= Solution
Solving the polarization quadratic gives all four signed roots:
$$
\boxed{\omega_{h,s}=\frac q2\left(hkd_i+s\sqrt{4+k^2d_i^2}\right),\qquad h,s=\pm1.}
$$
A sign-independent description is the pair of positive frequency magnitudes
$$
\boxed{\Omega_{\rm fast,slow}=|k_\parallel|v_A\left[\sqrt{1+\frac{k^2d_i^2}{4}}\ \pm\ \frac{kd_i}{2}\right].}
$$
For wavelengths long compared with the <ion skin depth>, $kd_i\ll1$, both magnitudes approach $|k_\parallel|v_A$. The Hall correction is then small and the usual <Alfvén wave> dispersion $\omega^2=k_\parallel^2v_A^2$ is recovered. In an incompressible system the two transverse polarizations include the shear and <pseudo-Alfvén wave> sectors; away from this limit Hall coupling selects circular helicity combinations. A small parallel wavenumber alone does not remove the relative Hall splitting: the expansion parameter is the full $kd_i$.
Back to article page
