 Solution

Let $z$ increase upwards and let $g>0$ denote the downward gravitational acceleration. The environment obeys hydrostatic equilibrium, $dp/dz=-\rho g$. Displace a small parcel by $\zeta$ while it remains in pressure balance with its surroundings and exchanges negligible heat. For fixed composition and adiabatic index $\gamma$, its density change at constant specific entropy satisfies $d{\rho }_{\mathrm{par}}/dp=\rho /(\gamma p)$. At the displaced position, its density excess over the environment is thereforeThe buoyancy acceleration is $\ddot{\zeta }\simeq -g\Delta \rho /\rho$. HenceStability requires $N^2>0$, equivalentlyAn upward-displaced parcel is then denser than its surroundings and falls back; a downward-displaced parcel is lighter and rises. Equality gives neutral stability in this parcel model. For a perfect gas with constant composition, $ds=c_{v}d\ln (p/{\rho }^{\gamma })$ gives $N^2=(g/c_p)ds/dz$, so a rising specific entropy is stable. This is the parcel form of the Schwarzschild criterion. Rapid pressure equilibration, negligible heat diffusion during displacement and no composition-driven buoyancy are the assumptions behind this test.