.
e.g. Adiabatic compression as a method of heating a plasma
Suppose a plasma is trapped by a magnetic field. If the magnetic field is increased
then v⊥ increases. Collisions will distribute this extra energy. The plasma is heated.
There are two other invariants. They are illustrated by the following example.
(1) µ = constant
(2) Longitudinal (or second) adiabatic invariant
J=
∫
over a path
back and forward
between mirrors
v ⋅ dl = constant
So if the location of the mirrors changes slowly with time, due to the solar wind, this
remains constant.
(3) Third adiabatic invariant
The guiding centre may precess going from one field line to another. But the field
lines all lie on a flux surface - a barrel-shaped surface such that the enclosed flux is
constant.
Exercises
B = constant, uniform
1.
Calculate cyclotron frequencies and Larmor radii for
(i) 18 keV deuteron in a fusion reactor. B = 5.7 T
(ii) 5 eV electron in a plasma CVD source. B = 200 gauss.
(iii) 10 keV electron in the earth’s magnetic field. B = 0.5 gauss.
E constant, uniform. B constant, uniform.
2.
In a low temperature plasma device called a magnetron, B is typically 300
gauss, the potential difference V is 500 V over 2 mm in the region of interest and the
E is perpendicular to the B.
Estimate the drift velocity of the electrons.
B = constant, non-uniform. Case (a):
3.
(a) In a magnetic mirror where the magnetic field is B0, the trajectory makes
an angle θ0 with the magnetic field line.
B0
Show that reflection occurs where the magnetic field is B =
.
sin 2θ 0
(b) Suppose now Bmax is the maximum value of the magnetic field.
B0
then there is no reflection. In a magnetic mirror
Bmax
device, this would mean the particle was not trapped - it would be lost. This angle
defines a cone in velocity space - the loss cone
Show that if θ 0 < sin −1
4.
In our gyrotron millimetre-wave source most of the electrons travel from the
electron gun through the resonant cavity. But some electrons are reflected.
At the electron gun (z = −0.30 m) the magnetic field is 0.52 T, at the resonant
cavity (z = 0 m) it is 12.0 T.
Consider a particular electron that is reflected at the cavity. It leaves the gun
with an energy of 10 keV. Its guiding centre is 3.00 mm from the axis.
Calculate
(i) v⊥ at the cavity,
(ii) v⊥ and hence vz at the gun,
(iii) the distance to the guiding centre from the axis at the
cavity,
(iv) the Larmor radius at the cavity,
(v) the Larmor radius at the gun.
B = constant, non-uniform. Case (b):
5.
In a small experimental plasma device a toroidal B is produced by uniformly
winding 120 turns around a toroidal vacuum vessel, and passing a current of 250 A
through it. The major radius is 0.6 m.
A plasma is produced in hydrogen by a radiofrequency field. The electron
temperature is 80 eV and the ion temperature is 10 eV. The two temperature
distributions are Maxwellian. The plasma density at the centre of the vessel is 1016
m−3.
(i) Calculate the B field at the centre of the vessel
(ii) Calculate the total drift for both ions and electrons at the centre of the
vessel. On a sketch, show the directions of these drifts.
3
R 
6.
The earth’s magnetic field, in the equatorial plane, is B = 3 × 10  e  T .
 r 
At 5 Re in one of the Van Allen radiation belts, the electrons have an energy of
30 keV and the protons an energy of 1 eV.
−5
(i) Calculate the total drift for both protons and electrons. On a sketch show
the directions of these drifts.
(ii) If the plasma density is 105 m−3, calculate the ring current density.
Re = 6.37 × 106 m.
7.
For the Van Allen belt in Exercise 6, estimate the following times.
(i)
cyclotron period,
(ii)
time between mirror reflections,
(iii) time to drift once around the earth.
Summary of chapter
1. E = Ex$ . Constant acceleration along x.
2. B = B$z . Helical motion along z.
(angular) cyclotron frequency ω c =
Larmor radius rL =
qB
m
v⊥
ωc
3. E constant, B = B$z .
(i) if E = E$z gyration plus acceleration along z.
(ii) if E = Ex$ gyration plus E × B drift along y. vd =
4. B non-uniform.
Case (a):
E
B
1 2
mv ⊥
2
leads to
= constant, this quantity is an adiabatic invariant.
B
For magnetic mirror calculations use this and conservation of energy.
Case (b):
Constant drifts along y.
 v⊥0 2
2
±α 
+ vz0 
 2

vd =
ωc
where α =
∇B
, positive ions drift in the direction of −∇B × B .
B
You should be able to
Do calculations of cyclotron frequency, Larmor radius.
Recognise when E × B, gradient and curvature drifts occur. Describe what is
happening and carry out calculations. The derivations will not be examinable.
Describe magnetic mirror and carry out calculations.
Apply to toroidal plasmas and earth’s radiation belts.
PLASMA PHYSICS
III. FLUID DESCRIPTION OF A PLASMA
Fluid mechanics
Liquids and gases can be characterized by the physical quantities; density ρ, pressure
p, velocity v and temperature T.
In Fluid Mechanics the fluid is treated as a continuous medium. We look at what
happens to large numbers of molecules. This is a macroscopic approach as distinct
from the microscopic approach in Chapter II.
The key equations deal with mass, momentum and energy.
We obtain the equations either by Method 1 where we consider a fluid particle, or by
Method 2 where we look at some property carried along by the fluid through a fixed
volume.
Mass
Here we use Method 2.
mass flowing into the infinitesmal volume in ∆t
∂


= ρvx ∆t ∆y∆z −  ρvx + ( ρvx )∆x ∆t ∆y∆z +K


∂x
= −∇ ⋅ ( ρv) ∆x∆y∆z ∆t
increase in mass in ∆t
=
∂
( ρ ∆x∆y∆z) ∆t
∂t
Since mass is conserved, these are equal, so
∂ρ
+ ∇ ⋅ ( ρv ) = 0 .
∂t
This is the equation of conservation of mass or the Equation of continuity.
(For liquids and often for gases we make the approximation that the fluid is
incompressible i.e., ρ does not change and this equation reduces to ∇ ⋅ v =
0.
We usually express this equation in integral form and use it to solve problems like:
v1 A1 = v2 A2 + v3 A3 )
Momentum
This time we use Method 1.
First some mathematics. How does the scalar property T of a fluid particle change as
the fluid goes from point P at (x, y, z) to point P′ at (x+∆x, y+∆y, z+∆z) in time ∆t?
T = T ( x, y, z, t )
∂T
∂T
∂T
∂T
∆T =
∆x +
∆y +
∆z +
∆t
∂x
∂y
∂z
∂t
Divide throughout by ∆t and let ∆t → 0.
dT ∂T
=
+ v ⋅ ∇T
dt
∂t
dT/dt is called the time derivative following the motion. It is the total change in T as
the fluid particle passes P. It is equal to (i) the change in T because T at P is a
function of time plus (ii) the change in T because the particle is moving past P at
velocity v and T varies with position. (This last change is known as the convective
change).
Simple example. A river. Velocity of water is 10 km/day.
T is the temperature. At P, a fixed thermometer indicates T increases 0.2°C / day.
Near P, T increases in the direction of the flow 0.07°C / km.
Consider a thermometer drifting with the water. It measures the total rate of change
in T.
dT ∂T
∂T
=
+u
dt
∂t
∂x
= 0.2° C / day + 10 km / day × 0.07° C / km
= 0.9° C / day.
Momentum equation
Let T = vx.
dvx ∂vx
=
+ v ⋅ ∇vx
dt
∂t
Now dvx/dt is the acceleration ax, and by Newton’s 2nd Law Fx = max.
Write m = ρ∆x∆y∆z.
Define force per unit volume
fx =
Then
Fx
∆x∆y∆z
∂vx
+ ρv ⋅ ∇vx = fx , or
∂t
∂v
ρ
+ ρv ⋅ ∇v = f
∂t
ρ
In fluid mechanics, f is separated into three parts:
(i) force due to pressure fpressure = −∇p
(ii) force due to shear stresses. This gives a viscosity term.
(iii) body forces e.g., gravity fgravity = ρg
This leads to the Navier-Stokes equation
ρ
∂v
+ ρv ⋅ ∇v = −∇p + µ∇ 2 v + ρg .
∂t
(We usually express this equation in integral form and use it to solve problems like:
− ρv 1 v1 A1 + ρv 2 v2 A2 = F )
Energy
We can take v ⋅ of the momentum equation, integrate and get Bernouilli’s equation
which is just a form of the conservation of mechanical energy equation in a form
suitable for fluid mechanics.
(We usually express this equation in integral form and use it to solve problems like:
1 2
ρv + p + ρgh = constant)
2
Plasma
A plasma is an electrical fluid. We must add to the list of physical quantities; charge
density σ, current density j, electric field E and magnetic field B.
Continuity and momentum
ρ
∂ρ
+ ∇ ⋅ ( ρv ) = 0 .
∂t
∂v
+ ρv ⋅ ∇v = nq( E + v × B) − ∇p
∂t
p here is isotropic. More generally, the pressure term is
∇ ⋅ P where P is a pressure
tensor. The off-diagonal terms would be associated with viscosity.
Suppose we have a fully-ionized plasma of (a single variety of singly-charged, to keep
things simple) ions and electrons. It is a mixture of two fluids. An ion fluid and an
electron fluid.
For ions, the momentum equation is
∂v
ρ i i + ρ i v i ⋅ ∇v i = ni e( E + v i × B) − ∇pi − ρ iυ ie (v i − v e )
∂t
and for electrons it is
∂v
ρ e e + ρ e v e ⋅ ∇v e = − nee(E + v e × B) − ∇pe − ρ eυ ei ( v e − v i )
∂t
A force term due to collisions between the ions and the electrons has been added.
The νs are called collision frequencies for momentum transfer. Since momentum must
be conserved in collisions,
ρ iν ie ( vi − v e ) + ρ eν ei (v e − v i ) = 0 or
ρ iν ie = ρ eν ei .
In a less than fully-ionized plasma there would be equations for the neutrals as well.
In general,
ρα
∂ρ α
+ ∇ ⋅ (ρ α vα ) = 0
∂t
∂v α
+ ρ α v α ⋅ ∇v α = nα qα (E + v α × B) − ∇pα − ∑ ρ α υ αβ v α − v β
∂t
β
(
)
where α = i, e or n.
The neutral fluid interacts with the others only via collisions. The ion and electron
fluids interact via fields even in the absence of collisions.
Energy
It is difficult to include collisions. We normally use an equation of state instead.
If isothermal conditions apply, use
p = nkT ;
if adiabatic conditions apply use
pρ −γ = constant ,
c
5
where γ = p = for a monotomic gas .
cv 3
Both choices lead to ∇p = U 2 ∇ρ where U is a sound speed.
You do. Differentiate these expressions and obtain
isothermal sound speed
adiabatic sound speed
U2 =
U2 =
p
ρ
, and
γp
.
ρ
We have a self-consistent problem here. E and B depend on the charges in the plasma
and how they move; and how they move depends on E and B.
We have more than enough equations at this point to solve for the 16 unknowns ni,
ne, vi, ve, pi, pe, E and B. We have continuity (1 for ions and 1 for electrons),
momentum (3 components for ions and 3 for electrons), energy (1 and 1), Maxwells
equations (8), a total of 18. We can drop two Maxwells equations ∇ ⋅ E and ∇ ⋅ B .
They can be obtained by taking ∇ ⋅ of the other Maxwells equations. This leaves 16
equations in 16 unknowns.
We usually let ni = ne = n and avoid using ∇ ⋅ E =
σ
. This is the so-called plasma
ε0
approximation, closely related to the idea of quasineutrality.
E × B and diamagnetic drifts
Consider the ion momentum equation. Simplify it by ignoring time variations and
collisions. The equation becomes
0 = ni e(E + v i × B) − ∇pi .
Take equation × B, use v = v ⊥ + v
v i⊥ =
and rearrange
E × B ∇pi × B
−
.
B2
ni eB2
You do. Show this.
The first term is the E × B drift of the guiding centres we obtained in the single
particle approach.
The second term is called the diamagnetic drift. It does not involve any motion of the
guiding centres. There is no drift equivalent to this in the single particle approach.
You can see how it arises from the sketch.
There will be a net vi to the right.
The diamagnetic drift is in the opposite direction for electrons. The ion and the
electron drifts combine to give a diamagnetic current, to the right in the sketch.
The direction of the current is such as to reduce the magnetic field - hence
"diamagnetic".
e.g., in a cylindrical plasma with a density gradient radially inwards.
Plasma as a single fluid
We can define a density for the plasma as a whole,
ρ = ∑ ρα
α
and a velocity for the plasma as a whole,
ρ v
v=∑ α α
α
ρ
Continuity equation
Add the equations of continuity of all the species
∂ρ
+ ∇ ⋅ ( ρv ) = 0 .
∂t
Continuity of charge
1
1
× the ion continuity equation −
× the electron continuity equation.
mi
me
Use charge density σ = en1 − ene , current density j = eni v i − ene v e .
Take
∂σ
+ ∇⋅ j = 0 .
∂t
Momentum
Add the momentum equations
But
(i) linearize. This means neglect any quadratic terms in v. This is a
considerable
simplification.
(ii) use
total pressure
p=
pα
∑
α
momentum is conserved so
∑ ∑ ραυ αβ (vα − v β ) = 0 .
α
β
∂v
ρ = σE + j × B − ∇p
∂t
At this point let ni = ne = n. The Continuity of charge equation is no longer required
and σE in the Momentum equation can be dropped.
ρ
∂v
= j × B − ∇p
∂t
Generalized Ohm’s Law
For an ordinary conductor, E = ηj , where η is the resistivity. (Recall V = RI, hence El
= RjA so η = RA/l.)
For a plasma, take
1
1
× ion momentum equation −
× electron momentum
mi
me
equation.
But
(i) linearize
(ii) use
ni = ne = n
me << mi
Some useful manipulations include showing that ρv = ρivi + ρeve and j = en(vi - ve)
lead to v i
= v+
me
m
j and v e = v − i j ,
eρ
eρ
νie << νei leads to a collision term nνei(ve−vi) which we write as
η=
meν ei
is a constant of proportionality.
ne 2
−
ne
ηj . Here
me
me ∂j
1
1
= E+v×B−
j × B + ∇pe − ηj
2
ne ∂t
ne
ne
The j × B term is known as the Hall term.
You do. Refer to a textbook on the Hall effect in a solid material to see the connection.
Approximations
Various approximations are generally made.
1. Quasineutral approximation. ni = ne = n
2. Steady-state or very slow time variations.
∂v ∂j
,
are negligible.
∂t ∂t
3. Cold plasma. ∇pi , ∇pe are negligible.
Under these approximations, Maxwell’s ∇ × B equation becomes
∇ × B = µ0 j ,
the momentum equation reduces to
j× B = 0 ,
and the generalized Ohms law reduces to
0 = E + v × B − ηj .
The set of equations including this approximation is known as the
magnetohydrodynamic or MHD equations. These are standard tools for treating large
scale plasma motion.
4. Infinite conductivity. Resistivity can be neglected and
0 = E+v×B.
When this is true the magnetic field lines can be regarded as being frozen into the
plasma. We talk about the ideal MHD equations.
Confinement of a plasma
There are two ways in which we might discuss plasma confinement.
The first way is in terms of forces.
Consider a plasma in a cylinder where the plasma density falls off from a maximum at
the centre to zero at the wall.
The momentum equation
0 = j × B − ∇p
says that if there is equilibrium the pressure forces and the j × B forces balance.
The second more picturesque way is in terms of magnetic pressures and magnetic
tensions.
Substitute for j using ∇ × B = µ 0 j , rearrange to obtain

