Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 3 / 39D / b

In coordinates moving with the background fluid, $x^{\prime }=x-Ut$. The stationary hill profile becomesso in the convention $e^{i(k{x}^{\prime }-\omega t)}$ its frequency is $\omega =-kU$; its physical angular frequency has magnitude $kU$.

The boundary condition gives complex amplitude $w_0=Uk{h}_0$. Substitution of the internal gravity wave forminto the internal-wave equation givesSince $|\omega |=Uk$,For $k<N/U$, choose the sign of $m$ that gives upward energy propagation; with $\omega =-Uk$, this is $m>0$.

Horizontal momentum and incompressibility giveThereforeFor upward radiation take $\omega =-Uk$ and $m=\sqrt{N^2/U^2-k^2}>0$, obtaining

If $k>N/U$, then $m=i\alpha$ with $\alpha >0$ chosen so that the disturbance decays as $z\to \infty$. The pressure amplitude is then ninety degrees out of phase with $w$, since $\hat{p}/\hat{w}$ is purely imaginary. Hence

The mountain-wave cutoff follows because the hills force intrinsic frequency $Uk$, whereas internal gravity waves require $|\omega |\le N$. Thus $k_c=N/U$ separates long hills, which radiate vertically propagating waves and carry energy upward, from short hills, whose response is evanescent and carries no mean vertical energy flux.