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I've read a lot of answers to the questions why the sky is blue. However all the answers I found contain mostly qualitative analysis: Rayleigh scattering is changing the direction of blue light, so there is more blue light coming to the eye along the line of sight than the red one.

However these explanations raise additional questions.

First of all, the scheme of only single scattering seems to be an oversimplification: the light direction should be changed more than once. Can we prove that this is negligible by calculation, or is it not negligible? Does this change the analysis?

Further, the explanation says nothing about the exact amount of the blue light being scattered when looking into a particular direction. Assuming the sun is in zenith, it follows from symmetry that the color of the sky in the directions having the same zenith angle must be the same, but closer to the horizon the way of the scattered light differs a lot from the rays coming near zenith — so is it possible to derive theoretically a formula which would predict the color of sky given the azimuth angle and the position of the Sun (at least in a simple geometrical setup when the Sun is in zenith)? It's not clear why the color should not rapidly change from near blue at horizon to almost red near the Sun position: after all, the atmosphere is thicker along the lines going closer to horizon! The sky seems to be more uniformly blue than the typical explanation suggests.

Further, it follows from the usual explanation that blue light is partially reflected back into the space. Due to this, about half of all scattered light should be lost, so the total amount of red light coming from sun should be greater than the amount of blue light, which seems to contradict the observable reality. Does it?

I'm mainly interested in quantitative analysis, not the observations or qualitative considerations.


I've read the answers to this question and know that the physiology of the eye comes additionally into play, but let's neglect this for the sake of simplicity.

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    $\begingroup$ The sky really isn't uniformly blue. Check again ;) Especially at low sun angles. Might be hard to notice if you look at midday and the sun is overhead. If you have a pair of polarized glasses or a polarizing filter for a camera you can enhance the effect $\endgroup$ – Stian Yttervik Jun 23 at 11:09
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    $\begingroup$ @StianYttervik Exactly. Human perception is a miracle -- it corrects little imperfections even before the second bottle of wine ;-). $\endgroup$ – Peter - Reinstate Monica Jun 23 at 11:12
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    $\begingroup$ I've never experienced the Sun at zenith, but I've also never seen a uniformly clear sky. It's quite easy to see that the shade of blue varies with both zenith and azimuth angles, both with the eyes and in any photograph that includes the sky. $\endgroup$ – gerrit Jun 23 at 12:48
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    $\begingroup$ Why does the sky look black in pictures taken from the summit of everest?: "In shots looking outward from the summit there's a very interesting effect of having a black sky above with a blue sky lower down near the horizon." $\endgroup$ – Keith McClary Jun 24 at 3:04
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    $\begingroup$ Why do you think the Sun appears yellow? If you remove the atmosphere, it's very, very white. Mind, Rayleigh and Mie scattering is not the only thing going on. As for multiple scattering, don't forget that the scattering has the same chance of happening the second time, but for far fewer photons. So if we imagine a 1% scattering change, the second scattering will only be 1% of the 1%. There are cases where that matters to the human eye, but it generally needs the sky to already be very dark, e.g. when the Sun is under the horizon. $\endgroup$ – Luaan Jun 24 at 6:51
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First of all, the scheme of only single scattering seems to be an oversimplification: the light direction should be changed more than once. Can we prove that this is negligible by calculation, or is it not negligible?

This is an oversimplification, but for a clear sky at daytime it's not too wrong. See the following comparison of an atmosphere model computed with only single scattering and that including 4 orders of scattering (basically, 4 direction switches per light ray). The projection here is equirectangular, so you can see all the directions in one picture.

enter image description here

This becomes a much more problematic simplification when the sun is under the horizon, particularly noticeable under the belt of Venus, where the Earth's shadow is located:

enter image description here

Assuming the sun is in zenith, it follows from symmetry that the color of the sky in the directions having the same zenith angle must be the same, but closer to the horizon the way of the scattered light differs a lot from the rays coming near zenith — so is it possible to derive theoretically a formula which would predict the color of sky given the azimuth angle and the position of the Sun (at least in a simple geometrical setup when the Sun is in zenith)?

