Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 315 / 1

Treat the morning and evening halves of the day-night terminator as independent, isothermal, hydrostatic atmospheres with the same reference radius $R_0$, reference pressure $P_0$, composition, gravity $g$, and extinction coefficient ${\kappa }_{\lambda }$. Their atmospheric scale heights areFor $H_i\ll R_0$, the slant optical depth of half $i$ at tangent altitude $z$ is approximatelyThe standard isothermal effective altitude is thereforeup to a wavelength-independent choice of reference radius. The two semicircular limbs add in projected area, so the exoplanet transmission spectrum isEquivalently, $R_{\mathrm{tr}}^2=[(R_0+z_m{)}^2+(R_0+z_e{)}^2]/2$. Differentiating with respect to $\log {\kappa }_{\lambda }$ givesThus a homogeneous retrieval measures, to leading order,provided the opacity and composition do not themselves differ between the two limbs.

Along a ray, the radiative transfer equation has the formal solution of the radiative transfer equationfor each isothermal region in local thermodynamic equilibrium, neglecting scattering. For a self-luminous atmosphere with no incident intensity from below at the relevant photosphere, and in the optically thick limit, $I_{\nu ,i}=B_{\nu }(T_i)$.

The observed radiative flux is the projected-disc integralThe inner region occupies projected fraction $f={\sin }^2{\theta }_T$, henceFor finite optical depths, each Planck law function in this formula is replaced by its corresponding emergent intensity above.

A homogeneous interpretation assigns the planet a brightness temperature $T_b$ satisfyingAt a fixed wavelength let $x=hc/(\lambda k_B)$ andInverting the Planck law givesAt $\lambda =20\mu \mathrm{m}$, $x=719.4\mathrm{K}$. With ${\theta }_T=\pi /4$, $T_1=1500\mathrm{K}$, and $T_2=1000\mathrm{K}$, one obtainsThe Rayleigh-Jeans law would give the nearly identical area-weighted estimate $1250\mathrm{K}$.