 Radiation sign for an inertial wave

For fixed vertical wavenumber $k_z$, the inertial-wave dispersion relation is ${\omega }^2=4{\Omega }^2{k}_z^2/(k_x^2+k_z^2)$. Its horizontal group velocity is the derivative displayed above, opposite in sign to $\omega k_x$. In a half-channel $x>0$, an outgoing propagating mode therefore requires $\omega k_x<0$. With time factor $e^{-i\omega t}$ and $q=|k_z|\sqrt{4{\Omega }^2/{\omega }^2-1}$, choose $e^{-i\mathrm{sgn}(\omega )qx}$. Above the inertial-frequency cutoff, the horizontal wavenumber is imaginary and the appropriate radiation condition instead selects the exponentially decaying solution. Energy transport, rather than the direction of phase travel alone, determines the outgoing sign.