= Solution
\b[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> gives
$$
\boxed{\frac{D\phi}{Dt}=\frac{d}{dt}\phi(\mathbf X(t),t)
=\frac{\partial\phi}{\partial t}+\mathbf u\cdot\nabla\phi.}
$$
The <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.
\b[Momentum balance.] For an <inviscid flow>, <pressure> is the only surface stress. A small material volume $dV$ has net pressure <force> $-\nabla p\,dV$, 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>,
$$
\boxed{\rho\frac{D\mathbf u}{Dt}
=\rho\left[\partial_t\mathbf u+(\mathbf u\cdot\nabla)\mathbf u\right]
=-\nabla p.}
$$
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.
\b[Changing to rotating coordinates.] In a <rotating frame> with constant <angular velocity> $\boldsymbol\Omega$, the lab acceleration is the rotating-frame <material acceleration> plus $2\boldsymbol\Omega\times\mathbf u+\boldsymbol\Omega\times(\boldsymbol\Omega\times\mathbf r)$. Therefore
$$
\rho_0\left[\partial_t\mathbf u+(\mathbf u\cdot\nabla)\mathbf u
+2\boldsymbol\Omega\times\mathbf u\right]
=-\nabla p-\rho_0\boldsymbol\Omega\times(\boldsymbol\Omega\times\mathbf r).
$$
For uniform rotation,
$$
\boldsymbol\Omega\times(\boldsymbol\Omega\times\mathbf r)
=-\nabla\!\left(\frac12|\boldsymbol\Omega\times\mathbf r|^2\right).
$$
With constant $\rho_0$, this gives the <Euler equation in a uniformly rotating frame>,
$$
\boxed{\rho_0\left[\partial_t\mathbf u+2\boldsymbol\Omega\times\mathbf u
+(\mathbf u\cdot\nabla)\mathbf u\right]
=-\nabla\!\left[p-\frac{\rho_0}{2}|\boldsymbol\Omega\times\mathbf r|^2\right].}
$$
The term subtracted from $p$ is the equilibrium centrifugal contribution. Dividing the bracket by $\rho_0$ defines a reduced pressure $\Pi=p/\rho_0-|\boldsymbol\Omega\times\mathbf r|^2/2$, with units $\mathrm{m^2\,s^{-2}}$. The <Coriolis acceleration> remains a velocity-dependent term.
\b[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 are
$$
\partial_t\mathbf u+2\boldsymbol\Omega\times\mathbf u=-\nabla\Pi',\qquad
\nabla\cdot\mathbf u=0.
$$
Taking the <curl> removes the pressure <gradient>. The identity
$$
\nabla\times(\boldsymbol\Omega\times\mathbf u)
=\boldsymbol\Omega(\nabla\cdot\mathbf u)-(\boldsymbol\Omega\cdot\nabla)\mathbf u
=-(\boldsymbol\Omega\cdot\nabla)\mathbf u
$$
gives the <inertial-wave vorticity equation>,
$$
\boxed{\partial_t(\nabla\times\mathbf u)-2(\boldsymbol\Omega\cdot\nabla)\mathbf u=0,
\qquad \nabla\cdot\mathbf u=0.}
$$
\b[Plane-wave frequencies.] Let $\mathbf k\ne0$ and use the given <plane wave> convention. <Incompressible flow> requires $\mathbf k\cdot\mathbf u_0=0$, and the <vorticity> equation gives
$$
\omega\,\mathbf k\times\mathbf u_0
=2i(\boldsymbol\Omega\cdot\mathbf k)\mathbf u_0.
$$
For 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 that
$$
\boxed{\omega^2=\frac{4(\boldsymbol\Omega\cdot\mathbf k)^2}{k^2}
=4\Omega^2\frac{k_z^2}{k^2},\qquad \omega=\pm2\Omega\frac{k_z}{k}.}
$$
This 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.
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