Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 345 / 1

With buoyancy perturbation $b=-g{\rho }^{\prime }/{\rho }_0$, the inviscid Boussinesq equations areForincompressibility is automatic. Define the Jacobian determinantThe material derivative is $D/Dt={\partial }_t+J(\psi ,\cdot )$. Taking the $y$ component of the curl of momentum gives the exact finite-amplitude system

If the disturbance amplitude is small enough that each Jacobian is asymptotically smaller than its corresponding time derivative, the system can be linearized. Eliminating $b$ then gives

For a single phase $\varphi =kx+mz-\omega t$, every disturbance field is a function of $\varphi$ alone. Its velocity is tangent to the phase surfaces:Consequently both nonlinear Jacobians vanish exactly, irrespective of amplitude. Substitution givesA nonzero amplitude therefore requires the internal gravity wave dispersion relationThe wave is an exact nonlinear solution of the inviscid Boussinesq equations. The parameter $ϵ$ merely sets its amplitude; it does not change its frequency or waveform. Very large $ϵ$ can nevertheless invalidate the Boussinesq model or make the total stratification locally overturn.

At $O(ϵ\gamma )$, the slowly varying wave envelope obeyswhere the group velocity obtained from $\omega =Nk/(k^2+m^2{)}^{1/2}$ isIt satisfies $\mathbf{k}\cdot {\mathbf{c}}_g=0$. If initially $\mathbf{k}\cdot \hat{\nabla }\tilde{\psi }=0$, the envelope structure varies across the wavevector and translates unchanged in the direction ${\mathbf{c}}_g$ at speedThus internal-wave crests propagate with phase velocity parallel to $\mathbf{k}$, while the packet and its energy propagate perpendicular to $\mathbf{k}$ with the group velocity.