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

dvxdvydvz around velocity (vx,vy,vz) is proportional to exp −

1

22mv

kTdv dv dvx y z.

We can construct the Maxwellian velocity distribution function f(v)

( )f v nm

kT

mv

kT=

−

2

1

23

2

2

πexp ,

and use it to calculate average values.

e.g., Calculate the particle density

( )n f v dv dv dvx y z= ∫

calculate the mean velocity

( )u v= ∫1

nf v dv dv dvx y z = 0

calculate the mean speed

( )vn

vf v dv dv dvkT

mx y z= =∫1 8

π

calculate the rms speed

( )vn

v f v dv dv dvkT

mv

kT

mrms x y z rms2 21 3 3= = =∫ so

You do. Show the average kinetic energy per particle is E kTav = 3

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,

−×≅T

U

n

T

n

n i

in

i exp1032

3

27

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 self-sustaining. 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

Eb

p

Eexp

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.

One electron leaving the cathode becomes eηV electrons arriving at the anode.

And eηV −1 positive ions heading back towards the cathode.

One ion produces γ electrons.

So one electron gives rise to ( )γ ηe V −1 electrons.

For breakdown, this must be ≥ 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

j aTe

kT= −

2exp

φ

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

A e A A2+ ∗+ → +

is the fastest recombination process in a weakly-ionized gas like a glow discharge Radiative recombination

A e A h+ ∗+ → + ν 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 r from the charge is

φπε

= q

r4 0

.

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

factor becomes exp −+

1

22mv q

kTdv dv dvx y z

φ. φ depends on position so the

probability depends on position.

The particle density is given by ( )n f v dv dv dvx y z= ∫ so nq

kT∝ −

exp

φ

for electrons n ne

kTe = − −

0exp

φ

for ions (we will suppose they are singly-ionized) n ne

kTi = −

0exp

φ

Gauss’ Law can be written as

∇ =.Eσε 0

φ−∇=E so

−∇ =2

0

φ σε

.

This is Poisson’s equation.

The charge density is σ φ φ= − + = − + −

en en en

e

kT

e

kTe i 0 exp exp .

Assume that this potential term is very small, kTe <<φ

σ φ φ φ≅ − +

+ −

= −en

e

kTen

e

kT

n e

kT0 00

2

1 12

.

I am going to use spherical coordinates (and assume spherical symmetry)

≡∇dr

dr

dr

d

r

φφ 22

2 1.

Poisson’s equation becomes

kT

en

dr

dr

dr

d

r 0

202

2

21

εφφ −=

−

with solution

φπε ε

= −

q

r

r

kT

n e

4

20 0

02

exp .

The potential falls away exponentially.

Call λε

D

kT

n e= 0

02 the Debye length then

φπε λ

= −

q

r

r

D4

2

0

exp .

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 D>> λ , and

(ii) there must be enough electrons to produce shieldingND >>> 1 , where ND 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 n n ne i≅ ≅ 0 .

If we put an electrode into the plasma, it becomes shielded by a sheath ot thickness ≈ λ D .

)min K,in ( m 0.69 3-e

eD nT

n

T=λ

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.

Equation of motion for the slab of electrons is F md x

dt=

2

2

mass of electrons in slab m = meneLA What is the force on the electrons? We have two oppositely charged sheets facing

each other. The electric field between them is E s=σε 0

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 =−n LAeen x

ee

ε 0

.

Equation of motion becomes

n LAmd x

dtn LAe

en x

d x

dt

n e

mx

e e ee

e

e

2

20

2

2

2

0

= −

= −

ε

ε

or

with solution ( )x A tpe= cosω ,

where the (electron) plasma frequency is

ωεpe

e

e

n e

m=

2

0

.

This is a ‘natural’ frequency for the plasma.

f n npe e e= −8 98 3. Hz ( in m ).

We encounter ωpe when discussing wave propagation.

You do. Show e

Dpe m

kT=λω .

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 ionosphere van Allen belts aurora solar wind solar corona interstellar medium 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 = 1017 m−3 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

Debye lengthλ εD

kT

n e= 0

02

,

plasma frequencyωεpe

e

e

n e

m=

2

0

.

The derivations will not be examinable. In subsequent sections you will be expected to use Maxwell’s equations in differential form.

∇ ⋅ =

∇ × = −

∇ ⋅ =

∇ × = +

E

EB

B

B jE

σε

∂∂

µ ∂∂

0

0 2

0

1

t

c t

where, in cartesian coordinates

∇ = + +.E∂∂

∂∂

∂∂

E

x

E

y

E

zx y z . ( )∇ × = −E

xz yE

y

E

z

∂∂

∂∂

, etc.

and in cylindrical coordinates,

( ).

11

z

EE

rr

rE

rzr

∂∂+

∂∂

+∂

∂=⋅∇φ

φE

( )

zφrE ˆ1

ˆˆ1

∂∂

−∂

∂+

∂∂

−∂

∂+

∂∂

−∂∂

=×∇φφ

φφ rzrz E

r

rE

rr

E

z

E

z

EE

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 E v B= + ×q . E is measured in V m−1. B in T (often in gauss. 10000 gauss = 1 T)

1. E = constant, uniform Suppose E x= E$ . The Lorentz force equation becomes

dv

dt

q

mE

dv

dtdv

dt

x

y

z

=

=

=

0

0.

This describes a constant acceleration along x.

2. B = constant, uniform SupposeB z= B$ . Here is how we might produce a uniform magnetic field.

dv

dt

q

mv B

dv

dt

q

mv B

dv

dt

xy

yx

z

=

= −

=

(1)

(2)

(3)0.

Take d

dtof (1), substitute using (2)

d v

dt

q

m

dv

dtB

q

m

q

mv B Bx y

x

2

2= =

.

Write

ω c

q B

m=

the (angular) cyclotron frequency or gyrofrequency. (Note the symbol Ω is often used.)

d v

dtvx

c x

2

22 0+ =ω .

Similarly, d v

dtvy

c y

2

22 0+ =ω .

The solutions can be written as v v t

v v tx c

y c

= −=

⊥

⊥

sin

cos

ωωm

(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

xv

t r t

yv

t r t

cc L c

cc L c

= =

= =

⊥

⊥

ωω ω

ωω ω

cos cos

sin sinm m

rv

Lc

= ⊥

ω 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 z= B$ .

( )

( )

dv

dt

q

mE v B

dv

dt

q

mE v B

dv

dt

q

mE

xx y

yy x

zz

= +

= −

=

(1)

(2)

. (3)

(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

d v

dtv

E

Bx

c x cy

2

22 2+ =ω ω and

d v

dtv

E

By

c y cx

2

22 2+ = −ω ω ,

with solutions

v v tE

B

v v tE

B

x cy

y cx

= − +

= −

⊥

⊥

sin

cos

ω

ωm .

