Lecture course in PLASMA PHYSICS

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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 - 19 magnitude of charge on electron e = 1.60 × 10 C - 31 mass of electron m = 9.11 × 10 kg e - 27 mass of hydrogen ion m = 1.67 × 10 kg i - 23 - 1 Boltzmann’s constant k = 1.38 × 10 J K 8 - 1 velocity of light in vacuum c = 3.00 × 10 m s - 12 - 1 permittivity of vacuum ε = 8.85 × 10 F m 0 - 7 - 1 permeability of vacuum μ = 4π × 10 H m 0 - 19 1 eV = 1.60 × 10 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 - 3 - 3 - 3 6 - 3 plasma density n (m ) (often cm . 1 cm = 10 m ) temperature T (K or more conveniently eV) Temperature and energy For an ideal gas in thermal equilibrium, the probability that velocity lies in the range 1  2 mv   2 dv dv dv around velocity (v ,v ,v ) is proportional to exp - dv dv dv . x y z x y z   x y z kT     We can construct the Maxwellian velocity distribution function f(v) 1  2 3 mv   2 m   2 f v = n exp - , ( )       2πkT kT     and use it to calculate average values. e.g., Calculate the particle density n = f (v)dv dv dv x y z ∫ calculate the mean velocity 1 u = vf v dv dv dv = 0 ( ) x y z ∫ n calculate the mean speed 1 8kT v = vf (v)dv dv dv = x y z ∫ n πm calculate the rms speed 1 3kT 3kT 2 2 v = v f (v)dv dv dv = so v = rms x y z rms ∫ n m m 3 You do. Show the average kinetic energy per particle is E = kT av 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 E = 3 eV.) av Formation of a plasma Ionization at high temperatures We have said that a sufficiently hot gas becomes a plasma. Atoms in a gas have a spread in thermal energy and they collide with each other. Sometimes there is a collision with high enough energy to knock an electron out of the atom and ionize it. Energy must exceed ionization potential, 13.6 eV for hydrogen. In a cold gas such collisions are very infrequent, in a hot gas more likely. From the Maxwellian velocity distribution function we can derive the Saha equation which gives the fraction of ionization we can expect in a gas in thermal equilibrium at temperature T, 3 2 n U T   i 27 i ≅ 3×10 exp -   n n T   n i where n is the ion density, n is the neutral particle density and U is the ionization i n i energy. (In this expression U and T are both in eV.) i 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. - 15 very small currents ( ≈ 10 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. - 1 We can plot Townsend’s ionization coefficient, η (in V ), the number of ionizing collisions caused by an electron as it falls through a p.d. of 1 V - 1 - 1 Note the units. 1 V/cm. mmHg or 1 V/cm. torr = 0.75 V m Pa .   p p η = a exp - b   E  E η depends on E/p ( a common parameter in gas discharge work – it allows scaling), the gas. η is low at low pressures because an electron encounters hardly any atoms. η is low at high pressures because elastic collisions are more frequent and it is more difficult for the electron to gain sufficient energy to ionize. The figure shows the Penning effect in a gas mixture where argon is ionized by metastable excited neon atoms. Photoionization is not important in this case but might provide the initial ionization to start things off. The second important process is positive ions bombarding the cathode and knocking off electrons. This is described by γ, the secondary ionization coefficient, defined as the number of electrons knocked off the cathode by a single positive ion. γ depends on E/p, the gas and the cathode material (and the state of the cathode surface, whether it is a pure metal or has an oxide layer). Consideration of these two processes allow us to estimate the p.d. for breakdown. Suppose the p.d. between the electrodes is V. ηV One electron leaving the cathode becomes e electrons arriving at the anode. ηV And e - 1 positive ions heading back towards the cathode. One ion produces γ electrons. ηV So one electron gives rise to γ e - 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 - 8 - 6 4 weakly-ionized, fraction ionized is 10 to 10 , T is 10 K but T and T are 300 K. e i n 