Past exam of the mathematics course of the University of Cambridge 2002 iii Paper 41 4 Solution Created 2026-10-03 Updated 2026-10-14
Work at fixed enclosed mass , so the perturbations are fluid Lagrangian perturbations. The equilibrium acceleration vanishes. Varying the factor in the gravitational force contributes , while varying the factor multiplying the acceleration gives no first-order term because the background acceleration is zero. With , 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 . Use and the stellar hydrostatic equation to turn the mass-coordinate momentum equation into . 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 . The self-adjoint differential equation becomesHence and in the requested form. Multiplying by a real eigenfunction and integrating gives the boundary term . It vanishes for a regular centre and bounded admissible surface displacement and derivative, since . 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 by parts. Its boundary term again vanishes. The remaining integrand isSince , one has . Substitution proves the positive-square radial pulsation energy identityIf 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 somewhere inside the star.
For the opposite necessary condition, test the quotient with a nonzero constant . It has andIf everywhere and is strictly below 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 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 throughout, solves the pulsation equation with . The star is marginally stable in this radial sense, so the strict 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 everywhere and exceeds it on a positive-pressure interval, the square identity forces for every nonzero eigenfunction: zero energy would first force constant and then force that constant to vanish. If everywhere with strict inequality on an interval, the constant trial proves instability. When both types of region occur, somewhere does not alone prove instability, and 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.