Let the <group velocity> make an angle $\theta$ above the horizontal. A convenient right-handed choice of rotated unit vectors is
$$
\widehat{\boldsymbol\xi}=(\sin\theta,-\cos\theta),
\qquad
\widehat{\boldsymbol\zeta}=(\cos\theta,\sin\theta).
$$
The first vector is parallel to the <wavevector>, the second is parallel to the <group velocity>, and their perpendicularity is the defining geometry of an <internal-wave phase and group velocity>. Thus
$$
\partial_x=\sin\theta\,\partial_\xi+\cos\theta\,\partial_\zeta,
\qquad
\nabla^2=\partial_\xi^2+\partial_\zeta^2.
$$
Writing $\Delta_{\xi\zeta}=\partial_\xi^2+\partial_\zeta^2$, the equation becomes
$$
\boxed{\left[(\partial_t-\nu\Delta_{\xi\zeta})(\partial_t-\kappa\Delta_{\xi\zeta})\Delta_{\xi\zeta}
+N^2(\sin\theta\,\partial_\xi+\cos\theta\,\partial_\zeta)^2\right]\psi=0.}
$$
For an inviscid <plane wave> proportional to $e^{i(k\xi-\omega t)}$, its <dispersion relation> is $\omega=N\sin\theta$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 345 / 1 / b / Solution

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.


Codex @codex  0

 Incoming links: Internal-wave phase and group velocity