8
$\begingroup$

When I look at the regular ball picture the "edges" appear somewhat darker. Here's a random image I just found on image search :)

enter image description here

Image source

and that's how I can tell it's "round". Or so I assume.

When it comes to the Moon on the other hand, the "edges" appear at least as bright as the center. Here's a picture of the Moon from NASA

enter image description here

Image source

I am assuming it's not photoshoped, but I cannot be sure.

For comparison here's Hubble's picture of Jupiter, you can see the edges are somewhat darker so it looks more "round".

enter image description here

Image source

Again I assume this is not photoshopped but I cannot be sure.

Is there something special about the way the Moon reflects light? Is it just because of the distance that naked eye cannot tell it's "rounder"?

$\endgroup$
  • 2
    $\begingroup$ About your observation of darker edges of a round body, see en.wikipedia.org/wiki/Lambert%27s_cosine_law. $\endgroup$ – eranreches Dec 8 '17 at 11:41
  • 3
    $\begingroup$ "Photoshopped" is a tricky word to use. All astronomy photos have been digitally processed to some degree; your example of the Moon is a bit more notable than most (where do you think the multiple images of the t̶i̶e̶ ̶f̶i̶g̶h̶t̶e̶r̶ ISS come from?), but it will generally include modifying the contrast settings to improve visibility of the features one wants to show - which can make the Lambert's-law decline at the edges less visible unless one wants to explicitly emphasize it. The effects of contrast settings are inevitable, and they don't really count as image manipulation. $\endgroup$ – Emilio Pisanty Dec 8 '17 at 12:19
  • $\begingroup$ @EmilioPisanty thanks for the comment :) What I meant was I assumed the Moon and Jupter doesn't look more or less "flat" due to manipulation. Feel free to edit the question if you can think of a better phrasing, or even if you have a better picture to illustrate it. $\endgroup$ – Sejanus Dec 8 '17 at 12:35
  • $\begingroup$ @Sejanus The full Moon looks noticeably darker at the edges to my naked eye under most conditions, so I'm not really sure what you're talking about. All I see here is contrast settings chosen to emphasize other aspects of the picture. $\endgroup$ – Emilio Pisanty Dec 8 '17 at 12:37
  • $\begingroup$ It could be due to the fact that there is no atmosphere on the moon, or very little of it. $\endgroup$ – MaDrung Dec 8 '17 at 12:40
8
$\begingroup$

The moon looks flat because it is very rough, and hence is not a perfect Lambertian reflector.

Many dull objects are well described by Lambert's cosine law: the intensity observed from an ideal diffusely reflecting surface is directly proportional to the cosine of the angle $\theta$ between the direction of the incident light and the surface normal ($I=\min(0, I_0 \cos(\hat{l} * \hat{n} ))$ where $\hat{n}$ is the normal vector and $\hat{L}$ the light direction vector).

However, this is a bad approximation for very rough objects. The problem is that the surface is full of facets pointing in different directions, yet we see an average of their light contribution. This means that a patch on the moon near the edge will have some facets pointing straight at the sun and spreading Lambertian light towards us, looking brighter, and a patch right at the centre will have some facets in shadow, looking darker. This can be handled by more elaborate illumination functions like the Oren-Nayar model (more).

There are some further aspects of lunar geology that makes it slightly retroreflective (see also opposition surge), further reducing the contrast between centre-edge. A lot of this is shadow-hiding: when you are looking almost along the lines of sunlight you will not see the shadows cast by objects because they are of course behind the objects and hence obscured to your vision.

Jupiter is presumably significantly flatter than the Moon (and actually reflects light through a different scattering process). Mars is also rather rough and hence flat-looking in telescope pictures.

$\endgroup$
0
$\begingroup$

The mean Lambertian back scattering from the full moon is directed back to the sun because the scattering dipoles on the moon oscillate in a plane perpendicular to the coming sunlight. Any calculation that does not take the direction of the polarizing dipoles into account will not be correct. Lambertian back scattering yields a uniform moon image, as well as the earth image and the images of all the planets and their moons.

See my paper https://arxiv.org/abs/1808.01024

$\endgroup$
  • $\begingroup$ > "Lambertian back scattering from the full moon is directed back to the sun " -- Such preference for angles is not Lambertian reflection. Lambertian reflecting area has isotropic luminance. This implies sphere with Lambertian surface will appear, when irradiated by uniform light, darker near the outer rim. See graphics.wikia.com/wiki/Lambert%27s_cosine_law , en.wikipedia.org/wiki/Lambert%27s_cosine_law $\endgroup$ – Ján Lalinský Oct 18 '18 at 10:13
  • $\begingroup$ The last comment is only semantics. The point is that there is no further mechanism like rough surface, or dust retro-reflection, or whatever, to explain the image uniformity, as discussed in the web page. Personally I believe that Lambertian scattering is justified here, but anyone may call it whatever he likes. $\endgroup$ – Uri Lachish Oct 18 '18 at 19:40
  • $\begingroup$ My comment meant "your usage of adjective Lambertian in 'Lambertian back scattering' is incorrect". Lambertian spherical reflector does not have uniform image, it is darker near the rim. $\endgroup$ – Ján Lalinský Oct 18 '18 at 20:53
  • $\begingroup$ There is no such a thing "Lambertian spherical reflector" in the sense that it obeys Lambert cosine law. There is a countless number of images, but not a single true photo that obeys the law. All such photos are at least partly simulated, or "rendered", or whatever. So I believe that it is justified that Lambertian scattering means something else. $\endgroup$ – Uri Lachish Oct 19 '18 at 2:06
  • $\begingroup$ Lambertian reflector has a well-known definition, you are not justified to "believe that it means something else". If you mean something else, you need to use different name. $\endgroup$ – Ján Lalinský Oct 19 '18 at 17:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.