= Solution
A localized monochromatic cylinder emits four narrow beams, one in each quadrant, forming a St Andrew's cross. If $\alpha$ is the beam angle to the horizontal,
$$
\boxed{\sin\alpha=\omega/N};
$$
equivalently, its angle $\theta=\pi/2-\alpha$ to the vertical obeys $\cos\theta=\omega/N$. As $\omega$ rises from zero to $N$, the beams rotate from horizontal toward vertical. For $\omega>N$ no freely propagating internal wave exists and the response is evanescent.
The wavevector and <phase velocity> are parallel. The dispersion relation is homogeneous of degree zero in $(k,m)$, so Euler's theorem gives $\mathbf k\mathbin{\cdot}\mathbf c_g=0$: the <group velocity> is perpendicular to both the wavevector and phase velocity. Energy travels along the beams in the group-velocity direction.
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