Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 321 / 2

The first equation is vertical hydrostatic equilibrium. In a thin disk, stellar gravity has vertical component $-{\Omega }^{2}z$, giving $dP/dz=-\rho {\Omega }^{2}z$. The second is local energy conservation: Keplerian shear has $rd\Omega /dr=-3\Omega /2$, so viscous dissipation per unit volume isThe third equation is radiative diffusion. With radiation energy density $a_{\mathrm{R}}{T}^4=4\sigma T^4/c$, an optically thick medium carrieswhich rearranges to the stated gradient. Thomson scattering makes $\kappa$ approximately independent of density and temperature. The final equation adds perfect-gas pressure $k\rho T/({\mu }_m{m}_p)$ and radiation pressure $a_{\mathrm{R}}{T}^4/3=4\sigma T^4/(3c)$.

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The alpha disk prescription isEquivalently, the turbulent shear stress is of order $\alpha P$. Turbulent eddies cannot remain coherent on scales much larger than the vertical thickness $H$, and strongly supersonic eddies shock and dissipate, so their speed is at most of order $c_s$. The product of the largest plausible speed and length is therefore $c_{s}H$. Magnetorotational turbulence supplies a physical mechanism for correlated radial and azimuthal motions and magnetic stresses, while $\alpha$ parametrizes their uncertain efficiency.

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Let $\beta =P_{\mathrm{gas}}/P_{\mathrm{rad}}$. Constancy of $\beta$ givesThe total pressure isConsequently, if $T=T_{0}f$, then $\rho ={\rho }_0{f}^3$ and $P=P_0{f}^4$, whereSubstitution in hydrostatic balance shows that $f^{\prime }=-2z/H^2$ and

Finally,Absorbing the numerical factor $4/3$ and this dimensionless integral into $A$ gives the requested form

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The vertically averaged alpha prescription givesRadiative diffusion over thickness $H$ givesAt fixed $\Sigma$, part (c) gives $\beta H{T}_0^3=\text{constant}$.

In the gas-pressure limit $\beta \gg 1$, hydrostatic balance gives $H\propto T_0^{1/2}$ and hence $\beta \propto T_0^{-7/2}$. ThereforeA temperature increase raises cooling faster than heating, so this equilibrium is thermally stable.

In the radiation-pressure limit $\beta \ll 1$, the two structural relations give $H\propto T_0^4$ and $\beta \propto T_0^{-7}$. HenceHeating now rises faster, so this equilibrium is thermally unstable.

More generally, combining the structural relations givesBecause $\beta$ decreases monotonically with $T_0$,The ratio has one minimum, at $\beta =2/3$, and therefore a horizontal equilibrium level can be crossed at most twice. Thus there are at most two thermal equilibria: a stable gas-pressure branch and an unstable radiation-pressure branch.

Solved by gpt-5.6-sol high.