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Let the thermal-wind basic velocity beso that $f{U}_z=b_x=M^2$. Assume constant $f,M,N$, hydrostatic perturbations, $y$ independence, and normal modes proportional to $e^{i(kx+mz-\omega t)}$. The linearized equations areIncompressibility gives $w=-(k/m)u$. Eliminating $v,p,b^{\prime }$ yieldsThis quadratic in $k/m$ is positive for every orientation precisely whenOtherwise some disturbances grow monotonically through symmetric instability.

For the stable case, the minimum occurs atConstant-phase lines have slope $dz/dx=-k/m=-M^2/N^2$, exactly the slope of the basic isopycnals $M^{2}x+N^{2}z=\text{constant}$. The minimum-frequency displacement follows an isopycnal, allowing buoyancy and Coriolis restoring forces to oppose one another. Its frequency is below the inertial frequency $|f|$ of an unstratified rotating fluid.

If $\theta$ is the counter-clockwise angle of the wavevector from the horizontal, $k/m=\cot \theta$, andFor ${\omega }^2>f^2$, define $D=[M^4+N^2({\omega }^2-f^2){]}^{1/2}>M^2$. Taking $k>0$, the incident wave with group velocity down and right hasA horizontal reflection preserves $k$ and $\omega$ but selects the other vertical wavenumber, soThe incident and reflected phase lines have slopes $-(M^2+D)/N^2$ and $(D-M^2)/N^2$, respectively. Their group velocities are perpendicular to these phase lines: the incident ray points down-right and the reflected ray up-right. The isopycnal slope $-M^2/N^2$ lies between the two phase-line orientations.