I understand that the blue colour of the sky is because of the scattering of blue light by molecules in earth's atmosphere. The scattering appears to be happening from molecules that are far above in the earth's atmosphere. What about the scattering that happens because of molecules near the surface of the earth? Why can I not see the blue light scattered by molecules closer to the earth?
[Physics] Why can’t I see the blue color scattered by the lower atmosphere of the earth
atmospheric scienceraman-spectroscopyscatteringspectroscopyvisible-light
Related Solutions
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.
Well, the Earth is kind of blue when viewed from far away:
In the picture above, you see that the "edge" of the Earth is more bluish, which is due to the reason you give in your answer; here we see through a deeper layer of the atmosphere, so we see more blue light.
But the light that does reach Earth's surface is reflected back through the atmosphere, so this is seen as well. And since after all non-blue photons outnumber the blue ones, we see what's underneath the surface.
Best Answer
If I understand you right, you're referring to the phenomenon seen in this picture (from the first Google hit), that near the horison the color of the sky is more light-blue (not exactly white):
Rayleigh scatteringThe scattering in the atmosphere is for a large part Rayleigh scattering off of nitrogen and oxygen molecules, which are much smaller than the wavelength $\lambda$ of the light, and which has a preference for forward and backward scattering. If the Sun is more or less above you, the horison is roughly at 90º, which is the direction on which scattering is least likely.
Rayleigh scattering has an huge wavelength dependence of $\sim 1/\lambda^4$, meaning that blue light is scattered much more efficiently than red light. This is the reason for the blue sky in the first place, and the reason for the red sun at sunset.
Mie scatteringHowever, close to the ground — which is what you look at when you look at the horison — you have more water droplets which have a size of the order of the wavelength. In this case, the scattering is not Rayleigh scattering, but the more general solution of Mie scattering, which is not strongly wavelength dependent. Thus, all the colors are scattered more or less equally, producing a whitish glare.
(My first answer was not good, and did not take into account the Mie scattering. I'm sorry.)