B2  (B ⋅∇)B
∇ p +
.
=
2µ 0 
µ0

If we have a straight cylinder of plasma, the rhs is zero and
B2
p+
= constant .
2µ 0
B2
is the magnetic pressure.
2µ 0
B2
The magnetic tension =
N m−2. If the field lines are straight, this is in the
p is the particle pressure,
µ0
direction of the field lines, if curved there is a
related to this component.
⊥ component. The term on the rhs is
Near the centre n is larger, so p is larger and B is smaller (this is plasma
diamagnetism); near the wall n is smaller, so p is smaller and B is larger.
Compare this with a gas filling a cylindrical vessel.
Pinch effect
Axial current jz heats plasma and azimuthal Bθ due to this current confines the
plasma. Nice idea as a way of achieving fusion conditions but extremely unstable.
In this analysis you will be working in cylindrical coordinates and will make
the assumption that there are only radial variations.
The plasma has radius R. The total current is I.
(a) Suppose the current is distributed uniformly throughout the plasma.
Show I = jzπR2 .
µ Ir
1 dB z
and Bθ ( r ) = 0 2 .
2πR
µ 0 dr
(c) Write the radial component of the momentum equation 0 = j × B − ∇p ,
substitute.
Show
2
µ I 2r
Bz 
d 
=− 0
.
p
+
dr 
2 µ 0 
2π 2 R 4
(b) Use ∇ × B = µ 0 j and show jθ (r ) = −
(d) Finally, integrate over the range 0 to R and show
Bz2 ( R)
Bz2 ( 0) µ 0 I 2
= p( 0) +
− 2 2.
2µ 0
2µ 0
4π R
I = 0. This is like the case discussed above.
If I is large enough Bz(R) < Bz(0) and plasma is squashed. (This is similar to the effect
we see in the force between parallel currents demonstration and the squashing of a
lightning conductor.)
Problems with the pinch
(i)
Linear pinch with electrodes at the ends. There will be cooling at the ends.
To overcome this problem:
Bend the cylinder into a torus so there are no ends.
(ii)
There are instabilities.
Bz = 0. There is the sausage instability (b) and the kink instability in (c). The
regions where Bθ is stronger are shown. In these regions the magnetic pressure on the
plasma is larger and the instability becomes worse.
To overcome these instabilities:
(i) Apply a Bz. The field lines are frozen-in. In (b) the magnetic pressure is
increased where the plasma is being squeezed thus opposing the squeezing. In (c) the
magnetic tension is increased tending to straighten out the kink.
(ii) Use a vessel with conducting walls. If the field is applied suddenly, the
field lines are squashed against the wall increasing the magnetic pressure there.
The sausage instability is an example of an interchange instability. An example from
fluid mechanics is the Rayleigh-Taylor instability where one fluid is floating on a
second, less dense fluid. Another example from plasma physics is the flute instability.
Tokamaks and stellarators
The toroidal coil windings provide a toroidal B. Recall that B is larger the closer you
get to the centre of the torus and this ∇B causes drifts of the ions and electrons. The
drifts are in opposite directions so an E builds up causing an E × B drift. The plasma
moves out to the walls and cools.
In a tokamak, there is a plasma current around the torus so the lines of B are twisted
helically.
The E is shorted out. In a tokamak, this current also provides some heating.
In a stellarator, there is a helical coil winding to provide the twist. There is no need for
a current in the plasma, except perhaps to heat it.
Exercises
There is a standard way of checking out the relative importance of the terms in these
equations in order to make approximations. Choose a scale length L so that any space
∂u u
≈ and choose a scale time τ so that any time
derivative can be written as
∂x L
∂u u
derivative can be written as
≈ .
∂t τ
e.g., ∇ × E = −
∂B
E B
L
becomes ≈ . Write ≈ V a velocity, so E ≈ BV .
∂t
L τ
τ
1.
Examine the single fluid momentum equation. Show the terms are in the ratio
2
nme vth2 e
V
jBτ me vth e
nmi : jB:
or 1:
:
.
τ
L
nmi V mi V 2
This suggests that the ∇p term could be ignored if the plasma is cool so we can use
jBτ
V≈
.
nmi
2.
Examine the Generalized Ohms Law.
(a) Show the terms are in the ratio
2
1
1
1 vth e
ν ei
:11
::
:
:
.
2
2
ω ceω ciτ
ω ciτ ω ceτ V ω ceω ciτ
(b) Which terms are important or can be neglected if
1
1
1
1
(i) τ >>
, (ii) τ ≈
, (iii) τ ≈
and (iv) τ <<
?
ω ci
ω ci
(c) When can the ηj term be dropped?
ω ce
ω ce
1
3.
Show diamagnetic drift << E × B drift when τ >>
4.
Show that the quasineutrality approximation is good if
ω ci
.
ω ceω ci
<< 1 , i.e., if the
ω 2pe
density is sufficiently large
(a) Look at the term σE in the momentum equation. Compare it with j × B .
(b) Look at the
∇⋅ j.
∂σ
term in the continuity of charge equation. Compare it with
∂t
You will need to use ∇ ⋅ E =
5.
σ
.
ε0
(a) Show that for a cylindrical plasma in equilibrium,
µ I 2r
B2 
d 
 p + z  = − 0 2 4 .
dr 
2µ 0 
2π R
(b) Suppose
the plasma radius is R = 0.1 m,
the current density jz is uniform,

r2 
p( r ) = p( 0) 1 − 2  , p(0) = 100 kPa,

R 
Bz(0) = 1 T.
Draw graphs of p(r), Bθ(r) and Bz(r) for currents (i) I = 10 kA and (ii) I = 300
kA.
6. MHD generators and propulsion
The current produced by the moving plasma can deliver power to a load.
Compare the MHD generator with a conventional generator. In the latter, the fluid
energy is converted to kinetic energy of a conductor. The conductor is moving in a
magnetic field so the kinetic energy is converted into electrical energy. In the MHD
generator the intermediate step is eliminated leading to increased efficiency.
A conducting fluid flows (in the x-direction) in a magnetic field (in the zdirection). The generated voltage appears on the electrodes.
(i) Use generalized Ohms law to obtain an expression for the potential
difference between the electrodes. Suppose there is no load connected. Note that the
because of the electrodes jx = 0.
Working in reverse, applying a voltage to the electrodes, gives a plasma
propulsion system.
(ii) Use the MHD momentum equation to get an expression for the force on
the plasma.
Summary of chapter
There are two approaches:
1 Different species.
You should be able to use the continuity, momentum and energy equations.
∂ρ α
+ ∇ ⋅ (ρ α vα ) = 0
∂t
ρα
∂v α
+ ρ α v α ⋅ ∇v α = nα qα (E + v α × B) − ∇pα − ∑ ρ α υ αβ v α − v β
∂t
β
p = nkT or pρ −γ = constant
2 Single fluid.
(
)
You should be able to use the continuity, momentum, generalized Ohms law
and energy equations.
∂ρ
∂σ
+ ∇ ⋅ ( ρv) = 0 or
+ ∇⋅ j = 0
∂t
∂t
∂v
ρ = j × B − ∇p
∂t
me ∂j
1
1
j × B + ∇pe − ηj
= E+v×B−
2
ne ∂t
ne
ne
The derivations will not be examinable.
B2
Describe E×B drift, diamagnetic drift, magnetic pressure
.
2µ 0
Describe plasma confinement, pinch effect and instabilities and how they
might be overcome.
Describe tokamaks and stellarators.
Note
The equations for ions and electrons can be derived from kinetic theory
considerations. This approach starts with the Boltzmann equation which describes
how the particle distribution function varies in time and space.
PLASMA PHYSICS
IV. DIFFUSION
Collision parameters
Suppose electrons are incident on a slab area A, thickness dx, containing neutral
atoms, density nn. The atoms are regarded as solid spheres cross-sectional area σ.
n Adxσ
= nnσdx .
The fraction of the slab blocked by atoms is n
A
Flux Γ is defined as Γ = nv .
Suppose incident flux of electrons is Γ(x).
Flux emerging from the other side Γ(x+dx) = Γ(x)(1 − nnσdx)
Rewrite this as
dΓ
= − nnσΓ
dx
with solution
−
x
Γ = Γ( 0) e
or Γ ( 0) e
.
λm is the mean free path, the distance for the flux to fall to 1/e of its initial value.
1
λm =
.
nnσ
− nnσx
λm
The mean time between collisions for particles with velocity v is τ =
The collision frequency ν =
1
τ
λm
v
.
. (Strictly, we should average over the velocity
distribution.)
You do. Show λ m =
v
ν
.
Diffusion and mobility
Suppose the plasma density is higher near the centre of the plasma vessel. Ions and
electrons will collide with each other and with neutral particles and diffuse out to the
ends or the walls. Suppose there is an electric field. The ions and electrons will move
but the neutral particles will not. Diffusion and mobility leads to the loss of ions and
electrons from the plasma.
Diffusion
First, consider a gas of two kinds of neutral particles, A and B. The B particles are in
the minority.
There are two ways we might look at diffusion.
(i)
Suppose the background of A particles is uniform, but the density of the B
particles is not. The B particles collide with the A particles until the non-uniformity is
smoothed out.
The continuity and momentum equations for the B particles are
∂ρ B
+ ∇. ( ρ B v B ) = 0
∂t
ρB
∂v B
+ ρ B v B ⋅ ∇v B = −∇p B − ρ BAυ BA ( v B − v A )
∂t
Let us make some simplifications.
0
1
The non-uniformity is small so n B = n B + n B , where 0 indicates the uniform,
1
constant part and indicates a small first-order part that varies in space and time.
0
0
There are no zero order drifts so v A = 0 and v B = 0 .
We will use
p B = n B kTB .
Note that collisions between particles of the same type do not contribute to diffusion.
To first order, the equations become
∂n 1B
+ n B0 ∇ ⋅ v 1B = 0
∂t
(1)
∂v 1B
kTB
=−
∇n1B − ν BA v 1B
0
∂t
mB nB
Take
(2)
∇ ⋅ of (2) and substitute using (1).
∂n 1B
= DB ∇ 2 n 1B
∂t
kT
is the diffusion coefficient.
mν
1 ∂ 2 n1B
(Note a
term has been dropped.)
ν BA ∂t 2
The units of D are m2 s−1.
D=
The meaning of D
(a)
In terms of a scale length L, any initial non-uniformity is smoothed out in a
L2
time T ≈
.
D
(b)
D≈
2
vrms
ν
. Then D ≈
λ2m
. i.e., the diffusion coefficient is based on a random
τ
walk with a step equal to the mean free path between collisions.
You do. Show this.
(ii)
A steady state where there is a density gradient of the B particles. There will
clearly be a steady flow or flux of these B particles.
The momentum equation is (we will drop the subscripts and the superscripts)
0=−
kT
∇n
∇n − νv or v = − D
.
mn
n
The flux of the B particles is
Γ = nv = − D∇n .
This equation Γ = − D∇n is known as Fick’s Law.
Mobility
Next, suppose the B particles are electrons and there is an E field. The A particles are
still neutrals.
The momentum equation for electrons is
0 = −enE − ∇pe − me nνv e
Rearrange
ve = −
e
kT ∇n
E− e
meν
meν n
or
v e = − µ e E − De
∇n
n
e
is the mobility. µ and D are known as transport coefficients.
mν
The units of µ are m2 V−1 s−1.
µ=
These drift velocities are << the random velocities of the particles.
Free diffusion is driven by the density gradient and drift is driven by the electric field.
Weakly-ionized plasma, no magnetic field
Ambipolar diffusion
In a plasma there are ions and electrons. The electrons tend to diffuse more rapidly
than the heavier ions. If this results in ne being different from ni, an E field is
established. This E field accelerates the ions and slows down the electrons, so, to a
good approximation, they diffuse together. (This is the first key idea concerning
diffusion of a weakly-ionized plasma.)
Write down the equations for ions and electrons
∇n
n
∇n
v i = µ i E − Di
.
n
v e = − µ e E − De
Remember that in a weakly-ionized plasma νe and νi are electron-neutral and ionneutral collision frequencies. Electron-ion collisions can be ignored.
Set the electron flux equal to the ion flux
Γ = − nµ e E − De ∇n = nµ i E − Di ∇n .
Solve for E and substitute to obtain
Γ = − Da ∇n .
where Da =
µ e Di + µ i De
is the ambipolar diffusion coefficient.
µe + µi
From above µ e =
e
meν e
, µi =
e
kT
and ν ∝ v ∝
, so µ i << µ e .
miν i
m
 T 
Da ≅ 1 + e  Di .
Ti 