If we neglect non-uniformity of the atmosphere with latitude and longitude, this scenario will lead to the colors independent of azimuth. It's not quite clear what you mean by "position of the Sun" though, if you already put it into zenith. Also, if by "derive theoretically a formula" you mean some closed-form expression, then it's unlikely, given that the atmosphere is not a simple distribution of gases and aerosols. But it's possible to calculate the colors numerically, and the above pictures demonstrate this calculation done by my (work in progress) software, CalcMySky.

It's not clear why the color should not rapidly change from near blue at horizon to almost red near the Sun position: after all, the atmosphere is thicker along the lines going closer to horizon!

It shouldn't be bluer at the horizon than at the zenith. After all, you have relatively small thickness near zenith, which makes most of the light scattered to you not too extincted due to Beer-Lambert law, while near the horizon the thickness is much larger, and the light scattered into the observer, in addition to becoming bluer due to Rayleigh scattering depending on wavelength, becomes also redder due to extinction along this long path. The combination of this bluing and reddening effects gives a color closer to white (which you can see in the daytime simulation above), or reddish-orange (in the twilight).

Further, it follows from the usual explanation that blue light is partially reflected back into the space. Due to this, about half of all scattered light should be lost, so the total amount of red light coming from sun should be greater than the amount of blue light, which seems to contradict the observable reality.

Yes, the Earth indeed looks bluish from space, so the total radiation incoming from above should be redder at the ground level than at the top of atmosphere. But this is modified by the ozone layer, without which we'd have sandy color of twilight instead of blue. See for details the question Why is there a “blue hour” after the “golden hour”?

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    $\begingroup$ So you basically mean that the formula can be replaced with computation/modelling? If so, could you please shortly add some words about your mathematical model? $\endgroup$ – Vlad Jun 23 at 12:00
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    $\begingroup$ About sun in zenith: I used that assumption only for the geometrical setup explanation. Of course I'd prefer the computation in the general case for arbitrary position of Sun. $\endgroup$ – Vlad Jun 23 at 12:09
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    $\begingroup$ @Vlad for a very good explanation of the calculations including single and multiple scattering please see E. Bruneton's paper Precomputed Atmospheric Scattering. My model is heavily based on this one. Even more information on calculations, as well as inclusion of ozone absorption, you can find in this update to the original example that came with the paper. $\endgroup$ – Ruslan Jun 23 at 12:15
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    $\begingroup$ Nice article! I assume it's worth mentioning in the answer itself, as the comments are often ignored by the majority of readers. $\endgroup$ – Vlad Jun 23 at 12:31
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    $\begingroup$ This could answer (+1) could be further improved by adding a photo of the sky from which the colour variations are immediately apparent. $\endgroup$ – gerrit Jun 23 at 12:51
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Short explanation is this. Red light comes directly from Sun almost un-scattered or scattered to small degree. And when blue light enters atmosphere it gets scattered by air molecules a lot in each direction, thus according to Huygens-Fresnel principle making each point in atmosphere as a secondary sources of blue light. These blue light sources adds-up along the direction of view, which in the end increases intensity of blue waves, compared to the red ones which reaches us only directly from Sun. So speaking by analogy, Earth atmosphere acts as a some sort of optical lens, focusing blue light towards direction of view. Schematics :

enter image description here

Of course this is a bit oversimplified, because blue light is scattered in ALL directions across the air. You can imagine a thousands of blue light bulbs turned-on in the sky. Maybe this would be better analogy, because each point in air acts as an ambient light source for blue waves.