The path of an electron is a combination of uniform circular motion plus a drift, called an E B× drift.

( )v x yE B

E B y xBE E

B× = − = ×12

$ $ .

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 vE

BE B× = .

Here is an example where theE B× drift can cause a plasma to rotate.

4. B = constant, non-uniform. We will consider two distinct cases.

Case (a):

( )B B zz = +0 1 α

B B x

B B y

x

y

= −

= −

0

0

2

2

α

α

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

dt

q

mv B v B

q

mv B v B z v B y

dv

dt

q

mv B v B

q

mv B x v B v B z

dv

dt

q

mv B v B

q

mv B y v B x

xy z z y y y z

yz x x z z x x

zx y y x x y

= − = + +

= − = − − −

= − = − +

0 0 0

0 0 0

0 0

2

2

2 2

α α

α α

α α

We will write v v v= +0 1where 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 v0 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 v v x y zx y z

0 0 0 0 0 0, , , , , ; then we write down the first order terms, i.e., the terms in v1and

those containing α; then substitute for v v v x y zx y z0 0 0 0 0 0, , , , , .

This gives

dv

dt

q

mv B v t B v t v B r t

dv

dt

q

mv B r t v B v t B v t

dv

dt

q

mv t B r t v t B r t

xy c z z L c

yz L c x c z

zc L c c L c

11

00

00 0

0

10

01

00

00

10

00

0

2

2

2 2

= + +

= − − −

= +

⊥

⊥

⊥ ⊥

cos sin

cos sin

sin sin cos cos

ω α α ω

α ω ω α

ω α ω ω α ω

which can be written as

dv

dtv v v t t v v t

dv

dtv v t v v v t t

dv

dtv

xc y z c c z c

yc c x z c c

z

11 0 0 0

10 1 0 0

10 2

2

2

2

= ± − −

= ±

= −

⊥

⊥ ⊥

⊥

ω α ω ω α ω

α ω ω α ω ω

α

cos sin

cos sin

z0

z0

m m

upper sign ions, lower sign electrons, with solutions

v v v t t v v t t

v v v t t v v t t

v v t

x c c c

y c c c

z

1 0 0

1 0 0 2

1 0 2

2 2

2 2

2

= − −

= ±

= −

⊥ ⊥

⊥ ⊥

⊥

α ω ω α ω

α ω α ω ω

α

z0 2

z0

z0

z0

cos sin

cos sinm

.

Adiabatic invariant 1

22mv

B

⊥is a constant of the motion or adiabatic invariant. Adiabatic carries the idea of

slowly-changing.

You do. Show this is true to first order in α. Start with ( ) ( )v v v v vx x y y⊥ = + + +2 0 1 2 0 1 2and

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 ΦM BA= is constant. It follows that the magnetic moment of the gyrating charge is constant. The magnetic

moment is defined as µ = iA where iq

t= , where q is the charge and t is the time for

one gyration and A is the area encircled by the orbit.

µ πω

π= =⊥e

rmv

B

c

L2

1

22

2

.

So µ is constant. Magnetic mirror As an electron spirals into a higher B region, v⊥ increases and rL decreases. Since the

total energy 1

22mv is a constant, vz must decrease. Eventually vz = 0 and the electron

reverses direction. It has been reflected by a magnetic mirror.

e.g., magnetic mirror used to trap plasma in an experimental device.

To do a magnetic mirror calculation use

1

22mv

B

⊥ = constant and conservation of

energy 1

2

1

22 2mv mvz⊥ + = constant.

Case (b):

( )B B x

B B zz

x

= +=

0

0

1 αα

You do. Show Maxwell’s equations are satisfied. B x0α describes the gradient.

B z0α describes the curvature. In the derivations below, the curvature terms are

underlined.

( )

( )( )zBv

m

q

dt

dv

xBvBvzBvm

q

dt

dv

xBvBvm

q

dt

dv

yz

xxzy

yyx

α

αα

α

0

000

00

−=

−−=

+=

Proceed as before.

ttvvdt

dv

ttvvtvdt

dv

tvvdt

dv

czcz

ccxczcy

cycx

ωαω

ωωαωαω

ωαω

cos

cossin

cos

001

201201

22011

⊥

⊥

⊥

=

±±=

−±=

m

upper sign ions, lower sign electrons, with solutions

( )tttvv

v

vvt

vv

tvtv

v

cccc

zz

c

z

cc

cy

zcc

x

ωωωω

α

ωα

ωαω

ωα

αωω

α

sin1cos

2cos2

2

2sin2

001

2020201

20

201

−−=

±±=

+−=

⊥

⊥⊥

⊥

m

Consider vy

1 . There are constant drift terms.

± ⊥αωv

c

02

2 due to the gradient

± αωvz

c

02

due to the curvature.

They combine to give

v

vv

d

z

c

=

± +

⊥α

ω

0 2

02

2.

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, 0B

Bz∇=α .

(ii) From the sketch above, α = 1

Rc

You do. Show this.gradient drift

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 BB ×∇− or BR ×c .

e.g. in a toroidal magnetic field

r

NiB

πµ2

0=

so Br

Br

≈ ∇ ≈ −1 12

and . i.e., ∇B increases as you go radially in towards the axis.

In this caseα = ∇ = −B

B r

1.

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 = ⋅ =∫ v dlover a path back and forwardbetween mirrors

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.

Show that reflection occurs where the magnetic field is BB= 0

20sin θ

.

(b) Suppose now Bmax is the maximum value of the magnetic field.

Show that if θ 01 0< −sin

B

Bmax

then there is no reflection. In a magnetic mirror

device, this would mean the particle was not trapped - it would be lost. This angle defines a cone in velocity space - the loss cone 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.

6. The earth’s magnetic field, in the equatorial plane, is BR

re= ×

−3 10 5

3

T .

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. (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 x= E$ . Constant acceleration along x. 2. B z= B$ . Helical motion along z.

(angular) cyclotron frequencyω c

q B

m=

Larmor radius rv

Lc

= ⊥

ω

3. E constant, B z= B$ . (i) if E z= E$ gyration plus acceleration along z.

(ii) if E x= E$ gyration plus E B× drift along y. vE

Bd =

4. B non-uniform. Case (a):

leads to

1

22mv

B

⊥= constant, this quantity is an adiabatic invariant.

For magnetic mirror calculations use this and conservation of energy. Case (b):

Constant drifts along y.

v

vv

d

z

c

=± +

⊥α

ω

0202

2

where α = ∇B

B, positive ions drift in the direction of −∇ ×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

( )

( )

= − +

+

= −∇ ⋅

ρ ρ ∂∂

ρ

ρ

v t y z vx

v x t y z

x y z t

x x x∆ ∆ ∆ ∆ ∆ ∆ ∆

∆ ∆ ∆ ∆

K

v

increase in mass in ∆t

( )= ∂∂

ρt

x y z t∆ ∆ ∆ ∆

Since mass is conserved, these are equal, so

( )∂ρ∂

ρt

+ ∇ ⋅ =v 0 .

