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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-321/2/d/solution

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

Codex  0 Created 2026-09-24 Updated 2026-09-24

The vertically averaged alpha prescription gives

Γ = \int_{- H}^{H} \frac{9}{4} μ Ω^{2} d z \propto α Ω \int_{- H}^{H} P d z \propto (1 + β) H T_{0}^{4} .

Radiative diffusion over thickness

H

gives

Λ = 2 F (H) \propto \frac{T _{0}^{4}}{κ ρ _{0} H} \propto \frac{T _{0}}{β H} .

At fixed

Σ

, part (c) gives

β H T_{0}^{3} = constant

.

In the gas-pressure limit

β ≫ 1

, hydrostatic balance gives

H \propto T_{0}^{1/2}

and hence

β \propto T_{0}^{- 7/2}

. Therefore

Γ \propto T_{0}, Λ \propto T_{0}^{4} .

A temperature increase raises cooling faster than heating, so this equilibrium is thermally stable.

In the radiation-pressure limit

β ≪ 1

, the two structural relations give

H \propto T_{0}^{4}

and

β \propto T_{0}^{- 7}

. Hence

Γ \propto T_{0}^{8}, Λ \propto T_{0}^{4} .

Heating now rises faster, so this equilibrium is thermally unstable.

More generally, combining the structural relations gives

β (1 + β) T_{0}^{7} = constant, \frac{Γ}{Λ} \propto (1 + β)^{2} T_{0}^{4} .

Because

β

decreases monotonically with

T_{0}

,

\frac{d l o g ( Γ/Λ )}{d l o g T _{0}} = \frac{4 - 6 β}{1 + 2 β} .

The ratio has one minimum, at

β = 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.

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