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 - 3 - 1 4 fraction ionized is 10 to 10 , T and T are 10 K e i types of arcs thermionic arc - emission of electrons is due to cathode being heated by the large current of ions bombarding it. Cathode must withstand very high temperature, e.g., carbon, tungsten. This arc is self-sustaining. (Emission of electrons from a hot surface is described by eϕ   2 j = aT exp-    kT where ϕ is the work function.) e.g., carbon and tungsten arc lamps thermionic arc with cathode heated by external source - non-self-sustaining. field emission arc - emission of electrons is due to very high E at cathode. e.g., mercury arc lamp, mercury arc rectifier. metal arc - heating the cathode vapourizes the metal. high-pressure arc p 1 atm; low pressure arc 1 atm. (v) Processes of deionization Dissociative recombination + A + e → A + A 2 is the fastest recombination process in a weakly-ionized gas like a glow discharge Radiative recombination + A + e → A + hν is not important for electron removal but may be important for light emission. Diffusion to wall is slower in a well-developed discharge. Three-body electron-ion recombination + A + e + e → A + e is main process in high density, low temperature laboratory plasmas. Debye length and plasma frequency A plasma has a characteristic length and a characteristic time. Screening of electrostatic fields This leads to the Debye length λ . D First, consider a positive charge q all by itself. The potential at a distance r from the charge is q ϕ = . 4πε r 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 1  2  mv + qϕ   2 factor becomes exp - dv dv dv . ϕ depends on position so the   x y z kT     probability depends on position. qϕ   The particle density is given by n = f (v)dv dv dv so n ∝ exp-  x y z ∫   kT - eϕ   for electrons n = n exp -   e 0   kT eϕ   for ions (we will suppose they are singly-ionized) n = n exp -   i 0   kT Gauss’ Law can be written as σ ∇.E = ε 0 E = -∇ ϕ so σ 2 -∇ ϕ = . ε 0 This is Poisson’s equation. eϕ - eϕ   The charge density is σ = - en+ en= en - exp + exp  . e i 0   kT kT Assume that this potential term is very small, eϕ kT 2 eϕ eϕ 2n e ϕ     0 σ ≅ - en 1 +  + en 1 -  = - . 0 0     kT kT kT I am going to use spherical coordinates (and assume spherical symmetry) 1 d dϕ   2 2 ∇ ϕ ≡ r .   2 r dr dr   Poisson’s equation becomes 2 1 d dϕ 2n e ϕ   2 0 - r = -   2 dr dr ε kT r   0 with solution     q r   ϕ = exp - .   4πε r ε kT 0 0   2 2n e   0 The potential falls away exponentially. ε kT 0 λ = Call the Debye length then D 2 n e 0   q 2r ϕ = exp - .   4πε r λ   0 D 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. L λ (i) the system must be large enough , and D (ii) there must be enough electrons to produce shielding N 1 , where D N is the number of electrons in a Debye sphere. D Suppose there is a local concentration of charge. If plasma dimensions are much greater than λ , then on the whole plasma is still neutral (we can describe the plasma D as quasineutral) and we can take n ≅ n ≅ n . e i 0 If we put an electrode into the plasma, it becomes shielded by a sheath ot thickness ≈ λ . D T -3 λ = 69.0 m (T in K, n in m ) D e n e Plasma oscillations This leads to the plasma frequency ω . pe 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. 2 d x Equation of motion for the slab of electrons is F = m 2 dt mass of electrons in slab m = m n LA e e What is the force on the electrons? We have two oppositely charged sheets facing σ s each other. The electric field between them is E = where σ is the surface s ε 0 charge density or charge /area. σ = en xA/A so the restoring force is equal to charge of electrons in slab × electric s e field, so, en x e force on electrons = - n LAe . e ε 0 Equation of motion becomes 2 d x en x e n LAm = - n LAe or e e e 2 dt ε 0 2 2 d x n e e = - x 2 dt ε m 0 e with solution x = A cos ω t , ( ) pe where the (electron) plasma frequency is 2 n e e ω = . pe ε m 0 e This is a ‘natural’ frequency for the plasma. - 3 f = 8.98 n Hz (n in m ) . pe e e We encounter ω when discussing wave propagation. pe kT You do. Show ω λ = . pe D m e We can now use λ and ω to classify plasmas. D pe 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 5 1 atm = 1.013 × 10 Pa = 760 mmHg = 760 torr 1 micron = 1 mtorr Temperature and energy 6 2. Calculate v for protons and electrons at 10 K. rms 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 n = e 19 - 3 - 3 n = 10 m , T = 100 eV and volume of plasma = 1 m . How much would the i 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. 