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Choose $y$ northward and $z$ upward in the Northern Hemisphere. In a locally linear Boussinesq approximation, write the background buoyancy as $\bar{b}=b_{\ast }+By+N^{2}z$, with horizontal buoyancy gradient $B<0$ and positive squared buoyancy frequency $N^2$. The stable density stratification resists a purely vertical displacement, but the isopycnals slope upward poleward:The geopotential surfaces are horizontal in this local approximation. Their inclination differs from that of the isopycnals, which is the relevant baroclinicity.

Exchange two equal small fluid elements, each of volume $V$, between positions separated by $(\delta y,\delta z)$. The adiabatic parcel exchange has no thermal diffusion, so each parcel retains its buoyancy. The background buoyancy difference is $\Delta \bar{b}=B\delta y+N^2\delta z$. Because the variable-density contribution to gravitational potential energy is $-{\rho }_{0}Vbz$, the change in the pair's gravitational potential energy isFor a poleward-rising exchange, set $\delta z=s\delta y$ with $s>0$. ThenThe trajectory lies between a geopotential surface and an isopycnal. The poleward-rising parcel starts more buoyant than the parcel it replaces, and the equatorward-sinking parcel starts less buoyant: the buoyancy does positive work on both. This releases available potential energy into kinetic energy, allowing eddies to amplify if the full rotating dynamics supports the displacement and dissipation does not remove the gain.

A trajectory steeper than the isopycnals, $s>s_{\rho }$, instead costs gravitational potential energy; an equatorward-rising trajectory also costs it. Motion exactly along a geopotential surface or an isopycnal gives zero change in this ideal exchange. At fixed $\delta y$, the largest release occurs at $s=s_{\rho }/2$, since $N^{2}s(s-s_{\rho })$ is minimized there.

This is an energetic possibility argument for baroclinic instability, not a sufficient normal mode instability criterion. In particular, unrestricted parcel motion must not be assumed in a rotating fluid: conservation of angular momentum can impose an additional restoring effect. Symmetric instability has a separate Ertel potential vorticity criterion; ordinary Eady instability can occur without violating that criterion.

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.