To add to this lively discussion, I think it is obvious that the very selective spectral response of the eye is a key factor here. However, we might ask the question whether an ideal detector with a perfectly flat spectral response would "see" the sky as blue or violet. I'm not sure this would the case, at least under all atmospheric conditions.

To help quantify the issue, I generated the skylight spectral irradiance for two extreme conditions using realistic atmospheric data measured with AERONET sunphotometers and my SMARTS code. First for Mauna Loa (3397m amsl) under very clear conditions that should induce a dark-blue sky (Angstrom's beta = 0.005) on 5 Jan 2009. Second, for Kuwait Airport (sea level) under hazy conditions, conducive of a whitish sky (beta = 0.667) on 1 May 2009. Both are for an average sun position (Z=45°). From the plot below, it is obvious that there is relatively more violet at Mauna Loa than in Kuwait under such conditions. However, to grasp the complete picture, I think a more elaborate analysis using the CIE chromaticity system would be in order.

It is explained on the basis of the sensitivity of our eye's color receptors. Our eye is more sensitive to blue than violet or other UV radiations. Another point is that the emission intensity of solar spectra is not constant through out the spectra.

Thank you for your response. However, I beleive that a quantative investigation as well as experimental measurements are essential to approve the points you mentioned

I think we can get enough information on this from the existing literature. Here is one paper http://ajp.aapt.org/resource/1/ajpias/v73/i7/p590_s1

An explanation why sky is blue was given as below:

"Human color vision and the unsaturated blue color of the daytime sky"

Glenn S. Smith ,Am. J. Phys., Vol. 73, No. 7, July 2005

However,if the atmosphere spectrum obeys Rayleigh scattering then it should be seen violet. Does the spectrometry of atmosphere exhibit that violet color is dominant?

If our eye is more sensitive to blue than violet ,why we can see the violet color in the sky during rainbow?

@Parviz. Because violet is spatially (angularly) separated in a rainbow: the retina cones detecting violet and blue are spatially separated. For the sky colour the same detectors are at work for both blue and violet and since they are more sensitive in the blue part of the spectrum the sky is blue.

From the solar spectrum below (taken from the french version of wikipedia : https://fr.wikipedia.org/wiki/Spectre_solaire

one can see that violet and UV light from the Sun are much more absorbed by the atmospere as blue is absorbed. This is mainly due to the ozone layer in the stratosphere (https://en.wikipedia.org/wiki/Ozone_layer). So violet is more scattered by the air molecules but it is also more absorbed by these molecules. This contributes to the blueness of the sky.

Charles. Thank you for your comments. According to Rayleigh scattering and the solar spectrum still sky color my be heavy blue and not sky blue!

@Parviz. When you climb up mountains, by sunny days :-) you see a heavy blue sky because there is less absorption of the high frequencies due to a thinner atmosphere.

Charles. How do you interpret the white color of the clouds( may be due to Mie -scattering)? Why Rayleigh scattering does not work along the distance between cloud and our eye?

Yes, particles in clouds are much larger than air molecules. Rayleigh scattering for molecule is a limit to the general scattering law due to Lorentz and Mie. Rayleigh scattering is wavelength dependant (omega^4) but isotrope, while Mie for clouds is wavelength independent but takes place mostly towards the forward direction.

You can consider the thickness of the atmosphere to be of the order of 100 km while the altitude of clouds is only less than 10 km. So, yes there is Rayleigh scattering between the clouds and our eyes but not enough (by a factor of 10 at least) to be observable.

Hope it helps.

Charles. What a nice dialogue it is and thank you indeed for your wise response. How do you explain red color in the sunset and how do you model the red atmosphere of the Mars?

Charles. I agree that cloud looks like white according to your comments, but why moonlight (color of the moon ) is white? It pretends that Rayleigh scattering does not work well comparing the clouds?

The color is always a balance between scattering (Mie theory) and absorption. Both opticla properties have to be taken into account to understand the color of environmental objects.

For the clouds you are at the limit of the theory, water droplet ar big compared to visible wavelenght, so they scatter exactly what they receive, that is white light.

The same observation can be done for pure sea water, in abscence of anyother optical significant matter in water (such as phytoplankton) sea water should be violet, and in fact it is, as observed in the south pacific gyre during summer, the equivalent of a desert... very beautifull image to see, purple water in the middle of the ocean: http://www.insu.cnrs.fr/expeditions-et-campagnes/biosope/l-eau-de-mer-la-plus-pure-au-monde.

As soon you add phytoplankton, it increases the absorption in the blue, so decreases the reflectance (color proportionnal to scattering and inversely proportionnal to absoprtion) in this part of the spectra (water aer no more purple). The more phyto you add, the more absorption in the blue,/green (the peak is shifted from 443 to about 500 nm), and the more scatter you have in the grren. So you will have greener water.

THe color in the sunset is due to the longer path of the sun ray, which then have more chance to be scattered by aerosls, which scatter light following Mie theory (assuming that they are spherical), that is on average following a lambda-1 spectral law. For big aerosols you will have an almost flat spectral law. So, even if you still have more blue light scattered, you will have more orange/red photons compared to a pure sky with no aerosols.

BEst

hubert

Dear Hubert

According to your comments, Blue-Green color is found to have longer penetration and minimum attenuation in the oceanic water. The other lights are strongly absorbed in deep layers such that oceans usually look like blue-green.

