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Assume a nonrotating, nonmagnetic equilibrium, inviscid adiabatic perturbations, and no dissipation. Let $g={\varphi }^{\prime }(R)>0$, so hydrostatic equilibrium is $p^{\prime }=-\rho g$. Use $\mathbf{\xi }{e}^{i\omega t}$ for the fluid displacement. To avoid a source ambiguity, write Eulerian fluid perturbations as $p_1,{\rho }_1,{\varphi }_1$, and use $\Delta$ for quantities following the displaced material element, as in Eulerian and Lagrangian fluid perturbations. They are related by $\Delta p=p_1+{\xi }_R{p}^{\prime }$ and similarly for mass density and potential.

The printed identity needs two corrections: its compression denominator must be $\gamma p$, not $\gamma \rho$, and the pressure and potential appearing in this form of the identity must be Eulerian fluid perturbations. The printed denominator has the wrong dimensions for an energy density. The claimed Lagrangian interpretation does not produce the displayed buoyancy-compression split; the following derivation specifies the consistent version.

Linearized mass conservation, the adiabatic relation, and the Poisson equation giveConsequently $p_1=-\gamma p\nabla \cdot \mathbf{\xi }-{\xi }_R{p}^{\prime }$. Linearizing the Euler equations for an inviscid fluid about the static star gives the linear adiabatic stellar oscillation equationsRequire regularity at the center, a vanishing Lagrangian surface pressure, and the usual matching to a decaying exterior gravitational potential. Because $p$ and $\rho$ vanish at the surface, the pressure boundary term vanishes for regular displacements; compactly supported trials also suffice below. The perturbed Newtonian gravitational potential solves Laplace's equation outside the star and must be included there when its field energy is computed.

Multiply the momentum equation by $-{\mathbf{\xi }}^{\ast }$ and integrate. The pressure term is $-{\int }_V{p}_1\nabla \cdot {\mathbf{\xi }}^{\ast }dV$. Let $d=\nabla \cdot \mathbf{\xi }$. The adiabatic relation and hydrostatic equilibrium implyThe pressure and equilibrium-gravity terms then combine pointwise aswhere $N^2=g[(1/\gamma )p^{\prime }/p-{\rho }^{\prime }/\rho ]$ is the stellar buoyancy frequency. The potentially troublesome cross terms cancel exactly.

For the perturbed self-gravity term, integration by parts givesThus the corrected stellar displacement energy identity isThe right side is a real quadratic form. With these conservative boundary conditions, the force operator is a self-adjoint operator. Its Rayleigh quotient characterizes squared mode frequencies, and a negative admissible trial quotient proves an unstable spectral component; it does not require the trial displacement itself to be an exact eigenfunction.

For a generic compressive disturbance of scale $\ell$ in a region with slowly varying equilibrium coefficients, estimate ${\rho }_1\sim \rho \xi /\ell$, $p_1\sim \gamma p\xi /\ell$ and ${\varphi }_1\sim 4\pi G{\rho }_1{\ell }^2$. The self-gravity and compression energy densities have orderswith adiabatic sound speed $c_s^2=\gamma p/\rho$. Under the plausible stellar estimate $c_s^2\sim G\bar{\rho }{a}^2$, global modes can have comparable gravity and pressure contributions, while their ratio for small-scale compressive modes is of order $(\rho /\bar{\rho })(\ell /a{)}^2$. Thus perturbations with $\ell \ll c_s/\sqrt{4\pi G\rho }$ admit the Cowling approximation. This estimate compares two nonzero compressive terms; it cannot alone justify neglecting gravity for a trial deliberately chosen to have $p_1=0$.

For an explicit convective trial, choose $c<d<e<b$ and a nonzero smooth bumpLet $Y$ be a normalized spherical harmonic of degree $\ell \ge 1$, let $L=\ell (\ell +1)$, and let ${\nabla }_{\Omega }$ be the unit-sphere gradient. The pressure-free localized stellar convective trial isIt is supported away from the center and surface, so it satisfies every needed mechanical boundary condition. Since ${\Delta }_{\Omega }Y=-LY$,Therefore $p_1=-\gamma p\nabla \cdot \mathbf{\xi }+\rho g{\xi }_R=0$ exactly. Its buoyancy energy is ${\int }_d^e\rho N^2{R}^2|f{|}^{2}dR<0$, whereas its kinetic norm ${\int }_d^e\rho R^2(|f{|}^2+L|h{|}^2)dR$ is positive. In the Cowling approximation this gives a negative Rayleigh quotient, and hence $N^2<0$ in an interior shell implies convective instability. Retaining perturbed self-gravity only adds a nonpositive term, so the same trial proves instability in the full conservative problem too.

There is also a quantitative small-scale check for this very trial. Its mass density amplitude is ${\rho }_1=(\rho N^2/g)fY$, independent of $\ell$. Write ${\rho }_1=s(R)Y$ and ${\varphi }_1=F(R)Y$. The radial Poisson equation givesBy the Cauchy-Schwarz inequality, $A\le 4\pi G(\int R^4|s{|}^{2}dR{)}^{1/2}(A/L{)}^{1/2}$, and so the angular-degree bound on perturbed stellar self-gravity isIt tends to zero with horizontal scale $R/\ell$, while the strictly negative buoyancy integral is fixed. This justifies the small-scale neglect of gravity even though the trial's compression term vanishes.