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 Incoming links: Affine quadrupole mode of a star

Let the equilibrium radius be

a_{0} = b_{0} = c_{0}

and

2 T_{0} = Ω^{2} a_{0}^{2}

. Linearizing the

a

\frac{δ a ¨}{a _{0}} = Ω^{2} (- 2 \frac{δ a}{a _{0}} + \frac{δ T}{T _{0}}),

with cyclic analogues. The temperature scaling gives

\frac{δ T}{T _{0}} = - (γ - 1) (\frac{δ a}{a _{0}} + \frac{δ b}{a _{0}} + \frac{δc}{a _{0}}) .

For the affine breathing mode of a star, all three fractional axis changes equal

q

. Then

\overset{q}{¨} + (3 γ - 1) Ω^{2} q = 0, ω = Ω 3 γ - 1 .

This is the homologous compressional mode, which changes volume, density, and temperature.

For either independent affine quadrupole mode of a star, the three fractional changes sum to zero. Then

δ T = 0

and

\overset{q}{¨} + 2 Ω^{2} q = 0, ω = 2 Ω .

These two degenerate modes deform the sphere into an ellipsoid while preserving its volume to first order.

The additional acceleration is

- \nablaΨ = 2 Ψ_{0} (x, y, - 2 z) cos (Ω_{o} t) .

Its three fractional-axis forcing terms are therefore proportional to

(1, 1, - 2)

and have zero sum. Consequently it has no projection on the affine breathing mode of a star, which is not forced at linear order.

It lies entirely in the affine quadrupole mode of a star subspace. Put

\frac{δ a}{a _{0}} = \frac{δ b}{a _{0}} = q, \frac{δc}{a _{0}} = - 2 q .

Then

\overset{q}{¨} + 2 Ω^{2} q = 2 Ψ_{0} cos (Ω_{o} t),

so away from resonance

q (t) = \frac{2 Ψ _{0}}{2 Ω ^{2} - Ω _{o}^{2}} cos (Ω_{o} t)

up to free oscillations. The tidal forcing resonates with the quadrupole mode when

Ω_{o} = 2 Ω

; in the ideal undamped model the resonant amplitude grows secularly.