Radial Pulsations

Radial pulsation theory asks whether a spherical relativistic star returns to hydrostatic equilibrium after an infinitesimal, adiabatic compression. The answer is encoded in a self-adjoint eigenvalue problem: positive squared frequencies describe oscillations, while a negative fundamental eigenvalue describes exponential growth. At the boundary between those cases the fundamental mode is a static zero mode, so its frequency must pass through zero—not merely become small by numerical accident.

Helpful background. Stress–Energy and Conservation supplies the fluid conservation laws; the Einstein equation supplies the stellar background; Relativistic Matter and Compact Objects orients the material models; Black-Hole Perturbation Theory supplies eigenvalue language; Gravitational Radiation and Waveforms clarifies radiative versus spherical modes; and Computational Relativity supplies numerical boundary-value methods.

The radial displacement problem

Let the equilibrium star have

$$ds^2=-e^{2\nu(r)}dt^2+e^{2\lambda(r)}dr^2+r^2d\Omega^2, \qquad e^{-2\lambda}=1-\frac{2Gm(r)}{r},$$

with pressure $p(r)$, total energy density $\varepsilon(r)$, and surface $p(R)=0$. Consider a smooth, isotropic perfect-fluid interior with no internal phase boundary or surface stress. The time coordinate is normalized at infinity. Take a harmonic Lagrangian radial displacement

$$\delta r(t,r)=\xi(r)e^{-i\omega t}.$$

The perturbation is adiabatic on the oscillation timescale. Its material input is therefore the perturbative adiabatic index

$$\Gamma_1=\frac{\varepsilon+p}{p} \left(\frac{\partial p}{\partial\varepsilon}\right)_{s,Y_i},$$

where the derivative holds fixed entropy per particle and any composition variables that are frozen during a cycle. This $\Gamma_1$ need not equal the derivative along a beta-equilibrated background if reactions are slow. Confusing the two changes the restoring force.

Eliminating the metric and pressure perturbations in favor of $\zeta=r^2e^{-\nu}\xi$ gives the Sturm–Liouville form

$$\frac{d}{dr}\left(P\frac{d\zeta}{dr}\right) +\left(Q+\omega^2W\right)\zeta=0,$$

where

$$P=\frac{\Gamma_1p}{r^2}e^{\lambda+3\nu}, \qquad W=\frac{\varepsilon+p}{r^2}e^{3\lambda+\nu}>0,$$

and the remaining background coefficient is

$$Q=\frac{e^{\lambda+3\nu}}{r^2} \left[\frac{p'^2}{\varepsilon+p}-\frac{4p'}{r} -8\pi G e^{2\lambda}p(\varepsilon+p)\right].$$

The pressure-gradient terms include the change in gravitational force during compression; omitting $Q$ would leave a different acoustic problem. Chandrasekhar derives the pulsation equation, surface condition, variational principle, and orthogonality in Chandrasekhar 1964, §IV, Eqs. (59)–(62), p. 424. His metric exponents $\lambda_0,\nu_0$ equal $2\lambda,2\nu$ here. Writing the equation this way exposes the decisive sign: with positive weight $W$, the eigenvalues $\omega_n^2$ are real and ordered.

Center, surface, and orthogonality

Regularity at the center requires a finite physical displacement. Since $\xi\sim r$, the transformed variable obeys $\zeta\sim r^3$. At the material surface the Lagrangian pressure perturbation must vanish,

$$\Delta p(R)=0.$$

In these variables its interior expression is

$$\Delta p=-\frac{\Gamma_1p\,e^\nu}{r^2}\zeta', \qquad P\zeta'=-e^{\lambda+2\nu}\Delta p.$$

This makes the surface condition a condition on the weighted derivative, not merely the unperturbed fact $p(R)=0$. It also displays the boundary term needed for self-adjointness. We require $P\gt0$ in the open interior, finite weighted norm, and endpoint behavior making that term vanish. A sharp phase boundary needs extra matching conditions determined by its conversion rate; the smooth-star problem below cannot simply be integrated through it.

For an ordinary surface where $p\to0$ smoothly, this is the natural endpoint condition of the Sturm–Liouville problem; for a self-bound star with nonzero surface density, it must be imposed with the appropriate surface matching rather than inferred from $p=0$ alone.

Two modes with distinct eigenvalues satisfy

$$\int_0^R W(r)\,\zeta_m(r)\zeta_n(r)\,dr=0 \qquad (m\ne n).$$

For this self-adjoint problem the fundamental eigenfunction has no interior node, and the $n$th overtone has $n$ nodes. This ordering follows from the Sturm–Liouville oscillation theorem with the stated endpoint domain; it does not transfer to general radiating, rotating, or dissipative mode problems. These facts give stringent numerical checks: a purported fundamental mode with a node, a complex eigenvalue in the conservative problem, or loss of weighted orthogonality signals inconsistent boundary conditions or discretization.

Why the fundamental mode reaches zero

Multiplying the pulsation equation by $\zeta$ and integrating by parts yields the variational quotient

$$\omega^2[\zeta] = \frac{\displaystyle \int_0^R\left(P\zeta'^2-Q\zeta^2\right)dr} {\displaystyle \int_0^R W\zeta^2\,dr}.$$

The boundary term vanishes because of regularity and $\Delta p(R)=0$. The lowest eigenvalue is the minimum of this quotient. Along a smooth one-parameter family of equilibria, the coefficients vary continuously; the first transition from a positive to a negative minimum must therefore occur at $\omega_0^2=0$. The corresponding displacement is the tangent to a nearby equilibrium only when the sequence holds the appropriate material invariants fixed.

This is the precise limitation behind the familiar mass-maximum rule. For a cold, nonrotating, barotropic sequence with a single central-density parameter, the first maximum of gravitational mass normally coincides with the first radial zero mode. Rotation, multiple conserved charges, phase transitions, nonbarotropic composition, or disconnected stable branches can break that shortcut. The eigenproblem, not the shape of one mass–radius plot, is the independent stability test.

Dimensions provide another check. In units with $c=1$, $\omega$ has dimension $L^{-1}$; a Newtonian weak-field limit must recover the classical radial pulsation equation, and a homologous trial displacement must reproduce the $\Gamma_1=4/3$ marginality of a simple Newtonian model after relativistic corrections are removed.

Check the boundary condition

A numerical mode has $\zeta\propto r^2$ near the center and gives a small eigenvalue. Can it establish the onset of instability?

Solution

No. Regularity requires $\xi\sim r$, hence $\zeta=r^2e^{-\nu}\xi\sim r^3$. An $r^2$ behavior makes the physical displacement singular or direction-dependent at the center. The small eigenvalue can therefore be a boundary-condition artifact. Enforce the regular series, verify $\Delta p(R)=0$, check node count and convergence, and only then interpret a convergent $\omega_0^2\to0$ as a stability threshold.

References

• Chandrasekhar, Subrahmanyan. “The Dynamical Instability of Gaseous Masses Approaching the Schwarzschild Limit in General Relativity.” Astrophysical Journal 140 (1964): 417–433. DOI.