Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 74 / 1 / a

Use the inviscid, nonrotating Boussinesq approximation, with a stable background mass density $\bar{\rho }(z)$ and reference density ${\rho }_{\ast }$. Write $B=-g{\rho }^{\prime }/{\rho }_{\ast }$ for the perturbation buoyancy and $\pi =p^{\prime }/{\rho }_{\ast }$ for the kinematic pressure. The buoyancy frequency is $N_0^2=-(g/{\rho }_{\ast }){\bar{\rho }}_z>0$. Dropping products of perturbations in the Boussinesq equations gives the Linearized Boussinesq equationsThe last equation expresses incompressible flow; the second follows by advecting the background mass density gradient. Neglecting rotation and viscosity is part of this internal gravity wave model.

For a plane internal gravity wave proportional to $\exp [i({\mathbf{k}}_h\cdot {\mathbf{x}}_h+mz-\omega t)]$, put $\kappa =|{\mathbf{k}}_h|>0$ and $K^2={\kappa }^2+m^2$. Eliminating the horizontal velocity, pressure and buoyancy from the linear equations givesFor example, the horizontal momentum and continuity equations give $\pi =-\omega mw/{\kappa }^2$; substituting $B=-i{N}_0^{2}w/\omega$ into vertical momentum yields the displayed dispersion relation. Thus an IGW has $0<|\omega |\le N_0$ in this model.

On the positive-frequency branch, the phase velocity normal to a constant-phase plane and the group velocity areConsequently $({\mathbf{k}}_h,m)\cdot {\mathbf{c}}_g=0$, and phase and group velocity are perpendicular. Equivalently, the dispersion relation is homogeneous of degree zero in the wave vector, so differentiating with respect to its scale proves the same orthogonality. Energy travels with the group velocity, along the phase planes. At the degenerate limit $m=0$, $\omega =N_0$ and the group velocity vanishes; the orthogonality statement then has this limiting interpretation.