Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 321 / 3 / b

After a radial Fourier transform, Poisson's equation away from the sheet becomesThe decaying, reflection-symmetric solution is $A{e}^{-|k||z|}$. Integrating Poisson's equation across $z=0$ gives the derivative jumpso $-2|k|A=4\pi G\tilde{\Sigma }$. Thus the midplane potential is

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Write the perturbations as $\delta \Sigma =\sigma$, radial and azimuthal velocities $u,v$, and pressure $\delta P=p$, all proportional to $e^{ikx+st}$. Linearization in the shearing sheet givesandThe continuity and pressure equations implyEliminating $\sigma ,u,v,p,\varphi$ then yields

When $\tau |s|\ll 1$, temperature relaxes during a disturbance and the pressure response is isothermal, $p/\sigma \simeq c_0^2$. When $\tau |s|\gg 1$, relaxation is negligible and the response is adiabatic, $p/\sigma \simeq \gamma c_0^2$, so the effective sound speed is $\sqrt{\gamma }{c}_0$. At finite relaxation time the phase lag between compression and pressure also damps stable waves.

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At marginal stability write $s=i\omega$ with real $\omega$. The imaginary part of the pressure factor isFor $\gamma >1$, $\tau >0$, and nonzero $k$, the dispersion relation can have zero imaginary part only when $\omega =0$. Thus every marginal mode hasThe marginal equation is consequentlyDefining the isothermal Toomre parameter $Q_0=c_0\Omega /(\pi G{\Sigma }_0)$, its two roots areThey exist when $Q_0\le 1$, and the interval between them is unstable when $Q_0<1$. Finite thermal relaxation therefore leaves the onset criterionequal to the isothermal criterion, regardless of $\gamma$ and the nonzero value of $\tau$.

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For $\tau \Omega =ϵ\ll 1$, expand the pressure factor:For a stable density wave defineThe dispersion relation becomesPerturbing the two isothermal roots givesThus the leading amplitude-damping rate is

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