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Let $m(r)$ be the background enclosed mass, so $g=Gm/r^2$ and $m^{\prime }=4\pi r^2\rho$. The background hydrostatic equilibrium is $p^{\prime }=-\rho g$. Use a Lagrangian fluid displacement $\xi ={\xi }_r$ of each spherical mass shell; $\Delta$ denotes its Lagrangian perturbation. Conservation of shell mass givesand the adiabatic pressure law gives the Lagrangian pressure perturbationThis retains perturbed self-gravity: a shell keeps its enclosed mass while its gravitational acceleration changes with its radius.

In mass coordinates, the exact radial shell equation is $\ddot{r}=-4\pi r^2\partial p/\partial m-Gm/r^2$. Its first-order perturbation isHydrostatic equilibrium gives ${\partial }_{m}p=-g/(4\pi r^2)$, and ${\partial }_m\Delta p=(\Delta p{)}^{\prime }/(4\pi r^2\rho )$. Thus the perturbation of the pressure-force geometry contributes another $2g\xi /r$, in addition to the gravity term:Put $\eta =\xi /r$, so ${\xi }^{\prime }+2\xi /r=r{\eta }^{\prime }+3\eta$. Expanding the derivative, using constant $\gamma$ and $p^{\prime }=-\rho g$, yields the radial stellar pulsation equation:The coefficient $4$ includes both geometric pressure and self-gravity effects; dropping the gravitational perturbation would not give this equation.

For a normal mode $\xi (r,t)=\xi (r)e^{-i\omega t}$, multiply by $\rho r^2{\xi }^{\ast }$ and integrate. Since ${\xi }^{\ast }/r={\eta }^{\ast }$, integration by parts givesRegularity at the center requires $\xi =O(r)$ with finite $\eta$, so the lower endpoint vanishes. For an isolated star take $p(R)=0$ and the free-surface condition $\Delta p=0$, with a regular finite-energy displacement; these eliminate the upper boundary term. HenceThese boundary conditions give the self-adjoint radial Sturm-Liouville problem behind the weighted stellar pulsation Rayleigh quotient.

Use the homologous trial fluid displacement ${\xi }_r=r$, or $\eta =1$. It is regular at the center, and $\Delta p=-3\gamma p$ vanishes at the free surface. The derivative term in the Rayleigh quotient is zero, leavingThe ratio is a positive weighted average of $g/r$. Since the central mass density is its maximum,For $\gamma >4/3$ multiplication preserves this inequality, proving the central-density upper bound for a radial stellar frequency:The full positive quadratic form also shows radial stability for constant $\gamma >4/3$ under these assumptions. A uniform-density equilibrium has constant $g/r$, making the homologous displacement an exact mode and saturating the bound.

If $\gamma <4/3$, the same trial value is strictly negative. Therefore ${\omega }_{\min }^2<0$: the star has an exponentially growing radial instability, with growth rate $\sqrt{-{\omega }_{\min }^2}$. One must use this negative trial quotient, rather than carrying the central-density inequality across the sign change unchanged. At the limiting value $\gamma =4/3$, the homologous displacement has zero restoring force and is a neutral radial mode in this constant-exponent model.