= Solution
Let primes denote radial derivatives of the equilibrium and put $\Delta u=u'+2u/r$. The linearized continuity, adiabatic pressure and <Poisson equation> are
$$
\partial_t\delta\rho=-\frac1{r^2}\partial_r(r^2\rho u)=-\rho\Delta u-\rho'u,
\qquad \partial_t\delta p=-u p'-\gamma p\Delta u,
\qquad \partial_r(r^2\partial_r\delta\Phi)=4\pi G r^2\delta\rho.
$$
Take the time derivative of the <Poisson equation> and substitute continuity. Integration in radius gives
$$
r^2\partial_r\partial_t\delta\Phi=-4\pi G r^2\rho u+C(t).
$$
Centre regularity excludes a changing point mass at the origin, so $C(t)=0$ and
$$
\boxed{\partial_r\partial_t\delta\Phi=-4\pi G\rho u.}
$$
The perturbed <self-gravity> must be retained; this calculation does not make the <Cowling approximation>.
The radial linear momentum equation is $\rho\partial_tu=-\partial_r\delta p-\delta\rho\Phi'-\rho\partial_r\delta\Phi$. Differentiate it in time and use the preceding equations:
$$
\rho\partial_t^2u=\partial_r(\gamma p\Delta u)+\partial_r(up')+(\rho\Delta u+\rho'u)\Phi'+4\pi G\rho^2u.
$$
The equilibrium has $p'=-\rho\Phi'$ and $\Phi''+2\Phi'/r=4\pi G\rho$. The remaining terms simplify as
$$
\partial_r(up')+(\rho\Delta u+\rho'u)\Phi'+4\pi G\rho^2u
=-\frac{2up'}r+\rho u(4\pi G\rho-\Phi'')=-\frac{4up'}r.
$$
Consequently the radial <stellar oscillation> equation is
$$
\boxed{\rho\partial_t^2u=\partial_r(\gamma p\Delta u)-\frac{4u}r p'.}
$$
The derivative acts on the full variable coefficient $\gamma(r)p(r)$; discarding $\gamma'$ would change the result.
For $u=r\xi(r)e^{-i\omega t}$, $\Delta u=(3\xi+r\xi')e^{-i\omega t}$. Substitute and multiply by $-r^3$. Combining the two terms proportional to $\xi'$ into a total derivative gives the <radial stellar pulsation equation>,
$$
\boxed{\omega^2\rho r^4\xi=-\frac{d}{dr}\left(\gamma p r^4\xi'\right)+r^3\frac{d}{dr}\bigl((4-3\gamma)p\bigr)\xi.}
$$
Set $P=\gamma p r^4$, $W=\rho r^4$ and $Q=r^3[(4-3\gamma)p]'$. The <Sturm-Liouville operator> is
$$
L\xi=\frac1W\bigl[-(P\xi')'+Q\xi\bigr],\qquad
\langle\eta,\xi\rangle_W=\int_0^{R_s}W\eta^*\xi\,dr.
$$
For regular physical perturbations, $\xi$ is finite at the centre and $P\xi'\to0$ at the free surface. For a nonzero-frequency mode the <fluid displacement> is $u/(-i\omega)$, so vanishing <Lagrangian pressure perturbation> is equivalent to $\gamma p(3\xi+r\xi')\to0$. With $p(R_s)=0$ and finite $\xi$, this gives $P\xi'\to0$. 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 gives
$$
\langle\eta,L\xi\rangle_W-\langle L\eta,\xi\rangle_W
=\left[P(\eta'^*\xi-\eta^*\xi')\right]_0^{R_s}=0.
$$
Thus 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>,
$$
\boxed{\omega_0^2=\inf_{\xi\ne0}\frac{\displaystyle\int_0^{R_s}\left[\gamma p r^4|\xi'|^2+r^3\bigl((4-3\gamma)p\bigr)'|\xi|^2\right]dr}{\displaystyle\int_0^{R_s}\rho r^4|\xi|^2dr}.}
$$
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, and
$$
\int_0^{R_s}r^3[(4-3\gamma)p]'dr
=\left[r^3(4-3\gamma)p\right]_0^{R_s}-3\int_0^{R_s}r^2p(4-3\gamma)dr
=-3\int_0^{R_s}r^2p(4-3\gamma)dr.
$$
The denominator $\int\rho r^4dr$ is positive. Hence
$$
\boxed{\int_0^{R_s}r^2p(4-3\gamma)dr>0\ \Longrightarrow\ \omega_0^2<0\ \Longrightarrow\ \text{radial instability}.}
$$
This <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 $\gamma>4/3$, $Q=(4-3\gamma)r^3p'>0$ in a normally stratified <hydrostatic equilibrium>, so the quotient is positive and the star is radially stable.
Back to article page
