Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 65 / 1 / b

Use the nonrotating linear Boussinesq approximation with reference density ${\rho }_0$ and $N^2=-(g/{\rho }_0)d\bar{\rho }/dz>0$:Let $K_h^2=k^2+l^2>0$, $K^2=K_h^2+m^2$ and $E=e^{i(kx+ly+mz-\omega t)}$. The horizontal momentum equations give $u^{\prime }=k{p}^{\prime }/({\rho }_0\omega )$ and $v^{\prime }=l{p}^{\prime }/({\rho }_0\omega )$. Continuity requires $k{u}^{\prime }+l{v}^{\prime }+m{w}^{\prime }=0$. Thus the internal-wave polarization isPhysical perturbations are the real parts. The density is in temporal quadrature with the vertical velocity, because buoyancy responds to the vertical fluid displacement. Substitution into the vertical momentum equation determines the internal gravity wave dispersion relation,A nontrivial propagating wave requires a nonzero horizontal wave vector and a nonzero frequency. The formulas are not to be divided by $K_h^2=0$: incompressibility excludes a nonzero oscillatory vertical velocity for a purely vertical wave vector.

For the positive-frequency internal gravity wave branch, $\omega =N{K}_h/K$. Differentiating with respect to the components of the wave vector gives the group velocity,The wavefront-normal phase velocity is ${\mathbf{c}}_p=\omega \mathbf{k}/K^2$. Directly,The negative-frequency branch changes both propagation signs but not their orthogonality. Another proof is that $\omega$ is homogeneous of degree zero in the wave vector, so Euler's homogeneous-function identity gives $\mathbf{k}\cdot {\nabla }_{\mathbf{k}}\omega =0$. Energy propagates along constant-phase surfaces, rather than normal to them. When $m=0$, $\omega =N$ and ${\mathbf{c}}_g=0$; the scalar-product result remains true, but there is no nonzero group-velocity direction to describe geometrically.