 Solution

Take the inviscid limit of the Navier-Stokes equation, neglect gravity, and use ideal magnetohydrodynamics. The governing equations areThe continuity equation also gives $D_t\rho =0$. Write the first-order magnetic perturbation as $\mathbf{b}={\mathbf{B}}^{\prime }$ and linearize about the uniform stationary background:For a nonzero-wavenumber plane wave proportional to $e^{i(kz-\omega t)}$, the two divergence constraints imply $v_z=b_z=0$. The longitudinal momentum equation then gives $P^{\prime }=0$ for that mode, and ${\partial }_t{\rho }^{\prime }=0$ gives ${\rho }^{\prime }=0$ for nonzero $\omega$. Thus the propagating Alfvén wave is transverse. Its two independent transverse directions have the same dispersion relation.

The Linear Alfvén wave equations reduce toEliminating either amplitude givesFor $B_0>0$, the branch traveling along $+z$ has $\omega =k{v}_A$ and $\mathbf{b}=-\sqrt{{\mu }_0{\rho }_0}\mathbf{v}$. The opposite branch reverses that amplitude sign. Finite viscosity would additionally damp the motion; it is neglected in obtaining the real ideal dispersion relation.