You do. Show this.
If Ti = Te,
Example
Let us apply
Da ≅ 2 Di
∂n
= Da ∇ 2 n to a plasma slab where the initial electron density profile is
∂t
shown in the sketch.
We will treat this as a 1-dimensional problem so
∂n
∂ 2n
= Da 2 .
∂t
∂x
This can be solved by separation of variables. Write n = f(x)g(t) and substitute.
This leads to
t
t
 x 
 x 
−
−
T
T
 + Be sin 
.
n = Ae cos
 Da T 
 Da T 
Now substitute the initial conditions. This gives
t
−
πx
L2
T
n = n 0 e cos  where T =
.
 L
πDa
The electron density profile remains the same but the peak decreases exponentially
with time.
You do. Show this.
Weakly-ionized plasma in a magnetic field.
First consider electrons.
Start with the momentum equation,
0 = −en( E + v × B) − ∇pe − m e nνv e
The z-component equation yields the same µe and De as before.
The x- and y-component equations are
De
n
D
v y = − µ e Ey − e
n
v x = − µ e Ex −
This yields
v ⊥ = − µ ⊥e E − D⊥e
where µ ⊥e =
µe
De
and D⊥e =
.
2 2
1 + ω ceτ
1 + ω 2ceτ 2
∂n ω ce
−
vy
∂x ν
∂n ω ce
+
vx
∂y ν
∇n v E × B + v dia
+
2 2
τ
n
1 + ω ce
If ω 2ceτ 2 << 1, B is small and has little effect on diffusion.
If ω 2ceτ 2 >> 1. B is large and diffusion across B is retarded.
This is the second key idea. Mobility and diffusion across the magnetic field
are smaller. In this case
λ2m
λ2m
rLe2
τ
D⊥e ≈ 2τ 2 ≈
≈
2
2
ω ceτ
τ
 v   λm 
,
   
 rLe   v 
i.e., the diffusion coefficient is based on a random walk with a step equal to the
Larmor radius.
Diffusion of ions and electrons is ambipolar but is complicated. Whether the diffusion
of a particle is primarily along z or prependicular depends sensitively on the plasma
boundaries.
Fully ionized plasma in a magnetic field
Collisions between electrons and ions. Can derive an expression for
an expression for the resistivity
η || = 5.25 × 10 −5
Z lnΛ
and
3
Te 2
It is usually adequate to take ln Λ ≈ 10 .
η ⊥ = 2η ||
υ ei
and obtain
(Te in eV).
Note that η is independent of n, decreases rapidly as Te increases.
Start with the single fluid equations (so diffusion in this case is automatically
ambipolar)
0 = j × B − ∇p
0 = E + v × B − ηj
Multiply the second by × B and substitute using the first.
v⊥ = −
η⊥
∇p − v E × B = −
η ⊥ nk (Te + Ti ) ∇n
B
B2
So we can define a D⊥ for a fully-ionized plasma.
Compare this with
2
D⊥ for a weakly-ionized plasma. It is ∝
n
− v E×B .
1
, but is ∝ n as well,
B2
decreases with T, and is automatically ambipolar.
Comment
Laboratory verification of the 1/B2 dependence proved elusive. The experimental
results were better described by the empirical formula
D⊥ =
1 kTe
.
16 eB
This was called Bohm Diffusion, usually many orders of magnitude larger.
Anomalous losses due to oscillations and asymmetries are responsible.
Exercises
1.
A positive column of a glow discharge in helium at p = 1 torr, no magnetic
field.
The helium and the helium ions are at room temperature, the electrons have an
energy of 2 eV.
(i) Estimate nn.
(ii) Estimate ve rms, vi rms.
The electron-neutral collision cross-section is about 5 × 10−20 m2.
(iii) Estimate λm.
(iv) Estimate νen.
(v) Hence estimate De, and µe.
For the ions, Di is around 0.02 m2 s−1 and µi is around 1 m2 V−1 s−1.
(vi) Estimate Da.
(vii) If the plasma density is 1016 m−3, the axial electric field is 10 kV m−1 and
the column diameter is 1 cm, estimate the current.
2.
Calculate the resistivity of a plasma. Take n = 1019 m−3 and Te = 104, 105,
6
7
10 , 10 , 108, 109 K.
3.
= 1 T.
Consider a plasma of thermonuclear interest. n = 1019 m−3, Te = 100 keV, B
(i) Calculate η. High temperature plasmas are good conductors and ohmic
(P = I2R) heating is no longer useful. Compare your value with the resistivity of
stainless steel.
(ii) Compare the classical diffusion coefficient D⊥ and the Bohm diffusion
coefficient for this plasma.
Summary of chapter
Definitions include cross-section σ , mean free path λ m , mean time between
collisions τ , collision frequency ν , flux, transport coefficients, diffusion coefficient
D, Fick’s Law, mobility µ , ambipolar diffusion.
You should be able to
Do calculations of diffusion and mobility coefficients.
neutral particles
kT
λ2m
D=
D≈
τ
mν
e
µ=
mν
weakly ionized plasma
µ D + µ i De
Da = e i
µe + µi
in magnetic field
r2
D⊥e ≈ Le
τ
PLASMA PHYSICS - KEY IDEAS – 1
ionization at high T
ionization in E (gas discharge)
linearize - drop vv terms
ni = ne
often steady state
often cold plasma - all of above,
MHD eqs.
often infinite conductivity - all of
above, ideal MHD eqs
characterize by n, T, B ...
magnetic pressure
characteristic length - Debye length
shielding
sheaths
pinch plasmas
current heats
magnetic field confines
but instabilities
I Introduction
plasmas everywhere
characteristic time - oscillation at plasma
frequency
II Motion of ions and electrons in E and B
fields
1. Treat single ions and electrons
in E, linear acceleration
in B, gyration
cyclotron frequency
Larmor radius
in E and B, E×B drift
in non-uniform B
(a) magnetic mirror
use adiabatic invariant
(b) gradient and curvature
drift (e.g., toroidal plasma, earth’s B)
toroidal plasmas
ion & electron drift ⇒ E ⇒ E×B drift
to wall
tokamak - toroidal current provides
helical twist in B (also some heating)
stellarator - helical coil (or figure 8)
provides twist
MHD generator
vplasma, B ⇒ E
and propulsion
E, B ⇒ vplasma
IV Diffusion
diffusion - collisions smooth out nonuniformities
III Fluid description of a plasma
mobility - how they move in E
2. Treat plasma as ion fluid + electron fluid +
neutral fluid
ambipolar diffusion
ions & electrons diffuse together
add σ, j, E, B to fluid equations
continuity eq.
momentum eq.
energy eq eq. of state
diffusion in strong B - Larmor radius replaces
mean free path
get new result - diamagnetic drift
3. Treat plasma as single fluid
continuity of mass eq.
continuity of charge eq.
momentum eq.
Generalized Ohm’s Law
maths: must simplify
PLASMA PHYSICS
V. WAVES IN PLASMAS − 1
Wave equation
Start with Maxwell’s curl equations,
∇×E = −
∂B
∂t
1 ∂E
c 2 ∂t
The wave fields, which we shall write as E1, B1 to show that we are treating them as
first order quantities, and their (first order) effects on particle densities and particle
velocities all show an e j ( k⋅r −ωt ) variation. k is the wave vector and ω is the (angular)
frequency.
∇ × B = µ0 j +
Maxwell’s equations can then be written,
jk × E 1 = jω B 1
jk × B1 = µ 0 j1 − j
ω
c
2
E1 .
Waves in a vacuum
You do.
Show that for a vacuum the only solution describes
(i) electromagnetic waves,
(ii) with their fields perpendicular to k, i.e., transverse waves,
(ii) with k =
ω
.
c
(There are no currents.)
One approach is to take k along z, say. Write out the component
Maxwell’s equations. Eliminate B1, leaving equations in E1.
Phase velocity
v ph ≡
ω
k
where c is the velocity of light in a vacuum.
=c
Waves in a plasma, no magnetic field
The ion and electron momentum equations are
∂v
ρ i i + ρ i v i ⋅ ∇v i = ni e( E + v i × B) − ∇pi − ρ iυ ie (v i − v e )
∂t
and
ρe
∂v e
+ ρ e v e ⋅ ∇v e = − ne e(E + v e × B) − ∇pe − ρ eυ ei ( v e − v i ) .
∂t
Write n = n0 + n1 and v = v0 + v1, where n1 and v1 show the e j ( k⋅r −ωt ) variation.
Substitute and consider only terms to first order. But before we do we will make some
simplifications.
(i) an infinite, uniform plasma,
(ii) no drifts, i.e., v0 = 0. So to first order terms like v.∇v = 0 ,
(iii) cold plasma, so terms like ∇p = 0 ,
(iv) no collisions so last terms are zero,
(v) no steady magnetic field, so B0 = 0.
Then to first order the momentum equations are
n 0 mi ( − jωv 1i ) = n 0 eE1
n 0 me ( − jωv 1e ) = − n 0 eE1 .
It follows that v1i = −
me 1
v e so ion motions are small. We shall neglect them.
mi
v 1i ≅ 0
e
E1 .
ωm e
The electrons move but the ions remain at rest in the background.
v 1e = − j
Using the definition of current density, we have
n0e2 1
j = en ( v − v ) ≅ j
E.
ωm e
Take k along z, say. Write out the component Maxwell’s equations. Eliminate B1, j1,
leaving equations in E1.
1
0
1
i
1
e
The E1z equation gives
ω2 =
n0e2
. (1)
ε 0 me
The other equations give
ω2
n0e2
k = 2 −
. (2)
c
ε 0 me c 2
2
Plasma oscillations
Equation (1) describes oscillations, not waves. These are called plasma oscillations.
n0e2
Recall ω 2pe =
. So the oscillation frequency is just the plasma frequency
ε 0 me
ω = ω pe .
Transverse electromagnetic waves
Equation (2) describes electromagnetic waves where the fields are perpendicular to
the direction of propagation, i.e., transverse electromagnetic waves. An equation like
this relating k and ω is called a dispersion relation.
ω 2 − ω 2pe
2
k =
c2
This will describe propagating waves as long as the rhs is > 0; i.e., ω > ω pe . If
ω < ω pe we say the wave is cutoff.
Critical density
Suppose we are carrying out a wave propagation experiment at a fixed frequency ω.
ne 2
If n is too high then, because ω 2pe =
, ω 2pe will be too large and the wave will be
ε 0 me
cutoff.
Define critical density as
nc =
If n > nc the wave is cutoff.
ε 0 m eω 2
e2
.
Suppose we have a plane wave normally incident on a slab of plasma. We can
calculate the powers transmitted and reflected at the boundary as a function of n. If
n > nc the wave does not penetrate; all the power is reflected.
Alternatively, we can define a cutoff frequency f c = 8.98 n .
Phase velocity
We can plot vph as a function of ω. v ph =
At a cutoff v ph = ∞ .
c
ω 2pe
1− 2
ω
Note vph > c! Is this a problem? No, recall that energy is transmitted at the group
velocity
dω
vg =
.
dk
c2
You do. Show that in this case, v g =
which is less than c.
v ph
We can think of a plasma as being like a dielectric medium with a refractive index
c
µ=
.
v ph
You do. Show that in this case
ω 2pe
n
µ2 = 1− 2 = 1− .
nc
ω
Relax some of the simplifications made earlier, one by one.
(i) Allow a collision frequency.
Waves in a plasma, no magnetic field, with collisions
vi is still small so the collision term in the electron momentum equation becomes
−n 0 meν ei v1e
1
the expression for j becomes
j1 = j
and
k =
2
n 0 e2
E1
me (ω + jν ei )
ω (ω + jν ei ) − ω 2pe
c2
.
The wave is attenuated.
(ii) Allow a finite electron temperature.
Waves in a plasma, no magnetic field, with finite electron
temperature
∇pe = Ue2 ∇ρ e = Ue2 jne1me k
∇ρ e = me ∇ne and it is easy to show that ∇ne = jne1 k .
The expression for j1 becomes
n 0 e 2 1 eUe2 1
E −
ne k
meω
me
instead of electron oscillations, now have a wave with dispersion relation
ω 2 − ω 2pe − k 2Ue2 = 0 .
j1 = j
(iii) Allow a steady magnetic field.
Waves in a plasma, with magnetic field
We will suppose the steady magnetic field is in the z-direction, B 0 z$ .
The electron momentum equation, to first order yields the following three component
equations for v 1e .
ω
e
E x1 − j ce v ey1
vex1 = − j
ωm e
ω
vey1 = − j
ω
e
E y1 + j ce vex1
ωm e
ω
vez1 = − j
e
E z1
ωm e
eB0
where ω ce =
is the electron cyclotron frequency.
me
From these, we obtain the three components for j1
j x1 = j
ω ce e 2 n 0
e2n0
1
E
+
Ey1
x
2
2
 ω ce



ω
ωme 1 − 2 
ω 2 me 1 − ce2 
 ω 
 ω 
ω ce e 2 n 0
e2n0
1
j =j
Ey −
E1x
2
2
 ω 
 ω 
ω 2 me 1 − ce2 
ωme 1 − ce2 
 ω 
 ω 
1
y
jz1 = j
e2n0 1
E
ωme z
We can think of the plasma as being like a conducting fluid.
j1 = σ ⋅ E1 where σ is the conductivity tensor defined as

ε 0ω 2pe
 j
 ω 2ce 

 ω 1 − ω 2 

ε 0ω ceω 2pe
σ =  −
 ω2 
 ω 2 1 − ce2 
ω 




0

ε 0ω ceω 2pe
ω 

ω 
ε 0ω 2pe


2
ce
2


ω ce2 

ω2 
ω 2 1 −
j
ω 1 −
0






0 


ε 0ω 2pe 

j
ω 
0
But, as we said above, we can think of a plasma as a dielectric medium. What is the
connection between these two pictures?
Recall
∇ × B = µ0 j +
1 ∂E
∂D
= µ0
and D = ε ⋅ E .
2
c ∂t
∂t
Substitute for j1 and show

ε = ε 0I +

j

σ .
ωε 0 
Instead of deriving dispersion relations for waves in any arbitrary direction we will
find them for the two simplest cases: propagation along B0 and perpendicular to B0.
Waves in a plasma. Propagation along B0
Take k along z, the direction of B0. Write out the component Maxwell’s equations.
Eliminate B1, j1, leaving equations in E1.
k 2 Ex =
k Ey =
2
ω2
ω2
c
2
ω2
c
2
Ex +
Ey +
ω 2pe
ω ce 

