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A lunar phase is the appearance of the sunlit portion of the Moon as seen from Earth. Outside a lunar eclipse, the distant Sun illuminates approximately one hemisphere of the Moon, but an observer sees a changing overlap of that hemisphere with the hemisphere facing Earth. The lunar phase therefore changes even though synchronous rotation keeps approximately the same near side of the Moon facing Earth. ^[ref].

The cycle runs through new Moon, waxing crescent, first quarter, waxing gibbous, full Moon, waning gibbous, last quarter and waning crescent. “Waxing” means that the lunar illuminated fraction is increasing; “waning” means it is decreasing. A crescent has less than half its apparent disk sunlit, and a gibbous lunar phase has more than half. First and last quarter each show half a disk: “quarter” refers to progress around the cycle, rather than a quarter of the disk being lit. ^[ref].

An instructive circular orbit model puts Earth at the origin, the Sun far away along $+x$, and the Moon at $R(\cos \theta ,\sin \theta )$. The sunward position $x>0$, $\theta =0$, gives new Moon; the opposite position $x<0$, $\theta =\pi$, gives full Moon. Counterclockwise motion gives first quarter at $\theta =\pi /2$ and last quarter at $\theta =3\pi /2$. The Sun's direction is fixed during this snapshot of the cycle. Every small globe in the orbital view is illuminated on its $+x$ half; the adjacent apparent disks are separate views from Earth, rather than globes viewed from above.

The disk views choose north up, with the upward direction perpendicular to the orbital plane. In this convention waxing lunar phases are bright on the right and waning lunar phases on the left. Local horizon orientation and viewing from different locations on Earth can rotate the apparent disk; its lit edge always points toward the projected direction of the Sun. The lunar illuminated fraction is unaffected by this rotation.

The useful angle is the phase angle $\alpha =\angle SME$, with $S$ the Sun, $M$ the Moon and $E$ the observer at Earth's center: its vertex is at the Moon. Full Moon has $\alpha =0$, quarter has $\alpha =\pi /2$, and new Moon has $\alpha =\pi$ in the ideal model. By contrast, the elongation $ϵ=\angle SEM$ has its vertex at Earth. For a distant Sun, its rays are parallel, soHere $k$ is the lunar illuminated fraction: a fraction of projected disk area, not a fraction of the Moon's total surface area or its received radiant flux relative to full Moon. The phase angle and unsigned elongation each lie between $0$ and $\pi$, so either alone gives the illuminated amount without distinguishing waxing from waning. The orbital direction resolves that ambiguity.

The lunar illuminated fraction has an elegant geometric derivation. Model the Moon as a sphere of unit radius and view it along the $z$ direction. Take the Sun's unit direction to be $(\sin \alpha ,0,\cos \alpha )$. A visible surface point above apparent disk coordinates $(x,y)$ has $z=\sqrt{1-x^2-y^2}$. Direct illumination requires the outward normal's dot product with the sunward direction to be positive. The day-night terminator therefore obeys $x\sin \alpha +z\cos \alpha =0$. At a fixed $y$, set $q=\sqrt{1-y^2}$; the visible day-night terminator is $x_t=-q\cos \alpha$, and the illuminated strip extends from $x_t$ to the right edge $q$. Integrating its width givesThe integral is the area of half a unit circle. Thus a three-dimensional visibility problem reduces to a circle's area and a cosine. The derivation assumes parallel sunlight, a distant observer and a sphere, and ignores a lunar eclipse. Real observed brightness additionally depends on surface scattering of light and observing geometry, so $k=1/2$ does not promise half the full-Moon radiant flux.

For a worked example, elongation $ϵ=60^{\circ }$ gives phase angle $\alpha =120^{\circ }$ and $k=(1+\cos 120^{\circ })/2=1/4$: a crescent with $25\%$ of its apparent disk sunlit. At elongation $120^{\circ }$, the phase angle is $60^{\circ }$ and $k=3/4$: a gibbous lunar phase. At $90^{\circ }$, $k=1/2$. In each case the same $k$ occurs once while waxing and once while waning in the ideal cycle.

Ordinary lunar phases are not caused by Earth's shadow. A lunar eclipse occurs only when the Moon actually enters that shadow near full Moon. The orbital inclination of the Moon is about $5^{\circ }$ relative to Earth's orbital plane around the Sun; most full Moons pass above or below the shadow. This is why full Moon does not imply a lunar eclipse every month. ^[ref].

