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I know the moon is not uniformly grey, it has details, craters, it's not just a colored uniform circle from earth, however, when in a full moon, the intensity of the light received from different parts of the "disc" seems very similar.

Let's define the "border" of the moon as the geographical region that, during a full moon as seen from earth, corresponds to the, as seen from the earth, external close the perimeter zone of the territory of the moon. At those places the sun is close to the horizon from their perspective.

This seems to contradict a piece of intuition I have: from the point of view of the moon, the "borders" (what we see in the border of the disc at full moon) are at "sunset" or "sunrise" so, they have less light. As normal experience of living in the earth, the most light and heat you receive from the sun is at noon when it's at cenit, and when the sun starts inclining towards the horizon, the amount of light per meter squared decreases, and thus the floor shines dimmer. That's why I expect the "borders" of the moon to be dimmer than the center during a full moon.

Hypothesis to solve my confusion: Actually, the borders indeed receive less light from the sun because they are tilted with respect to it, but then again, in full moon they are tilted with respect to us also! So to us even if each meter squared receives and thus reflects less light, it is tilted to us it covers a smaller solid angle, so in turn it seems brighter. And since it's the same factor, it should compensate exactly.

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  • $\begingroup$ Here is the same question on another Q&A site $\endgroup$ – M. Enns Oct 23 '16 at 2:51
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    $\begingroup$ The surface roughness of the moon plays an important role in how we see it. An interesting anecdote: the observation that the light reflected by the moon was much like light reflected from a rough surface and not like light reflected from a smooth surface led Galileo Galilei to the conclusion that the moon was just like Earth, and not made of some "celestial" material as Aristotelian philosophers believed at the time. This argument can be found in Dialogue Concerning the Two Chief World Systems. $\endgroup$ – valerio Oct 23 '16 at 13:52
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    $\begingroup$ Vision, like our other senses, are not perceived linearly. Just because something looks uniform to our eyes does not mean that it is. Always something to consider when dealing with our sense. $\endgroup$ – HopelessN00b Oct 25 '16 at 5:01
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    $\begingroup$ @fishinear The linked picture clearly shows that the brightness doesn't drop. The sun is not "directly" behind the photographer but illuminates the Moon from slightly left of the observer. So you don't see a full moon but a ~98/100 (descendant, if the pic wasn't mirrored or turned up-side-down) moon. If the radiance would drop at the "edges" the reduced brightness would also be visible at the left (or western) side of the Moon - which can't be detected at all. $\endgroup$ – klanomath Oct 25 '16 at 16:42
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    $\begingroup$ @fishinear No it doesn't convince me - but it would be a better example to support your thesis. Here the sun is slightly below the observer and the angle between sun and observer is smaller than in the other picture - no brightness drop in the south (except the different surface material in the Tycho halo). $\endgroup$ – klanomath Oct 25 '16 at 17:32
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If the moon were a uniform sphere, it would indeed appear dimmer at the edges of a full moon.

Surface roughness is the main reason the edges are not as dim as the sphere model predicts. When viewing a full moon from earth, the light is coming from nearly behind us. That means the edges of the full moon are illuminated by sun rays nearly parallel to the lunar surface. Now consider what on the moon those rays are going to hit. Just like here on earth at sunrise or sunset, vertical objects facing the sun will be better illuminated. When the surface has roughness, more of the glancing sunlight will hit more vertical parts of the surface.

What we are seeing at the edges of a full moon are predominantly the walls of craters, sides of rocks, and the like that are oriented towards the sun and therefore towards us. This more flat-on average orientation reflects significantly more light than a smooth-sphere surface would.

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    $\begingroup$ If the Moon were a uniform sphere and an ideal Lambertian reflector, the edges wouldn't appear dimmer. That's the key message of Lambert's cosine law. If the Moon were an ideal (non-diffuse) reflector it would be completely dark irradiated from one point light source (except one point if the source is in the observer's hemisphere and not directly behind the observer). $\endgroup$ – klanomath Oct 24 '16 at 20:32
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    $\begingroup$ if the moon was a perfect sphere and lambertian it would be darker at the edge, see www1.cs.columbia.edu/CAVE/projects/oren/oren.php $\endgroup$ – Brice Oct 25 '16 at 0:50
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    $\begingroup$ @klanomath - The "equal brightness effect" you link to describes how the brightness of parts of the object don't depend on the angle from which it's viewed. The brightness does, however, depend on the angle to the light source, and so the edges -- which on a uniform sphere are not pointing towards the sun -- are indeed dimmer in Lambert's model. $\endgroup$ – Jibb Smart Oct 25 '16 at 3:12
  • $\begingroup$ @JibbSmart I have to gather 10 points in physics.se somewhere to be allowed to answer this question ;-). You are right with the radiance intensity but not with the radiance. And only the radiance is relevant in the context of the question. $\endgroup$ – klanomath Oct 25 '16 at 4:03
  • $\begingroup$ @klanomath Fair enough :P I can see how it might be interpreted differently, but my experience with Lambertian diffuse model comes from how it's used in video games (ubiquitously for many years), which definitely dims on faces not pointing directly towards the light source (as per Brice's link). The games and film industries could have it wrong and be stuck with our interpretation by convention -- that kind of thing is not unheard of -- but I'm not inclined to think so. $\endgroup$ – Jibb Smart Oct 25 '16 at 5:48
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I've wondered the same thing, why the edges of the full Moon don't look darker, and why the terminator for a quarter-Moon phase doesn't look dimmer than the point at the edge of the Moon that is opposite the Sun. It isn't a foreshortening effect, as in your last hypothesis, that actually doesn't work. If the Moon's surface were uniformly bright in all directions, like a ping-pong ball perhaps, then the dusk and dawn regions, which get less illumination per square meter, should indeed look darker. It has to be something about the roughness of the Moon that is not uniformly bright in all directions. The foreshortening effect, whereby we see larger areas at the edge of the Moon, just doesn't work like that, we see the brightness of the surface not the flux per unit area from the surface (and brightness is per solid angle, so it's the solid angle we are seeing that counts, not the Moon surface area within the pixel).