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:

v A v A v A1 1 2 2 3 3= + )

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= , , ,

∆ ∆ ∆ ∆ ∆TT

xx

T

yy

T

zz

T

tt= + + +∂

∂∂∂

∂∂

∂∂

Divide throughout by ∆t and let ∆t → 0.

dT

dt

T

tT= + ⋅ ∇∂

∂v

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

dt

T

tu

T

x= +

= ° + × °= °

∂∂

∂∂

0 2 10 0 07

0 9

. .

.

C / day km / day C / km

C / day.

Momentum equation Let T = vx.

dv

dt

v

tvx x

x= + ⋅ ∇∂∂

v

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

fF

x y zxx=

∆ ∆ ∆

Then

ρ ∂∂

ρv

tv fx

x x+ ⋅ ∇ =v , or

ρ∂∂

ρv

v v ft

+ ⋅ ∇ =

In fluid mechanics, f is separated into three parts: (i) force due to pressure f ppressure= −∇

(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∂∂

ρ µ ρv

v v v gt

p+ ⋅ ∇ = −∇ + ∇ +2 .

(We usually express this equation in integral form and use it to solve problems like:

− + =ρ ρv v F1 1 1 2 2 2v A v A )

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

22ρ ρv p gh+ + = constant)

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∂

ρt

+ ∇ ⋅ =v 0 .

( )ρ∂∂

ρv

v v E v Bt

nq p+ ⋅ ∇ = + × − ∇

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

( ) ( )ρ ∂∂

ρ ρ υii

i i i i i i i ie i etn e p

vv v E v B v v+ ⋅ ∇ = + × − ∇ − −

and for electrons it is

( ) ( )ρ ∂∂

ρ ρ υee

e e e e e e e ei e itn e p

vv v E v B v v+ ⋅ ∇ = − + × − ∇ − −

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 ν ρ νρ ν ρ ν

i ie i e e ei e i

i ie e ei

v v v v− + − ==

0 or

.

In a less than fully-ionized plasma there would be equations for the neutrals as well. In general,

( )∂ρ∂

ραα αt

+ ∇ ⋅ =v 0

( ) ( )ρ ∂∂

ρ ρ υαα

α α α α α α α αβ

αβ α βv

v v E v B v vt

n q p+ ⋅ ∇ = + × − ∇ − −∑

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,

where = for a monotomic gasγc

cp

v

= 5

3.

Both choices lead to ∇ = ∇p U2 ρ where U is a sound speed.

You do. Differentiate these expressions and obtain

isothermal sound speed Up2 =ρ

, and

adiabatic sound speed Up2 = γρ

.

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 B and .

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 ∇⋅ =Eσε 0

. This is the so-called plasma

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 = + × − ∇n e pi i iE v B .

Take equation × B, use v v v= +⊥ and rearrange

vE B B

ii

iB

p

n eB⊥ = × − ∇ ×2 2

.

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,

vv= ∑

ρρ

α α

α

Continuity equation Add the equations of continuity of all the species

( )∂ρ∂

ρt

+ ∇ ⋅ =v 0 .

Continuity of charge

Take 1

mi

× the ion continuity equation − ×1

me

the electron continuity equation.

Use charge density σ = −en ene1 , current densityj v v= −en eni i e e .

∂σ∂t

+ ∇⋅ =j 0 .

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 p= ∑ αα

momentum is conserved so ( )ρ υα αβ α ββα

v v− =∑∑ 0 .

ρ ∂∂

σvE j B

tp= + × − ∇

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 Bt

p= × − ∇

Generalized Ohm’s Law For an ordinary conductor, E j= η , where η is the resistivity. (Recall V = RI, hence El

= RjA so η = RA/l.)

For a plasma, take 1

mi

× ion momentum equation − ×1

me

electron momentum

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 tov v j v v jie

eim

e

m

e= + = −

ρ ρ and ,

νie << νei leads to a collision term nνei(ve−vi) which we write as − ne

me

ηj . Here

η ν=

m

nee ei

2 is a constant of proportionality.

m

ne t ne nepe

e2

1 1∂∂

ηjE v B j B j= + × − × + ∇ −

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 jt t

, are negligible.

3. Cold plasma. ∇ ∇p pi e, are negligible.

Under these approximations, Maxwell’s ∇ × B equation becomes

∇ × =B jµ0 ,

the momentum equation reduces to j B 0× = ,

and the generalized Ohms law reduces to 0 E v B j= + × −η .

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 jµ0 , rearrange to obtain

( )∇ +

=

⋅∇p

B2

0 02µ µB B

.

If we have a straight cylinder of plasma, the rhs is zero and

pB+ =

2

02µ constant.

p is the particle pressure, B2

02µis the magnetic pressure.

The magnetic tension = B2

0µ N m−2. If the field lines are straight, this is in the

direction of the field lines, if curved there is a ⊥ component. The term on the rhs is related to this component.

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 j Rz= π 2 .

(b) Use∇ × =B jµ0 and show ( )dr

dBrj z

0

1

µθ −= and ( )B rIr

Rθµπ

= 022

.

(c) Write the radial component of the momentum equation 0 j B= × − ∇p , substitute. Show

42

20

0

2

22 R

rIBp

dr

d z

πµ

µ−=

+ .

(d) Finally, integrate over the range 0 to R and show ( ) ( ) ( )B R

pB I

Rz z2

0

2

0

02

2 220

0

2 4µ µµπ

= + − .

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

derivative can be written as∂∂u

x

u

L≈ and choose a scale time τ so that any time

derivative can be written as ∂∂ τu

t

u≈ .

e.g., ∇ × = − ≈EB∂

∂ τt

E

L

B becomes . Write

LV

τ≈ a velocity, so E BV≈ .

1. Examine the single fluid momentum equation. Show the terms are in the ratio

nmV

jBnm v

Lie th e

τ: :

2

or 12

2: :

jB

nm V

m

m

v

Vi

e

i

th eτ.

This suggests that the ∇p term could be ignored if the plasma is cool so we can use

VjB

nmi

≈ τ.

2. Examine the Generalized Ohms Law. (a) Show the terms are in the ratio

111

1 12

2

2ω ω τ ω τ ω τν

ω ω τce ci ci ce

th e ei

ce ci

v

V: : : : : .

(b) Which terms are important or can be neglected if

(i) τω

>> 1

ci

, (ii) τω

≈ 1

ci

, (iii) τω

≈ 1

ce

and (iv) τω

<< 1

ce

?

