Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 79 / 1

Assume small-amplitude, nonrotating, inviscid Boussinesq approximation motion about a fluid at rest, with stable background stratification. In either uniform layer let $b=-g{\rho }^{\prime }/{\rho }_{\ast }$ be the buoyancy perturbation and $N^2=-(g/{\rho }_{\ast }){\rho }_0^{\prime }(z)>0$. The linear equations areFor a plane internal gravity wave proportional to $e^{i(kx+mz-\omega t)}$, continuity gives $u=-mw/k$, and horizontal momentum gives $p={\rho }_{\ast }\omega u/k$. Substituting these into vertical momentum and buoyancy evolution givesChoose $k>0$, $\omega >0$. Its group velocity isThus downward energy propagation has $m>0$. If $\theta$ is the acute angle between the ray and the vertical in its direction of propagation, then $\tan \theta =|m|/k$, soThe phase velocity is parallel to $(k,m)$, while ${\mathbf{c}}_g\cdot (k,m)=0$: phase and group velocities are perpendicular, and their vertical components have opposite signs. Propagating incident waves require $0<\omega <N_1$; both layers support propagating waves only when $0<\omega <\min (N_1,N_2)$. At or above $N_2$, the lower layer cannot carry a propagating downward internal wave.

For the amplitude convention used below, $\eta$ multiplies the unit particle-displacement vector along the ray, with positive horizontal component. That vector is $(\sin \theta ,-\cos \theta )$ for a downward wave and $(\sin \theta ,+\cos \theta )$ for an upward wave. Stating this polarization convention is important when assigning a sign to a reflection coefficient.

Time and horizontal translational invariance at the interface conserve $\omega$ and $k$. For every component, horizontal momentum givesThe background density has the same value on both sides, so pressure continuity implies continuity of horizontal velocity. Normal velocity is also continuous: no fluid accumulates at the interface.

Use the particle-displacement amplitudes and polarization directions defined in part (a). For downward incident and transmitted waves and an upward reflected wave, continuity of $u$ and $w$ givesSolving the two equations proves the internal-wave transmission across a buoyancy-frequency jump formulaThe reflected wave has the same frequency and wavelength as the incident wave and opposite vertical wavenumber. The transmitted wave hasIts group velocity is downward at angle ${\theta }_2$; the formula in part (a), with $N_2,m_t$, gives its speed.

If instead one uses signed vertical-displacement amplitudes, the reflected coefficient has the opposite sign: $(m_1-m_2)/(m_1+m_2)$. This follows because the reflected polarization's vertical component changes sign, and is not a difference in the physical reflected field. As an energy check, the reflected fraction is $[(m_1-m_2)/(m_1+m_2){]}^2$ and the transmitted fraction is $4m_1{m}_2/(m_1+m_2{)}^2$; they sum to one.

When $N_2<\omega <N_1$, take $m_t=-i\kappa$, $\kappa =k\sqrt{1-N_2^2/{\omega }^2}$, so $e^{i{m}_{t}z}$ decays as $z\to -\infty$. The transmitted field is evanescent, carries no mean vertical energy flux, and the reflected wave has unit amplitude magnitude. A real lower-layer ray angle and wavelength are then inapplicable.

The rigid bottom reflects the lower-layer wave. Its repeated returns to the interface create an internal-wave cavity response containing both upward and downward components in $-H<z<0$, while the upper layer contains the imposed incident wave and one net reflected wave.

Let the lower downward and upward particle-displacement coefficients be $D$ and $U$, multiplying $e^{i{m}_{2}z}$ and $e^{-i{m}_{2}z}$ respectively. Their vertical displacement has opposite polarization signs:Zero normal velocity at $z=-H$ gives $U=Dq$, where $q=e^{-2i{m}_{2}H}$. Interface continuity consequently becomesFor the specified angles, $\sin {\theta }_2/\sin {\theta }_1=1/\sqrt{3}$ and $\cos {\theta }_2/\cos {\theta }_1=\sqrt{3}$. ThusThe dependence on bottom depth is thereforeIn particular,Here $k=\pi /{\lambda }_i$ and $m_2=k/\sqrt{3}$, so the positive depths giving this maximum areThe lower vertical-displacement standing-wave envelope is $2\cos {\theta }_2|D||\sin [m_2(z+H)]|$, while its horizontal-displacement envelope has the complementary cosine form. These distinguish the total standing disturbance from the amplitude of either traveling component. The traveling amplitude has minimum $|{\eta }_i|/\sqrt{3}$ at $m_{2}H=(n+1/2)\pi$.

Since $|q|=1$, $|-1+2q|=|2-q|$. Hence the net upward-propagating wave in the upper layer has amplitude magnitude $|{\eta }_i|$ for every $H$, with the phase given above. At an enhancement maximum, $q=1$ and ${\eta }_r={\eta }_i$. No energy escapes through the bottom, so complete eventual reflection is consistent with energy conservation.