 Solution

Let $a_{\xi }=\mathbf{\xi }\cdot \nabla \Psi$, $d_{\xi }=\nabla \cdot \mathbf{\xi }$ and ${\rho }_{\Psi }=d\rho /d\Psi$. Since $\nabla p=-\rho \nabla \Psi$, the pressure and mass density perturbations are $\delta p_{\xi }=\rho a_{\xi }-\gamma p{d}_{\xi }$ and $\delta {\rho }_{\xi }=-\rho d_{\xi }-{\rho }_{\Psi }{a}_{\xi }$. Integrating the pressure-gradient term by parts in the mass density-weighted inner product givesThe boundary term vanishes for regular admissible displacements because $p,\rho$ vanish there and $\delta p=\rho a-\gamma pd$. Completing the pressure square giveswhereThis effective-potential stratification coefficient is real. All coefficients of the bilinear form are real, so $⟨\eta ,\mathcal{F}\xi ⟩=⟨\mathcal{F}\eta ,\xi ⟩$: the operator is symmetric on the stated boundary domain, giving the usual self-adjoint realization of the stellar normal-mode problem.

Set $\eta =\xi$ and use $\mathcal{F}\xi =-{\omega }^2\xi$. The Cowling energy principle for a rotating barotropic star isThe Rayleigh quotient $Q[\xi ]/\int \rho |\xi {|}^2$ tests stability. If $Q\ge 0$ for every admissible displacement, no mode has ${\omega }^2<0$, so there is no exponentially growing mode. If an admissible trial displacement has $Q<0$, the Rayleigh-Ritz variational principle puts negative spectrum below zero; in the usual discrete stellar mode problem this gives a mode with $\omega =i\sigma$, $\sigma >0$. Equality allows neutral modes, rather than establishing strictly positive frequencies. A locally negative coefficient alone is not a complete instability proof: a trial function must also control the pressure and other positive terms.