My question is a duplicate of this; Clarification on Rayleigh scattering causing various sky colors. The accepted answer from the link above says that at sunset the scattering occurs farther away and does not reach the observer, which is unsatisfactory and vague to me. (not even sure if it's the right and correct answer)

It doesn't quite make sense to me that at noon, we see scattered blue light whereas, at sunset, we see the sunlight itself minus the scattered blue light. At noon or at sunset, we are not seeing the sun, but the scattered sunlight indirectly. The logic should apply both cases equally and should imply that at sunset we see bluer sky than at noon because of more scattering.

Please let me know if this duplicate question was unnecessary because the accepted answer from the link above was enough to answer the question.

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    $\begingroup$ Perhaps you could slightly rephrase a bit. Saying that the answer "doesn't quite make sense" when it's you who just didn't understand it sounds quite arrogant. Your comments under the answers also give me the same impression. $\endgroup$ – YiFan Jan 5 at 4:20
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    $\begingroup$ Does this answer your question? Clarification on Rayleigh scattering causing various sky colors $\endgroup$ – Nij Jan 5 at 8:47
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    $\begingroup$ @YiFan I am not sure where you sensed the arrogance but you should check again what I was referring to with "doesn't quite make sense". Try finding the subject of the sentence. $\endgroup$ – Seung Jan 5 at 14:38
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    $\begingroup$ The OP here has made it clear it's a duplicate, and that they just don't like the answers there. The appropriate action in that circumstance is to place a bounty on the original question seeking greater clarity in the answers, not to post the same question again. @Ruslan $\endgroup$ – Nij Jan 5 at 20:25
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    $\begingroup$ Expecting people to make an effort is deliberately unwelcoming to those who aren't going to effectively contribute. Even if one can't bounty the question, it likely would have sufficed to comment on an answer that isn't quite clear, or find the chat. You could have easily clicked the username as well, to see they have multiple profiles several months old - they are not unfamiliar with Stack Exchange. @Ruslan $\endgroup$ – Nij Jan 5 at 22:51

From one of your comments to other answers:

I think the sunlight doesn't lose its blue that much on the way to the point where red scattering dominates.

Why wouldn't it be able to?

Rayleigh scattering has cross section proportional to $\lambda^{-4}$. The spectrum of light illuminating the volume scattering it gets multiplied by $\lambda^{-4}$, which indeed makes smaller wavelengths amplified more than larger ones. This indeed makes the scattered light, when seen immediately after scattering, bluer.

But on the other hand, this same scattering mechanism removes corresponding amount of power from the light that hasn't gotten scattered and continues propagating forwards. What we now have is this factor of the scattering cross section getting into the exponent of the Beer-Lambert law. In the limit of large distance $d$, the factor $\lambda^{-4}\exp(-d \lambda^{-4})$ has a reddening effect, not bluing. Here the distance you have to take into account is the sum of 1) distance traversed before scattering and 2) distance traversed to you after scattering.

Now, even if you take into account double scattering, which would give you a second factor of $\lambda^{-4}$, you'll still get $(\lambda^{-4})^2\exp(-d_\Sigma \lambda^{-4})$ where $d_\Sigma$ is the sum of distances traversed by the sunlight 1) before first scattering, 2) between first and second scatterings and 3) after second scattering until it got to you. This is even larger distance, and exponential function in the above mentioned expression, again, is much more selective by wavelength than the simple power in the factor it's multiplied by.

The result of this reddening, apart from the orange solar disk, is the phenomenon of Belt of Venus — the area of the sky on the opposite of the solar azimuth, which has purple-red color.

There's also a peculiarity of the Earth atmosphere that fools humans into thinking that the blue sky at day and blue sky at twilight have the same reason. Actually, if the atmosphere had no ozone, twilights also wouldn't even be as blue as they are. Beer-Lambert law would have much greater effect. On the Earth the twilight is blue because the sunlight, traveling in the upper parts of the atmosphere, gets absorbed in the red part of the spectrum by the Chappuis absorption band of ozone. If not the ozone layer, twilights would actually have sandy-brown color.


