Phases, elongation and viewing geometry

A phase describes a view of illumination

A spherical body lit by a distant source has an approximately illuminated hemisphere. An observer generally sees a different hemisphere. The visible overlap of these two regions changes with the source–body–observer geometry, producing the familiar sequence of phases.

The terminator separates the illuminated and unilluminated regions on the body. In this simple model, it is a geometric boundary, not the edge of another body’s shadow. Eclipse shadowing is a separate alignment effect and is excluded from the phase calculations below.

Near side and far side refer to orientation relative to an observer. Day and night refer to illumination. Synchronous rotation can keep broadly the same lunar side facing Earth while that side undergoes changing illumination. The far side is therefore not a permanently dark hemisphere.

Here the sphere, source directions and observer directions are ideal geometric objects. We neglect an atmosphere, surface relief, finite-source effects and light-travel-time corrections. The examples are for reasoning from diagrams and archived data rather than locating an object near the Sun.

Name the vertex before naming the angle

Label the Sun S, a target P and the observer O. The phase angle α is the angle S–P–O: its vertex is the target, and its sides point from the target toward the Sun and observer. In the ideal model, α = 0 degrees gives a fully illuminated apparent disk and α = 180 degrees gives an unilluminated apparent disk.

Solar elongation ε is the angle S–O–P. Its vertex is the observer, and its sides point toward the Sun and target. It describes their angular separation on the observer’s sky rather than how the illuminated hemisphere is presented at the target.

These two angles can belong to one triangle while taking different values. The phrase ninety degrees is incomplete unless the relevant vertex and directions are identified. Even a correct numerical angle becomes misleading if moved from P to O in a diagram.

The original figure uses the same length scale along its two coordinate axes. Its enlarged point markers and any phase-disk insets do not represent physical body sizes. Lines identify the geometric directions; they do not turn the drawing into a full apparent-position calculation.

An illuminated fraction measures area in a defined projection

For the ideal sphere seen as a circular disk, the projected illuminated area divided by the full disk area is k = (1 + cos α)/2. This fraction is dimensionless. It refers to the apparent projected area, not the fraction of the whole spherical surface in daylight.

Angles 0, 60, 90, 120 and 180 degrees give k values 1, 0.75, 0.5, 0.25 and 0. The sphere remains approximately half illuminated by the distant Sun throughout; what changes is how that illuminated hemisphere projects toward this observer.

The fraction k is not a reflected-brightness law. Parts of a real illuminated disk contribute different amounts depending on incidence, viewing direction, surface scattering, roughness and material. Observer distance and incident irradiance also affect the received flux.

At k = 0 the model gives no projected directly sunlit area. It does not establish zero radiation of every kind from a real target. Thermal emission, atmospheric effects or additional illumination would require a model that explicitly includes them.

A right angle at the target need not be a right angle on the sky

Choose arbitrary length units L₀ and a coplanar instantaneous diagram. Set S = (0, 0), O = (1, 0), and give an inner circular target orbit radius 0.7 L₀. One tangent point is P = (0.49, 0.499900) L₀, with the last coordinate rounded.

The Sun–target distance is 0.7 L₀. The target-to-observer direction is perpendicular to the Sun–target radius at this tangent point, so α = 90 degrees at P. Its ideal projected illuminated fraction is therefore 0.5.

At O, the triangle instead gives sin ε = 0.7/1. The angle is approximately 44.43 degrees, rather than 90 degrees. This is the maximum elongation of a point on that fixed inner circle from the stipulated observer location in this instantaneous geometry.

The numbers are not an ephemeris for a named planet or date. Eccentric or inclined paths, a moving observer and apparent-position corrections require their own calculation. A two-dimensional classroom triangle cannot provide a real observing schedule.

Similar sky separations can hide different illumination

In a point-geometry diagram, put the inner target between S and O. Its elongation is zero, while the target sees Sun and observer in opposite directions: α = 180 degrees. Put the target beyond S on the same line and elongation is again zero, while α = 0 degrees.

Thus identical elongation can accompany very different ideal phases. These collinear limiting diagrams neglect finite body sizes and occultations; they establish a distinction between angles rather than a claim that either configuration is directly observable.

Actual prediction services specify the conventions of their reported quantities. The JPL Horizons manual distinguishes observer-centred elongation from a target-centred Sun–target–observer angle and describes associated apparent-coordinate corrections. Our instantaneous Euclidean triangle retains its separate model definition.

To compare measured brightness across epochs, preserve the illumination and viewing geometry as well as wavelength and calibration. The next lesson asks which physical region supplies that light when a surface, atmosphere or both contribute.

Worked examples

Try the reasoning

1. Where is the vertex of solar elongation?

   Show the answer and reasoning

   At the observer.

   Its two directions point from the observer toward the Sun and target.

2. Does a quarter-illuminated projected disk mean only one quarter of the sphere is in daylight?

   Show the answer and reasoning

   No.

   Under distant illumination roughly half the sphere is sunlit; projection toward the observer changes the visible fraction.

3. Must an ordinary crescent be caused by Earth’s shadow?

   Show the answer and reasoning

   No.

   The changing view of the illuminated hemisphere produces ordinary phases; eclipse shadowing is separate.

4. Does one elongation value uniquely determine the target’s phase angle?

   Show the answer and reasoning

   No.

   The collinear inner-target examples have the same zero elongation but opposite limiting phase angles.

What to carry forward

• Phase angle and elongation describe different vertices of the source–target–observer geometry.
• Projected illuminated area and received reflected flux require different calculations.
• A diagram’s geometry and a service’s corrected apparent quantities must retain their stated conventions.