Derive the MHD Wave Dispersion Relations for Warm Plasma

ITEM#1. Derive the MHD wave Dispersion Relations for warm plasma.

 1)  U  0 (continuity)  t

       2)   uu   -   j  B (momentum)   t   B  3)  -   E (Faraday’s Law)  t    4)   B  o j (Ampère’s Law, without displacement current)    5) B  0 (Divergenceless B )    6) E U  B  0 (Ohm’s Law)

      U    0 7)      (entropy conservation)   t   p     For linear waves: B  B  b,     ,      while assuming we are in the plasma frame, U=0 and E, u are 1st order quantities. Assuming background field B and density are constant and spatially uniform, j is also a first order quantity, whereas  t  B t  0 .

From (3) and (6) we get:  b   8)    u  B t  while from (5) we get b  0. This is a boundary (initial) condition because if we take    the divergence of (8) we obtain b  0 i.e., if b  0 initially, it must be zero t

  p  2 always. The equation of state results from (7) and since p    p   p  cs       p     0 p  Therefore,  2  and , the sound speed, is set by the initial conditions. t  cs     Thus we can express p as a function of . We are left with 7 unknowns: , u, B and 7 equations (1), (2), and (3), after replacing (6) into (3), so we can solve the system. The linearized equations are:

1     u  0 (1)’’ t  dv       p  ( b) B  0 (2)’’ dt b       (u  B). (3)’’ t

Assume 1-dimensional waves propagating in the x-direction, and a background field at an angle  to xˆ . Without loss of B  generality we take the ( xˆ , B ) plane to contain the zˆ -axis. z Retaining only derivatives in x, we obtain:  B  u    x  0 (1 II)  t x    u   bB 0 Bx b    xˆ  xˆ  (2 II)  t x x  x y   0 b U  U x  II  Bx    (3 )  t x  x  x Note equation (2II) is derived by linearizing the force from the total field, expressed as:          2      B  B  -B 20  BB 0   2bB 20  Bb 0    realizing that B is uniform and b 2 , bb are second-order quanities.

Assuming an exponential form of waves, eikxt  , we obtain the equations relating the   i ;  x  ik. amplitudes by replacing  t Dropping the “i”, these result in: III  - k u x  0 (1 )    III u - k xˆbB Bxb 0  k  0 (2 )    III b  kBxu  u x B  0 (3 )

2 III We can substitute  with cs  and then relate  to u x from (1 ). We thus get: k   c 2 u . Then (2 III) becomes, after division by k : s  x       bB  B b  k ˆ  x 2 IV u  x     cs u x  0 (2 ) k   0   0   

2 whereas (3III) becomes, after division by kB:   b    B u  u B  0 (3IV) k B x x

We now recall   B 2 B B  Bcos, 0, Bsin and  V 2 , Bˆ  , and get, after multiplying (2IV ) with  :   A  k  0  B   2     bBˆ    b   u  xˆ   cos  V 2  xˆc 2u  0 V  k       A s x (2 )   k B   k B      b   ˆ     cos u  u x B 0. k  B     (3V)

We now substitute (3V) into (2V) and divide by

2  k cs VA . We use the symbols X 0  , X s  and get VA VA 2   ˆ ˆ  2 X  u  xˆux  cos u  B cos u x B  cos u xˆcs ux  0 2  2  ˆ 2  ˆ  X  u  xˆux 1 X s  cos u  B cos   u  u x cos  B  0

Now take the x, y, and z components:

2 2  ˆ 2 2 xˆ : X  u x  u x 1 X s  cos u  B cos  u x  u x cos   0

2 2 2  X  1 X s  u x  u x cos   u z sin cos  0

2 2 2  X   X s  sin  u x  sin cos u z  0

2 2 yˆ : X  u y  cos  u y  0

2 2 zˆ : X  u z  cos  u z  u x sin cos  0

For the above system of equations to have solutions arbitrary in ux, uy, and uz, the determinant must be zero:

