Internal-wave phase and group velocity

For a uniformly stratified fluid, internal-wave phase velocity is parallel to the wavevector while group velocity is perpendicular to it. Energy from a localized monochromatic source propagates along four beams forming a St Andrew's cross.

Choose $k,m>0$, $K=(k^2+m^2{)}^{1/2}$ and the positive angular frequency $\omega =Nk/K$. Then the phase velocity is $Nk(k,m)/K^3$ and the group velocity is $N(m^2,-km)/K^3$. Their dot product vanishes and their sum is horizontal with magnitude $N/K$. Draw the group velocity from the tip of the phase velocity: the two form the perpendicular sides of a right triangle. The horizontal hypotenuse is their sum, not the horizontal phase velocity $\omega /k$.

For a propagating, non-cutoff two-dimensional internal gravity wave proportional to $e^{i(kx+mz-\omega t)}$, incompressibility gives $U=-mW/k$. Thus the oscillatory velocity field is perpendicular to the wavevector and parallel to the group velocity, whereas the phase velocity is parallel to the wavevector.

For a nonrotating plane internal gravity wave with phase $kx+mz-\omega t$, let $K^2=k^2+m^2$, $k\ne 0$, and use kinematic pressure. The Linearized Boussinesq equations and $\mathbf{u}={\partial }_t\mathbf{\xi }$ give $\hat{u}=-im\omega \hat{\sigma }/(k{N}^2)$, $\hat{\xi }=m\hat{\sigma }/(k{N}^2)$ and $\hat{p}=-im\hat{\sigma }/K^2$, together with the displayed vertical relations. Buoyancy and displacement field are in phase or opposite phase; velocity and pressure are in quadrature with them. Incompressibility makes both the oscillating velocity and displacement field perpendicular to the wavevector.

For a two-dimensional internal gravity wave with phase $kx+\int m(z)dz-\omega t$, a constant-phase line at fixed time has slope $dz/dx=-k/m(z)$. Its normal spacing from the next crest is $2\pi /(k^2+m^2{)}^{1/2}$; the wavevector $(k,m)$ is normal to the line. The intrinsic energy ray is tangent to these phase lines, while mean-flow advection changes the laboratory ray direction.