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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.

Let $L_{\perp }=k_{\perp }^{-1}$ and let $U$ be the common size of horizontal and vertical velocity perturbations, as found in part b. The linear inertial wave time and nonlinear advection time areThe critical balance in a rapidly rotating fluid is thereforeThe Rossby number is of order $ϵ$. Horizontal geostrophic flow balances the leading pressure gradient against Coriolis acceleration, soThese are dimensionally comparable statements, rather than comparisons of a dimensional frequency with a dimensional velocity. Equivalently, use $L_{\perp }$ and $(2\Omega {)}^{-1}$ as units. ThenThus rotation stays $O(1)$ in these units, while the motions evolve on the slower $O({ϵ}^{-1})$ time. Vertical advection $u_{\parallel }{\partial }_z$ is smaller than horizontal advection by $ϵ$.

Write ${\mathbf{u}}_{\perp }={\mathbf{u}}_{\perp }^{(0)}+{\mathbf{u}}_{\perp }^{(1)}+\cdots$, with ${\mathbf{u}}_{\perp }^{(1)}=O(ϵU)$, and $p=p_0+p_1+\cdots$. At leading horizontal order,Define the stream function $\Phi =p_0/(2\Omega )$, including its horizontally uniform part in this convention. ThenThe next-order incompressibility condition is not zero horizontal divergence:This term has size $ϵk_{\perp }U$ and is essential: its product with the large rotation rate is a leading contribution to the vertical vorticity equation.

To see that contribution explicitly, write $D_{\perp }={\partial }_t+{\mathbf{u}}_{\perp }^{(0)}\cdot {\nabla }_{\perp }$. At the first horizontal inertial order,Taking its vertical curl eliminates $p_1$. The two-dimensional advection identity and the two divergence orders giveEquivalently the ageostrophic flow satisfieswhose divergence is $-D_{\perp }\zeta /(2\Omega )$. This explicitly verifies that the next-order divergence has the required sign.

The vertical momentum equation, at the same slow order, isFinally ${\mathbf{u}}_{\perp }^{(0)}\cdot {\nabla }_{\perp }f=\{\Phi ,f\}$, where the canonical two-dimensional Poisson bracket is $\{f,g\}=f_x{g}_y-f_y{g}_x$. The reduced equations for an anisotropic rotating fluid areThe omitted vertical-advection, ageostrophic-advection and relative-vorticity stretching terms are smaller by $O(ϵ)$ in the equations retained here. Linearizing these equations recovers ${\omega }^2=4{\Omega }^2{k}_{\parallel }^2/k_{\perp }^2$, the anisotropic limit of part a.