Another possible effect to bear in mind is the way the eye sees contrast-- it tries to limit small contrasts when there are much larger contrasts in the field of view. So it's possible the surface is not as uniformly bright as it looks. I don't know about that, so let's say for the purposes of argument that the full Moon is uniformly bright, even though the edges get less solar illumination per square area. That would require that the brightness sent out by the Moon's surface is not uniform, but rather peaks along the direction that the sunlight comes from. That peak would have to be especially pronounced when the sunlight comes from a steep angle.

There is actually independent evidence that the brightness of the Moon does indeed peak along the direction of the incident sunlight-- it is well known that the full Moon gives the Earth about ten times more illumination than the quarter phase Moon does, even though the illuminated area is only doubled. It's a roughness effect. Many road signs are intentionally built to do the same thing, so they reflect light preferentially back toward oncoming headlights. The Moon is acting like a spherical version of those road signs.

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    $\begingroup$ This is the best answer so far. There is a real problem expalining the apparent uniform brightness and it is not resolved by the concentrating effect of the angles near the edges, as the OP and one of the other responsders suggest. I think you are probably right about the psycho/optical effect of brightness levelling within the human eye. $\endgroup$ – Marty Green Oct 23 '16 at 4:22
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    $\begingroup$ The key word is Opposition surge or opposition effect $\endgroup$ – user27542 Oct 23 '16 at 6:15
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    $\begingroup$ It seems unlikely to be an effect of the eye. Photographs don't exhibit this effect either: images.google.com/images?q=full%20moon — using an eyedropper tool in case someone claims that the same effect exists when looking at the photograph! $\endgroup$ – Gnubie Oct 24 '16 at 16:37
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For the full moon, the answer in the OP is clearly correct, as others have confirmed: the brightness is related to the solar energy per square foot, and when viewing from the front, the square feet increase as you go towards the edge at the same rate that the solar energy per foot fades. This assumes that reflections are omnidirectional (as with matte dust rather than a polished sphere), but this seems a reasonable assumption about moondust.

There remains the question that others have asked: if the above is correct, and brightness scales down from the center of the side facing the sun ("midday region"), down to the edges ("twilight region"), then why is the cutoff we see when looking at the half moon so very sharp? Shouldn't it gradually fade off?

My suspicion is that it's a brightness/saturation thing.

Here's Cassini image of Saturn's moon, Rhea:

Cassini image of Saturn's moon, Rhea [via http://photojournal.jpl.nasa.gov/catalog/PIA07606]

Here's our own moon:

Half moon [via https://www.emaze.com/@ALRIZCWQ/David-Moon]

Notice how the former, much further away from the sun, has a far slower cutoff?

I believe it's because the very brightest point is far dimmer sunlight: it's about ten times further from the sun, which means, by the inverse square law, that it's about 100 times dimmer out there.

So I think it's just that our vision is saturated. If we take that lunar picture and desaturate it by increasing contrast by 65% while reducing brightness by 70%, we get...

Half moon desaturated

Which is much more like the first, I feel, though it's clear that a lot of the levels have been lost.

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    $\begingroup$ I think OP is asking about a full moon. $\endgroup$ – JonathanReez Oct 23 '16 at 10:55
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    $\begingroup$ This is an excellent answer with good documentation as to the saturation effect. Even though it is written with regard to the half moon, the logic applies equally well to the full moon. $\endgroup$ – Marty Green Oct 23 '16 at 14:38
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    $\begingroup$ @MartyGreen Does it? $\endgroup$ – Peter A. Schneider Oct 24 '16 at 13:15
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    $\begingroup$ The geometry is very different for a half-moon. Because our point of view is different, different parts of the moon's surface appear compressed. $\endgroup$ – Jyrki Lahtonen Oct 25 '16 at 6:43
  • $\begingroup$ Sorry, and good call. I should have explained my rationale for displaying the half-moon rather than the full. I'll edit. $\endgroup$ – Dewi Morgan Oct 25 '16 at 18:35
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The sun illuminates every particle of the moon that it "sees" with the same intensity. During a full moon we see each particle of the moon from the same angle and therefore we see each particle eliminated with the same intensity. The relative macro slope of the moon's surface has no influence on the brightness of each particle.

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protected by Qmechanic Oct 23 '16 at 8:30

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