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    $\begingroup$ And a down-vote is for ... ? $\endgroup$ – Agnius Vasiliauskas Jun 23 at 8:35
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    $\begingroup$ Perhaps because this is only a qualitative consideration? (Countered the downvote with an upvote.) $\endgroup$ – Vlad Jun 23 at 9:30
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    $\begingroup$ I wish I could vote twice for hand sketches. Always better for explaining. $\endgroup$ – Stian Yttervik Jun 23 at 11:10
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    $\begingroup$ If the blue light is focused on me, why does my neighbor 100m down the road also see a blue sky? Shouldn't there be less blue light available for his sensors, given that it's preferentially arriving at my sensors instead? $\endgroup$ – Daniel Wagner Jun 25 at 14:31
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    $\begingroup$ @DanielWagner "Of course this is a bit oversimplified, because blue light is scattered in ALL directions across the air." Agnius has already conceded that this model is inaccurate in precisely the way you describe. $\endgroup$ – jpaugh Jun 25 at 18:37
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Here are some answers, albeit back-of-the-envelope.

At a reasonably good site with a low amount of atmospheric aerosols and dust, the "extinction" is about 0.3 magnitudes per airmass at 400 nm, in astronomers units, compared to about 0.1 mag/airmass at 550 nm and about 0.04 mag/airmass at 700 nm.

What this means is that if light travels through the atmosphere at zenith, then a factor of $10^{-0.3/2.5}=0.758$ of blue light makes it to the ground, compared with a factor of 0.912 for green light and 0.963 for red light. Most of the remainder will be Rayleigh scattered (although there is some component from atmospheric absorption and scattering by aerosols in these numbers).

From this you can see that multiple scattering cannot be negligible for blue light, because at least a quarter of it is scattered by just travelling through the minuimum possible amount of air between space and the observer.

The next point: yes, it is possible to calculate the spectrum of the daylight sky given the appropriate atmospheric conditions (the run of density with height) and the aerosol content (the latter is important because the dependence of the scattering cross-section on wavelength is much more uniform than for Rayleigh scattering). Is there a simple formula - no. An example of where detailed calculations have been set out in great detail can be found here.

Then, why doesn't the sky become red near the Sun? Why would it? Red light is not effectively scattered, so red light that is emitted by the Sun does not get scattered towards the observer. On the other hand if you look directly towards the Sun (please do not do this) then blue light is preferentially scattered out of the direct sunlight, and indeed the Sun is "redder" than it would appear from space (plot below).

The only source of illumination from directions that are not towards the Sun are from scattered light. If we ignored multiple scattering and aerosols then that scattered light would have a spectrum that was proportional to the illuminating light multiplied by the Rayleigh scattering cross-section. The illuminating light does get progressively redder as the zenith angle increases (because the illuminating beam has to travel further and deeper through the atmosphere), so you would expect a whiter colour near the horizon, transitioning to a deeper blue higher above the horizon. However, this is not a very strong effect because only a quarter of blue light is scattered per airmass (and the eye has a pseudo-logarithmic response to spectral flux). Note though that in practice aerosols are not absent and that scattering from aerosols and particulates has some concentration in the forward scattering direction, which messes up this simple prediction, by making the sky whiter near to the Sun. Multiple scatterings also make the sky whiter near to the horizon because some of the blue light coming from that direction is then scattered out of the line of sight.

This is perfectly illustrated by a calculated sky image that shows the separate contribution of Rayleigh and aerosol (Mie) scattering (taken from this website, which does quantitative calculations, but which does not take account of multiple scattering). The sky is quite white near the horizon, then becomes a deeper blue at higher angles and is finally quite white again near to the Sun because of Mie scattering. Rayleigh + Mie scattering