When you look at the sun overhead (not advisable) you see the white light from the sun with a little bit of the blue scattered horizontally. The blue sky is blue light scattered from sunbeams going elsewhere. When the sun is near your horizon, its light passes through a much greater distance of dense atmosphere, and most of the blue is lost from the beam to give other people a blue sky.

  • $\begingroup$ I understand from your answer that the light scattered farther does not reach me. But surely the sunlight that scatters close to me should make the sky blue. Your answer also doesn't explain the red sunset. $\endgroup$ – Seung Jan 4 at 19:46
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    $\begingroup$ @Seung "the sunlight that scatters close to me should make the sky blue" It does! Look to your left/right and you'll see blue. $\endgroup$ – Lightness Races with Monica Jan 5 at 3:23
  • $\begingroup$ @Seung The particles in the air scatter both the red and the blue end of the spectrum, but the blue light scatters more. Scattering sends light in random directions, and what you see as the blue of the sky are those rays that happen to hit your eye. During daytime, there's also direct sunlight which remains white-yellowish, and includes a fair ammount of the blue light that didn't scatter. In the evening, more of the blue component gets scattered (contributing to the blue sky color), and so direct sunlight appears reddish, but the red bleeds out a bit because it scatters too, to a lesser deg $\endgroup$ – Filip Milovanović Jan 5 at 11:02

Considering 11 km as the thickness of the atmosphere that scatters the blue light, when the sun is just at the horizon, the length travelled by the sun light is:

$$l = \sqrt{(6371+11)^2 - 6371^2} = 374 km.$$

When that rays come to the place where we are looking at the horizon, much of the blue components were already scattered along the long journey through the atmosphere.

We see light deprived from bluish frequencies, what look reddish to our eyes.

  • $\begingroup$ Your math does not make sense, but this idea was my original speculation, but I threw it away because Rayleigh scattering scatters away blue very little amount. $\endgroup$ – Seung Jan 4 at 19:31
  • $\begingroup$ I forgot the sqrt... $\endgroup$ – Claudio Saspinski Jan 4 at 19:37
  • $\begingroup$ Ah it makes more sense. Anyway, I think the sunlight doesn't lose its blue that much on the way to the point where red scattering dominates. $\endgroup$ – Seung Jan 4 at 19:49
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    $\begingroup$ @Seung : The scattering rate is 5%/km for light near 400 nm. So at a depth of 11 km (noon), $(1-0.05)^{11} \approx 57\%$ of the blue light has not been scattered -- that is about half of the 400 nm rays from the sun that reached the upper atmosphere are still travelling, undeflected by Rayleigh scattering when they reach your eye. At 374 km (sunset), $(1-0.05)^{374} \approx 5 \times 10^{-7}\%$ of the blue light is still travelling directly to you from the Sun. The first half of the light was scattered in the first 11 km, more than 374 - 11 = 363 km away from you. $\endgroup$ – Eric Towers Jan 5 at 10:17

The Rayleigh scattering cross-section from nitrogen molecules is around $1.6\times 10^{-30}$ m$^2$ at 400 nm (blue light). At sea level, there are about $2\times 10^{25}$ molecules per cubic metre. The mean free path of a blue photon is therefore 31 km. Since the Earth's atmosphere has a scale height of around 8 km, then the vast majority of photons travelling from the Sun at zenith, directly towards the ground, make it to sea level. Hence the colour of the Sun is changed very little when it is overhead.

It is also the case that because the mean free path of a blue photon is greater than the atmospheric scale height, that most photons that are scattered, are only scattered once before reaching the ground, because the location where they are scattered is closer than their mean free path to the observer.

On the contrary, when we look towards the horizon, the path length through the atmosphere is drastically increased. When the Sun is 5 degrees above the horizon (about 20-30 minutes before sunset) the "airmass" that photons must pass through is increased by a factor of 10. So now there are roughly 3 mean free paths of nitrogen in the way of direct photons from the Sun and there is therefore almost no chance of a blue photon traversing this without being scattered. Further, any blue photons that are scattered through a small angle are also unlikely to make it to the observer, because they will be multiply scattered. The result is that blue photons are almost absent from a direct line towards the Sun.


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