# Trying to understand lunar phases

Complete novice here. I'm trying to understand what appears to be two approaches to quantifying lunar phases. Have I got this right?

1. Named lunar phases can be understood as the Moon's ecliptic longitude minus the Sun's ecliptic longitude. These are, for example, exactly $$0^{\circ},90^{\circ},180^{\circ}$$ and $$270^{\circ}$$ respectively for a new moon, first quarter, full moon, and last quarter.

2. Illuminated fraction of the Moon's disc uses the formula$$k=\frac{1+\cos i}{2},$$where $$i$$ is the Moon's phase angle (angle from the Sun to the Moon to the Earth).

Phase angles of $$0^{\circ},90^{\circ},180^{\circ}$$ and $$270^{\circ}$$ are only approximately equal to (respectively) a new moon, first quarter, full moon, and last quarter. This is because the Moon's orbital plane is slightly inclined to the ecliptic plane.

Is this correct?

• The phase angle and the named phases are exact by definition. It is the illuminated fraction which is approximate when using that formula. So the first quarter generally will not have 50% illumination. Nov 2, 2022 at 2:11

Yes, that's correct. The phases of the Moon are defined in terms of the Moon's elongation: the difference in (geocentric) ecliptic longitudes of the Moon and Sun, with New Moon at 0°. The ecliptic latitude is ignored.

From Wikipedia Lunar Phases

There are four principal (primary/major) lunar phases: the new moon, first quarter, full moon, and last quarter (also known as third or final quarter), when the Moon's ecliptic longitude is at an angle to the Sun (as viewed from the centre of the Earth) of 0°, 90°, 180°, and 270°, respectively.

— Seidelmann, P. Kenneth, ed. (1992). Explanatory Supplement to the Astronomical Almanac

However, that article then contradicts itself by saying

Each of these phases appears at slightly different times at different locations on Earth.

which doesn't make sense, given the geocentric definition stated by Dr Seidelmann.

Here's a plot, created using JPL Horizons, of the Moon's ecliptic elongation for the first synodic month of 2022 (with a 6 hour time step), taken from my answer to a similar question. As you mention, to calculate the illuminated fraction of the Moon we need to use the true 3D Sun-Moon-Earth angle. Horizons provides that angle, but in the range 0° to 180°. Here's the corresponding plot, over the same time span as the plot above. To compare these values we can transform the elongation angle by subtracting it from 180° and taking the absolute value. When we subtract the Sun-Moon-Earth angle from the transformed elongation angle, we get a plot like this (for 2021, with a 1 day time step). The difference is usually quite small, except near the New and Full Moons, when they can differ by a couple of degrees. I assume the variation from month to month depends on how close the Moon is to the ecliptic at New & Full Moon, but I haven't investigated that.

All of these plots were created for a geocentric observer. The actual observed angles for an observer on Earth's surface will be slightly different. The Moon is relatively close to the Earth, so parallax effects are relatively large. And of course we should also make adjustments for atmospheric refraction, especially when the Moon is near the horizon.

• Excellent answer. There's a nice online quote: "On Earth, we never see a perfectly full Moon, since the true phase angle we see is in the order of 5 degrees. With a zero degree phase angle the Moon would be in Earth’s shadow, and we would experience a total lunar eclipse. Apollo astronauts reported that a true full Moon is about 30% (0.2 magnitudes) brighter than what we see here on Earth." Nov 2, 2022 at 13:40
• Good analysis, but the phases are defined by the geocentric ecliptic longitude, not the elongation. They will be slightly different due to the moon's inclination, and that's what the OP was asking about. Nov 2, 2022 at 14:10
• @Greg I've edited my answer to emphasise that the official definition of the lunar phases is based on the geocentic ecliptic longitudes of the Sun & Moon. As I'm sure you're aware, Dr Seidelmann was a prominent member of many IAU committees, and a USNO director. Nov 2, 2022 at 19:33
• @PM2Ring The Explanatory Supplement doesn't have the offending Wikipedia comment about appearing at different times, instead actually immediately follows with "Because the time are determined from the geocentric coordinates, they are independent of location on the Earth". But I don't think the Wikipedia version is contradictory, the phases do appear at different times on the surface of the Earth, they just aren't the official times. So it could be worded better. Nov 3, 2022 at 2:15
• @PM2Ring Actually, if you’re finicky, lunar phases do appear at a slightly different times from one place to the other on the Earth, as the terminator would not be exactly facing each part of the globe at the same time. This is due to parallax, which affects the lunar longitude topocentrically. Nov 4, 2022 at 5:09

Yes and no; the Moon’s orbital plane is indeed tilted with respect to the ecliptic plane, but it doesn’t affect the phase angle. So, New Moon is exactly 0°, First Quarter is exactly 90°, Full Moon is exactly 180°, and Last Quarter is exactly 270° (technically, those phases last only a split second as far as it is “exactly”).

The orbit’s tilt will play on which part of the Moon is lit when it’s, e.g., 50% illuminated as seen from the Earth: for example, if the Moon is North of the ecliptic, then at First Quarter, the North Lunar Pole will be in the dark and the South Lunar Pole will be lit, but the point on the equator that’s at exact longitude 0° (not counting libration) will be on the terminator (the line between the dark and lit regions).

• My definition of phase angle is the angle from the Sun to the Moon to the Earth. That is not the same as the difference between the Moon and Sun's ecliptic longitude. Therefore, how can the phase angle at first quarter, for example, be (as you say) 90 degrees? Nov 3, 2022 at 9:08
• Because, e.g., first quarter does NOT happen at Δλ = 90°, but at a slightly smaller angle due to finite distances to the Moon and to the Sun. However, at that moment, the Moon has exactly half its visible surface (as seen from the Earth) illuminated, so should you divide the visible surface from 0° (e.g.) on the left to 180° on the right, then the center of the disk would be at 90° and the terminator would also be at 90° at the center of the disk. If the Moon then has a latitude β ≠ 0°, then either its north pole or its south pole would be in the shade and the other one, lit. Nov 4, 2022 at 5:06