= Solution
For $\mathbf u_0=-3\Omega x\mathbf e_y/2$, advection vanishes and the <Coriolis acceleration> exactly cancels $3\Omega^2x\mathbf e_x$, so constant $P_0$ and $\theta_0=0$ complete the equilibrium.
Write perturbations $(\mathbf v,p,\vartheta)e^{i\mathbf k\cdot\mathbf x-i\omega t}$ with $\mathbf k=(k_x,0,k_z)$. Incompressibility gives $\mathbf k\cdot\mathbf v=0$. It also makes the quadratic terms $\mathbf v\cdot\nabla\mathbf v=i(\mathbf k\cdot\mathbf v)\mathbf v$ and $\mathbf v\cdot\nabla\vartheta$ vanish exactly; the background-advection terms vanish because $k_y=0$. The amplitude equations are
$$
-i\omega v_x=-\frac{ik_xp}{\rho_0}+2\Omega v_y,
\qquad
-i\omega v_y=-\frac12\Omega v_x,
$$
$$
-i\omega v_z=-\frac{ik_zp}{\rho_0}-N^2\vartheta,
\qquad
-i\omega\vartheta=v_z,
\qquad
k_xv_x+k_zv_z=0.
$$
Eliminating the amplitudes yields the <inertia-gravity wave> dispersion relation
$$
\boxed{\omega^2=\frac{k_z^2}{k^2}\Omega^2
+\frac{k_x^2}{k^2}N^2,
\qquad k^2=k_x^2+k_z^2}.
$$
For $N^2=0$, $\omega=\pm\Omega k_z/k$, so the <group velocity> is
$$
\boxed{\mathbf c=\nabla_{\mathbf k}\omega
=\pm k^{-3}[\mathbf k\times(\boldsymbol\Omega\times\mathbf k)]}.
$$
It is perpendicular to $\mathbf k$ and hence to the <phase velocity>. An <inertial wave> packet transports energy along beams lying in its phase surfaces.
If $N^2>0$ and $k_x/k_z\to0$, incompressibility suppresses vertical motion and $\omega\to\Omega$: horizontal epicyclic motion and rotation dominate. If $k_z/k_x\to0$, radial motion is suppressed and $\omega\to N$: vertical buoyancy oscillations dominate. Intermediate ratios give hybrid <inertia-gravity waves>.
If $N^2<0$, exponential growth occurs exactly when
$$
\boxed{N^2k_x^2+\Omega^2k_z^2<0
\quad\Longleftrightarrow\quad
\frac{k_x^2}{k_z^2}>\frac{\Omega^2}{-N^2}}.
$$
Only modes with sufficiently large radial wavenumber permit enough vertical displacement for unstable buoyancy to overcome rotational restoration; the other orientations remain stabilized by the Coriolis force.
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