Past exam of the geophysical fluid dynamics course of the University of Oxford / 2024 / 1 / b / Solution

Use a plane wave with phase $kx+mz-\omega t$, and take $N>0$, $k,m\ne 0$. The positive-frequency branch of the given hydrostatic dispersion relation is $\omega =N|k|/|m|$. The wavevector is $\mathbf{K}=(k,m)$. Differentiating with respect to its components gives the group velocity, which carries the wave packet and its energy:The normal phase velocity, which follows a crest in its normal direction, isConsequently,Thus the group velocity lies along the constant-phase lines $z=-(k/m)x+\mathrm{constant}$, perpendicular to the phase velocity. For $k,m>0$, crests move upward and rightward, whereas energy moves downward and rightward. Choosing $m<0$ gives upward and rightward energy propagation and downward and rightward normal phase propagation. Conjugate or oppositely directed components supply the other outgoing directions from a localized source.

The speeds areThe hydrostatic limit is $|k/m|\ll 1$. In that limit energy propagates almost horizontally, with $|c_{g,x}|=N/|m|$ and $|c_{g,z}|=N|k|/m^2$, while normal phase propagation is almost vertical and has speed approximately $N|k|/m^2$. The horizontal speed at which a crest intersects a fixed height is $\omega /k=c_{g,x}$; this is not the horizontal component of the normal phase velocity vector. For example, $N=10^{-2}{\mathrm{s}}^{-1}$, vertical wavelength $2\pi /|m|=10\mathrm{km}$ and horizontal wavelength $2\pi /|k|=1000\mathrm{km}$ give $|c_{g,x}|\simeq 15.9\mathrm{m}{\mathrm{s}}^{-1}$ and $|c_{g,z}|\simeq 0.159\mathrm{m}{\mathrm{s}}^{-1}$.

These are local speeds in a uniform medium, without a background current. The non-rotating approximation requires $|f|\ll \omega$, while the hydrostatic approximation requires $\omega \ll N$. At lower frequencies, rotation changes the dispersion relation to that of inertia-gravity waves.