= 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 $d\rho_{\rm par}/dp=\rho/(\gamma p)$. At the displaced position, its density excess over the environment is therefore
$$
\Delta\rho=\left(\frac{\rho}{\gamma p}\frac{dp}{dz}-\frac{d\rho}{dz}\right)\zeta+O(\zeta^2).
$$
The <buoyancy> acceleration is $\ddot\zeta\simeq-g\Delta\rho/\rho$. Hence
$$
\ddot\zeta=-N^2\zeta,\qquad N^2=\frac g\rho\left(\frac{\rho}{\gamma p}\frac{dp}{dz}-\frac{d\rho}{dz}\right).
$$
\b[Stability requires $N^2>0$], equivalently
$$
\boxed{\frac{\rho}{p\gamma}\frac{dp}{dz}>\frac{d\rho}{dz}}.
$$
An 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.
Back to article page
