( ). One set of terms has a ω in

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1 Laptag Class Notes W. Gekelan Cold Plasa Dispersion relation Suer Let us go back to a single particle and see how it behaves in a high frequency electric field. We will use the force equation and Maxwell s equations. The high frequency field will be that of a wave in the plasa. The high frequency field is E ( t) = E e t. The frequency can be as high as the cyclotron frequency. The force law is dv dt = q ( E e t + v B ). Let v= v c + v E e t, where vc does not depend on ω. The force law gives us: dv c + v E e t = q E e t + v c B + v E Be t dt front of the all and an e t we have equations: dv (I) c dt + = q ( v c B ) v E e t = q E e t + q v E Be t. One set of ters has a ω in dependance, the other does not; in fact The first is the usual cyclotron otion equation, we know the answer( see appendix ). The second ay be re-written as () (+ q B ) v E = q E. Now ultiply equation II by the operator (- q B ) (- q B ) (+ q B ) v E = q (- q B ) E. Let us now see what the left hand side is (- q B ) (+ q = ω ve q B ) v E = ω ve q B B ve Bi ve B + q B ve

2 Equating both sides ω ve q ( ω c ) v E q ( BivE ) B + q B ve = q (- q Bi ve B = q (- q B ) E This ay be written as B ) E ; q B = The next step is to break the velocity into coponents perpendicular and parallel to the agnetic field. First for the parallel case. The parallel case B i v E = Bv E v E = v E + v E ( ω c ) v E q B ve = q E, B E is to B (3) v q E = -i E The parallel coponent of v oscillates as if B was ω iπ not there but the oscillation is out of phase by 9 degrees ( i = e ). For the perpendicular coponent ( ω c ) v E = q (- ω c ) E, ω c = q B (4) v E = q (- ) Note this has a resonance at the cyclotron frequency. This is an operator equation of the for v = A where A is a coplex operator, which could be a tensor. Now let us further break down the perpendicular velocity and electric field (which is that of the wave) into two coponents each rotating around the agnetic field in opposite directions. v = v L + v R (5) = + E R. Using (4) as a guide + (i ω c ) E ; E ( ) R (i ω c ) E ( ) ; E => E e t. Let us now assue the agnetic field is constant and is in the z direction. = et ˆr + i e t ˆθ ; E R = e t ˆr i e t ˆθ ; B = B ˆk Re( E L ) = cos ωt ) ˆθ = ˆr + Re(i cos( ωt) + isin(ωt) cos ( ωt ) ˆr E sin(ωt)e ˆθ

3 This is an electric field vector that rotates clockwise around the agnetic field. This is the sae direction that an ion will take so the EL field can resonate with the ion gyro otion. The ER field will resonant with the electrons as it will rotate in the counterclockwise direction. If we re-write the electric field in the perpendicular direction for ion otion as: = E x î + ĵ ω C = ω C ˆk for, = E xî + ĵ + i ˆk E x î + ĵ î + iĵ = E ie x y the tie dependence is still inside E in the above. If we put the tie dependence back Re î + iĵ et = cos( ωt)î sin ( ωt ) ĵ which is a unit vector spinning in the L direction. Now substitute the rotating vectors into equation (IV) first for EL then for ER. (-ω c ) E L = (- ω c ) + (i ω c ) E ( ) = { E c E ( ω c ) E ( ) (i ) = i (ω C E + i ω c E ω c = i(ω C ( note E c = ) Then the operator equation (4) is (6) v L = qi (ω + ) E L ω c v R = qi ( + ω) = qi ( ) This ay be written as a tensor for the rotating electric field in the frae of the rotating particle (7) v= v L v R v ω c = νe = iq + ω ω } E R E 3

4 Note that in this rotating frae the obility tensor is a diagonal. Now using the definitions for ER, EL etc in the notes we can transfor back into the xyz syste (see appendix A) to get: (8) v x v y v z ω c ω c = q ω c ω c i ω What s the next step? Fro the particle velocity we can get currents. W can put these into Maxwell s equations and fro this we can derive a dispersion relation is we assue wave solutions. Note there is only the velocity of single particles in the wave that we considered so the plasa ust be cold. Consider waves of the for e i ( k ir ωt ). This eans that we have Fourier analyzed the solution t the wave equation. (See the notes on Fourier analysis) Here is the strategy. We know the relationship between the velocity and electric field. They are related by a tensor (9) v= ν E The tensor is called the obility tensor. In the rotating frae!! is diagonal (9Rot) νe = iq E x E z + ω ω In rectangular coordinates!! has ore coponents and fro (8) 4