3 2 2 2 X   X s  sin  0 sin cos    2 2 Det 0 X   cos  0   0  2 2  sin cos 0 X   cos  

We start from the middle element of the determinant: 2 2 2 2 2 2 2 2 2 X   cos  X   X s  sin  X   cos   sin  cos  0

2 2 4 2 2 2 2  X   cos  X   X  Xs 1 X s cos  0  b   Note solutions to quadratic: ax2+bx + c = 0 are: x  ,   b 2  4ac. 2c 2 2 2 In our case,   xs 1  4cos , and solutions are :

2 2 1  2 2 2 2 2  X   X F ,S  X s 1 X s 1  4X s cos   2   2 2 2 2 expanding X s  cs VA and multiplying by VA , we get

2 2 2 2 2    2 1 2 2 2 2 2 2 2 2 X  VA  X F ,S VA     VF ,S  cs VA  cs VA   4cs VA cos   k  2

2 2 2 2 2 2 and the MHD dispersion relation so far is: X   cos   X   X F X   X s   0 Note that we had 7 equations but we have now 6 solutions, i.e., two waves propagating in +/- directions for each mode: Fast, Alfvén (otherwise called Intermediate) and Slow. th 2 III What happened to the 7 solution? When we substituted   cs u x / k from (1 ) from the continuity equation intto the momentum equation, we assumed that /k0. This is not necessarily true. The trivial solution /k=0 also satisfies all MHD equations and in particular in that case all perturbations in equations (1 III)-(3 III) except the density are zero, and the wave is “propagating” neither in the positive nor negative direction, but is degenerately at rest in the frame of the plasma. Small perturbations in density and   pressure can have a spatial variation but no temporal variation since u  b  0 . III Specifically, (1 ) is satisfied because =ux=0 even though   0 . This “wave” still satisfies the entropy conservation equation (7) and is called the isentropic wave. Combining the above solution with the solution to (2 III) we obtained for F,I,S waves we 2 2 2 2 2 2 have the 7 solutions to the MHD equations: X  X   cos   X   X F X   X S   0 . Multiplying back with the assumed non-zero Alfvén speed we obtain:

2 2 2 2 2 2 2 V  V VA cos   V VF  V VS   0

Division by VA in the determinant was not mandatory, but only a convenience. If it is not done the derivation follows the same path and results in the same solution but

4 encompasses the limiting case where VA=0. In that the MHD equations result in 3 waves: ordinary sound waves (the limiting case of Fast waves) and the entropy wave.

5 ITEM#2. Prove that p and B are in opposite phase for slow wave, whereas in  bBˆ B phase for fast wave. Note,   relative compressional perturbation. B B V Recipe: Start from x-component of momentum equation, (2 ) and replace u x with . Then show that p and B are opposite sign for slow waves and same sign for fast waves.

  ˆ 2     bB    b   k u  xˆ    cos  V 2  xˆc 2u  0 V        A s x 2   k  B   k B   2   bBˆ   B ˆ u    c 2      . x -component: x  k   s      k  B  k B

Replacing ux with , we use continuity equation,

III    u  (1 ), i.e., x  k 

  c 2    Since s  , we get

2 2    2  B  cs B     cs     2  k  2 2 cs B   k  cs B

2 Now we have to show that   c 2 is positive for fast waves and negative for slow  k  s waves. We use the dispersion relation:

2   c 2   k  s 1 2 2 2 2 2 2 2 2 2  cs VA  2cs  cs VA   4cs VA cos    2 

 2 2 2 2 2 2 2 2  1 2VA  Cs  Cs VA   4Cs VA cos   . Note that:  

2 2 2 2 2 2 2 2 2 2 2 2 2 A  VA  Cs   VA  Cs   4Cs VA  VA  Cs   4Cs VA cos   B i.e., A < B because cos 2  1. Therefore, A ± B will be positive for fast waves (+) and negative for slow waves (-) and as a resultp, B will have the same sign for fast waves and opposite sign for slow .