−
E
+
j
Ey 

x
2


ω


ω
c 2 1 − ce2 
ω 

ω 2pe
ω


 − E y − j ce E x 

ω
 ω 
c 2 1 −

ω 

2
ce
2
(1)
(2)
ω 2pe
( − Ez ) (3)
c2
c2
Equation (3) describes plasma oscillations at frequency ω = ωpe with E1 along k and
B0.
0=
Ex +
Equations (1) and (2) have a solution if their determinant is zero, i.e., if
k −
2
ω2
c
2
+
ω 2pe
j
 ω 
c 2 1 −

ω 

2
pe
2
ω 2pe
ω ce
−j
2
 ω pe  ω
c 2 1 − 2 
ω 

ω ce
 ω  ω
c 2 1 −

ω 

ω 2pe
ω2
k −
2
ω 2pe
2
pe
2
c2
+
 ω 2pe 
c 1 − 2 
ω 

2
Multiply this out
ω2 −
k2 =
ω 2pe
ω
1 ± ce
ω
c2
This describes two waves with E1 perpendicular to k and B0.
=0
Phase velocity
We can plot vph as a function of ω.
You do. Find expressions for ω1 and ω2.
It is easy to show that Ey1 = m jEx1 so the waves are circularly-polarized.
The wave corresponding to the upper sign has its E vector rotating opposite to the way
the electrons gyrate. This is called the left-hand circularly-polarized (LHCP) wave;
the wave corresponding to the lower sign has its E vector rotating in the same way the
electrons gyrate. This is called the right-hand circularly-polarized (RHCP) wave.
Note that these definitions of handedness are different from those used in Optics. The
LHCP and RHCP waves are the characteristic waves for propagation along the
magnetic field.
Resonances
Resonances occur when v ph = 0 .
At a resonance, the present approach breaks down. This is easy to see. Consider the
term E + v × B . This could be written

k × E1 
E 1 + v 1 × ( B 0 + B1 ) = E 1 + v 1 ×  B 0 +


ω 
To first order, we have been ignoring the last part and writing
a resonance,
v ph =
ω
k
E1 + v 1 × B 0 . But at
is very small so this simplification may break down.
For propagation along B0 there is a resonance for the RHCP wave when ω = ωce. Note
that if the magnetic field is high, the RHCP wave always propagates. In this regime it
is known as the whistler wave.
The refractive index is
ω 2pe
2
µ 2L = 1 − ω
.
ω ce
R
1±
ω
ω 2pe
ω
If we let X = 2 and Y = ce this can be written more compactly as
ω
ω
X
µ 2L = 1 −
.
Y
1
±
R
CMA diagram
You have seen plots of vph vs. ω. The CMA diagram is another graphical
representation of the same information.
Note the relation between the vph vs. ω graph and the CMA diagram.
Waves in a plasma. Propagation perpendicular to B0.
This time take k along x. B0 is still along z. Write out the component Maxwell’s
equations. Eliminate B1, j1, leaving equations in E1.
Proceeding as before
0=
ω2
c
2
k Ey =
2
k Ez =
2
Ex +
ω2
c
2
ω2
c2
ω 2pe
ω


 − E x + j ce E y 


ω
 ω 
c 2 1 −

ω 

Ey +
Ez +
2
ce
2
ω 2pe
ω


 − E y − j ce E x 

ω
 ω 
c 2 1 −

ω 

2
ce
2
ω 2pe
c2
( − Ez )
(1)
(2)
(3)
Equation (3) describes transverse electromagnetic waves with E1 along the zdirection, parallel to B0.
ω 2 − ω 2pe
k=
c2
This is the same as the dispersion relation for the wave when there is no magnetic
field. This wave is called the ordinary (O) wave.
Equations (1) and (2) yield
ω 2pe
ω −
ω 2ce
1− 2
ω − ω 2pe
2
k2 =
c2
.
This wave is called the extraordinary (X) wave.
The O and X waves are the characteristic waves for propagation perpendicular to the
magnetic field.
Phase velocity
You do. Find expressions for ω1 and ω2.
ω 2pe
ω
The refractive indices, if we let X = 2 and Y = ce , are
ω
ω
µ O2 = 1 − X and µ 2X = 1 −
CMA diagram
X
.
Y2
1−
1− X
ω 2pe + ω 2ce is known as the upper hybrid frequency.
Waves in a plasma. Propagation at an arbitrary angle to B0.
X
µ2 = 1−
1
2
2
Y sin θ   Y 2 sin 2 θ 
2
2
1−
± 
 + Y cos θ 
2(1 − X )   2(1 − X ) 

2
2
where θ is the angle between k and B0. This is one of Appleton’s equations for the
case where collisions are neglected. The other equations describe the polarization of
the waves. In general E1 will have a component parallel to k.
Ion motions
Is v 1i still always negligible in the presence of a steady magnetic field?
The momentum equations are
n 0 mi ( − jωv 1i ) = n 0 eE 1 + n 0 ev 1i × ( B 0 z$ )
n 0 me ( − jωv 1e ) = − n 0 eE 1 − n 0 ev 1e × ( B 0 z$ )
As an example, one of the components of
j
vix1 =
v 1i is
ω e
e
E x + 2ci E y
ωm i
ω mi
ω ci2
1− 2
ω
Clearly this becomes very important when ω ≈ ω ci .
.
e.g., for propagation along B0, the dispersion relation is

 1
m
1
ω 2 − ω 2pe 
+ e
 1 ± ω ce mi 1 m ω ci

ω
ω
k2 =
2
c





The effects can be seen by comparing the phase velocity and CMA diagrams below
with those above.
Exercises
Waves in a plasma, no magnetic field
1. Plasma diagnostics
Calculate nc for the standard microwave frequencies used in the Plasma
Physics Department; (i) 10 GHz, (ii) 35 GHz and (iii) 110 GHz.
2. Space Shuttle reentry
The Space Shuttle communicates with the ground using three frequency
bands; UHF 259.7 - 296.8 MHz, S-band 1.7 - 2.3 GHz and Ku-band 15.25 - 17.25
GHz. During reentry there is a communications blackout because of ionization of the
air around the spacecraft. This lasts for about 15 minutes as the Shuttle descends from
80 km down to 50 km.
Estimate the minimum plasma density.
Waves in a plasma, with magnetic field
3.
Consider a plasma slab that has a density profile like:
A wave, frequency ω, is launched into it from the outside. What is the
maximum density that the plasma can have so that the wave will pass through
(i) if B0 = 0,
(ii) if wave is LHCP along B0,
(iii) if wave is RHCP along B0,
(iv) if wave is O perpendicular to B0,
(v) (HARDER) if wave is X perpendicular to B0?
Waves in a plasma. Propagation along B0
4. Plasma diagnostics - Faraday Rotation
A linearly-polarized laser beam enters a uniform plasma slab, thickness L,
parallel to the magnetic field.
Obtain an expression for the Faraday Rotation of the beam in terms of B0, n
and L.
Assume ωce, ωpe << ω as it would be for a laser beam.
Waves in a plasma. Propagation perpendicular to B0.
5. Plasma diagnostics
An O wave microwave beam is transmitted through a plasma slab, thickness L,
density n << nc everywhere.
Obtain expressions for the phase change of the wave across the slab
(i) if the density is uniform across the slab, and
(ii) (HARDER) if the density profile is a parabola.

4x 2 
The equation for the parabola is n = n max 1 − 2  .

L 
In this case express the phase change in terms of the average density.
6. Reflection of radio waves by the ionosphere
Suppose waves are being transmitted vertically upwards from a point on the
earth where the earth’s magnetic field is horizontal. Take the value of the field to be
3 × 10−5 T.
(i) Calculate the frequencies for reflections of O waves from each layer.
(ii) Calculate frequencies of reflection of X waves from the F2 layer.
You will need the following:
O wave cutoff is given by ω = ω pe and
X wave cutoff by ω 2 = ω 2pe + ωω ce .
Compare the values of ωce and ωpe. Does this allow you to simplify the latter
expression?
(iii) Compare with an actual measurement.
The vertical axis is the time delay for the reflected signal but it has been
relabelled effective height; the horizontal axis is frequency.
7. Electron cyclotron heating of a plasma
The waves will be launched from outside the plasma and be completely
absorbed in the region where ω ≈ ω ce . (When we allow heating at a harmonic of
the ωce, take the velocities of the electrons and relativistic effects into account, the
condition generalizes to
ω=
nω ce
γ
+ k|| v|| .)
Possible heating strategies can be checked using the CMA diagram.
For a tokamak plasma,
B≈
1
, where r is the major radius. So the
r
resonance will be in the region shown below. The field is higher on the inside and
lower on the outside.
The magnetic field is in the toroidal direction.
Here are some possible strategies
(i) Launch O waves from the high-field side to fundamental
resonance,
(ii) Launch O waves from the low-field side to fundamental
resonance,
(iii) Launch X waves from the high-field side to fundamental
resonance,
(iv) Launch X waves from the low-field side to fundamental
resonance,
(v) Launch O waves from the high-field side to second harmonic,
(vi) Launch O waves from the low-field side to second harmonic,
(vii) Launch X waves from the high-field side to second harmonic,
(viii) Launch X waves from the low-field side to second harmonic.
Plot each of these on a CMA diagram and comment.
Each plot will start on the
ω 2ce
ω2
axis and head towards the resonance. The
variation in B is small so there will be either the fundamental or second harmonic but
not both. The first question is; can the resonance be reached?
(HARDER) Calculate the maximum value of
ω 2pe
ω ce2
in each case.
Note that (i) n = 0 outside the plasma, (ii) only one resonance, fundamental
or second harmonic, will be present (Why?).
Waves in a plasma. Propagation at an arbitrary angle to B0.
8. Whistlers
Whistlers are radio signals in the audio-frequency range that "whistle". A
lightning stroke excites a pulse that travels from one hemisphere to another and back
again. The wave travels in a duct of enhanced electron density that follows the
magnetic field lines.
Simple theory.
Take the equation for propagation at an arbitrary angle and apply the
quasilongitudinal approximation. θ is sufficiently small that
X
µ2 ≅ 1−
1 ± Y cos θ
Above the F2 layer, X > 1+Y so, since the whistler travels well above the ionosphere,
only the minus sign is of interest.
In fact it is reasonable, particularly for low frequencies, to simplify this further
to
µ2 ≅
X
.
Y cos θ
Under these approximations,
(i) find vph
(ii) find vg.
1
.
f
(iv) (HARDER) The correct definition of group velocity is
(iii) show delay time ∝
vg =
∂ω
∂ω
∂ω
$z .
x$ +
y$ +
∂kx
∂ky
∂kz
Apply this definition to the whistler dispersion equation to obtain θr
the ray direction. Then by calculating the maximum value, show that this direction
always lies within about 20° of the direction of B0, i.e., the whistler wave is guided by
the field line to within this angle.
When the quasilongitudinal approximation is used, nose whistlers are
predicted.
Note that in laboratory situations, RHCP waves propagating where ωce > ω are
known as whistlers or helicon waves.
Ion motions
9. Lower hybrid resonance
The X wave, with ion motions, has the dispersion relation
c2k 2
ω2
where
= 1−
ω 2p
ω 2 (ω ce − ω ci )
ω − ω ceω ci + 2
ω p − ω 2 + ω ceω ci
2
2
ω 2p = ω 2pe + ω 2pi .
What is the resonant frequency when the density n = 0 and when
10. Alfven waves
n= ∞?
(i) Show that when ωce < ωpe, the dispersion relation for propagation along
v ph = c
B0 that includes ion motions yields
(ii) Define the Alfven speed
VA =
ω ceω ci
ω 2pe
B
µ0ρ
as
ω → 0.
. Show vph = VA.
Summary of chapter
Definitions include phase velocity, group velocity, dispersion relation, refractive
index, cutoff, resonance, critical density, plasma oscillations, characteristic wave.
Dispersion relations.
no magnetic field
k2 =
ω 2 − ω 2pe
c
2
or µ 2 = 1 − X
with magnetic field
LHCP and RHCP waves:
ω 2pe
ω −
ω
1 ± ce
ω or µ 2 = 1 − X
k2 =
L
2
2
c
R
O and X waves:
ω2 −
k =
2
ω 2 − ω 2pe
and k 2 =
1± Y
ω 2pe
ω2
1 − 2 ce 2
ω − ω pe
c2
X
µ O2 = 1 − X and µ 2X = 1 −
Y2
1−
1− X
c
2
,or
You should be able to
Handle the first order approach to obtaining these dispersion relations. I would
expect you to derive the wave properties from Maxwell’s equations and the
momentum equation for simple cases.
Do calculations of critical density.
Deduce cutoffs and resonances.
Know the effects of including collisions, finite temperature.
Do calculations relating to plasma diagnostics (phase change and Faraday
rotation).
Describe reflection of radio waves by the ionosphere, electron cyclotron
heating of a plasma and whistlers.
PLASMA PHYSICS
VI. WAVES IN PLASMAS − 2
Effects of geometry and boundaries
Helicon waves
Helicon waves have a role in industrial plasmas. They are whistler waves in a
cylindrical plasma.
The method of calculating the properties of these waves is outlined below
Suppose that fields, etc., in a cylindrical geometry vary as
m integer.
Maxwell’s equations
∇×E = −
∇ × B = µ0 j+
f ( r)e
j ( mφ + k z z −ωt )
,
∂B
∂t
1 ∂E
c 2 ∂t
must be expanded using the cylindrical polar coordinate identities.
The electron momentum equation to first order is,
n 0 me ( − jωv 1e ) = − n 0 e( E 1 + v 1e × B 0 )
Use
ω 2pe >> ω 2
equation for
Writing
and
ω ce >> ω . Eliminate variables and finally arrive at a Bessel
1
z
B which has a solution
Bz1 = AJ m ( k ⊥ r ) .
k ⊥ = k sin θ leads to the earlier dispersion equation for the whistler wave.
k 2c2
ω
2
=
ω 2pe
ω2
ω ce
cos θ
ω
or
µ2 =
X
.
Y cos θ
Boundary condition. At the conducting wall, the tangential electric field must be
 mJ m (k ⊥ r )

