Inertial wave

An inertial wave is a rotating-fluid wave restored by the Coriolis force. Its frequency depends on the angle between its wavevector and the rotation axis.

For fixed vertical wavenumber $k_z$, the inertial-wave dispersion relation is ${\omega }^2=4{\Omega }^2{k}_z^2/(k_x^2+k_z^2)$. Its horizontal group velocity is the derivative displayed above, opposite in sign to $\omega k_x$. In a half-channel $x>0$, an outgoing propagating mode therefore requires $\omega k_x<0$. With time factor $e^{-i\omega t}$ and $q=|k_z|\sqrt{4{\Omega }^2/{\omega }^2-1}$, choose $e^{-i\mathrm{sgn}(\omega )qx}$. Above the inertial-frequency cutoff, the horizontal wavenumber is imaginary and the appropriate radiation condition instead selects the exponentially decaying solution. Energy transport, rather than the direction of phase travel alone, determines the outgoing sign.

Linear incompressible motion about rest in a frame rotating with $\mathbf{\Omega }=\Omega \mathbf{k}$ obeys ${\mathbf{u}}_t+2\mathbf{\Omega }\times \mathbf{u}=-\nabla P/\rho$. If $\zeta =v_x-u_y$ is its vertical vorticity, taking curl gives ${\zeta }_t=2\Omega w_z$, while eliminating pressure using continuity gives ${\partial }_t{\nabla }^{2}w+2\Omega {\zeta }_z=0$. Differentiating in time yields the displayed scalar equation. All three velocity components must be retained, even when the fields depend on only two spatial coordinates. The Coriolis acceleration generally generates a transverse velocity that is absent in a strictly two-component description.

For $\omega =\sigma 2\Omega k_{\parallel }/k$, differentiating the dispersion relation gives the displayed group velocity. It is perpendicular to the wavevector. If $|k_{\parallel }|\ll k_{\perp }$, energy propagates almost along the rotation axis while the phase velocity is almost horizontal. In an orthonormal basis ${\mathbf{e}}_1,{\mathbf{e}}_2$ normal to $\mathbf{k}$, with ${\mathbf{e}}_2=\hat{\mathbf{k}}\times {\mathbf{e}}_1$, the complex velocity is $A({\mathbf{e}}_1-\sigma i{\mathbf{e}}_2)$ and its real oscillation is circular. In the anisotropic limit the vertical and transverse-horizontal velocity amplitudes are comparable, while the horizontal component parallel to ${\mathbf{k}}_{\perp }$ is smaller by $|k_{\parallel }|/k_{\perp }$.

Insert a transverse plane wave $\mathbf{u}={\mathbf{u}}_0{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$ into the inertial-wave vorticity equation, with $\mathbf{k}\cdot {\mathbf{u}}_0=0$. It gives $\omega \mathbf{k}\times {\mathbf{u}}_0=2i(\mathbf{\Omega }\cdot \mathbf{k}){\mathbf{u}}_0$. Applying $\mathbf{k}\times$ again and using transversality gives the title dispersion relation. If $\theta$ is the angle to the rotation axis, $|\omega |=2|\Omega \cos \theta |$. At fixed direction the angular frequency does not depend on wavelength. Parallel wavevectors give the maximum $2|\Omega |$; perpendicular ones have zero frequency and belong to the nonoscillating balanced limit.

Linearize the Euler equation in a uniformly rotating frame about rest and impose incompressible flow, $\nabla \cdot \mathbf{u}=0$. Taking the curl eliminates the reduced pressure. For constant $\mathbf{\Omega }$, $\nabla \times (\mathbf{\Omega }\times \mathbf{u})=-(\mathbf{\Omega }\cdot \nabla )\mathbf{u}$, giving the displayed vorticity equation. Taking another curl and time derivative yields ${\partial }_t^2{\nabla }^2\mathbf{u}+4(\mathbf{\Omega }\cdot \nabla {)}^2\mathbf{u}=0$. These equations describe the restoring influence of background rotation on small disturbances without buoyancy or viscosity.

In a uniform-density incompressible flow inside a smooth rigid sphere rotating at angular speed $\Omega$, a normal mode with phase $e^{i\sigma t+im\varphi }$ has intrinsic frequency $s=\sigma +m\Omega$. For $s\ne 0,\pm 2\Omega$, eliminating velocity from the linearized Euler equations gives the displayed pressure equation, with $W=p^{\prime }/\rho$. Impermeability gives $R{W}_R+(2m\Omega /s)W+(1-4{\Omega }^2/s^2)z{W}_z=0$ on the sphere. There is no condition $W=0$: the rigid wall supplies variable normal pressure. These inertial waves are restored by Coriolis acceleration, and their pressure equation is hyperbolic for $|s|<2|\Omega |$.

For an axisymmetric normal mode in a rigid rotating sphere, substitute $W=z(A{R}^2+B{z}^2+C)$ into the pressure equation and its impermeability condition. With $\beta =1-4{\Omega }^2/{\sigma }^2$, the interior equation gives $2A+3\beta B=0$. The boundary polynomial gives $3\beta B=(2+\beta )A$ and $\beta C=-(2+\beta )A{r}_0^2$. A nontrivial oscillatory solution has $\beta =-4$, hence the displayed frequency and $B=A/6$, $C=-A{r}_0^2/2$. It satisfies both equations identically.