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.

Solution

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

{wiki}

In plane-parallel optical-depth coordinates without scattering,
$$
\mu\frac{dI_\nu}{d\tau_\nu}=I_\nu-S_\nu,
$$
with source function $S_\nu$.


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