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Treating the moon as an ideal, bright Lambertian reflector, I was trying to compare its efficiency as a light source to an equally-sized flat sheet of drywall. It's pretty straightforward to set the problem up, but I'm not so sure of my calculus. I got that the flat sheet of drywall would provide 50% more illumination in full moon conditions as compared to the spherical moon, treating both of them as smooth, diffuse lossless reflectors. I wonder if anyone wants to second-guess my math?

I talk about the problem on my blogsite here.

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Is there a particular piece of your work that you're not sure of? General work-checking questions are kind of discouraged... – David Z Jan 27 '12 at 6:11
Would it be a better question if I didn't mention that I'd tried to solve it already? – Marty Green Jan 27 '12 at 10:26
Nah, I wouldn't say so. I guess I shouldn't say that work-checking questions are discouraged, really... but I figure if you're asking for someone to check your work, you must have some reason to suspect that it's incorrect, and it'll make a better question if you identify that reason. Think of it like this: "how do I do it?" < "what's wrong with my work?" < "did I use this particular concept correctly?" – David Z Jan 27 '12 at 18:40
Reason? Not really. I just normally assume I'm going to do something wrong when things get complicated. – Marty Green Jan 27 '12 at 21:02
I hope you understand the moon is not a Lambertian reflector. It has a strong retro-reflection component, giving it a "flat" look during full moon. – Mike Dunlavey Jan 28 '12 at 2:44
up vote 1 down vote accepted

EDITED: Indeed, the flat Moon would provide 50% more illumination than the round Moon. I assumed that it is full moon now, so the Sun illuminates the Moon, and we are on axis $z$ connecting the Moon and the Sun (the origin is at the center of the Moon). If $\theta$ is the angle between axis $z$ and some direction, and illumination (total light energy per area - both incident and reflected, as there is no absorption) on the surface of the (round or flat) Moon equals $a$ for $\theta=0$. Then illumination on the surface of the round Moon for some $\theta$ equals $a\cos(\theta)$, as the same Sun light energy falls on a larger area. Therefore, the Lambertian light intensity in the direction $z$ will be proportional to $a\cos^2(\theta)$. Therefore, we need to compare an integral of $a$ over a unit circle with an integral of $a\cos^2(\theta)$ over the surface of the unit hemisphere. (Initially, I integrated $a\cos^2(\theta)$ over the unit circle, rather than the unit hemisphere, and offered a wrong result here).

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I think you have an extra factor of cos(theta) for the spherical moon. Yes, the Lambertian intensity give you a factor of cos(theta) but this only compensates for the greater area of the section. The real effect of the Lambertian profile is that the brightness looks the same regardless of viewing angle, so those factors cancel out. In other words, where you would have an effective brightness of 1/2 at 45 degrees off axis, I would have 71%. – Marty Green Jan 27 '12 at 10:23
I stand corrected. I should have integrated $\cos^2(\theta)$ over the surface of the round Moon, not over the unit circle. Indeed, the flat Moon would provide 50% more illumination than the round Moon. – akhmeteli Jan 27 '12 at 15:18
Very nice. If you edit your correction into your answer I will be glad to mark it as accepted. – Marty Green Jan 27 '12 at 21:01
Thank you, I have edited the answer. – akhmeteli Jan 28 '12 at 0:36
It's a small point, but when I edit a one of my answers I like to leave the wrong answer there and and the corrections under the title "EDIT:". I think the conversation makes more sense that way. But in any case thanks for working it out! – Marty Green Jan 28 '12 at 10:24

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