(c) When can the ηj term be dropped?

3. Show diamagnetic drift << E × B drift when τω

>> 1

ci

.

4. Show that the quasineutrality approximation is good if ω ω

ωce ci

pe2 1<< , i.e., if the

density is sufficiently large (a) Look at the term σE in the momentum equation. Compare it with j B× .

(b) Look at the ∂σ∂t

term in the continuity of charge equation. Compare it with

∇⋅ j .

You will need to use ∇⋅ =Eσε 0

.

5. (a) Show that for a cylindrical plasma in equilibrium,

42

20

0

2

22 R

rIBp

dr

d z

πµ

µ−=

+ .

(b) Suppose the plasma radius is R = 0.1 m,

the current density jz is uniform,

( ) ( )p r pr

R= −

0 1

2

2, p(0) = 100 kPa,

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 z-direction). 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∂

ραα αt

+ ∇ ⋅ =v 0

( ) ( )ρ ∂∂

ρ ρ υαα

α α α α α α α αβ

αβ α βv

v v E v B v vt

n q p+ ⋅ ∇ = + × − ∇ − −∑

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∂

ρt

+ ∇ ⋅ =v 0 or ∂σ∂t

+ ∇⋅ =j 0

ρ ∂∂v

j Bt

p= × − ∇

m

ne t ne nepe

e2

1 1∂∂

ηjE v B j B j= + × − × + ∇ −

The derivations will not be examinable.

Describe E×B drift, diamagnetic drift, magnetic pressure B2

02µ.

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.

The fraction of the slab blocked by atoms is n Adx

An dxn

n

σ σ= .

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

d

dxnn

Γ Γ= − σ

with solution

( ) ( )Γ Γ Γ= −−

0 0e or en x

x

n mσ λ .

λm is the mean free path, the distance for the flux to fall to 1/e of its initial value.

λσm

nn= 1

.

The mean time between collisions for particles with velocity v is τ λ= m

v.

The collision frequency ντ

= 1. (Strictly, we should average over the velocity

distribution.)

You do. Show λνm

v= .

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∂

ρBB Bt

+ ∇ =. v 0

( )ρ ∂∂

ρ ρ υBB

B B B B BA BA B Atp

vv v v v+ ⋅ ∇ = −∇ − −

Let us make some simplifications.

The non-uniformity is small so n n nB B B= +0 1 , where 0 indicates the uniform,

constant part and 1 indicates a small first-order part that varies in space and time.

There are no zero order drifts so v 0A0 = and v 0B

0 = .

We will use p n kTB B B= .

Note that collisions between particles of the same type do not contribute to diffusion. To first order, the equations become

∂∂n

tnB

B B

10 1 0+ ∇ ⋅ =v (1)

∂∂

νvvB B

B BB BA Bt

kT

m nn

1

01 1= − ∇ − (2)

Take ∇ ⋅ of (2) and substitute using (1).

∂∂n

tD nB

B B

12 1= ∇

DkT

m=

ν is the diffusion coefficient.

(Note a 1 2 1

2ν∂∂BA

Bn

t term has been dropped.)

The units of D are m2 s−1. The meaning of D (a) In terms of a scale length L, any initial non-uniformity is smoothed out in a

time TL

D≈

2

.

(b) Dvrms≈

2

ν. Then D m≈ λ

τ

2

. i.e., the diffusion coefficient is based on a random

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 v= − ∇ −kT

mnn ν or v = − ∇

Dn

n.

The flux of the B particles is Γ = = − ∇n D nv .

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 E v= − − ∇ −en p m ne e eν

Rearrange

v Eee

e

e

e

m

kT

m

n

n= − − ∇

ν ν or v Ee e eD

n

n= − − ∇µ

µν

= e

m is the mobility. µ and D are known as transport coefficients.

The units of µ are m2 V−1 s−1. 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

v Ee e eDn

n= − − ∇µ

v Ei i iDn

n= − ∇µ .

Remember that in a weakly-ionized plasma νe and νi are electron-neutral and ion-neutral collision frequencies. Electron-ion collisions can be ignored. Set the electron flux equal to the ion flux

Γ = − − ∇ = − ∇n D n n D ne e i iµ µE E .

Solve for E and substitute to obtain Γ = − ∇D na .

whereDD D

ae i i e

e i

=++

µ µµ µ

is the ambipolar diffusion coefficient.

From above µν

µνe

e ei

i i

e

m

e

m= =, and ν ∝ ∝v

kT

m, so µ µi e<< .

DT

TDa

e

ii≅ +

1 .

You do. Show this. If Ti = Te,

D Da i≅ 2

Example

Let us apply ∂∂n

tD na= ∇2 to a plasma slab where the initial electron density profile is

shown in the sketch.

We will treat this as a 1-dimensional problem so ∂∂

∂∂

n

tD

n

xa=2

2.

This can be solved by separation of variables. Write n = f(x)g(t) and substitute. This leads to

n Ax

D TB

x

D T

t

T

a

t

T

a

=

+

− −e cos e sin .

Now substitute the initial conditions. This gives

n nx

L

t

T=

−

0e cosπ

where TL

Da

=2

π.

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 E v B v= − + × − ∇ −en p m ne e eν

The z-component equation yields the same µe and De as before. The x- and y-component equations are

v ED

n

n

xv

v ED

n

n

yv

x e xe ce

y

y e ye ce

x

= − − −

= − − +

µ ∂∂

ων

µ ∂∂

ων

This yields

v Ev v

⊥ ⊥ ⊥×= − − ∇ + ++

µω τe e

E B dia

ce

Dn

n 1 2 2

where µ µω τ ω τ⊥ ⊥=

+=

+ee

cee

e

ce

DD

1 12 2 2 2 and .

If ω τce

2 2 << 1, B is small and has little effect on diffusion.

If ω τce2 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

Dv

r v

re

m

ce

m

Le

m

Le⊥ ≈ ≈

≈

λτ

ω τ

λτ

λ τ

2

2 2

2

2 2

2

,

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 υei and obtain

an expression for the resistivity

η ||

ln= × −5 25 10 53

2.

Z

Te

Λ and η η⊥ = 2 || (Te in eV).

It is usually adequate to take ln Λ ≈ 10. 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 j= + × −η

Multiply the second by × B and substitute using the first.

( )v v v⊥

⊥×

⊥×= − ∇ − = −

+ ∇ −η ηB

pnk T T

B

n

nE Be i

E B2 2.

So we can define a D⊥ for a fully-ionized plasma.

Compare this with D⊥ for a weakly-ionized plasma. It is ∝ 12B

, but is ∝ n as well,

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

DkT

eBe

⊥ = 1

16.