The phase cycle is a synodic month, about $29.5$ days, whereas the Moon's orbital period relative to distant stars is a sidereal month, about $27.3$ days. During one sidereal month, Earth advances around the Sun, so the Moon must move farther to repeat the sunward alignment. For coplanar circular orbits traversed in the same direction, subtract their mean motions: $n_{\mathrm{syn}}=n_{\mathrm{sid}}-n_{\mathrm{year}}$, with $n=2\pi /P$. HenceUsing $P_{\mathrm{sid}}=27.321661\mathrm{days}$ and $P_{\mathrm{year}}=365.256\mathrm{days}$ gives $P_{\mathrm{syn}}=29.530591\mathrm{days}$, consistent with the mean synodic month of $29.53059$ days. After one sidereal month, the sunward direction has advanced about $26.93^{\circ }$; catching up takes about $2.21$ more days. These are mean-period estimates: orbital eccentricity and changing geometry make individual synodic months vary. ^[ref]. ^[ref].

The lunar phase is also a practical clock for observing. Full Moon is opposite the Sun and rises roughly at sunset; first quarter is about a right angle from the Sun and is convenient in the evening, while last quarter is convenient before dawn. These are rough guides, not exact rise-time predictions for every place and season. ^[ref]. Historically, repeated lunar phases provided the month in lunar calendars, while half-lit disks gave Aristarchus of Samos a way to infer astronomical distances. ^[ref].

A natural interpretation of the short-wavelength peak is reflected starlight, while the longer-wavelength peak is planetary thermal radiation. A reflected spectrum approximately follows the stellar spectrum multiplied by the wavelength-dependent geometric albedo; thermal radiative flux approximately follows the planet's Planck function, modulated by molecular opacity. Thus two peaks need not represent two planetary surface temperatures.

For a quantitative estimate assume both are broad $F_{\lambda }$ peaks, reflection has a slowly varying geometric albedo, and star and planet have approximately blackbody spectral envelopes. Wien's displacement law then givesThe stellar estimate is compatible at order of magnitude with an old, relatively small main-sequence star; equal age does not mean equal temperature to the Sun. Assume the far-infrared signal is thermal and sufficiently long-wavelength that the Rayleigh-Jeans law applies to both bodies. The planet-star radius estimate in the Rayleigh-Jeans limit givesand thereforeThis is a small volatile-rich-planet-sized estimate, not a Jupiter-sized one. It is conditional: peaks caused by molecular windows, strongly chromatic reflection, or a spectrum expressed as $F_{\nu }$ rather than $F_{\lambda }$ do not support those two Wien estimates. Without $T_p/T_{\ast }$, the far-infrared ratio only fixes $R_p^2{T}_p/T_{\ast }$.

For a close-in hot planet around a small star, transit-based atmospheric observations are the natural route if the orbital geometry allows them. Exoplanet transmission spectra gain from the small stellar radius, with a limb signal scaling as $2R_{p}H/R_{\ast }^2$; they probe composition at the day-night terminator. The stated $10^{-3}$ thermal contrast also makes exoplanet secondary eclipse measurements a particularly useful route to the dayside exoplanet emission spectrum. Resolving such a close-in small planet by exoplanet direct imaging is much harder. Transit and secondary-eclipse spectroscopy are favoured for a transiting close-in interpretation; the supplied spectrum alone does not determine the orbital geometry or a unique best technique.

The two measurements sample different atmospheric regions and different radiative transfer geometries. An exoplanet emission spectrum probes the dayside; an exoplanet transmission spectrum probes the morning and evening limbs of the day-night terminator. Several alternatives to an exoplanet cloud deck can therefore suppress the transmission feature.

Temperature, mean molecular weight, limb chemistry, continuum opacity and the dayside temperature gradient are alternatives or degeneracies to test. In particular, the three-scale-height amplitude alone cannot establish different CO abundances between the hemispheres.

Three complementary methods available in the setting of this 2017 paper are:

Transmission probes the limb, eclipses probe the dayside, and direct imaging isolates spatially resolved planetary light. Radial velocities alone measure orbital motion rather than atmospheric composition; high-resolution planetary Doppler spectroscopy can instead detect atmospheric spectral lines if planetary light is isolated statistically.

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.