The <internal effective temperature of a planet> $T_{\rm int}$ parametrizes intrinsic cooling, while the <irradiation temperature> $T_{\rm irr}$ parametrizes incident stellar flux before the redistribution factor $f$. The quantity
$$
\gamma=\frac{\kappa_{\rm vis}}{\kappa_{\rm IR}}
$$
is the visible-to-thermal mean-opacity ratio, and $\tau$ is downward thermal <optical depth>. Under the Eddington two-stream boundary condition,
$$
a=\frac34,
\qquad b=\frac23,
\qquad\boxed{ab=\frac12},
$$
while the common semi-grey choice is $c=\sqrt3$.

For a young giant at $40\,\mathrm{au}$, stellar heating is weak and a still-large $T_{\rm int}$ dominates. Its pressure-temperature profile rises steadily inward, approximately as $T^4\propto b+\tau$, and joins a deep convective adiabat.

For a hot Jupiter close to a Sun-like star, $T_{\rm irr}\gg T_{\rm int}$. It has a broad, nearly isothermal irradiated radiative layer, followed by a deep rise where intrinsic flux and increasing opacity matter. If $\gamma>1$, absorption of starlight above the thermal photosphere can create an <atmospheric thermal inversion>.

For a temperate sub-Neptune around an M dwarf, irradiation and internal cooling can be more comparable. Its profile generally has a moderate radiative layer above a convective interior; near-infrared stellar radiation, molecular opacity, clouds, and hazes determine whether the upper profile is weakly inverted or decreases outward.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 315 / 1 / a / Solution

Without scattering and in <local thermodynamic equilibrium>, the <radiative transfer equation> has the emergent solution
$$
I_\nu(0,\mu)=\int_0^\infty
B_\nu[T(\tau_\nu)]e^{-\tau_\nu/\mu}
\frac{d\tau_\nu}{\mu}.
$$
The <Eddington-Barbier relation> gives the useful approximation
$$
I_\nu(0,\mu)\simeq B_\nu[T(\tau_\nu=\mu)].
$$
A molecular band has larger opacity than its neighboring continuum and therefore reaches optical depth unity at lower pressure.

If temperature decreases outward, the band samples cooler gas and appears in absorption. If the atmosphere is isothermal, both levels have the same source function and the feature disappears. If an <atmospheric thermal inversion> makes the upper layer hotter, the band appears in emission. For a weak separation of formation pressures,
$$
\Delta I_\nu\simeq
\frac{\partial B_\nu}{\partial T}
\frac{dT}{d\log P}\Delta\log P,
$$
which explicitly shows that feature sign and amplitude measure the vertical temperature gradient.

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 315 / 4 / b / Solution

{wiki=Inversion_(meteorology)}

An atmospheric thermal inversion is a region where temperature increases outward. In a strongly irradiated atmosphere it can arise when shortwave opacity exceeds thermal infrared opacity and stellar energy is absorbed high in the atmosphere.


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