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, 106, 107, 108, 109 K. 3. Consider a plasma of thermonuclear interest. n = 1019 m−3, Te = 100 keV, B = 1 T. (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

DkT

m=

ν D m≈ λ

τ

2

µν

= e

m

weakly ionized plasma

DD D

ae i i e

e i

=++

µ µµ µ

in magnetic field

Dr

eLe

⊥ ≈2

τ

PLASMA PHYSICS - KEY IDEAS – 1

I Introduction plasmas everywhere ionization at high T ionization in E (gas discharge) characterize by n, T, B ... characteristic length - Debye length shielding sheaths 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) III Fluid description of a plasma 2. Treat plasma as ion fluid + electron fluid + neutral fluid add σ, j, E, B to fluid equations continuity eq. momentum eq. energy eq eq. of state 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

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 magnetic pressure pinch plasmas current heats magnetic field confines but instabilities 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 non-uniformities mobility - how they move in E ambipolar diffusion ions & electrons diffuse together diffusion in strong B - Larmor radius replaces mean free path

PLASMA PHYSICS

V. WAVES IN PLASMAS −−−− 1

Wave equation Start with Maxwell’s curl equations,

∇ × = −

∇ × = +

EB

B jE

∂∂

µ ∂∂

t

c t0 2

1

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 ( )ej tk r⋅ −ω variation. k is the wave vector and ω is the (angular) frequency. Maxwell’s equations can then be written,

j j

j jc

k E B

k B j E

× =

× = −

1 1

10

12

1

ω

µ ω.

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 kc

= ω.

(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

vk

cph ≡ =ω

where c is the velocity of light in a vacuum.

Waves in a plasma, no magnetic field The ion and electron momentum equations are

( ) ( )ρ ∂∂

ρ ρ υii

i i i i i i i ie i etn e p

vv v E v B v v+ ⋅ ∇ = + × − ∇ − −

and

( ) ( )ρ ∂∂

ρ ρ υee

e e e e e e e ei e itn e p

vv v E v B v v+ ⋅ ∇ = − + × − ∇ − − .

Write n = n0 + n1 and v = v0 + v1, where n1 and v1 show the ( )ej tk r⋅ −ω 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 m j n e

n m j n e

i i

e e

0 1 0 1

0 1 0 1

− =

− = −

ω

ω

v E

v E .

It follows that v vie

ie

m

m1 1= − so ion motions are small. We shall neglect them.

v 0

v E

i

ee

je

m

1

1 1

≅

= −ω

.

The electrons move but the ions remain at rest in the background. Using the definition of current density, we have

( )j v v E1 0 1 10 2

1= − ≅en jn e

mi eeω

.

Take k along z, say. Write out the component Maxwell’s equations. Eliminate B1, j1, leaving equations in E1. The Ez

1 equation gives

ωε

20 2

0

= n e

me

. (1)

The other equations give

20

20

2

22

cm

en

ck

eεω −= . (2)

Plasma oscillations Equation (1) describes oscillations, not waves. These are called plasma oscillations.

Recall ωεpe

e

n e

m2

0 2

0

= . So the oscillation frequency is just the plasma frequency

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.

kc

pe22 2

2=

−ω ω

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.

If n is too high then, because ωεpe

e

ne

m2

2

0

= , ω pe2 will be too large and the wave will be

cutoff. Define critical density as

nm

ece=

ε ω02

2.

If n nc> the wave is cutoff.

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 nc = 8 98. .

Phase velocity

We can plot vph as a function of ω. vc

ph

pe

=

−12

2

ωω

At a cutoff vph = ∞ .

Note vph > c! Is this a problem? No, recall that energy is transmitted at the group velocity

vd

dkg = ω.

You do. Show that in this case, vc

vgph

=2

which is less than c.

We can think of a plasma as being like a dielectric medium with a refractive index

µ = c

vph

.

You do. Show that in this case

µωω

22

21 1= − = −pe

c

n

n.

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 me ei e0 1ν v

the expression for j 1 becomes

( )j E10 2

1=+

jn e

m je eiω ν

and

( )k

j

cei pe2

2

2=

+ −ω ω ν ω.

The wave is attenuated. (ii) Allow a finite electron temperature.

Waves in a plasma, no magnetic field, with finite electron temperature

∇ = ∇ =p U U jn me e e e e e2 2 1ρ k

eee nm ∇=∇ρ and it is easy to show that ∇ =n jne e1k .

The expression for j 1 becomes

j E k10 2

12

1= −jn e

m

eU

mn

e

e

eeω

instead of electron oscillations, now have a wave with dispersion relation ω ω2 2 2 2 0− − =pe ek U .

(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,B0

$z . The electron momentum equation, to first order yields the following three component equations for ve

1 .

v je

mE j v

v je

mE j v

v je

mE

exe

xce

ey

eye

yce

ex

eze

z

1 1 1

1 1 1

1 1

= − −

= − +

= −

ωωω

ωωω

ω

where ω cee

eB

m=

0

is the electron cyclotron frequency.

From these, we obtain the three components for j 1

j je n

m

Ee n

m

E

j je n

m

Ee n

m

E

j je n

mE

x

ece

xce

ece

y

y

ece

yce

ece

x

ze

z

12 0

2

2

12 0

22

2

1

12 0

2

2

12 0

22

2

1

12 0

1

1 1

1 1

=−

+−

=−

−−

=

ω ωω

ω

ω ωω

ω ωω

ω

ω ωω

ω

We can think of the plasma as being like a conducting fluid. j E1 1= ⋅σ where σ is the conductivity tensor defined as

σ

ε ω

ω ωω

ε ω ω

ω ωω

ε ω ω

ω ωω

ε ω

ω ωω

ε ωω

=

−

−

−−

−

j

j

j

pe

ce

ce pe

ce

ce pe

ce

pe

ce

pe

02

2

2

02

22

2

02

22

2

02

2

2

02

1 1

0

1 1

0

0 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 jE Dµ ∂

∂µ ∂

∂0 2 0

1

c t tand D E= ⋅ε .

Substitute for j 1 and show

ε εωε

σ= +

0

0

Ij

.

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 Ec

E

c

E j E

k Ec

E

c

E j E

cE

cE

x xpe

ce

xce

y

y ype

ce

yce

x

xpe

z

22

2

2

22

2

22

2

2

22

2

2

2

2

2

1

1

0

= +−

− +

= +−

− −

= + −

ω ωωω

ωω

ω ωωω

ωω

ω ω

(1)

(2)

(3)

Equation (3) describes plasma oscillations at frequency ω = ωpe with E1 along k and B0. Equations (1) and (2) have a solution if their determinant is zero, i.e., if

kc

c

j

c

j

c

kc

c

pe

pe

pe

pe

ce

pe

pe

ce pe

pe

22

2

2

22

2

2

22

2

2

22

2

22

2

2

22

2

1 1

1 1

0

− +−

−

−−

− +−

=

ω ωωω

ωωω

ωω

ωωω

ωω

ω ωωω

Multiply this out

kc

pe

ce

2

22

2

1=

−±

ωω

ωω

This describes two waves with E1 perpendicular to k and B0.

