I have searched Wikpedia and the Physics Stack Exchange archieves, and I cannot find the answer to these two related questions. If it is, please guide me to where this information is located.

  1. If Rayleigh scattering is the cause of the blue sky, it is my understanding that the density of the atmosphere must be very low. Rayleigh scattering in the lower and mid atmospheric regions must suppress the lateral scattering of blue light. If this was not true, then blue light would be scattered in the lower atmosphere and distanced mountains would then appear red (if I understand this correctly?). So here is my question: what regions of the atmosphere are doing the bulk of the blue light scattering that we actually see? Is it the stratosphere? The Kármán line? I have no idea.
  2. Is the only reason why we have a blue sky is because of Rayleigh scattering? That is, what role does the atmosphere play here? Since the atmosphere’s density is constantly fluctuating, wouldn’t this play a role in the blue sky picture as well?

Thanks in advance for any help.

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    $\begingroup$ Great question. I'm not guaranteeing this is the whole story, but for (1) perhaps someone should calculate what fraction of sunlight needs to be scattered to give the sky color. If it is small, say 1%, then you could have Rayleigh scattering of "mountain-light" at the same level without leaving the mountain looking red. $\endgroup$
    – user10851
    Commented Jun 28, 2013 at 5:53

1 Answer 1


1) Yes, in the real world, only a tiny portion of the light scatters by the Rayleigh scattering. This may be reinterpreted as the simple fact that generic places in the blue sky are far less bright than the Sun. It means that the generic places of the sky become blue but the Sun itself remains white. For the same reason, distant mountains keep their color. Also, the distant mountains don't increase the amount of blue light from other directions much simply because the intensity of light reflected from distant mountains into our eyes is vastly smaller than the intensity of light coming directly (or just with Rayleigh scattering) from the Sun to our eyes. And even if it were not smaller, e.g. when the Sun is right below the horizon and the mountains are needed, we won't be able to easily distinguish that the blue sky actually depends on the mountains.

Rayleigh scattering is caused by particles much smaller than the wavelength, i.e. individual atoms and molecules, so it doesn't really matter which of them they are. The rate of Rayleigh scattering is therefore more or less proportional to the air density which means that a vast majority of it occurs in the troposphere, especially the part closer to the surface.

2) The changing atmospheric density only impacts the angle of the propagation of the sunlight substantially if the atmospheric density changes at distance scales comparable to the wavelength. If the length scale at which the density changes is much longer than that, the impact on the direction of light is negligible and calculable by Snell's law.

If you ever watch Formula 1 races, you may see some fuzzy waving water-like illusion near the hot asphalt. This is indeed caused by density fluctuations caused by the variable heat near the asphalt (well, a campfire could have been enough instead of Formula 1). However, in this case the direction of light only changes slightly because the regions of hot and cold air are still much longer than the wavelength (half a micron or so).

If you think about ways how to get density fluctuations comparable to the wavelength which is really short, you will see that the source is in statistical physics and the naturally fluctuating air density due to statistical physics is actually nothing else than an equivalent macroscopic description of the Rayleigh scattering! When you calculate the Rayleigh scattering, you may either add the effect of individual air molecules; or you may directly calculate with a distribution of many air molecules and the source of the effect is that their density isn't really constant but fluctuates. So these two calculations are really equivalent. They are the microscopic and macroscopic description of the same thing, like statistical physics and thermodynamics.

If the blue light manages to come from a direction that differs from the direction of the source of light, the Sun, then – assuming that the atmosphere doesn't emit blue light by itself, and it doesn't (at least not a detectable amount of it) – it is scattering by definition. To get a substantial change of the direction, you need small particles, and that's by definition Rayleigh scattering. So there's no other source of the blue sky than the Rayleigh scattering – although the Rayleigh scattering may be described in several ways (microscopic, macroscopic etc.).

Well, there's also the Mie scattering – from particles much larger than the wavelength, especially spherical ones, like water droplets. However, for the Mie scattering to be substantial, you need a substantial change of the index of refraction $n$ inside the spheres, which is OK for water. Also, the Mie scattering is much less frequency-dependent (because $n$ only slightly depends on the frequency, nothing like the fourth power here) than the Rayleigh scattering so it doesn't influence the overall color much. Not only during the sunset, some grey-vs-white strips on the clouds near the horizon are caused by the Mie scattering. The Rayleigh scattering really has a monopoly on the substantial change of the color.

  • $\begingroup$ Even if the density fluctuation are long-wavelength, wouldn't you still expect the light to behave diffusively over long enough scales? The point is there would still be no real frequency dependence. Also by visual inspection it cannot be a very big effect. Probably easy to estimate the mean free path as well. $\endgroup$ Commented Jun 28, 2013 at 18:44
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    $\begingroup$ @Lubos, these are great answers. I expected the answer you gave for question (1) but the depth of answer (2) was fantastic – thanks. However, you now have raised new questions from your response. I will post this in the near future. $\endgroup$
    – Carlos
    Commented Jun 29, 2013 at 13:07
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    $\begingroup$ @Carlos Just so you are clear (although, given the thoughtfulness your questions bespeak, you likely know this already) Rayleigh scattering is the small particle limit of the full theory of scattering (Mie theory) - so they're not fundamentally different. Born and Wolf "Principles of Optics" treats them both in the same framework - Rayleigh scattering is the first order perturbation of an incident plane wave arising from the presence of any scatterer. But Lubos's answer is, as seemingly always, fantastic. $\endgroup$ Commented Aug 5, 2013 at 1:51
  • $\begingroup$ @WetSavannaAnimal: NO – I did not know that Mie scattering is the full theory. What level is Mie scattering at? I am an undergraduate and next quarter I will take graduate quantum mechanics. Would you recommend reading Born and Wolf even though the last edition was in 1970? I see that you are in optics, something that I am really interested in learning. $\endgroup$
    – Carlos
    Commented Aug 16, 2013 at 12:17
  • $\begingroup$ I loved Born and Wolf, but I am fifty years old. Its style and notation is decidedly dated: technical writing has come a long way since then and now we have the hyperlinked internet. I should think Mie scattering would be well within your capability: an easier approach is the multipole method rather than Born and Wolf's more first principles one. See section 7 of math-physics-tutor.com/web_documents/multipole.pdf I can't think of any other good references. $\endgroup$ Commented Aug 16, 2013 at 13:29

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