= Solution
The vertically averaged alpha prescription gives
$$
\Gamma=\int_{-H}^H\frac94\mu\Omega^2dz
\propto\alpha\Omega\int_{-H}^HP\,dz
\propto(1+\beta)HT_0^4.
$$
Radiative diffusion over thickness $H$ gives
$$
\Lambda=2F(H)\propto\frac{T_0^4}{\kappa\rho_0H}
\propto\frac{T_0}{\beta H}.
$$
At fixed $\Sigma$, part (c) gives $\beta HT_0^3=\text{constant}$.
In the gas-pressure limit $\beta\gg1$, hydrostatic balance gives $H\propto T_0^{1/2}$ and hence $\beta\propto T_0^{-7/2}$. Therefore
$$
\boxed{\Gamma\propto T_0,\qquad\Lambda\propto T_0^4}.
$$
A temperature increase raises cooling faster than heating, so this equilibrium is thermally stable.
In the radiation-pressure limit $\beta\ll1$, the two structural relations give $H\propto T_0^4$ and $\beta\propto T_0^{-7}$. Hence
$$
\boxed{\Gamma\propto T_0^8,\qquad\Lambda\propto T_0^4}.
$$
Heating now rises faster, so this equilibrium is thermally unstable.
More generally, combining the structural relations gives
$$
\beta(1+\beta)T_0^7=\text{constant},
\qquad
\frac{\Gamma}{\Lambda}\propto(1+\beta)^2T_0^4.
$$
Because $\beta$ decreases monotonically with $T_0$,
$$
\frac{d\log(\Gamma/\Lambda)}{d\log T_0}
=\frac{4-6\beta}{1+2\beta}.
$$
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.
Back to article page