Your final point, I've already covered. Yes, direct sunlight arriving at the surface of the Earth is "redder" than received at the top of the atmosphere. Plot below from https://commons.wikimedia.org/wiki/File:Solar_Spectrum.png

enter image description here

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    $\begingroup$ Multiple scattering is not needed to explain whitening of the sky near the horizon (extinction due to scattering must be take into account though). With multiple scattering, it becomes just more white, but the qualitative phenomenon is apparent even with single-scattering model. $\endgroup$ – Ruslan Jun 23 at 10:35
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    $\begingroup$ @Ruslan. How does the whitening work with single scattering? By single scattering do you just mean that you don't follow the multiply scattered photons. It seems that if you take into account "extinction due to scattering" that multiple scattering has taken place. $\endgroup$ – Rob Jeffries Jun 23 at 11:55
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    $\begingroup$ @Ruslan My answer already says that a deeper blue is expected higher up (and therefore whiter near to the horizon) because the illuminating light will be redder as a result of extinction. Is this what you mean by the extinction effect? $\endgroup$ – Rob Jeffries Jun 23 at 11:59
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    $\begingroup$ Yeah, I mean I'm sampling direct irradiance at the points on the light ray from observer (points of scattering), taking into account how much the light was extincted on the path between the Sun and the sampling point, and then between the sampling point and the observer. This is cheap enough to compute in real time. What I (and other implementers of calculations) call multiple scattering is simulation of scattering of a single "photon" happening more than once. While single scattering requires only one-dimensional quadrature per pixel, this one requires integration in 4D, 7D, 10D etc. spaces. $\endgroup$ – Ruslan Jun 23 at 12:09
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    $\begingroup$ @EricDuminil Not vintage xkcd: physics.stackexchange.com/questions/28895/… $\endgroup$ – Rob Jeffries Jun 23 at 14:49
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the light direction should be changed more than once.

Yes, that happens. The blue light that reaches you has probably been scattered several times. It makes no fundamental difference, it's still coming from everywhere and it's still blue.

It's not clear why the color should not rapidly change from near blue at horizon to almost red near the Sun position

It does change markedly. Dust and the long distance to the horizon make low-level light less pure and typically containing scattered direct green-and-red too, so it is more of a washed-out "sky blue" compared to the stronger, deeper blue of the zenith. But the human optical system compensates so the grading is often barely noticeable. Remember that the blue is scattered multiple times, so even close to the Sun the sky is still dominated by scattered blue.

about half of all scattered light should be lost, so the total amount of red light coming from sun should be greater than the amount of blue light

Astronauts have described the Earth seen from space as a "blue marble" so you are right on the money there. What we have evolved to think of as "white" is yellower than the Sun's actual color.

I'm mainly interested in quantitative analysis

You have asked a lot of qualitative questions though. Quantitative models depend heavily on the height of the Sun in the sky and the composition/pollution of the atmosphere both visible and within scattering range of the horizon and at different altitudes.

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but closer to the horizon the way of the scattered light differs a lot from the rays coming near zenith

The horizon has about 5 km of radius. The difference of the distance through the atmosphere in negligible between the center where the observer is and 5 km from him.

Further, it follows from the usual explanation that blue light is partially reflected back into the space

The blue light scatters to all directions, including to outer space. The argument for the lack of blue at sunset is different. The sunlight travels thousands of km through the atmosphere, scattering blue light during this path.

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    $\begingroup$ Well, this looks like qualitative reasoning. (Besides, rays which are close to horizontal direction clearly go longer way inside the atmosphere than the vertical ones, right?) $\endgroup$ – Vlad Jun 23 at 1:04
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    $\begingroup$ @vlad it depends. If the sun is at the zenith, and assuming 100 km for the atmosphere, the distance travelled for a ray to our horizon is $(100^2 + 5^2)^{1/2}$. Very small difference, comparing to 100 km. $\endgroup$ – Claudio Saspinski Jun 23 at 1:33
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    $\begingroup$ Your statement about horizon is very misleading. You only measure the distance from the 2  m high observer to the apparent horizon. But the atmosphere right above the horizon is much thicker along the line of sight, so luminance from that direction is much higher than from the ground at the horizon. $\endgroup$ – Ruslan Jun 23 at 6:12

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