5 (9Rect) ν = q ω c ω c i ω V and E are also related by Maxwell s equations. One of Maxwell s equation has the current in it. The current is related to the particle drift velocity. (Here we will assue that the current is carried by the electrons to siplify the equations. If you also include ion currents there are additional ters but the physics is the sae!) Now we will two relations between the current and electric field and these can be used to find the dielectric tensor. Of paraount utility in this unravelling is the plasa conductivity σ. It can be shown that if σ -> infinity the agnetic field lines are frozen into the plasa. If σ is not equal to infinity the plasa will dissipate heat like the wires in a toaster. We will consider a plasa with σ finite. The Maxwell equation we will use is () B == µ j + µ ε E t Next we will use Ohs law. The first tie you see Oh s law in high school it is written as V=IR. In icroscopic for the Voltage, Current and Resistance are represented by R ρ = σ V E and I J Therefore I = V R J = σ E The current and electric field need not be in the sae direction. Consider a high frequency field where ω is so large that the ions cannot respond. During an oscillation the electrons can drift in the EXB direction and cause an oscillating current. Thus a field in one direction can cause a current in another, As v = νe, j = nα q αvα = σ E. σ is the conductivity tensor. α =i,e 5

6 j x j y j z σ xx σ xy σ xz = σ yx σ yy σ yz σ zx σ zy σ zz E x E z For exaple for j z = σ zz E z this coponent of the conductivity tensor ediates the current in the z direction cause by an electric field in the z direction, but a current in the z direction can be caused by fields in the other two directions as well. In its ost general sense Oh s law is therefore: () J = σ E Now using Fourier analysis we tie variation of E is E = E ( r )e t Equation () becoes B = µ σ E + ε E t and then () B = µ ( σ ε )i E We can cobine the ters on the right hand side (3) B iσ = µ ε i E where ε = ε + ωε and define a new ter (4) iσ + ωε = κ But this not so new because! ust be related to the obility. This is because v = νe ; J = σ E and by definition J = nqv and Equating these to eliinate E (5) = inq ν + ωε = κ Here we ust be careful and we have to use a nuber to a tensor. when doing the addition since you cab t blithely add 6

7 also ω pe = nq ε is the definition of the plasa frequency. κ = inq ν + ωε = + inq ωε q i ω (6) κ = ω pe ω ce i ω pe ω ( ω ce ) i ω pe ω ω ce ω pe ω ce pe ω Appendix A 7

8 How do we go fro rotating coordinates back to rectangular ones? In the notes: (A) E L = ( E ie x y )( î + iĵ ) What about the other coponent? (A) E R = E i ω c E, substitute for in this E R = E xî + E ĵ i ω c E x î + ĵ y = E xî + E ĵ ˆk c E x î + ĵ y E r = E xî + E ĵ i ( E x ĵ î ) y = ( E x î iĵ + E y ĵ + iî ) î iĵ (A3) E r = E x + i Now with (A) and (A3) we have to find Ex, Ey in ters of the left and right coponents. The velocities have the sae for as the fields (A4) î + iĵ v L = v x iv y v R = v x + iv y î iĵ But v L = iq Therefore by substitution: ( v x iv y ) = iq (A5) v y = q ( E ie x y ) ( ) ( E ie x y ) iv x Next use the definition of vr v R = iq + ω E R 8

9 and using the definitions again ( v + iv x y ) = iq ( E + ie x y ) iv y = iq v y = q + ω ( ( E + ie x y ) v x (A6) ( ( E + ie x y ) + iv x Next equate the two (A5) and (A6) q ( ( E + ie x y ) + iv x = q ( ) ( E ie x y ) iv x iv x = q iv x = q v x = q ω c ( ) ( E ie x y ) + q ( ( E + ie x y ) + ( E x + q q E x + i Do the sae thing for the y coponent to get v y = q E x + q ω c Fro before the z coponent was siple v = v z = iq ω E z We can put this into atrix for ( + ( ) v x v y v z = q ( ω c ) ω c ω c ( ω c ) i ω E x E z 9