Eφ1 = 0 . Now Eφ1 ≈  k
+ k z J ' m (k ⊥ r ) so k ⊥ a must be a
k⊥r


 mJ m (k ⊥ a )

zero of  k
+ k z J ' m (k ⊥ a ) . We know the radius a so this limits the
k⊥a


possible values of k ⊥ .
zero, i.e.,
For a uniform density profile radius a = 50 mm, n = 3 × 1018 m−3, B = 500 gauss and f
= 27.12 MHz, some representative wave magnetic field profiles are sketched below.
MHD Waves
This time work with the low-frequency Maxwell’s equations
∂B
∇×E = −
∂t
∇ × B = µ0j
and the single fluid equations
∂ρ
+ ∇ ⋅ ( ρv ) = 0
∂t
∂v
ρ = j × B − ∇p
∂t
0 = E+v×B.
To first order these can be written
jk × E 1 = jωB1
jk × B1 = µ 0 j1
− jωρ 1 + ρ 0 jk ⋅ v 1 = 0
ρ 0 ( − jωv 1 ) = j1 × B 0 − U 2 jkρ 1
0 = E1 + v 1 × B 0 .
We will suppose the steady magnetic field is B0 z$ and
that k = k x x$ + kz z$ . If the angle between the direction
of propagation and the magnetic field is θ, then
kx = k sin θ, kz = k cos θ.
Eliminate ρ1, j1, E1, B1 leaving v1 and use Alfven speed VA =
The component equations are
1
1
v 1x = 2 k 2 VA2 v 1x + 2 U 2 ( k x v 1x + k z v 1z ) k x
ω
v 1y =
v z1 =
ω
1
ω
2
1
ω
2
k z2 VA2 v 1y
B0
µ0ρ0
.
(1)
(2)
U 2 ( k x v 1x + k z v 1z ) k z
(3)
Alfven wave
Equation (2) gives a wave, the Alfven wave
ω2
k
2
= VA2 cos 2 θ .
Properties of this wave
v1 is along y, from Equation (2)
v 1 × B 0 so is along x
1
B1 is along k × E so is along y,
E1 is along
i.e.,
B1 is perpendicular to the direction of propagation.
Phase velocity does not depend on frequency. It does depend on the direction
of propagation.
E1
Recall that for an electromagnetic wave in a vacuum, B1 ≈
. Here
c
E1
B1 ≈
. Since VA << c, the wave magnetic field is particularly important.
VA
Consider the special case θ = 0.
The waves bend the field lines. This transverse perturbation of the field lines leads to
the Alfven wave being called the shear wave. In a cylindrical geometry it is called the
torsional wave.
This wave is analogous to the transverse wave on a stretched string.
velocity of a wave on a stretched string
v ph =
τ
µ
, where τ is tension and µ is
mass per unit length.
velocity of Alfven wave
B0
v ph =
µ0ρ0
.
µ = ρA , where A is area, so the two expressions are equivalent if τ =
Recall that magnetic tension
=
B0
B0
2
µ0
A.
2
µ0
. This magnetic tension provides the restoring
force.
Fast and slow MHD waves
Equations (1) and (3) give two waves. The upper sign gives the fast MHD wave; the
lower sign the slow MHD wave.
ω2
(U
2
+ VA2 ) ±
(U
2
+ VA2 ) − 4U 2 VA2 cos 2 θ
2
=
k2
2
Note that in this limit the phase velocities do not depend on frequency.
Consider two special cases.
(i) Propagation along B0: ion acoustic wave and Alfven wave
θ = 0°.
The two solutions are
ω2
k2
and
ω2
k
2
= U2
(1)
= VA2
(2)
Equation (1) gives an acoustic wave, the ion acoustic wave.
Properties
v1 is along z
E1 = − v × B . But v1 and B0 are
parallel so there can be no E1.
1
B1
0
∝ k × E1
so there can be no B1.
So indeed it is an acoustic wave.
Equation (2) gives the Alfven wave again.
Properties
v1 is along x
v 1 × B 0 so is along y
1
B1 is along k × E so is along x
1
so B is perpendicular to the direction of propagation.
E1 is along
(ii) Propagation perpendicular to B0: compressional wave
θ = 90°.
ω2
k2
= U 2 + VA2
Properties
v1 is along x
v 1 × B 0 so is along y
1
B1 is along k × E so is along z
1
so B is in the direction of B0 and perpendicular to the direction of propagation
E1 is along
Other names for this wave are the fast wave, and if U is not neglected, the
magnetosonic or magnetoacoustic wave.
The waves are summarised on the phase velocity vs. angle of propagation diagram
below.
In laboratory plasmas, VA >> U; in space plasmas VA < U.
Exercises
Effects of geometry and boundaries
1.
In a uniform cylindrical plasma with a conducting wall, the magnetic field
components of the helicon wave are
 k mJ m ( k ⊥ r ) k z

Bz1 = AJ m ( k ⊥ r ), Br1 = jA
+
J m ' ( k ⊥ r ) ,
k⊥ r
k⊥

 k⊥
 k mJ m ( k ⊥ r ) k

Bφ1 = − A z
+
J m ' ( k ⊥ r )
k⊥ r
k⊥
 k⊥

each with a factor e
j ( mφ + k z z −ωt )
.
Consider a cross-section of the plasma. Sketch the field lines for m = 0 at
different instants.
MHD waves
2.
(i)
Write down the first-order equations (in cartesian coordinates)
corresponding to the cold plasma case where U = 0,
∂B
∇×E = −
∂t
∇ × B = µ0j
∂v
ρ = j× B
∂t
0 = E + v × B,
for the case where B is in the z-direction as usual, and k has both x and z
components.
0
(ii)
Find the dispersion relations for all the waves. Are any missing?
3. Lower hybrid waves
See if there is a dispersion relation for electrostatic waves propagating
perpendicular to B0. Electrostatic waves means no wave magnetic field, or
∇ × E = 0.
Use these equations
∂ρ e
+ ∇ ⋅ (ρ eve ) = 0
∂t
∂ρ i
+ ∇ ⋅ (ρ ivi ) = 0
∂t
∂v
ρ e e = − nee(E + v e × B)
∂t
∂v
ρ i i = ni e(E + v i × B)
∂t
and use ne = ni.
In fact, using these equations, there are no waves. There are oscillations at
the lower hybrid frequency
ω = ω ciω ce
.
4. Solid-state plasmas
Similarities to gaseous plasmas
When plasma criteria are satisfied, can observe something resembling
electrical discharges in gases, plasma confinement and waves.
Differences
density: below 1022 m−3 in weakly-doped or intrinsic semiconductors up to
1029 m−3 in metals.
temperature: usually in equilibrium with the host lattice, room temperature
down to liquid helium temperatures
carrier masses: The effective masses of the electrons may be two orders of
magnitude less than me; the effective masses of the holes are similar but not equal to
those of the electrons. These masses depend on the direction with respect to the
lattice.
dielectric constant: ε r can be very large, e.g., ≈ 100 for bismuth.
There are several kinds of solid-state plasma.
(1) compensated, where the numbers of mobile holes and mobile electrons are
equal, i.e., if ne = nh. Intrinsic semiconductors, semimetals (bismuth, antimony) and
certain metals (iron, tungsten).
(2) uncompensated, ne >> nh or nh >> ne and overall charge neutrality is
provided by the lattice of immobile ions. Doped or extrinsic semiconductors and other
metals (sodium, copper).
(3) Apply an external electric field strong enough to cause avalanche
breakdown.
(There are also liquid plasmas. For example, mercury, electrolytic solutions)
Waves in solid-state plasmas
The earlier equation for propagation along B0 that included ion motions can be
modified to


2
2
2


ω
ω
ω
pe
ph
pi
ω2 −
+
+

 1 ± ω ce 1 m ω ch 1 m ω ci 

ω
ω 
ω
k2 =
2
v
where the subscripts e, h and i refer to electrons, holes and lattice ions (this expression
assumes they are positive), respectively. The ion mass can be treated as infinite so the
last term is zero. In a solid we should use ε r ε 0 instead of ε 0 in calculating the plasma
frequencies and the velocity of light in the material is v =
1
µ 0ε rε 0
instead of c
where ε r is the dielectric constant.
Rearranging gives
k 2v2
ω2
ω 2pe
ω 2ph
ω 2ph 

