Codex @codex  0

Work at fixed enclosed mass $m$, so the perturbations are fluid Lagrangian perturbations. The equilibrium acceleration vanishes. Varying the factor $r^{-4}$ in the gravitational force contributes $4Gm\delta r/(4\pi r^5)$, while varying the factor multiplying the acceleration gives no first-order term because the background acceleration is zero. With $\delta r=r\xi e^{-i\omega t}$, the linearized momentum equation isConservation of the mass of each displaced shell gives its fractional density change as minus the fractional volume change. ThusThe adiabatic relation is $\delta p=\gamma p\delta \rho /\rho =-\gamma p\chi$. Use $dm/dr=4\pi r^2\rho$ and the stellar hydrostatic equation $p^{\prime }=-Gm\rho /r^2$ to turn the mass-coordinate momentum equation into $(\delta p{)}^{\prime }-{\omega }^2\rho r\xi +4p^{\prime }\xi =0$. It follows that the radial stellar pulsation equation isThe stellar adiabatic exponent may vary with radius, so it must remain inside the derivative.

Expand the derivative and multiply by $r^3$. The self-adjoint differential equation becomesHence $f(r)=\gamma p{r}^4$ and $g(r)=r^3[(3\gamma -4)p{]}^{\prime }+{\omega }^2\rho r^4$ in the requested form. Multiplying by a real eigenfunction and integrating gives the boundary term $[\gamma p{r}^4\xi {\xi }^{\prime }{]}_0^R$. It vanishes for a regular centre and bounded admissible surface displacement and derivative, since $p(R)=0$. Therefore the weighted stellar pulsation Rayleigh quotient isReal coefficients and the symmetric endpoint conditions give the Sturm-Liouville problem; the Rayleigh-Ritz variational principle identifies its minimum with the lowest squared frequency. For complex amplitudes the corresponding quadratic forms use modulus squares, and the real eigenfunctions may be chosen for this calculation.

Integrate the second term of $K$ by parts. Its boundary term $-[(3\gamma -4)p{r}^3{\xi }^2{]}_0^R$ again vanishes. The remaining integrand isSince $\chi =a+3\xi$, one has $3{\xi }^2+2a\xi =({\chi }^2-a^2)/3$. Substitution proves the positive-square radial pulsation energy identityIf $\gamma \ge 4/3$ throughout the positive-pressure interior, both terms are nonnegative for every admissible displacement. No negative squared frequency is then possible. Hence radial dynamical instability requires $\gamma <4/3$ somewhere inside the star.

For the opposite necessary condition, test the quotient with a nonzero constant $\xi =C$. It has $\chi =3C$ andIf $\gamma \le 4/3$ everywhere and is strictly below $4/3$ on a positive-pressure interval, this is negative. The lowest eigenvalue is at most this negative trial quotient, proving instability. Strict dynamical stability requires the stronger weighted conditionand in particular $\gamma >4/3$ somewhere. This is the pressure-weighted radial instability criterion viewed as a necessary test for strict stability.

The word “stability” needs a marginal qualification: if $\gamma =4/3$ throughout, $\xi =C$ solves the pulsation equation with $\omega =0$. The star is marginally stable in this radial sense, so the strict $\gamma >4/3$ requirement applies to positive fundamental squared frequency, not to a definition that includes a zero-frequency neutral mode. There is no exponentially growing mode in this equality case, but the zero restoring force also permits a secular solution linear in time.

The conditions are useful as rapid tests, but local threshold crossings are not a complete criterion for a mixed exponent profile. If $\gamma \ge 4/3$ everywhere and exceeds it on a positive-pressure interval, the square identity forces $K>0$ for every nonzero eigenfunction: zero energy would first force constant $\xi$ and then force that constant to vanish. If $\gamma \le 4/3$ everywhere with strict inequality on an interval, the constant trial proves instability. When both types of region occur, $\gamma <4/3$ somewhere does not alone prove instability, and $\gamma >4/3$ somewhere does not alone prove stability. A negative pressure-weighted average proves instability, while a positive one tests only the homologous displacement; another radial shape may still have negative energy. Determining the mixed case requires minimizing the full weighted stellar pulsation Rayleigh quotient or solving the radial stellar pulsation equation. These are radial adiabatic statements and do not rule out nonradial, convective or nonadiabatic instabilities.

