We left off trying to obtain the x, y-components of velocity associated with the wave, by use of Fourier transforms. The x-component is:

w^2 v_x – c_s^2 k^2 v_x + B_z k^2/ u_o rho_o [v_zB_x – v_xB_z] = 0

The y-component is decoupled from the others (x, z) and can be written:

w^2 v_y - B_x^2 k^2 v_y/ u_o rho_o = 0

or simply:

w^2 = [B_x^2/ u_o rho_o] k^2

where the quantity in brackets is the

**Alfven velocity**or alternatively written:

v(A) = [w/ k] = B_x / [u_o rho_o]^ ½

or

v(A) = B_o/ [u_o rho_o]^ ½

since B_o is in the x –direction

:

For completeness, we should be able to show the z-component equation is:

w^2 v_z - B_x k^2/ u_o rho_o [v_z B_x – v_x B_z] = 0

Here, let me back up and refer readers again to the basic wave equation one can obtain by getting the 2nd derivative of:

@v_1/ @t = -(c_s^2) grad p_1/ rho - 1/ u_o rho [B_o X Curl B_1]

One can then find the solution in terms of plane waves by assuming:

v_1 = v_1*[exp ik.x – iwt]

for which taking the second derivative, of v_1 with respect to t yields the original equation in w we found earlier. All of this the energetic reader should be able to work out, but most of it (after taking derivatives) reduces to brute force algebra!

For completeness, I need to note what happens when you solve the preceding (simultaneous) equations in x, and z.

w^4 + w^2[-c_s^2 k^2 - B_x^2 k^2/ u_o rho_o] + c_s^2 k^4 [B_x^2 / u_o rho_o] = 0

or:

w^4 – w^2(c_s^2 + v(A)^2)k^2 + c_s^2 v(A)^2 cos^2 (Θ) k^4 = 0

and finally,

w^2 = ½[(c_s^2 + v(A)^2k^2 +/- [(c_s^2 + v(A)^2 k^4 – 4 c_s^2 v(A)^2 cos^2(Θ) k^4]^ ½

Now, if one plots the preceding using for the vertical axis (c_s^2 + v(A)^2) and for the horizontal B_o (e.g. x) one will get what is called “Friedrich’s diagram” (see sketch image). It consists of

1)A smaller “dumb bell” or figure-8 shaped graph centered at the origin. This will be for what we call “slow mode” waves

2)A single larger lobe that envelopes the smaller right lobe of the dumb bell. This will be for Alfven waves proper.

3) A circle shaped graph surrounding both 1, 2 above. This will be for what we call the “fast MHD” mode.

The critical thing to note here is that the fast mode is the only MHD wave able to carry energy perpendicular to the magnetic field. This has important ramifications for solar flares, as well as magnetospheric effects (such as the aurora). Meanwhile, the phase velocity (w/k) of the slow mode wave perpendicular to the magnetic field is always zero. In the limit where the sound speed c_s^2 < < v(A)^2, and the Alfven speed v(A)^2 << c_s^2, the slow wave disappears. (Which one can easily validate and confirm for the equation in w^2)

Other properties, points to note:

-the velocity perturbation v_1 is orthogonal to B_o

-the wave is incompressible since DIV v_1 = ik.v_1 = 0

-the magnetic field perturbation (B_1) is aligned with the velocity perturbation. Since both are perpendicular to k and B_o

-the current density perturbation (J_1) exists as a current perturbation perpendicular to k and B_o e.g.

J_1 = k X Bo

- When c_s^2 << v(A)^2 the fast mode wave becomes a compressional Alfven wave. This has a group velocity equal to its phase velocity w/k

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