  ω 2pe
 m

= 1± 
−
−
 ω ce (ω ± ω ce ) ω ch (ω m ω ch )   ωω ce ωω ch 
Alfven waves
(i)
Show that if the plasma is compensated the last term vanishes.
Further, if ω ce or h >> ω then the dispersion relation simplifies to
1
1
1
= 2 + 2.
2
v ph v
VA
VA << v so
v ph = VA .
Damping is small if ω τ >> 1.
An early experiment looked at the propagation of Alfven waves in a small 4
mm diameter cylinder of bismuth. The bismuth was cooled to liquid helium
temperatures. The wave frequency was 16.25 GHz.
(ii)
Calculate the Alfven velocity.
Use n = 3.1 × 1023 m−3, effective mass for electrons = 0.080 me (multiply this
by 4.55 to take account of anisotropy), effective mass for holes = 0.068 me (multiply
by 1), B = 1 T,
(iii)
Check that ω ce or h >> ω .
Helicon waves
If the plasma is uncompensated, the last term does not vanish. If it is n-type ne
>> nh; if it is p-type nh >> ne. Again ω ce or h >> ω . Only the lower sign gives a wave.
v 2ph = v 2
ωω c
.
ω 2p
Damping is small if ω c τ >> 1 .
(iv)
Compare this with the whistler-helicon dispersion relations discussed earlier.
(v)
Show that you do not need to know the mass in this case.
One early experiment looked at helicon waves propagating along B0 in the
metal sodium. The sodium was cooled to liquid helium temperatures. They put their
sample in a 1 T magnetic field and looked at standing waves on a length of 4 mm.
(vi)
Predict the frequency of the lowest order.
Use n = 2.7 × 1028 m−3.
Another early experiment looked at helicon waves propagating parallel to B0
in the semiconductor indium antimonide using 9 GHz microwaves. The InSb sample
was at room temperature. They looked at standing waves on a length of 2 mm.
(vii)
Predict the magnetic field that gave the lowest frequency.
Use n = 1.2 × 1020 m−3.
Summary of chapter
helicon waves
You should be able to
Handle the first order approach to obtaining the dispersion relations.
Describe the perturbation of the magnetic field lines under shear and
compressional waves.
Describe effects of geometry and boundaries.
MHD waves dispersion relations
B0
Definitions include Alfven speed VA =
µ0ρ0
,
Alfven waves (also known as shear waves, torsional waves):
ω2
k2
Fast and slow MHD waves:
along B0, Alfven waves
= VA2 cos 2 θ
ω2
2
= VA2 and ion acoustic waves
ω2
2
= U2 .
k
k
⊥ B0, compressional waves (also known as fast waves, magnetosonic
or magnetoacoustic waves):
ω2
k
2
= U 2 + VA2 .
Note. There is a lot that has not been covered.
waves in non-uniform plasmas,
waves in gas mixtures,
waves in a current-carrying plasma,
reflection and refraction of waves,
kinetic effects that require consideration of the particle velocity distribution
functions,
non-linear effects,
shocks,
mode conversion ,
coupling between waves,
damping,
heating by waves.
PLASMA PHYSICS - KEY IDEAS – 2
cutoffs shift
resonance of RHCP at ω = ωce where
E1 rotation matches electron gyration
exactly
V Waves in plasmas - 1
derivations start with Maxwell’s curl equations
waves in vacuum
no j’s
get dispersion equation (relation between k
and ω) that describes:
transverse em waves
waves in plasma, no B
calculate j’s from ion and electron momentum
equations
ignore collisions
ignore T
no drifts
then,
ion velocity very small (electrons
move easily in wave, ions remain at
rest)
at a resonance vphase zero
perpendicular to B
get 2 characteristic waves:
both linearly-polarized
O wave: E1 parallel to B
X wave: E1 perpendicular to B
include + ions
get:
resonance of LHCP at ω = ωci
Know how to use information like ω << ωci,
ω >> ωpe etc. to simplify dispersion equations. (See for
example Faraday rotation, radio waves in ionosphere,
whistlers)
VI Waves in plasmas - 2
get 2 solutions:
oscillations at plasma frequency
transverse em waves
for propagation, k must have real part
propagation if ω high or n
small.
cutoff if ω low or n large.
(cutoff density nc)
at a cutoff vphase infinite
include boundaries
plasma is not infinite
Note: we are no longer talking about MHD
waves.
helicon waves
whistler waves in a cylindrical plasma
Next we use Maxwell’s curl equations and single fluid
equations to predict MHD waves
include collisions
k is complex
cutoff is not sharp
n<nc, propagation, some attenuation
n>nc, some propagation but high attenuation
Alfven waves
transverse perturbation of field lines
(like waves on a string)
include T
get:
parallel to B
Alfven wave
new electrostatic wave (E parallel to
k)
waves in plasma, with B
propagation of waves depends on direction
parallel to B
get 2 characteristic waves:
LHCP wave: E1 rotates in same
direction as + ions gyrate
RHCP wave: E1 rotates in same
direction as electrons gyrate
Fast and slow MHD waves
ion acoustic wave: finite T, pure
electrostatic wave
perpendicular to B
compressional wave: compresional
perturbation of magnetic field lines
PLASMA PHYSICS
VII. PLASMA DIAGNOSTICS
Electrostatic or Langmuir Probes
Probes are inserted into the plasma to measure Te and n.
The probes perturb the plasma electrically and may be, in the case of tokamak
plasmas, an intolerable source of impurities.
Single probe
Typically a tungsten probe in a glass tube.
Probe perturbs the plasma but the effects are small outside a thin sheath surrounding
ε 0 kTe
the probe. The sheath thickness is of the order of a Debye length λ D =
(see
n0 e 2
Chapter I) and is << R, the probe radius..
I-V characteristic
(i) VS is the space or plasma potential (the potential of the plasma in the absence of a
probe). There is no E. The current is due mainly to the random motion of electrons
(the random motion of the ions is much slower).
(ii) If the probe is more positive than the plasma, electrons are attracted towards the
probe and all the ions are repelled. An electron sheath is formed and saturation
electron current is reached.
(iii) If the probe is more negative than the plasma, electrons are repelled (but the
faster ones still reach the probe) and ions are attracted.
The shape of this part of the curve depends on the electron velocity distribution.
For a Maxwellian distribution with Te > Ti, the slope is
dI e I − I sat i
=
.
dV
kTe
(
)
(iv) VF is the floating potential (an insulated electrode would assume this potential).
The ion flux = the electron flux so I = 0.
(v) All the electrons are repelled. An ion sheath is formed and saturation ion current
is reached.
Collisions
If there are collisions, particles may have to rely on diffusion to treach the probe.
D
From before λ m ≈
. Collisions are important if λ m << R .
vrms
Magnetic field
The effect of a magnetic field is extremely complicated. The ions and electrons gyrate
and this affects their random motions and collisions. The behaviour of the probe will
depend on its orientation in the magnetic field. The magnetic field can be ignored if
rLe >> R.
Points to note
Swept Langmuir probes may give time behaviour of plasma parameters, but there are
difficulties.
Any secondary emission, photoemission will lead to errors
Care is required if distribution is non-Maxwellian. For instance if there is a drift.
Double probe
I-V characteristic
Assume symmetry.
The system “floats” and follows any changes in VS.
If V is slightly positive, there are more electrons reaching 1 and fewer reaching 2.
If V is very positive, saturation ion current into 2. The current into 1 cannot exceed
this value.
It can be shown that
I = Isat i tanh
eV
2 kTe
and the slope of the characteristic is
dI eIsat i
=
.
dV 2 kTe
Points to note
In a magnetic field, shadowing of the probes must be avoided.
Magnetic probes
Magnetic coil
(i) A small coil is inserted into the plasma and oriented to measure a particular
component of B or to pick up MHD waves.
Typically the small coil is in a glass tube – it has no electrical contact with the plasma.
V = NA
dB
.
dt
You do. Derive this.
Use integrator.
Vout =
NAB
.
RC
(ii) Flux loop or diamagnetic loop. A large coil surrounding the plasma vessel is used
to measure total magnetic flux.
Recall that if Iz = 0,
Bz ( r )
= constant .
2µ 0
The presence of the plasma acts to decrease the magnetic field inside it (hence
diamagnetic).
p( r ) +
2
If we measure Bz ( a) using a small coil and <Bz> using a diamagnetic loop then
B ( a) − < Bz > 2
< p>= z
2µ 0
gives an estimate of the total kinetic energy in the plasma (using p = nkT).
2
You do. Show this.
(iii) Monitor position of plasma in a tokamak
If the plasma moves up, the flux in the upper loop increases while the flux in the
lower loop decreases. Take difference and integrate.
If the plasma moves out, the flux in the outer solenoid increases while the flux in the
inner solenoid decreases. Take difference and integrate.
(iv) Loop voltage measurements
The emf induced in the loop equals the emf in the plasma loop. Measure plasma
current I independently and calculate the plasma resistivity using
R=
ηl
A
.
η
depends on electron temperature so Te can be estimated.
Rogowski coil
To measure I the total current. The large loop completely surrounds I. The
measurement is independent of how the current is distributed and it has the advantage
of making no electrical contact with the current being measured.
Note the return wire to minimize the effect of any flux threading the large loop.
dI
V = nAµ 0 .
dt
where n is the number of turns per unit length, A is the area of the small loop.
You do. Derive this. You have to assume B is uniform across A.
Use integrator.
The coil should be shielded to avoid electrostatic pickup from the plasma.
(Rogowski coil can be used as a current probe in non-plasma applications.)
Microwave interferometry
This is a non-perturbing way of measuring n. The interferometer shown is the
microwave version of the Mach-Zehnder interferometer. The optical path length in the
probe arm changes as the plasma density varies.
The waves in the plasma are usually the O waves whose refractive index
n
µ2 = 1 −
nc
depends only on the density and not on the magnetic field.
Field at detector due to signal passing through probe arm
(
E1 = A1 cos ωt + φ 1 + φ plasma
where
φ plasma =
∫ (k − k )dx , k
l
0
0
0
=
ω
c
,
k=µ
ω
c
)
.
Field at detector due to signal passing through reference arm
E2 = A2 cos(ωt + φ 2 ) .
The output of the square-law detector is
V = ( E1 + E2 )
2
(
)
(
)
= A12 cos 2 ωt + φ 1 + φ plasma + 2 A1 A2 cos ωt + φ 1 + φ plasma cos(ωt + φ 2 )
+ A22 cos 2 (ωt + φ 2 )
=
(
A12 + A12 cos 2 ωt + φ 1 + φ plasma
2
(
)
)
(
+ A1 A2 cos 2ωt + φ 1 + φ plasma + φ 2 + A1 A2 cos φ 1 + φ plasma − φ 2
+
A22 + A22 cos 2(ωt + φ 2 )
)
2
The capacitance of the detector shorts the microwave frequency signals to ground.
The remaining slowly-varying part is
A12
A22
V=
+ A1 A2 cos φ 1 + φ plasma − φ 2 +
.
2
2
(
)
Points to note
As the derivation shows, the interferometer measures average density along the chord.
To obtain a density profile would require measurements along many chords.
The longer the wavelength the better is the sensitivity to small densities but the poorer
is the spatial resolution.
Beam bending may occur.
Microwave reflectometry
Measures n. Different frequencies will be reflected from different layers in the
plasma.
Point to note
There is a problem if density is not monotonically increasing.
Laser interferometry
Measures <n>.
Two forms of interferometers are the Mach-Zehnder interferometer and the Michelson
interferometer.
Some lasers that have been used in the School of Physics are He-Ne laser (visible 633
nm, infrared 3.39 µm), CO2 laser (infra-red 10.6 µm), formic acid molecular vapour
laser (far-infrared 433 µm), HCN laser (far-infrared 337 µm).
Scattering of electromagnetic radiation from a plasma
(a) Thomson scattering
Scattering of laser light from electrons in the plasma to make a non-perturbing
measurement of Te.
The scattered wave satisfies
k s = k i + ∆k and ω s = ω i + ∆ω
Since ks = ki,
∆k = 2 ki sin
θ
2
.
Thomson scattering can be described classically. The principle is that the incident
wave accelerates the electrons; In the non-relativistic case,
e
v& = −
E i cos( k i ⋅ r − ωt ) ,
me
and because the electrons are accelerating, they radiate; At large distances, in the far
field,
e r × (r × v& )
Es = − 2
c
r3
where the quantities on the rhs are evaluated at retarded time.
(At high energies a quantum mechanical treatment is more appropriate. Compton
scattering.)
The ions are too massive to radiate - they still have an effect however.
The contributions from all the electrons must be combined.
1
<< 1 . There are two ways of looking at this case
∆k λ D
(a) ∆k is large so the resolution is high, individual electrons can be seen.
(b) λ D is large so plasma effects are unimportant.
In other words, the effects of individual electrons are important. The phases of
the wave arriving at the different electrons is random as will be the phases of their
α=
scattered waves. So the scattering is incoherent. The total scattered intensity is given
by adding the scattered intensities from each electron.
(If we think of
1
scale length
as a scale length, then α =
.)
∆k
Debye length
1
>> 1
∆k λ D
(a) ∆k is large so the resolution is low, only a ‘cloud’ can be seen,
(b) the scattering is from fluctuations in charge density (ion acoustic waves)
and is collective or coherent. The total scattered intensity is found by calculating the
total far field and squaring it.
α=
The intensity in the incoherent case is ∝ n and in the coherent case is much larger,
∝ n2 .
The intensity of the scattered radiation is given by
I (ω ) = I 0 re2 (1 − cos 2 φ ) nS( ∆k, ∆ω )
e2
is the classical radius of the electron
where I0 is the incident intensity, re =
4πε 0 me c 2
(which is very small), (1 − cos2φ ) is a geometrical factor in which φ is the angle
between Ei and ks, n is the plasma density and S( ∆k, ∆ω ) is the dynamic form factor.
From this expression we see
(i) that the scattered intensity is greatest for φ = 90°
So choose φ = 90°.
(ii) the scattered intensity is extremely small
So use an energetic pulsed laser (ruby laser).
(iii) the spectrum goes as
S( ∆k, ∆ω ) which depends on α.
S( ∆k, ∆ω ) has been plotted below for two values of
α.
S( ∆k, ∆ω ) has an electron term and an ion term. If α << 1, the ion term is
negligible and the shape is due to Doppler broadening. Electron thermal motion gives
a doppler shift and the width of the line gives Te. n could be obtained from the
absolute value of the scattered intensity.
If α ≥ 1 the electron peak can give n but at still higher α depends on n and Te. At α ≥
1 the ion term becomes important and, since the peak is interpreted as the existance
of an ion acoustic wave, depends on Te. Under different conditions the width can
depend on Ti or Te. The condition α ≥ 1 can be obtained by using a long wavelength
and/or a small scattering angle.
Points to note
Avoid parasitic light. But note the light of interest is at a slightly different wavelength
from the incident light. Use a beam dump (a near perfect absorber).
A full treatment must include magnetic fields and relativistic effects.
(b) Scattering from macroscopic density fluctuations
Scattering from waves, instabilities, turbulence.
Choice of frequency/wavelength. Assume you know what range of
frequencies/wavelengths you want to study.
Given λi, the minimum ∆λ that can be investigated is λi/2 (or maximum ∆k is
2ki).
If ∆λ is too large (or if ∆k is too small), the scattering angle θ is small and
spatial resolution is poor.
If ∆ω is too low (or the wavelength of the waves you are studying is too
large),
beam bending may occur (see above),
the width of the beam which is set by diffraction may be too large and spatial
resolution is poor.
You do. Find a reference to gaussian beams. What determines how the beam spreads
with distance
Electron cyclotron emission
Electron cyclotron emission from a single gyrating electron is at the electron
cyclotron frequency and its harmonics.
Electron cyclotron emission from the plasma.
(i) The spectrum is broadened principally because the magnetic field is not uniform.
e.g., A tokamak whose major radius R = 0.54 m, minor radius a = 0.10 m and
in which
the magnetic field B = 6.1 T.
At the inside wall of torus B = 7.2 T, at the outside wall B = 5.0 T.
You do. Show this. (See Chapter II) What range of frequencies might we expect for
each harmonic?
(ii) Intensity of the radiation.
At low frequencies, the plasma is optically thick. Any radiation is reabsorbed.
The plasma is like a black-body
ω 2 kTe
I (ω ) = I bb (ω ) =
8π 3 c 2
so Te can be estimated from the intensity.
At high frequencies, the plasma is optically thin. Reflections from the walls
become important.
1 − e −τ (ω )
I (ω ) = Ibb (ω )
1 − re −τ (ω )
where τ is the optical thickness and r is the reflection coefficient. Note that optical
thickness cannot be deduced from the earlier cold plasma dispersion relations.
(iii) Usually observe at 90° to the magnetic field. In this direction, X waves dominate.
The spectrum of electron cyclotron emission is obtained using Fourier transform
spectroscopy.
Suppose plasma emits a single frequency
Field at detector due to beam reflected off stationary mirror
E1 = Acos( kd − ωt ) .
Field at detector due to beam reflected off moving mirror
E2 = Acos( k ( d + x ) − ωt ) .
The output of the square-law detector is
( E1 + E2 ) 2 = 4 A 2 cos 2  k
x

 cos 2  k  d +

 
2
x

 − ωt 


2
x 1

= 4 A 2 cos 2  k  1 + cos2 k  d +
 2 2 
 
x

 − ωt  

2
The low-frequency part is
x
V = 2 A 2 cos 2  k  = A 2 (1 + coskx ) .
 2
Express this as
intensity
I = S( k )(1 + coskx ) .
The plasma emits a range of frequencies. Using a similar approach
∞
intensity
I ( x ) = ∫ S( k )(1 + coskx ) dk
0
S(k) is the spectrum.
In a measurement, I(x) is recorded for a range of x, the average value is subtracted
out leaving what is called the interferogram
∞
Int ( x ) = ∫ S( k ) coskx dk .
0
This is a Fourier transform. Carry out the inverse transform to obtain the spectrum
S(k)
S( k ) =
Int ( x ) coskx dx .
π∫
0
We do not have readings for x = 0 to
the instrument is limited to
∆k =
∞
2
π
∞ , only to xm. This means that the resolution of
xm
.
Plasma spectroscopy
Important processes that determine populations include
1 Radiative, involving photons
bound-bound transitions - absorption and (its inverse process) emission of
photons
A + hν ↔ A ∗
free-bound transitions - photoionization and recombination with the
emission of a photon
A + hν ↔ A + + e
2 Collisional, involving electrons. Since the processes involve electrons Te is
important.
bound-bound transitions - electron impact causing excitation or deexcitation
A + e ↔ A∗ + e
free-bound transitions - electron impact causing ionization or three-body
recombination
A + e ↔ A+ + e + e
Two models are of particular interest.
1.
Local thermal equilibrium (LTE)
High density, low temperature plasmas. Collisions excite and deexcite. The
populations of the energy levels is given by Boltzmann distribution for temperatute T.
For two levels n and m
nn
g −
= ne
nm gm
En − Em
kTe
where the ns are the number densities, the gs are the statistical
weights and the Es are the energies.
This can be generalized to give ratios of number densities of energy
levels for atoms in different states of ionization and as a special
case the Saha equation (see Chapter I).
2.
Coronal equilibrium. Like the sun’s corona.
Low density, high temperature plasmas. Collisions excite and photons
deexcite.
Since collision rate depends on density, which is low, the only way for a
downward transition is by spontaneous emission
A low density plasma is optically thin so any photons escape before they can
excite another atom and the only way for an upward transition is by collisions.
Most plasmas lie outside LTE. The coronal model is appropriate for tokamak plasmas
as long as there has been sufficient time for equilibrium to have been reached.
Coronal equilibrium may apply to lower energy levels but LTE may still apply to the
higher.
Line spectra
Neutral atoms and ions that still have bound electrons emit radiation whenever they
make a transition from a higher energy state to a lower one. This provides a way of
studying the working gas and any impurities in the plasma. However interpretation
(besides simply revealing which species are present) is generally very difficult. We
need to know the populations in the various possible states (complicated functions of
n and Te and the composition of the plasma) and understand the processes that
maintain them.
A** → A* + hν
Diagnostics frequently use a monochromator to measure the
absolute intensity of line
ratio of intensities of two lines
line shape and/or width
Line broadening
The width of spectral lines is principally due to
Doppler broadening - particle thermal motion gives a doppler shift. This would give
Ta or Ti.
Pressure broadening which includes collisional broadening and Stark broadening.
depends on the influence of nearby particles on the emitting atom.
Collisional broadening. Most of the time the atom radiates undisturbed,
occasionaly there is a collision and an abrupt phase change.
is the mean time between collisions.
∆fFWHM ≈
1
πτ
where τ
Stark broadening. The most important perturbing effect is the E field of
nearby atoms.
Instrumental broadening. The measuring instrument has a finite resolution.
Use convolution to combine the effects of the different kinds of broadening
Continuum spectra
Diagnostics may be based on
absolute intensity
ratio of intensities at two wavelengths
In visible, uv and x-ray.
bremsstrahlung
free-free transitions - interaction of electrons with ions of effective nuclear
charge Z.
emissivity (radiated power /unit volume /unit angular frequency /unit solid
angle)
j = 8.6 × 10
−53
2
1
e 2
n e ni Z T
−
−
e
hω
kTe
No bremsstrahlung from non-relativistic like particle interactions.
Radiation losses by a fusion plasma are worse if the impurities make the
Zeffective higher.
recombination radiation
A + + e → A∗
j is similar but the exponent is replaced by a series of terms with
−
Ge
hω −
Z2
n2
kTe
E1
where E1 is the ionization potential of hydrogen and G = 1 if
Z2
hω >> 2 E1 , otherwise 0.
n
Recombination radiation is less important than bremsstrahlung if
kTe > 3E1 Z 2 .
The combined spectrum is shown below.
Note
−
at low frequencies
1
2
j ≈ Te ,
the recombination edges. The highest frequency edge is when the
atom ends up in the ground state after the electron is captured.
Points to note
Need to calibrate setup to find its sensitivity at the wavelengths being used.
Need to be aware of: any lines in the wavelength range, radiation from vessel walls.
Exercises
Electrostatic or Langmuir probes
1.
The random ion current reaching the probe Ii remains appoximately equal to
the saturation ion current given by
kTe
I sat i ≈ − 0.57ne
A
mi
where A is the effective probe area.
The electron current is
e ( V − VS )
Ie = Isat e e
The total current is
kTe
where Isat e =
1
8kTe
ne
A.
4
πme
I = Ie + Ii .
(
)
e ( V − VF )