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 E jEy x

1 1= m 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 vph = 0 .

At a resonance, the present approach breaks down. This is easy to see. Consider the term E v B+ × . This could be written

( )E v B B E v Bk E1 1 0 1 1 1 0

1

+ × + = + × + ×

ω

To first order, we have been ignoring the last part and writing E v B1 1 0+ × . But at

a resonance, vkph = ω

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

µ

ωω

ωω

L

R

pe

ce

2

2

2

11

= −±

.

If we let X Ype ce= =ωω

ωω

2

2 and this can be written more compactly as

µ L

R

X

Y2 1

1= −

±.

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

1

1

2

2

2

22

2

22

2

2

22

2

22

2

2

2

= +−

− +

= +−

− −

= + −

ω ωωω

ωω

ω ωωω

ωω

ω ω

cE

c

E j E

k Ec

E

c

E j E

k Ec

Ec

E

xpe

ce

xce

y

y ype

ce

yce

x

z zpe

z

(1)

(2)

(3)

Equation (3) describes transverse electromagnetic waves with E1 along the z-direction, parallel to B0.

kc

pe=−ω ω2 2

2

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

kc

pe

ce

pe2

22

2

2 2

2

1

=

−−

−

ωω

ωω ω

.

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.

The refractive indices, if we let X Ype ce= =ωω

ωω

2

2 and , are

µ µO XXX

Y

X

221 1

11

= − = −−

−

and 2 .

CMA diagram

ω ωpe ce2 2+ is known as the upper hybrid frequency.

Waves in a plasma. Propagation at an arbitrary angle to B0.

( ) ( )

µθ θ θ

2

2 2 2 2 2

2 2

1

2

1

12 1 2 1

= −

−−

±−

+

X

Y

X

Y

XY

sin sincos

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 i

1 still always negligible in the presence of a steady magnetic field?

The momentum equations are

( ) ( )( ) ( )

n m j n e n e B

n m j n e n e B

i i i

e e e

0 1 0 1 0 1 0

0 1 0 1 0 1 0

− = + ×

− = − − ×

ω

ω

v E v z

v E v z

$

$

As an example, one of the components of v i1 is

v

je

mE

e

mE

ixi

xci

iy

ci

12

2

21

=+

−

ωωω

ωω

.

Clearly this becomes very important when ω ω≈ ci .

e.g., for propagation along B0, the dispersion relation is

k

m

m

c

pece

e

i ci

2

2 2

2

1

1

1

1=

−±

+

ω ω ωω

ωω

m

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.

The equation for the parabola is n nx

L= −

max 1

4 2

2.

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 ω ωω2 2= +pe 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 ωγ

= +n

k vce|| ||.)

Possible heating strategies can be checked using the CMA diagram.

For a tokamak plasma, Br

≈ 1, where r is the major radius. So the

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 ωω

ce2

2 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 ωω

pe

ce

2

2 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

µθ

2 11

≅ −±

X

Ycos

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

µθ

2 ≅ X

Ycos.

Under these approximations, (i) find vph (ii) find vg.

(iii) show delay time ∝ 1

f.

(iv) (HARDER) The correct definition of group velocity is

v x y zgx y zk k k

= + +∂ω∂

∂ω∂

∂ω∂

$ $ $ .

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

( )c k p

ce cice ci

p ce ci

2 2

2

2

22 2

2 2

1ω

ω

ω ω ωω ω ω

ω ω ω ω

= −

− +−

− +

where ω ω ωp pe pi2 2 2= + .

What is the resonant frequency when the density n = 0 and when n = ∞ ? 10. Alfven waves

(i) Show that when ωce < ωpe, the dispersion relation for propagation along

B0 that includes ion motions yields v cphce ci

pe

=ω ω

ω 2as ω → 0 .

(ii) Define the Alfven speed VB

A =µ ρ0

. 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

kc

pe22 2

2=

−ω ω or µ 2 1= − X

with magnetic field LHCP and RHCP waves:

kc

pe

ce

2

22

2

1=

−±

ωω

ωω or µ L

R

X

Y2 1

1= −

±

O and X waves:

kc

pe22 2

2=

−ω ω and k

c

pe

ce

pe2

22

2

2 2

2

1

=

−−

−

ωω

ωω ω

,or

µ µO XXX

Y

X

221 1

11

= − = −−

−

and 2

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 ( ) ( )f r j m k z tze φ ω+ −,

m integer. Maxwell’s equations

∇ × = −EB∂

∂t

∇ × = +B jEµ ∂

∂0 2

1

c t

must be expanded using the cylindrical polar coordinate identities. The electron momentum equation to first order is,

( ) ( )n m j n ee e e0 1 0 1 1 0− = − + ×ωv E v B

Use ω ωpe2 2>> and ω ωce >> . Eliminate variables and finally arrive at a Bessel

equation for Bz1 which has a solution

( )B A k rz m1 = ⊥J .

Writing k k⊥ = sinθ leads to the earlier dispersion equation for the whistler wave.

k cpe

ce

2 2

2

2

2

ω

ωω

ωω

θ=

cos or µ

θ2 = X

Ycos.

Boundary condition. At the conducting wall, the tangential electric field must be

zero, i.e., Eφ1 0= . Now

( ) ( )

+≈ ⊥

⊥

⊥ rkkrk

rkmkE mz

m 'JJ1

φ so k a⊥ must be a

zero of ( ) ( )

+ ⊥

⊥

⊥ akkak

akmk mz

m 'JJ

. We know the radius a so this limits the

possible values of k⊥ .

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

∇ × = −EB∂

∂t

∇ × =B jµ 0

and the single fluid equations

( )∂ρ∂

ρt

+ ∇ ⋅ =v 0

ρ ∂∂v

j Bt

p= × − ∇

0 E v B= + × .

To first order these can be written j jk E B× =1 1ω

jk B j× =10

1µ

− + ⋅ =j jωρ ρ1 0 1 0k v

( )ρ ω ρ0 1 1 0 2 1− = × −j U jv j B k

0 E v B= + ×1 1 0 . We will suppose the steady magnetic field is B0

$z and that k x z= +k kx 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 VB

A =0

00µ ρ

.

The component equations are

( )

( )

v k V v U k v k v k

v k V v

v U k v k v k

x A x x x z z x

y z A y

z x x z z z

12

2 2 12

2 1 1

12

2 2 1

12

2 1 1

1 1

1

1

= + +

=

= +

ω ω

ω

ω

(1)

(2)

(3)

Alfven wave Equation (2) gives a wave, the Alfven wave

ω θ2

22 2

kVA= cos .

