Past exam of the physics of atmospheres and oceans course of the University of Oxford / 2018 / 6 / b

Put $K=(k^2+m^2{)}^{1/2}$ and $P=p^{\prime }/{\rho }_0$. Acting on the Fourier mode, $D_0=-i(\hat{\omega }-Uk)=-i\omega$. Continuity and horizontal momentum give $u=-mw/k$ and $P=\omega u/k$. The buoyancy equation gives $g{\rho }^{\prime }/{\rho }_0=i{N}^{2}w/\omega$. Substitution into vertical momentum yieldsThe two signs describe the two frequency branches; when the intrinsic frequency is required positive, $\omega =N|k|/K$. For a stationary disturbance $\hat{\omega }=0$, so $\omega =-Uk>0$. With $U>0$ this requires $k<0$ and selects the minus sign, $\hat{\omega }=Uk-Nk/K$. The stationary dispersion condition isDifferentiate the unconstrained dispersion before imposing stationarity. The stationary mountain-wave group velocity is obtained from the ground-frame dispersion; before imposing stationarity it isOn the stationary locus this becomesDifferentiating after substituting $K=N/U$ as though all neighbouring modes were stationary would incorrectly erase the frequency gradients.

For upward energy propagation from the lower boundary, choose $m<0$, since $k<0$ and ${\hat{c}}_{gz}>0$. It requires real nonzero $m$, henceThe downstream ray inclination above horizontal isIt is fixed for a given monochromatic forcing, with the two signs of $m$ giving the upward/downward rays.

Frame qualification for the requested phase angle: the normal intrinsic phase velocity is ${\mathbf{c}}_p=\omega (k,m)/K^2$. Because $\omega =-Uk$, the boxed stationary group velocity is exactly $-{\mathbf{c}}_p$: its angle to intrinsic phase propagation is $\pi$, and it is perpendicular to the stationary phase lines. By contrast the intrinsic group velocity ${\mathbf{c}}_g=\widehat{{\mathbf{c}}_g}-U\hat{\mathbf{x}}$ satisfies ${\mathbf{c}}_g\cdot (k,m)=0$, so its angle to intrinsic phase propagation is $\pi /2$. In the ground frame the phase velocity of a strictly stationary pattern is zero and has no nonzero propagation direction. For example $k=-0.6N/U$, $m=-0.8N/U$ gives $\widehat{{\mathbf{c}}_g}=(0.36U,0.48U)$ and ${\mathbf{c}}_p=(-0.36U,-0.48U)$. These distinctions prevent applying the usual zero-flow perpendicularity rule to two velocities expressed in different frames.