dI e I − Isat i
(i) Show I = I sat i 1 − e kTe  and hence
=
or


dV
kT
e


d
e
ln I − Isat i =
.
dV
kTe
Hint: Use the definition of VF.
( (
))
(ii) Show that, for a hydrogen plasma, VS ≈ VF + 3.3
kTe
.
e
This is how VS might be found.
2.
Estimate Te and n for a hydrogen plasma from the (a) single probe and the (b)
double probe traces below. Take A = 5 mm2.
Microwave interferometry
3.
The output from the detector in a 35 GHz microwave interferometer is shown
below. The fringes in this trace occurred during he decay of the plasma.
(i) Plot <n> vs. t on a log-lin graph.
Take the path length l to be 150 mm and assume the density profile is
parabolic.
Hint: Each interference fringe corresponds to a 2π change in phase. You could
calculate densities at times when φplasma = Nπ.
(ii) How many fringes do you calculate to be between n = nc and n = 0?
Notice how attenuation due to collisions becomes important near cutoff.
Thomson scattering
8πre2
4.
The Thomson scattering cross-section σ =
= 6.65 × 10 −29 m 2 .
3
(i) Calculate the fraction of photons collected when the plasma density is 1020
m−3. Suppose the optics is looking at 1 cm of the total path through the plasma and
collects photons within a solid angle of 0.01 steradian.
(The fraction is very small.)
(ii) If the ruby laser wavelength is 694.3 nm and the pulse has an energy of 10
J, how many photons arrive at the photomultiplier.
Scattering from electron density fluctuations
5.
Suppose there is a spectrum of fluctuations in the plasma ranging in frequency
from 0 to 20 kHz. There are two ways this could be investigated.
1. Keep the geometry fixed, i.e., keep the scattering angle θ fixed. Change the
frequency of the incident wave.
2. Keep the frequency of the incident wave fixed and change the angle at
which the scattering is viewed. Get a k-spectrum.
In this example, the frequency of the incident waves was kept fixed at 35 GHz.
The frequency spectra obtained for different values of scattering angle θ are shown.
(i) Plot the spectrum of the fluctuations.
(ii) Plot the dispersion relation (ω vs. k) and estimate the velocity of the wave.
6.
On a log-log plot of n vs. Te (see Chapter 1), show where scattering from
density fluctuations might be studied with waves of (i) f = 100 GHz, and (ii) λ = 10.6
µm.
Two criteria must be satisfied.
1. the waves must be able to propagate. (So n < nc.)
2. in order to see collective behaviour λ >> λD. (Set the criterion as λ = 10 λD.)
Summary of chapter
Electron density measurements
electric probe
interferometry
reflectometry
absolute spectral intensities
spectral line broadening
Electron temperature and electron energy distribution
electrical conductivity
electric probe
Thomson scattering
electron cyclotron emission
ratios of spectral intensities
Ion temperature
spectral line broadening
Electric current, magnetic field
magnetic coils
Have not mentioned several diagnostics important for fusion studies
charged particle analysers
ion temperature by charge exchange with neutrals
neutron diagnostics
It is useful to have more than one way of measuring a particular plasma parameter.
PLASMA PHYSICS
VIII. PROCESSING PLASMAS
Introduction
Plasmas are used to manufacture semiconductors, to modify the surfaces of materials,
to treat emissions and wastes before they enter the environment, etc.
The plasma is a source of ions.
Industry wants
plasma devices that are simple and compact
that enable processing at high rates with high efficiencies
processing to be uniform over large areas
In order to produce the best plasma for the process in question,
we can control
size and shape of plasma device, gas mixture, ps, V, i, B, ω
which determine
ion and electron densities and temperatures, ion fluxes and energies.
Besides the plasma physics, there atomic and molecular processes within the volume
of the gas and on the surface to be understood as well.
Applications
Deposition by sputtering
target is source of coating material
DC sputtering: metals, the target is the cathode
RF sputtering: non-conducting materials
ion-beam sputtering:
reactive sputtering: e.g., TiN coatings for wear resistance, Ti target and N2 gas
substrate may be biassed so ion bombardment modifies the growing film
Deposition by Plasma-assisted CVD
CVD (chemical vapour deposition)
is a thermal process - the reaction between gas and hot surface.
requires high temperatures
Plasma-assisted (or plasma-enhanced) CVD
Electron bombardment of atoms and molecules in the plasma volume
results in excitation, ionization and dissociation thereby producing a
variety of chemically reactive species with vastly different properties
from their parent gas.
requires lower temperatures
e.g., TiN for wear resistance. A gas mixture of TiCl4, N2 and H2.
thermal CVD 900-1100 °C (above the softening temperature for steel),
plasma CVD 500 °C
e.g., Si3N4 for passivation layer in semiconductor manufacture. A gas
mixture of SiH4, N2, NH3. thermal CVD 900 °C, plasma CVD 300 °C.
e.g., diamond thin films. A diamond thin film is exceptionally hard,
low electrical conductivity, high thermal conductivity. Our experiment
uses a 2.45 GHz magnetron source, no magnetic field. The process
uses a 99% H2, 1% CH4 gas mixture at a pressure of 10’s of torr. The
film is grown on a silicon wafer which is subsequently etched away.
The individual diamonds are nm to µm in size depending on the detail
of the deposition process.
Etching
sputter etching
reactive ion etching
e.g., in semiconductor manufacture, reactive F radicals react with Si to
form volatile species that can be pumped out.
A generic plasma reactor for deposition and etching
DC discharges
Cathode sheath
At the cathode, ions accelerated across the sheath strike the cathode and cause (i)
secondary emission of electrons (essential to the maintenance of the discharge - see
Chapter I) and (ii) sputtering of material from the cathode. This material coats the
substrate.
Te >> Ti in the plasma.
However we will suppose the electron density in the sheath is small enough to ignore
and the ions have a small velocity
us =
kTe
as they enter the sheath. Small
mi
means that the kinetic energy of the ions as they enter the sheath is much less than the
kinetic energy they gain as they are accelerated across the sheath towards the
cathode.
If low pressure, there are no ion collisions.
The equation of continuity is
n( x ) v( x ) = ns u s (1).
The equation of conservation of energy is
1
2
mi v( x ) + eφ ( x ) = 0 (2).
2
Eliminate v(x) from (1) and (2) and obtain an expression for n(x). Substitute this into
Poisson’s equation
d 2φ( x)
en( x )
=−
2
dx
ε0
and solve.
Find that the potential across the sheath follows the the Child-Langmuir law
4
φ( x) ∝ x 3 ,
and the sheath thickness
3
 eφ  4
s ≈  s  λD,
 kTe 
so the thickness of this sheath could be hundreds of λD’s.
If high pressure, there are ion-neutral charge exchange collisions and both ions and
neutrals strike the cathode.
Instead of conservation of energy use
v = µE = − µ
The ion energy distribution function looks like:
dφ
where µ is the mobility.
dx
The low energy continuum is a result of ion collisions.
Sheath near a floating electrode or wall.
There are both rapidly-moving electrons and slowly-moving ions in the sheath.
The key equation expresses the fact that the net current (due to both ions and
electrons) to the floating electrode is zero (See Chapter VII, Exercise 1).
The sheath thickness is ≈ λ D and the p.d. across it is VF − VS ≈
kTe
which is
e
insufficient to accelerate the ions for sputtering.
Magnetron
The figure shows the planar magnetron.
(The figures do not show gas inlets, vacuum pumping ports or matching networks.)
The magnetron is used for sputter coating and metallization. It is capable of high
current densities - and fast processing.
The magnetic field (typically 0.02 T) confines the secondary electrons so a bright
plasma ring sits above the cathode. Ions however can reach the cathode and bombard
it.
RF discharges
Capacitively-coupled RF discharge also called RF diode
Under the applied RF voltage the plasma-sheath boundary at each electrode oscillates
up and down. The bulk of the plasma remains uniform.
This is the most common plasma source for materials processing. Low pressure
discharges can provide high ion energies for etching and high pressures discharges
can provide low ion energies for deposition.
The main plasma heating mechanism are:
Ohmic heating. P = I2R. In this case the RF current is capacitively-coupled
across the sheaths.
Stochastic heating. The sheath edges are oscillating up and down. Electrons
striking the sheath edge have their velocities changed and over 1 period, there is a net
gain in energy. The word stochastic refers to the probabilistic nature of the electron
collisions with the sheath.
Advantages:
simple construction,
no magnetic field required
Disadvantages:
Low ion flux and high ion energy (typically 100’s of eV), these cannot be
varied independently. If damage to the substrate is likely, processing must be
carried out slowly.
Voltage drop at sheath is sensitive to geometry (The total area of the grounded
surfaces is much greater than the area of the powered electrode. The sheath at
the powered electrode therefore has a smaller capacitance and hence a larger
voltage drop.)
Inductively-coupled RF discharge
The spiral coil is the primary, the plasma is the secondary.
A multipole magnetic field can be used to enhance the confinement of the plasma.
Advantages:
Ion energy can be controlled independently by applying capacitively-coupled
rf to bias the substrate.
The ion energies are much less, 10’s of eV and have a much milder action on
the substrate.
Disadvantages:
diameter/height is large making cooling and pumping difficult
non-uniform density profile (ring-shaped)
RF or microwave heated discharges
Here is an outline of how we might calcuate the power absorbed by the plasma.
Start with the electron momentum equation
∂v
ρ e e + ρ e v e ⋅ ∇v e = − ne e( E + v e × B) − ∇pe − ρ eυv e
∂t
In the last term we have assumed that velocities vi and vn are << ve.
This time, consider oscillations, not waves, so E = E 1 e − jωt x$ , write
n = n 0 + n 1 e − jωt and v e = v 1e e − jωt and consider only the first-order terms.
(i) No magnetic field.
−e
1
E1 .
m e − jω + ν
Power absorbed per unit volume
p = j⋅ E .
0
1
Now j x = − n evex .
Some care is required here. To calculate the power you must take the real part
of jx and multiply it by the real part of E. This gives the instantaneous power. The
time-averaged power absorbed (over one cycle) is
2
1 n 0 e 2 E1
ν2
p =
2 me ν ν 2 + ω 2
vex1 =
Note that if there are no collisions, there is no power absorbed.
(ii) magnetic field
1 n 0 e2 E1
p =
2 me ν
2
1