Properties of this wave

v1 is along y, from Equation (2)

E1 is along v B1 0× so is along x

B1 is along k E× 1 so is along y, 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.

Recall that for an electromagnetic wave in a vacuum, BE

c1

1

≈ . Here

BE

VA

11

≈ . Since VA << c, the wave magnetic field is particularly important.

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 vph = τµ

, where τ is tension and µ is

mass per unit length.

velocity of Alfven wave vB

ph =0

00µ ρ

.

µ ρ= A , where A is area, so the two expressions are equivalent if τµ

= BA

02

0

.

Recall that magnetic tension = B02

0µ. 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.

( ) ( )ω θ2

2

2 2 2 2 2 2 2 24

2k

U V U V U VA A A=

+ ± + − cos

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 ω 2

22

kU= (1)

and ω 2

22

kVA= (2)

Equation (1) gives an acoustic wave, the ion acoustic wave. Properties

v1 is along z

E1 = − v B1 0× . But v1 and B0 are parallel so there can be no E1.

B1 ∝ k E× 1 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

E1 is along v B1 0× so is along y

B1 is along k E× 1 so is along x so B1 is perpendicular to the direction of propagation. (ii) Propagation perpendicular to B0: compressional wave

θ = 90°. ω 2

22 2

kU VA= +

Properties

v1 is along x

E1 is along v B1 0× so is along y

B1 is along k E× 1 so is along z so B1 is in the direction of B0 and perpendicular to the direction of propagation

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

( ) ( ) ( )

( ) ( )

B A k r B jAk

k

m k r

k r

k

kk r

B Ak

k

m k r

k r

k

kk r

z m rm z

m

z mm

1 1

1

= =

= − +

⊥⊥

⊥

⊥ ⊥⊥

⊥

⊥

⊥ ⊥⊥

J , J

+ J ,

JJ

'

'φ

each with a factor ( )e j m k z tzφ ω+ − . 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,

∇ × = −EB∂

∂t

∇ × =B jµ 0

ρ ∂∂v

j Bt

= ×

0 E v B= + × ,

for the case where B0 is in the z-direction as usual, and k has both x and z components. (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∂

ρee et

+ ∇ ⋅ =v 0

( )∂ρ∂

ρii it

+ ∇ ⋅ =v 0

( )ρ ∂∂e

ee et

n ev

E v B= − + ×

( )ρ ∂∂i

ii it

n ev

E v B= + ×

and use ne = ni.

In fact, using these equations, there are no waves. There are oscillations at

the lower hybrid frequency ω ω ω= ci 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

kv

pe

ce

ph

ch

pi

ci

2

22 2 2

2

1 1 1=

−±

+ +

ωω

ωω

ωωω

ωωω

m m

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 εr 0 instead of ε 0 in calculating the plasma

frequencies and the velocity of light in the material is vr

= 1

0 0µ ε ε instead of c

where ε r is the dielectric constant. Rearranging gives

( ) ( )k v pe

ce ce

ph

ch ch

pe

ce

ph

ch

2 2

2

2 2 2 2

1ω

ωω ω ω

ωω ω ω

ωωω

ωωω

= ±±

−

−

mm

Alfven waves (i) Show that if the plasma is compensated the last term vanishes. Further, if ω ωce h or >> then the dispersion relation simplifies to

1 1 12 2 2v v Vph A

= + .

VA << v so v Vph A= .

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 ωce h or >> .

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 ωce h or >> . Only the lower sign gives a wave.

v vphc

p

2 22=

ωωω

.

Damping is small if ω τc >> 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

Definitions include Alfven speed VB

A =0

00µ ρ

,

Alfven waves (also known as shear waves, torsional waves): ω θ

2

22 2

kVA= cos

Fast and slow MHD waves:

along B0, Alfven waves ω 2

22

kVA= and ion acoustic waves

ω 2

22

kU= .

⊥ B0, compressional waves (also known as fast waves, magnetosonic

or magnetoacoustic waves): ω 2

22 2

kU VA= + .

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

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)

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 collisions

k is complex cutoff is not sharp n<nc, propagation, some attenuation n>nc, some propagation but high attenuation

include T

get: 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

cutoffs shift resonance of RHCP at ω = ωce where E1 rotation matches electron gyration exactly 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 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

Alfven waves transverse perturbation of field lines (like waves on a string)

Fast and slow MHD waves parallel to B

Alfven wave 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

the probe. The sheath thickness is of the order of a Debye length λ εD

ekT

n e= 0

02 (see

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

dV

e I I

kT

sat i

e

=−

.

(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.

From before λ mrms

D

v≈ . Collisions are important if λ m R<< .

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 IeV

kTsat ie

= tanh2

and the slope of the characteristic is dI

dV

eI

kTsat i

e

=2

.

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 NAdB

dt= .

You do. Derive this. Use integrator.

VNAB

RCout = .

(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,

( ) ( )p r

B rz+ =2

02µ constant .

The presence of the plasma acts to decrease the magnetic field inside it (hence diamagnetic). If we measure ( )B az using a small coil and <Bz> using a diamagnetic loop then

( )< > =− < >

pB a Bz z

2 2

02µ

gives an estimate of the total kinetic energy in the plasma (using p = nkT). 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 Rl

A= η

. 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.

V nAdI

dt= µ0 .

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

µ2 1= − n

nc

depends only on the density and not on the magnetic field.

Field at detector due to signal passing through probe arm

( )E A t1 1 1= + +cosω φ φ plasma

where ( )φ plasma = −∫ k k dxl

00, k

c0 = ω, k

c= µ ω

.

Field at detector due to signal passing through reference arm

( )E A t2 2 2= +cosω φ .

The output of the square-law detector is

( )( ) ( ) ( )

( )

( )

( ) ( )( )

V E E

A t A A t t

A t

A A t

A A t A A

A A t

= +

= + + + + + +

+ +

=+ + +

+ + + + + + −

++ +

1 2

2

12 2

1 1 2 1 2

22 2

2

12

12

1

1 2 1 2 1 2 1 2

22

22

2

2

2

2

2

2

2

cos cos cos

cos

cos

cos cos

cos

ω φ φ ω φ φ ω φ

ω φ

ω φ φ

ω φ φ φ φ φ φ

ω φ

plasma plasma

plasma

plasma plasma

The capacitance of the detector shorts the microwave frequency signals to ground.

The remaining slowly-varying part is

( )VA

A AA

= + + − +12

1 2 1 222

2 2cosφ φ φplasma .

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 k ks i= + ∆ and ω ω ωs i= + ∆

Since ks = ki,

∆k ki= 22

sinθ

.

