 Solution

The group velocity, which gives wave-packet and energy propagation, is the gradient of the dispersion relation:For $\mathbf{k}=(k_{\perp },0,k_{\parallel })$ this becomesThus for $ϵ=|k_{\parallel }|/k_{\perp }\ll 1$, energy travels almost along the rotation axis, with a small horizontal component. By contrast, the normal-to-phase-front phase velocity ${\mathbf{v}}_{\mathrm{ph}}=\omega \mathbf{k}/k^2$ is almost horizontal. Indeed $\mathbf{k}\cdot {\mathbf{v}}_g=0$, so these phase and group directions are perpendicular. Specifying which meaning of propagation is intended matters for an inertial wave.

For the wave polarization, incompressibility and the $y$ equation in part a giveThe horizontal component parallel to ${\mathbf{k}}_{\perp }$ is small, but $|{\hat{u}}_y|\simeq |{\hat{u}}_z|$: the other horizontal and vertical components are in quadrature. More exactly, introduce the orthonormal transverse basisThen $\hat{\mathbf{u}}=A({\mathbf{e}}_1-\sigma i{\mathbf{e}}_2)$. The physical velocity at a fixed point traces a circle in the plane perpendicular to $\mathbf{k}$; its handedness reverses between branches. In the anisotropic limit this plane is nearly vertical. The group velocity and polarization of an inertial wave therefore have quite different geometric roles.