Why is 8 am (just after sunrise) darker than 12 pm (noon)? Does it have something to do with the geometry of spheres? Or is it due to the atmosphere?


The main factor is geometry. The Sun doesn't change, and it emits a constant "radiant intensity", defined as "luminous flux per solid angle"


And the illumination depends on it. Imagine a disc on the ground. Of course it is much more illuminated if the source is opposite to it. Explicitly, the illumination is

$ \frac{dF}{dS_2}=\frac{Id\Omega}{dS_2}=\frac{I\cdot \cos\alpha_2}{r^2}$

And so the illumination directly depends on the cosine of the angle of incidence. That's why you want to sunbath perpendicularly.

Flux, irradiance and solid angle

Besides this, the atmospehre plays also a role. When the rays are not perpendicular, they must travel along more thickness, so they are very slightly mittigated (thats why you shouldn't sunbath too much at noon). But this effect is much more noticeable as for the scattering of wavelenghts. That's why sunset and sunrises look red. Check this image:


  • $\begingroup$ I'd make a distinction here between how bright the ground looks and how bright the sun (or sky) looks. Clearly the ground brightness changes because of flux. But if you were to stare straight at the sun (don't do this), it looks dimmer in the morning because of the atmosphere, and has nothing to do with the flux. $\endgroup$ – Jahan Claes Aug 30 '17 at 2:49
  • $\begingroup$ Oh, isn't it equivalent? After all if we can consider the atmospehre as a diffusor (right?), then the radiance $L_e$ it transmits is proportional to the irradiance on it. $\endgroup$ – FGSUZ Aug 30 '17 at 11:27
  • $\begingroup$ Certainly the brightness of the sun itself has nothing to do with the angle of the incidence of the light rays, it's entirely due to the thicker atmosphere. I have to think about the sky around the sun, though. $\endgroup$ – Jahan Claes Aug 30 '17 at 18:39
  • $\begingroup$ The brightness itself doesn't, but the flux does indeed depend on all that. If the brigness is the radiance $L_e$, then the flux is $\frac{L_e \cos (\alpha_1) \cos(\alpha_2) dS_1 dS_2}{r^2}$. When you stare at a light you see it brighter than if you look in other direction, it's also brighter if you are closer to it. At least that's how I saw it.... isn't it? $\endgroup$ – FGSUZ Aug 30 '17 at 19:07
  • $\begingroup$ I'm not sure I'm understanding what you're saying, so let's see if we can agree on this point: If there were no atmosphere, the sun would appear just as bright at 8 a.m. as at noon. The position of the sun in the sky has no effect on its brightness. It ONLY affects how brightly the sun illuminates the ground. Yes? $\endgroup$ – Jahan Claes Aug 30 '17 at 19:59

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