Thomson scattering can be described classically. The principle is that the incident wave accelerates the electrons; In the non-relativistic case,

( )&v E k r= − ⋅ −e

mt

ei icos ω ,

and because the electrons are accelerating, they radiate; At large distances, in the far field,

( )E

r r vs

e

c r= − × ×

2 3

&

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.

αλ

= <<11

∆k D

. There are two ways of looking at this case

(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

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

∆k as a scale length, then α = scale length

Debye length.)

αλ

= >>11

∆k 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. 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 I r nSeω φ ω= −02 21 cos ∆ ∆k,

where I0 is the incident intensity, re

m cee

=2

024πε

is the classical radius of the electron

(which is very small), ( )1 2− cosφ 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

( ) ( )I IkT

cbbeω ω ω

π= =

2

3 28

so Te can be estimated from the intensity.

At high frequencies, the plasma is optically thin. Reflections from the walls become important.

( ) ( )( )

( )I Ie

rebbω ωτ ω

τ ω= −−

−

−

1

1

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

( )E A kd t1 = −cos ω .

Field at detector due to beam reflected off moving mirror

( )( )E A k d x t2 = + −cos ω .

The output of the square-law detector is

( )E E A kx

k dx

t

A kx

k dx

t

1 2

2 2 2 2

2 2

42 2

42

1

21 2

2

+ =

+

−

=

+ +

−

cos cos

cos cos

ω

ω

The low-frequency part is

( )V A kx

A kx=

= +2

212 2 2cos cos .

Express this as

intensity ( )( )I S k kx= +1 cos .

The plasma emits a range of frequencies. Using a similar approach

intensity ( ) ( )( )I x S k kx dk= +∞

∫0 1 cos

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 kx dk=∞

∫0 cos .

This is a Fourier transform. Carry out the inverse transform to obtain the spectrum S(k)

( ) ( )S k Int x kx dx=∞

∫2

0π cos .

We do not have readings for x = 0 to ∞ , only to xm. This means that the resolution of

the instrument is limited to ∆kxm

= π.

Plasma spectroscopy

Important processes that determine populations include 1 Radiative, involving photons bound-bound transitions - absorption and (its inverse process) emission of photons

A h A+ ↔ ∗ν free-bound transitions - photoionization and recombination with the emission of a photon

A h A e+ ↔ ++ν 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

n

n

g

gn

m

n

m

E E

kTn m

e=−

−

e

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.

νhAA +→ ***

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. ∆fFWHM ≈ 1

πτ where τ

is the mean time between collisions.

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 n n Z Te i ekTe= × − −

−8 6 10 53 2

1

2. ehω

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

∗+ →+ AeA

j is similar but the exponent is replaced by a series of terms with

ekT

En

Z

G

12

2

e

−−

ωh

where E1 is the ionization potential of hydrogen and G = 1 if

12

2

En

Z>>ωh , otherwise 0.

Recombination radiation is less important than bremsstrahlung if

kT E Ze > 3 12 .

The combined spectrum is shown below.

Note

at low frequencies j Te≈−

1

2 ,

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 I i remains appoximately equal to the saturation ion current given by

I nekT

mAsat i

e

i

≈ − 0 57.

where A is the effective probe area. The electron current is

( )

I Ie sat e

e V V

kTS

e=−

e where I nekT

mAsat e

e

e

= 1

4

8

π.

The total current is I I Ie i= + .

(i) Show ( )

I I sat i

e V V

kTF

e= −

−

1 e and hence ( )dI

dV

e I I

kTsat i

e

=−

or

( )( )d

dVI I

e

kTsat ie

ln − = .

Hint: Use the definition of VF.

(ii) Show that, for a hydrogen plasma, V VkT

eS Fe≈ + 3 3. .

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

4. The Thomson scattering cross-section σ π= = × −8

36 65 10

229re . m2 .

(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 ukT

mse

i

= as they enter the sheath. Small

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 n us s= (1).

The equation of conservation of energy is

( ) ( ) 02

1 2 =+ xexvmi φ (2).

Eliminate v(x) from (1) and (2) and obtain an expression for n(x). Substitute this into Poisson’s equation

( ) ( )d x

dx

en x2

20

φε

= −

and solve. Find that the potential across the sheath follows the the Child-Langmuir law

( )φ x x∝4

3 ,

and the sheath thickness

se

kTs

eD≈

φ λ

3

4

,

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 Ed

dx= = −µ µ φ

where µ is the mobility.

The ion energy distribution function looks like:

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 V VkT

eF Se− ≈ which is

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

( )ρ ∂∂

ρ ρ υee

e e e e e e e etn e p

vv v E v B v+ ⋅ ∇ = − + × − ∇ −

In the last term we have assumed that velocities vi and vn are << ve.

This time, consider oscillations, not waves, so E E x= −1e j tω$ , write

n n n j t= + −0 1e ω and v ve ej t= −1e ω and consider only the first-order terms.

(i) No magnetic field.

ve

m jEex

e

1 11= −− +ω ν

.

Power absorbed per unit volume p j E= ⋅ .

Now j n evx ex= − 0 1 .

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

pn e

m

E

e

=+

1

2

0 2 12 2

2 2νν

ν ω

Note that if there are no collisions, there is no power absorbed. (ii) magnetic field

( ) ( )p

n e

m

E

e ce ce

=+ −

++ +

1

2

1

2

1

2

0 2 12 2

2 2

2

2 2νν

ν ω ων

ν ω ω

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 non-resonant 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 wave-particle 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 proton-proton 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 cross-section.

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

preaction = n2

4 <σv>E

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

pbrems = 1.6 × 10−40 neniZ2T

1

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

= × × ÷ =23

2

3n kT

nkTE

E

ττ

.

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, n Eτ > 1020 m−3 s, Similarly,

for D-D reactions,n Eτ > 1022 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)

PLASMA CURRENT

I(MA)

TOROIDAL FIELD B(T)

START DATE

CHS JAPAN 0.2 1 2

HELIOTRON E JAPAN 0.2 2.2 2 WENDELSTEIN 7-AS GERMANY 0.2 2 3.5

H-1 AUSTRALIA 0.2 1 1 LHD JAPAN 0.5-0.6 3.9 3 1998

WENDELSTEIN 7-X GERMANY 0.5 5.5 3 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 USA 40 10 GEKKO JAPAN 100 12 OMEGA USA 30 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

the containment time τ E

na≈

×2 10202 to show that the maximum temperature is far

below that for fusion. (Get resistivity from Chapter IV.) Inertial confinement fusion 3. The Lawson criterion n Eτ > 1020 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,

τ E

R

U≈ 1

4

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 BE× 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

1 2 1

2

1

ν∂∂

∂∂BA

B Bn

t

n

t<< can be written as νT >> 1 and this leads to λ m L<< . i.e., there

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 ∇ ⋅ =Eσε 0

to get an expression for

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 BE× 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.