 Solution

Choose the positive-frequency branch with $k>0$, let $K=|\mathbf{k}|$, and write $\mathbf{k}=K(\cos \theta ,\sin \theta )$. The internal gravity wave dispersion relation is $\omega =Nk/K=N\cos \theta$. The phase velocity and group velocity areThey are perpendicular: the phase advances normal to the crests while energy travels along them. For a general signed $k$, the positive frequency is $N|k|/K$ and the horizontal component of group velocity changes sign with $k$.

If $\omega >N$, there is no real wave vector satisfying this relation. For prescribed real horizontal wavenumber, $m^2=k^2(N^2/{\omega }^2-1)<0$, so a bounded boundary response is an evanescent wave, rather than a freely propagating plane wave.

For a single incompressible plane wave, internal-wave polarization gives ${\mathbf{u}}_1\cdot \mathbf{k}=0$. Its oscillatory displacement ${\mathbf{\xi }}_1$ is parallel to ${\mathbf{u}}_1$ and therefore also perpendicular to $\mathbf{k}$. Every spatial derivative of the plane-wave field is proportional to a component of $\mathbf{k}$, soThere is no leading Stokes drift for that single wave. The fluid parcels oscillate along straight lines within phase planes and never sample a spatial change of the wave along their displacement. This conclusion does not prohibit cross terms between different wave vectors.