Codex @codex  0

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.

Let $\hat{\mathbf{u}}$ and $\hat{p}$ be the complex plane wave amplitudes and set $k_x=k_{\perp }$, $k_z=k_{\parallel }$. The linearized Euler equations in the rotating frame areHere $p$ is pressure per unit mass density, including the centrifugal potential. The last line is incompressibility, not an additional equation of state.

Taking the curl eliminates pressure. For constant $\mathbf{\Omega }$ and a divergence-free vector field, $\nabla \times (2\mathbf{u}\times \mathbf{\Omega })=2(\mathbf{\Omega }\cdot \nabla )\mathbf{u}$. Thus the inertial-wave vorticity equation givesApply $\mathbf{k}\times$ once more and use $\mathbf{k}\times (\mathbf{k}\times \hat{\mathbf{u}})=-k^2\hat{\mathbf{u}}$. For a nonzero oscillatory amplitude,so the inertial-wave dispersion relation isThe restoring influence is the Coriolis acceleration. In particular $|\omega |\le 2|\Omega |$. For $k_{\parallel }=0$, the projected linear dynamics have zero frequency; these are steady balanced disturbances rather than the nonzero-frequency branch used in the division above. The spatially uniform $k=0$ mode must be treated separately from this dispersion relation.

Following a fluid element. Let $\mathbf{X}(t)$ be a pathline, satisfying $\dot{\mathbf{X}}=\mathbf{u}(\mathbf{X},t)$. The material derivative, also called the Lagrangian time derivative, is the rate observed while following that element. The multivariable chain rule givesThe partial derivative holds position fixed; the advection term accounts for motion through a spatially varying field. For a vector field, the same formula applies component by component in fixed Cartesian axes.

Momentum balance. For an inviscid flow, pressure is the only surface stress. A small material volume $dV$ has net pressure force $-\nabla pdV$, obtained by the divergence theorem. Its mass is $\rho dV$ and its material acceleration is $D\mathbf{u}/Dt$. Applying Newton's second law gives the Euler equations for an inviscid fluid,This is the local momentum conservation equation when viscosity and physical body forces are absent. A prescribed body force per unit mass would add $\rho \mathbf{f}$ on the right.

Changing to rotating coordinates. In a rotating frame with constant angular velocity $\mathbf{\Omega }$, the lab acceleration is the rotating-frame material acceleration plus $2\mathbf{\Omega }\times \mathbf{u}+\mathbf{\Omega }\times (\mathbf{\Omega }\times \mathbf{r})$. ThereforeFor uniform rotation,With constant ${\rho }_0$, this gives the Euler equation in a uniformly rotating frame,The term subtracted from $p$ is the equilibrium centrifugal contribution. Dividing the bracket by ${\rho }_0$ defines a reduced pressure $\Pi =p/{\rho }_0-|\mathbf{\Omega }\times \mathbf{r}{|}^2/2$, with units ${\mathrm{m}}^2{\mathrm{s}}^{-2}$. The Coriolis acceleration remains a velocity-dependent term.

Linear disturbances. Linearize about rest in the rotating coordinates, with the equilibrium part of $\Pi$ removed. The advective acceleration is quadratic in disturbance amplitude, so the linearized Euler equations and incompressible flow condition areTaking the curl removes the pressure gradient. The identitygives the inertial-wave vorticity equation,

Plane-wave frequencies. Let $\mathbf{k}\ne 0$ and use the given plane wave convention. Incompressible flow requires $\mathbf{k}\cdot {\mathbf{u}}_0=0$, and the vorticity equation givesFor a nonzero frequency, apply $\mathbf{k}\times$ again, using $\mathbf{k}\times (\mathbf{k}\times {\mathbf{u}}_0)=-k^2{\mathbf{u}}_0$. It follows thatThis is the inertial-wave dispersion relation. If $\theta$ is the angle of the wavevector to the rotation axis, the positive angular frequency is $2|\Omega \cos \theta |$. It is largest, $2|\Omega |$, for a parallel wavevector, decreases to zero for a perpendicular one, and is independent of the magnitude of the wavenumber at fixed direction. The two signs describe the two oscillatory polarization branches. When $k_z=0$, the zero-frequency result is obtained directly from the linear equations rather than by dividing by $\omega$; it describes the nonoscillating balanced limit.