UCLA ESS 200C - Homework#6 - D2623428

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UCLA ESS 200C - Homework#6

School name University of California, Los Angeles

Course Ess 200c- Space Plasma Physics

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200C, Winter 2009, Homework #6(1) Starting from the steady electron and ion momentum equations, including collisions, but ignoring pressure: e(E  ueB)  meenue0q(E  uiB)  miinui0derive the ionospheric Ohm’s law, as given in equations (7.35b), (7.36), (7.37), and (7.38) of Kivelson and Russell. In deriving these equations assume the magnetic field is uniform, and defines the z-axis of a Cartesian coordinate system. Also assume quasi-neutrality, in which case,j nequi eue . [Technical note, this Ohm’s law is derived based on the electric field as observed in the frame of the neutrals, which is why the velocity of the neutrals does not enter the collision frequency terms.](2) Derive the equation for the divergence of the field-aligned current, starting from the MHD momentum equation, assuming isotropic pressure:dudtj B  PStep 1: Take the dot product of B and the curl of the momentum equation to show thatB  B  j  j B22 B  dudtStep 2: Using Ampere’s law, without the displacement current to express the Lorenz force in terms of gradients of the magnetic field, and since j j B 0, show that:j B  B  j B22and hence from Step 1:B  jB  j B2 B  dudtStep 3:Take the vector cross product of B with the momentum equation to show that:jB2B dudt jB B  B PStep 4:Use the result from Step 3 to replace j in the right-hand side of the equation derived at the end of Step 2, gather terms involving B   and hence show that:B  jBB22B3B  B P 1B4B dudtB21B2B  dudt(Remember B22BB).Step 5:Making use of the cyclical permutation identity: A  B C B  C A C  A B , and since the Alfven speed is given by VA2B20, rearrange terms from Step 4 to show that:B  jBB22B3B  P B 1B4B dudtVA2VA21B2B  dudt(Note that  fA  fA  A f, when expanding the last term on the right-hand side of the equation derived in Step 5.Step 6:Almost done, we could stop at Step 5. However, it is useful to proceed further, given that vorticity is defined by  u. We need to evaluate the last term on the right-hand side of the equation derived in step 5. On expanding the total time derivative d dtand u  u (u), show thatB   dudtB t B   u   Expand the curl on the right-hand side, and noting that  0, getB  dudtB ddt B u  B    uFinally, in this step, since  (u)   0, show that B   (u)    B (u)   B  u  B    u 0and henceB  dudtB ddt   B u  B  u Step 7:Now invoke the frozen-in condition and Faraday’s law to show thatdBdt B  u  B u and henceB  dudtB ddt  dBdtStep 8:Gather the terms from Step 5 and 8 and get the final form:B  jBB22B3B  P B 1B4B dudtVA2VA21B2B ddt 

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