1
ν2
ν2


+
2 ν2 + ω −ω 2 2 ν2 + ω +ω 2 
(
(

ce )
ce ) 
ECR discharge (electron cyclotron resonance)
Note that in case (ii) above the power absorbed is large if the frequency is near
the electron cyclotron resonance frequency. ECR is an improvement over the nonresonant case.
The ECR discharge uses inexpensive 2.45 GHz magnetron microwave sources
that can deliver 0.3 to 6 kW. The microwaves are launched as RHCP waves into the
high-field region and are absorbed in the resonance region where ω = ωce. When fce =
2.45 GHz, the resonance magnetic field B = 0.0875 T. The plasma created in the
resonance region flows into the main chamber.
You do. Why does the ECR discharge use RHCP waves travelling from the high field
region? Refer to the CMA diagram in Chapter V.
Helicon plasma
RF driven antenna excites a helicon wave (see Chapter VI) and a resonant waveparticle interaction transfers energy to the plasma. The plasma flows into the main
chamber. This results in a high density plasma.
Vacuum arc
plasma is fully ionized
high deposition rate, low substrate temperature
greatest drawback is the formation of macroparticles. Use magnetic field filter.
PI3 Plasma immersion ion implantation
Ions from the plasma are accelerated by means of a series of negative high-voltage
pulses. The beam injected into the surface changes the atomic composition and the
structure near the surface.
semiconductor maunufacture; now routine
metallurgy; an emerging technology in for creation of new surface alloys - not
restriced to planar surfaces.
PLASMA PHYSICS
IX. FUSION PLASMAS
Fusion reactions
Fusion reactions are the source of energy in stars.
You do. Look up this topic in a Modern Physics textbook, in particular the protonproton chain and the carbon cycle.
Controlled thermonuclear reactions are a potential source of energy on earth.
The first fusion reactors will exploit the reaction
D + T → α(3.5 MeV) + n(14.1 MeV) (energy of reaction 17.6 MeV)
D or 2H is deuterium, T or 3H is tritium and α or 4He is an alpha particle. n is a
neutron and p is a proton.
This reaction has the lowest ignition temperature (about 4 keV). Ignition - when all
the energy in the α’s is sufficient to maintain the reaction.
In a magnetically-confined plasma, the charged α’s remain trapped long enough to
deliver most of their energy back into the plasma before escaping, the n’s escape
immediately.
Fuel is readily available.
D is a readily separable component of sea water. One hydrogen nucleus in
6700 is D.
T is regenerated when n’s are absorbed in the lithium blanket surrounding the
reactor vessel (tritium breeding).
n + 6Li → T + α
n + 7Li → T + α + n
Energy of n’s and α’s is converted in the blanket into heat which is carried away by a
suitable coolant to make steam for conventional electricity generation. One blanket
design uses vanadium alloy as the structural material to withstand the high radiation
environment and liquid lithium as both tritium breeder and coolant.
Another process is
D + D → T + p (4.0 MeV) or equally probably
D + D → 3He + n (3.3 MeV)
followed by
D + 3He → 4He + n (18.34 MeV)
In this process there is no need to manufacture T. However the ignition temperature is
much higher (about 35 keV)
To get a fusion reaction, the two nuclei have to get sufficiently close to each other for
the strong but short-range nuclear force of attraction to take over from the Coulomb
force of repulsion. The plasma will have a Maxwellian distribution and it is the fast
particles in the tail of the distribution which undergo fusion.
Cross-sections for D-T and other reactions. The D-T reaction has the highest crosssection.
A reactor must produce more power from the reaction than is required for heating the
plasma and operating the device.
Power produced per unit volume
n2
<σv>E
preaction =
4
where <σv> is an average over the distribution, E is the 17.6 MeV released in each
reaction.
Power lost per unit volume by bremsstrahlung
1
pbrems = 1.6 × 10−40 neniZ2 T 2
This result follows from the expression for emissivity in Chapter VII.
Keep the effective Z as low as possible to minimize power lost by bremsstrahlung.
Power lost per unit volume by escaping D and T ions
plost = energy density ÷ energy confinement time
3
3nkT
= 2 × n × kT ÷ τ E =
.
τE
2
The energy confinement time τE, how long it takes to cool down once the external
heating is switched off, is a measure of how effective the confinement is.
Breakeven is when the power output is sufficient to maintain the reaction. Let us
calculate a criterion for breakeven.
Assume that the external power input + power carried by the α’s produced in
the DT reaction replaces the power lost.
pext + pDT,α = pbrems + plost.
This external power comes from retrieving some of the power lost by bremsstrahlung,
escaping D and T ions and some of the power produced by the n’s. The efficiency η is
estimated to be about 0.3. So
pext = η( pbrems + plost + pDT,n).
Substitute using the earlier expressions and plot nτE vs T. There is a minimum
at about 30 keV. This minimum leads to the so-called Lawson criterion that
for D-T reactions, nτ E > 10 20 m−3 s,
Similarly,
for D-D reactions, nτ E > 10 22 m−3 s.
The temperature must be sufficiently high so an alternative criterion is that
for D-T reactions, nτET > 5 × 1021 m−3 s keV.
Of course, economic viability will eventually be the most important
consideration.
Ignition is when the power in the α’s is sufficient to balance the power lost by
bremsstrahlung and escaping hot D and T ions. This is more difficult to achieve.
Major problems
Plasma confinement - keeping the hot plasma out of thermal contact with the
vessel walls.
Plasma heating
Main candidates for controlled fusion
Magnetically-confined plasmas. Closed systems like tokamaks and
stellarators.
Inertially-confined plasmas. Laser fusion.
Magnetic confinement fusion
Plasma confinement
The magnetic field guides the charged particles and restricts their diffusion to the
walls.
See Chapter II.
Use a closed system, a torus rather than mirror to avoid end losses.
Twist magnetic field lines to avoid E×B drift.
tokamak: internal plasma current
stellarator: external helical conductors
Tokamak
In a tokamak the plasma is like secondary winding of transformer.
The current heats the plasma (Ohmic heating) and helps confine it. Cannot
analyse confinement and heating separately.
A steady current cannot be produced this way.
ITER (International Tokamak Experimental Reactor)
Stellarator can operate steady state.
Existing and projected large stellarators.
MACHINE
COUNTRY
MINOR
RADIUS
A(m)
MAJOR
RADIUS
R(m)
CHS
HELIOTRON E
WENDELSTEIN 7-AS
H-1
LHD
WENDELSTEIN 7-X
JAPAN
JAPAN
GERMANY
AUSTRALIA
JAPAN
GERMANY
0.2
0.2
0.2
0.2
0.5-0.6
0.5
1
2.2
2
1
3.9
5.5
PLASMA
CURRENT
I(MA)
TOROIDAL
FIELD
B(T)
2
2
3.5
1
3
3
START
DATE
1998
constr.
begun
Plasma heating
Ohmic heating
P = I2R. However this is far from sufficient to reach fusion temperatures.
Auxiliary heating is required. The main methods used at present and envisaged for the
future are:
Electron cyclotron resonance heating
heat electrons, collisions transfer energy to ions.
The system proposed for ITER uses a bank of millimetre-wave gyrotrons each
delivering a power of 1 MW cw at the fundamental frequency, 170 GHz, a total power
> 60 MW.
Neutral particle injection
neutralize accelerated D and T ions in a gas cell.
neutral particles can pass through the magnetic field into the plasma.
in plasma, energy is transferred in charge-exchange collisions with cold ions.
TFTR used 4 neutral-beam injectors with accelerating voltages of 110 kV
delivering 40 MW of power.
Ion cyclotron range-of-frequencies heating
DIII-D uses 4 MW of radio-frequency power in the 30-120 MHz range for
heating and current drive.
Current drive
The plasma in the tokamak is like the secondary winding of a transformer and the
current drops to zero at the end of the pulse.
The millimetre-wave power from the gyrotrons will also be employed to “push” the
electrons so they move in the same direction thereby maintaining the plasma current.
Impurities
Want to restrict impurities sputtered off the vessel wall to reduce bremsstrahlung
radiation losses.
Choose suitable wall materials (e.g., carbon has a low Z).
Use limiter to define the outer boundary of the plasma and keep it away from the
walls.
Use a magnetic diverter to divert particles into a separate chamber from which they
are pumped out.
Stability
The operation of present tokamaks is limited not by confinement but by disruptions if a certain maximum density is exceeded a MHD instability suddenly destroys
confinement.
Other approaches
spherical tokamak, reversed-field pinch, spheromak
fission-fusion hybrid.
Inertial confinement fusion
In laser fusion, the power from a bank of high-power pulsed lasers is focussed onto a
D-T target.
Application of the Lawson criterion shows that the energy required to initiate a
reaction is too high, we need to compress the fuel beyond solid densities. Need a
central hotspot 100-200× solid density at about 5 keV and the surrounding main fuel
region 1000-5000× solid density at a lower temperature.
Two kinds of targets are being studied:
Direct drive
A spherical target. The fuel is surrounded by a
spherical shell. The outer part is ablated and the
rest of the shell implodes towards the centre
compressing the fuel. This approach requires
uniform irradiation to avoid instabilities.
Indirect drive
The target is in a hohlraum (a radiation cavity) made of a high-Z material. The laser
beams strike the walls and are converted to x-rays. This gives a more uniform
implosion.
The lasers used in the present high-power experiments are Nd-glass lasers operating
in the infrared at 1.06 µm. The radiation is frequency-tripled to 351 nm.
Current installations
MACHINE
COUNTRY
ENERGY (kJ)
NO. OF BEAMS
NOVA
GEKKO
OMEGA
USA
JAPAN
USA
40
100
30
10
12
60
Experiments can achieve 1000× solid density, can produce neutrons.
The illustration shows the Nova upgrade.
NIF Proposed National Ignition Facility
1.8 MJ, 500 TW peak power, 20 ns pulse glass laser. 192 beams grouped into 12 lines.
Targets have been designed on the computer that should ignite under a 1.35 MJ pulse.
Other approaches
Inertial confinement fusion by light ions and heavy ions is also being studied.
Exercises
Fusion reactions
1.
Suppose the distance between the centres of the two nuclei must be within 3
fm for the nuclear force to be effective.
Use this to estimate the ignition temperature.
Comment. The temperature you obtain will be too high because:
the nuclei have a finite size,
quantum mechanical tunnelling through the Coulomb barrier can
occur,
the fast nuclei at the tail of the Maxwellian distribution are most
important.
Magnetic confinement fusion
2. Ohmic heating
(i) Find an expression for the pohmic, the power per unit volume for ohmic
heating of a toroidal plasma, in terms of resistivity, plasma current and plasma
dimensions.
(ii) Write down the expression for plost, the power lost per unit volume by the
escaping hot D and T ions, in terms of temperature and the energy confinement time.
(iii) The maximum temperature that can be reached by ohmic heating is set by
pohmic > plost.
Use the parameters for JET, I = 7 ×106 A, a = 1 m and the empirical value for
n
the containment time τ E ≈
a 2 to show that the maximum temperature is far
2 × 10 20
below that for fusion.
(Get resistivity from Chapter IV.)
Inertial confinement fusion
3.
The Lawson criterion nτ E > 10 20 m3 s and a T of 10 keV must be satisfied.
(i) Estimate n for solid hydrogen (Take ρ = 200 kg m−3).
Use this definition of the energy containment time τE, the time for the plasma
to expand freely,
1 R
τE ≈
4U
where R is the target radius and U is the sound speed (about 106 m s−1 for T = 10
keV).
(ii) Estimate R.
(iii) Estimate the energy required if all the atoms in a sphere of radius R were
to have an energy of 10 keV. This is much more than can be provided by present-day
lasers.
(iv) Suppose ρ is 100× higher.
Summary of chapter
D-T reaction, reactor basics, Lawson criterion.
Magnetically-confined fusion, confinement in tokamaks and stellarators,
methods of heating, current drive, impurities, stability.
Inertially-confined fusion, direct drive and indirect drive.
PLASMA PHYSICS
SOLUTIONS TO EXERCISES
Dr Ferg Brand
School of Physics, University of Sydney NSW 2006, AUSTRALIA
Ch I Ex 1
Ch I Ex 2
Ch I Ex 3
Ch I Ex 4
Ch I Ex 6
Ch I Ex 7
Ch I Ex 8
Ch II Ex 1
Ch II Ex 2
Ch II Ex 3
Ch II Ex 4
Ch II Ex 5
Ch II Ex 6
Ch II Ex 7
Ch II E × B drift. Derive a condition for the trajectory of the charge to look like
Ch III Ex 1
Ch III Ex 2
Ch III Ex 3
Ch III Ex 4
Ch III Ex 5
Ch III Ex 6
Ch IV Ex 1
Ch IV Ex 2
Ch IV Ex 3
Ch IV Using the notions of scale length L and scale time T (the distance and time for
a significant change in density, particle velocity, etc), show that the assumption that
∂n 1B
1 ∂ 2 n1B
<<
can be written as νT >> 1 and this leads to λ m << L . i.e., there
ν BA ∂t 2
∂t
must be many collisions over L.
Ch IV The electric field set up because the ions and electrons diffuse at different rates
can be neglected if the mobility term << the diffusion term.
Show that this leads to L << λD. (Hint. Use ∇ ⋅ E =
σ
to get an expression for
ε0
E.)
Since λD is usually very small, this condition is rarely satisfied
Ch V Ex 1
Ch V Ex 2
Ch V Ex 3
Ch V Ex 4
Ch V Ex 5
Ch V Ex 6
Ch V Ex 7
Ch V Ex 8
Ch V Ex 9
Ch V Ex 10
Ch VI Ex 1
Ch VI Ex 2
Ch VI Ex 3
Ch VI Ex 4
Ch VII Ex 1
Ch VII Ex 2
Ch VII Ex 3
Ch VII Ex 4
Ch VII Ex 5
Ch VII Ex 6
Ch VIII Derive the Childs-Langmuir law equations for the potential across the sheath
and the sheath thickness, for the case where there are no ion collisions.
Ch VIII Derive the expression for the time-averaged absorbed power for the case
where there is a magnetic field.
Ch IX Ex 1
Ch IX Ex 2
Ch IX Ex 3
Physics 3 & 4 PLASMA PHYSICS
Assignment 1
Due Monday 7 Aug 2000
Normal
1
Ch I Ex 7
2
Ch I Ex 8
3
Ch I Locate a book or website that illustrates the colours in a glow
discharge for different gases.
4
Ch II Ex 3
5
Ch II Ex 6
Advanced
1
Ch I Ex 7
2
Ch I Ex 8
3
Ch II E × B drift. Derive a condition for the trajectory of the charge to
look like
4
Ch II Ex 3
5
Ch II Ex 6
6
Ch III Ex 5
Physics 3 & 4 PLASMA PHYSICS
Assignment 2
Due Monday 28 Aug 2000
Normal
1.
Ch V Ex 3 Parts (i) to (iv)
2.
Ch V Ex 4
3.
Ch V Ex 5
4.
Ch V Ex 6 But this time calculate the frequencies of reflection of X waves
from the F1 layer. (Take the density to be 4.5×105 cm−3.)
Advanced
1.
Ch V Ex 3 Parts (i) to (v)
2.
Ch V Ex 4
3.
Ch V Ex 5
4.
Ch VI Ex 2
Physics 3 & 4 PLASMA PHYSICS
Assignment 3
due Monday 16 Oct 00
Normal
1.
Ch VII Ex 2
2.
Ch VII Ex 3
3.
See Ch VIII page 3. Derive the Childs-Langmuir law equations for the
potential across the sheath and the sheath thickness for the case where
there are no ion collisions.
Advanced
1.
Ch VII Ex 2
2.
Ch VII Ex 3
3.
See Ch VIII page 7. Derive the expression for the time-averaged
absorbed power for the case where there is a magnetic field.
PLASMA PHYSICS
EXAMINATION PAPER, November 2000
Time allowed 1 ½ hours.
Plasma physics formula sheet is included.
Candidates may bring in a ‘cheat sheet’ – one sheet of
A4 paper, handwritten on one side only.