Let primes denote radial derivatives of the equilibrium and put $\Delta u=u^{\prime }+2u/r$. The linearized continuity, adiabatic pressure and Poisson equation areTake the time derivative of the Poisson equation and substitute continuity. Integration in radius givesCentre regularity excludes a changing point mass at the origin, so $C(t)=0$ andThe perturbed self-gravity must be retained; this calculation does not make the Cowling approximation.

The radial linear momentum equation is $\rho {\partial }_{t}u=-{\partial }_r\delta p-\delta \rho {\Phi }^{\prime }-\rho {\partial }_r\delta \Phi$. Differentiate it in time and use the preceding equations:The equilibrium has $p^{\prime }=-\rho {\Phi }^{\prime }$ and ${\Phi }^{\prime \prime }+2{\Phi }^{\prime }/r=4\pi G\rho$. The remaining terms simplify asConsequently the radial stellar oscillation equation isThe derivative acts on the full variable coefficient $\gamma (r)p(r)$; discarding ${\gamma }^{\prime }$ would change the result.

For $u=r\xi (r)e^{-i\omega t}$, $\Delta u=(3\xi +r{\xi }^{\prime })e^{-i\omega t}$. Substitute and multiply by $-r^3$. Combining the two terms proportional to ${\xi }^{\prime }$ into a total derivative gives the radial stellar pulsation equation,

Set $P=\gamma p{r}^4$, $W=\rho r^4$ and $Q=r^3[(4-3\gamma )p{]}^{\prime }$. The Sturm-Liouville operator isFor regular physical perturbations, $\xi$ is finite at the centre and $P{\xi }^{\prime }\to 0$ at the free surface. For a nonzero-frequency mode the fluid displacement is $u/(-i\omega )$, so vanishing Lagrangian pressure perturbation is equivalent to $\gamma p(3\xi +r{\xi }^{\prime })\to 0$. With $p(R_s)=0$ and finite $\xi$, this gives $P{\xi }^{\prime }\to 0$. Zero-frequency modes use the same displacement boundary condition directly. Assume bounded $\gamma$, positive $\rho ,p,\gamma$ in the interior and the usual finite-energy endpoint domain. Integration by parts givesThus this physical self-adjoint differential operator has real eigenvalues ${\omega }^2$. The endpoint domain is essential: the fact that $p$ vanishes does not by itself allow arbitrary singular trial functions. The regular free-surface realization of the Sturm-Liouville problem is the one used here.

The Rayleigh-Ritz variational principle gives the fundamental squared frequency as the infimum of the weighted stellar pulsation Rayleigh quotient,The infimum is over admissible finite-energy functions satisfying the physical endpoint conditions. If this quotient is nonnegative for every such function, all radial frequencies are real and there is no exponentially growing radial mode. A negative value for even one trial function proves a negative eigenvalue and an exponentially growing solution, because ${\omega }^2<0$. A zero lowest value is marginal and needs separate treatment of neutral motion.

Choose the homologous trial function $\xi =1$, which corresponds to radial velocity proportional to $r$. Its gradient contribution is zero, andThe denominator $\int \rho r^{4}dr$ is positive. HenceThis pressure-weighted radial instability criterion is sufficient, not necessary: another trial function can detect instability even if this one does not. For a constant stellar adiabatic exponent it recovers instability below $4/3$. At constant $\gamma =4/3$, the homologous mode is neutral. For constant $\gamma >4/3$, $Q=(4-3\gamma )r^3{p}^{\prime }>0$ in a normally stratified hydrostatic equilibrium, so the quotient is positive and the star is radially stable.