25 - 3 Use n = n + n = 3 × 10 m (about what it is at room temperature, 1 total i n 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, T , T , fractional ionization e i 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, n = e 17 - 3 10 m and diameter about 100 mm, (i) calculate λ , D (ii) calculate N the number of electrons in a Debye sphere. D 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 n by assuming the solid is sodium and that each e atom contributes one electron. Summary of chapter You should be able to Describe ionization by electrons in a gas discharge, the role of positive ions, deionization. Do calculations of ε kT 0 Debye lengthλ = , D 2 n e 0 2 n e e plasma frequencyω = . pe ε m 0 e The derivations will not be examinable. In subsequent sections you will be expected to use Maxwell’s equations in differential form. σ ∇ ⋅ E = ε 0 ∂B ∇ × E = - ∂t ∇ ⋅ B = 0 1 ∂E ∇ × B = μ j + 0 2 c ∂t where, in cartesian coordinates ∂E ∂E ∂E ∂E ∂E y y x z z ∇.E = + + . ∇ × E = - , etc. ( ) x ∂x ∂y ∂z ∂y ∂z and in cylindrical coordinates, ∂E 1 ∂(rE ) 1 ∂E ϕ r z ∇ ⋅ E = + + . r ∂r r ∂ϕ ∂z ( )  ∂E   ∂ rE  1 ∂E ∂E ∂E 1 ∂E   ϕ ϕ z r z r  ˆ ˆ  ˆ ∇ × E = - r + - φ + - z       r ∂ϕ ∂z ∂z ∂r r ∂r ∂ϕ       Note that we use σ for (volume) charge density to distinguish it from ρ for mass density. PLASMA PHYSICS II. MOTION OF IONS AND ELECTRONS IN E AND B FIELDS We consider the paths of ions and electrons in E and B fields for some simple cases. The Lorentz force on a point charge is F = q(E + v × B) . - 1 E is measured in V m . B in T (often in gauss. 10000 gauss = 1 T) 1. E = constant, uniform Suppose E = Ex . The Lorentz force equation becomes dv q x = E dt m dv y = 0 dt dv z = 0. dt This describes a constant acceleration along x. 2. B = constant, uniform Suppose B = Bz . Here is how we might produce a uniform magnetic field. dv q x = v B (1) y dt m dv q y = - v B (2) x dt m dv z = 0. (3) dt d Take of (1), substitute using (2) dt 2 dv d v q q q y   x = B =  v B B . x 2   dt m dt m m Write q B ω = c m the (angular) cyclotron frequency or gyrofrequency. (Note the symbol Ω is often used.) 2 d v x 2 + ω v = 0 . c x 2 dt Similarly, 2 d v y 2 + ω v = 0 . c y 2 dt The solutions can be written as v = - v sinω t x ⊥ c v = mv cosω t y ⊥ c (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 v ⊥ x = cosω t = r cosω t c L c ω c v ⊥ y = m sinω t = mr sinω t c L c ω c v ⊥ r = is called the Larmor radius, radius of gyration, or gyroradius. L ω c 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 v , this z-component of the motion is unchanged. The charge moves z in a helical path. 3. E constant, uniform. B constant, uniform. Suppose B = Bz . dv q x = E + v B (1) ( ) x y dt m dv q y = E - v B (2) ( ) y x dt m dv q z = E . (3) z dt m (3) gives constant acceleration along z. You do. Suppose E has a z-component only. Describe the motion and sketch the path for this case. (1) and (2) are manipulated as before, they give 2 E d v y x 2 2 + ω v = ω and c x c 2 dt B 2 d v E y 2 2 x + ω v = - ω , c y c 2 dt B with solutions E y v = - v sinω + t x ⊥ c B E x v = mv cosω t - . y ⊥ c B The path of an electron is a combination of uniform circular motion plus a drift, called an E × B drift. 1 E × B v = E x - E y= . ( ) E×B y x 2 B B Note that the drift term is independent of the charge and its sign, so all the charges will drift together. The paths are cycloids. E If E⊥B then v = . E×B B Here is an example where the E × B drift can cause a plasma to rotate. 4. B = constant, non-uniform. We will consider two distinct cases. Case (a):

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