= Solution
For the positive-frequency branch and $K^2=k^2+m^2$, the wavefront-normal <phase velocity> and the <group velocity> are
$$
\boxed{\mathbf c_p=\frac{\omega}{K^2}(k,0,m),\qquad
\mathbf c_g=\frac{N^2-f^2}{\omega K^4}(km^2,0,-mk^2).}
$$
Their scalar product vanishes since $k(km^2)+m(-mk^2)=0$. Equivalently, the frequency is homogeneous of degree zero in the <wave vector>, and hence $\mathbf k\cdot\nabla_{\mathbf k}\omega=0$.
The vertical sign relation is
$$
\boxed{c_{pz}c_{gz}=-\frac{(N^2-f^2)k^2m^2}{(k^2+m^2)^3}.}
$$
It is negative for the usual geophysical ordering $N>|f|$ and nonzero $k,m$. Thus vertical phase and energy propagation are opposite under that ordering; negative-frequency waves obey the same product relation. The PDF does not state this ordering, so its unconditional direction assertion needs qualification. For example, $N=1$, $f=2$, $k=m=1$ gives positive vertical components on the positive-frequency branch. If $N=|f|$ the <group velocity> vanishes, and in axial limiting directions one vertical component can be zero. Orthogonality is valid throughout, while strictly opposite nonzero vertical components need the stated nondegeneracy and $N>|f|$.
Back to article page
