Stationary topographic internal gravity wave ($\omega =0,\hat{\omega }=-kU$)

A fixed terrain obstacle forces a zero laboratory-frequency internal gravity wave through the kinematic boundary condition $w=U{h}_x$. Its intrinsic frequency is $-kU$. Propagation, radiation and the transient arrival of its permanent component are distinct questions.

Define intrinsic positive frequency $\omega$ and ground frequency $\hat{\omega }=Uk+\omega$ for a nonrotating internal gravity wave. Stationarity with $U>0$ selects $k<0$ and $\sqrt{k^2+m^2}=N/U$. Differentiating the full dispersion before imposing that condition gives the displayed ground-frame group velocity. Upward energy propagation requires $m<0$ and $|k|<N/U$. Its ray inclination is ${\cos }^{-1}(U|k|/N)$. The ground-frame group velocity is opposite the normal intrinsic phase velocity; the intrinsic group velocity is perpendicular to that phase velocity. A strictly stationary ground-frame phase pattern has zero phase velocity, so mixing frames can produce a false angle claim.

For a sinusoidal small-amplitude bed, constant current $U>0$ and buoyancy frequency $N$, put $m=\sqrt{N^2/U^2-k^2}>0$. The linear vertical velocity amplitude is $W(z)=ikU{\eta }_0\sin [m(H-z)]/\sin (mH)$. The upward-wave coefficient is $A_{+}=-kU{\eta }_0{e}^{-imH}/[2\sin (mH)]$. At $mH=n\pi$, the prescribed nonzero bed motion and homogeneous rigid-lid condition are incompatible with a bounded stationary inviscid solution. With angle $\theta$ between intrinsic group velocity and vertical, $\tan \theta =m/k$, giving $kH\tan \theta =n\pi$.

For a localized terrain source in uniform flow, the stationary phase method applied to $kx+m(k)z$ gives phase $\Phi =(N/U)\sqrt{x^2+z^2}$ on the downstream branch. Permanent phase lines are circular arcs, whereas an individual monochromatic component has straight phase lines. Reflected arcs use an image source; transmitted phase is continuous across the velocity interface.