Add a globe's light to make a phase curve

Projected area supplies the weight

For latitude $\beta$ and longitude $\lambda$, a patch subtends solid angle $d\Omega=\cos\beta\,d\beta\,d\lambda$ at the planet's centre. Its physical area is $R_p^2d\Omega$. Let $\mu_o$ be the cosine of the angle between its outward normal and the direction to the observer. A visible patch has positive $\mu_o$, and its apparent area is smaller than its physical area by that factor.

Use wavelength $\Lambda$, distinct from longitude $\lambda$. Let $I_\Lambda$ be spectral intensity and $D$ the system's distance. Summing visible, unocculted patches gives

Why divide total surface emission by pi? Intensity is per solid angle and per projected area. Integrating a constant intensity over an outward hemisphere gives its emitted flux multiplied by the cosine projection. In polar angle $\theta$, that angular integral is $2\pi\int_0^{\pi/2}\cos\theta\sin\theta\,d\theta=\pi$. The same integral across a uniformly bright observed sphere returns its projected disk area.

The planet-to-star ratio cancels distance when both are in the same system. A uniform bolometric blackbody planet therefore gives the squared radius ratio times the fourth power of its temperature ratio to the star. A nonuniform planet requires the actual integral, with the fourth power applied before adding patches.

Reading: Nicolas B. Cowan and Eric Agol, A Model for Thermal Phase Variations of Circular and Eccentric Exoplanets · Rodrigo Luger et al., STARRY: Analytic Occultation Light Curves

Follow the orbit before masking an eclipse

Adopt an observer along negative global z, with sky x right and sky y up. The circular orbit is inclined by $i$. Orbital phase $q=0$ is primary transit and $q=1/2$ geometric secondary conjunction. The projected separation alone does not tell which body is in front; the line-of-sight coordinate supplies that information.

In the substellar coordinates used previously, the observer longitude is $\pi-2\pi q$ and its latitude is $\pi/2-i$. Each change of orbital phase therefore changes the viewing weights. At primary transit the planet may cover the star, but the planet's own emission remains visible. At secondary eclipse the foreground star removes planetary contributions wherever the projected disks overlap.

A numerical mask must use both projected overlap and front/back depth. It leaves the stellar heating term in the thermal equation unchanged. The star is not between itself and the planet during secondary eclipse; it is between the planet and us.

For a uniformly bright planet, a partial eclipse is simply the overlap area of two circles times constant intensity. That supplies a check independent of the transported temperature map. A nonuniform map weights the hidden areas differently. Ingress and egress can then measure how brightness varies across the occultor’s path.

Reading: Rodrigo Luger et al., STARRY: Analytic Occultation Light Curves · Taylor J. Bell et al., Nightside clouds and disequilibrium chemistry on the hot Jupiter WASP-43b

Reflection has a second cosine

The illuminating cosine $\mu_s$ measures how directly a patch faces the star. It is different from the observing cosine. An ideal Lambert surface reflects with the same intensity into every outward direction. To return a fraction $A_B$ of incident power, its local bolometric intensity must be

The visible hemisphere integral includes both the illumination factor and the projected area toward us. For phase angle $\alpha$, the angle between star and observer as seen from the planet, the normalized result is

The thermal model uses this bolometric reflection as a separate ideal channel. A general atmosphere's Bond albedo does not specify its reflectance at every wavelength. It would be incorrect to add an arbitrary fraction of this reflected curve to a measured mid-infrared band and call the result a calibrated detector prediction.

Reading: Nikku Madhusudhan and Adam Burrows, Analytic Models for Albedos, Phase Curves, and Polarization of Reflected Light from Exoplanets · Nicolas B. Cowan and Eric Agol, The Statistics of Albedo and Heat Recirculation on Hot Exoplanets

A longitude map becomes a convolution

Now set the orbit edge-on, omit occultations and assume isotropic thermal intensity. Take orbital angle $\xi=2\pi(q-1/2)$, so the observer faces longitude $-\xi$. In the projection, one latitude cosine comes from surface area and another from the viewing angle. Define a latitude-weighted longitude brightness $J(\lambda)$ that includes those factors and the chosen flux normalization.

To see what it preserves, expand the map in cosine and sine modes. A mode's multiplier is the integral of that mode times the kernel over the visible half-circle. Direct integration gives

The third harmonic disappears completely in this geometry. For example, a positive map with uniform brightness one and another with brightness $1+0.1\cos(3\lambda)$ give exactly the same phase curve. Positivity alone does not select one. A smoothness preference chooses among solutions by adding a preference for gentler spatial variation.

Reading: Nicolas B. Cowan and Eric Agol, Inverting Phase Functions to Map Exoplanets · Rodrigo Luger et al., STARRY: Analytic Occultation Light Curves

Worked example of a missing mode

Suppose a normalized curve is a constant plus first and second harmonics, as in the real-data tutorial. Dividing its constant by two and its cosine amplitudes by the corresponding nonzero $K_m$ gives one longitude map. The sine coefficients also change sign. This operation recovers only the represented modes.

For a numerical example, take the explicitly synthetic flux ratio below, with no sine terms. Its constant gives a map mean of 0.0015. Dividing the first cosine by $K_1$ gives $0.002/\pi$; dividing the second by $K_2$ gives 0.0003.

Since every cosine lies between minus one and one, this map is everywhere above $0.0015-0.002/\pi-0.0003\simeq0.000563$. Adding $10^{-4}\cos3\lambda$ therefore leaves it positive. The added mode has zero visibility multiplier, so both maps return the same entire curve, not merely the same maximum and minimum.

The supplied WASP-43b reduction yields a positive baseline longitude map for that conditional fit. Adding a third cosine with amplitude forty percent of its smallest sampled brightness keeps the example maps positive while leaving their ideal edge-on phase curves unchanged. These are brightness maps, not a conversion to local gas temperature and not two independently solved heat-transport fields.

Eclipse observations can distinguish some spatial differences as a stellar limb changes which subset of the disk remains visible. In a finite latitude-independent basis through harmonic order six, the sampled phase-only operator has four vanishing singular directions even at the observed inclination near eighty-two degrees. Including eclipse samples gives a nonzero response in all thirteen selected directions. A general two-dimensional map has additional degrees of freedom, so that finite-basis result does not establish general map uniqueness.

Reading: Nicolas B. Cowan and Eric Agol, Inverting Phase Functions to Map Exoplanets · Rodrigo Luger et al., STARRY: Analytic Occultation Light Curves

Average through the exposure

An instrument averages its signal over a time window. If a harmonic of angular frequency $\omega$ is averaged over a centred uniform exposure of duration $\Delta t$, direct integration gives

For the nominal MIRI integration of about ten seconds and an orbit of about nineteen and a half hours, this correction is tiny for the first orbital harmonic. It can matter more during ingress, where brightness changes quickly. The next calculation retains actual sample midtimes and gaps, distinguishes the rounded exposure approximation, and propagates uncertainty through the resulting curve.

Reading: Taylor J. Bell et al., Nightside clouds and disequilibrium chemistry on the hot Jupiter WASP-43b · Rodrigo Luger et al., STARRY: Analytic Occultation Light Curves