Regards

Parviz

According to black body radiation, dominant color of sun is yellow -orange . Why its reflectance from the moon resembles to be white ?!

In addition to the Rayleigh Scattering law, two parameters need to be factored in: (a) the spectral distribution of the solar radiation, with maximum at approx 557 nm, tapering off towards the blue and the violet, and (b) the spectral sensitivity of the human ocular response. Both help explain why we see 'blue' and not 'violet', which otherwise would have been much stronger, according to Rayleigh inverse quartet law.

George. Thank you for your comments. In order to see sky blue, another fact my be the volumetric property of Rayleigh scattering, while Rayleigh scattering is not isotropic and angularly forward scattering (teta=0 degree) and backward scattering(teta=180 degree) are dominant.

This is a problem related to the wave-particle interaction & scattering theory. Scattering is a function of size of the particle, wavelength of incident radiation, angle of incident radiation and absorption cross section. That is why clouds will be white-grey sort of depending on their constituents and density (Mie scattering), while dark blue when the scattering is by nitrogen molecules (clear sky time). When sun is setting angle comes into the play and you observe long wavelength of incident radiation dominating.

What we observe is depends on the detector sensitivity. Our eyes are sensitive in visible region hence mostly we observe blue (in day) and red (in twilight). For more clarification you may read any standard book on scattering theory which are easily available in internet.

Though earlier good responses are given by others, I hope that this simple explanation will be of help to you.

Taori, Thank you for your explanation . Can you reply me the following question :"According to black body radiation, dominant color of sun is yellow -orange . Why does its reflectance from the moon resemble to be white ?!".

The color you see depends on the average temperature of the black body radiation (for finer details read blackbody radiation). Solar surface has about 6000 K temperature which according to Weins law shall result in visible wavelengths. Moon absorbs solar radiation and re-radiates in long-waves which is seen by us as White.

Hope this clears your doubt.

Taori.Thank you for your answer. I believe that sun wavelengths longer than yellow-orange may exhibit relatively red light from the moon based on your comment (according to the relative spectral abunance ). Otherwise, moon soil may absrorb not only visible light but also UV spectral region of the solar irradiation ( moon has no atmosphere to absorb UV ).Therefore, moon does not reflect sunlight whereas moonlight arises from the fluoresence of soil by aborption of UV (and shorter visible wavelengths) regarding the sun dominant spectral region (yellow-orange), then moon emits longer waves respect to UV such that the mean spectral irradtiance covers whole visible lights ( white). Nevertheles, I think this should be checked by inspection of the fine spectra of the moonlight and sunlight. However , I am not sure yet whether moon reflects light or re-emit the absorbed light at longer wavelengths(fluoresence)?!

After NASA missions to the moon and lunar explorations in 1970's, the compositions of the moon soil should be reported. Are there articles available on the moon compositions, its optical properties, particularly the fluoresence property ?

Red and orange tinted Moon, as seen from Earth during a lunar eclipse, where the Earth comes between the Moon and Sun, my be an interesting evidence that if earth blocks most of solar UV radiation (SUVR) then lunar rocks and lunar soil re-radiate at longer visible spectal range while SUVR normally is the main source that moon resemles white.

Because the sun light has stronger spectral power density in blue range compared to violet one.

According to Sun radiation spectrum , the ratio of blue range intensity to violet one is less than 1.5, while Rayleigh cross-section indicates that the quartet walength ratio of violet to sky blue , i.e. (450 nm/400 nm)^4 =1.6, is greater than 1.6. It benefits on behalf of violet and means violet may be stronger overall ?!

If we integrate AM1.5 spectrum between 450-480 nm (blue region, the result is 50.2 W/m2 reaching ground level) and also between 380-450 nm (violet, 71.7 W/m2 reaching ground), there will be about 1.67 times more power within the violet range. But the ather factors are the eye's sensitivity being greater for the blue region, and the possibility that most substances in the atmosphere (like water vapour) absorb UV/violet stronger than blue. Maybe that is why the sky looks more like the violet and blue colours are somewhat mixed on very dry and clear days.

Dear mikhail

1) According to rough calculation on Sun radiation spectrum integrating over blue and violet range separately, violet emission power exceeds blue one as we expect.

2) Regarding the Rayleigh scattering cross-section, the violet is more abundant in atmosphere than sky- blue photons as we expect.

3) Items 1 and 2 emphasize the dominant attendance of violet photons in number density and based on scattering.

4) The higher absorption rates of UV/violet respect to blue means that in dry atmosphere, the contribution of violet is dominant, however sky does not look like violet even in dry atmoshere?!

I agree, the answer may then be more related to the eye sensitivity curve

Can anyone add comments on the sensitivity of human eye to verify our present discussion ?

Definitely, the response of the human visual system plays an important role. Please see an attached article by Glenn S. Smith who convincingly demonstrated that "the average observer sees daytime skylight as blue because the response of the eye to the spectrum of skylight is the same as that to a mixture of monochromatic blue light ~spectral blue light and white light, that is, it is the same as the response to unsaturated blue light"

To add to this lively discussion, I think it is obvious that the very selective spectral response of the eye is a key factor here. However, we might ask the question whether an ideal detector with a perfectly flat spectral response would "see" the sky as blue or violet. I'm not sure this would the case, at least under all atmospheric conditions.

To help quantify the issue, I generated the skylight spectral irradiance for two extreme conditions using realistic atmospheric data measured with AERONET sunphotometers and my SMARTS code. First for Mauna Loa (3397m amsl) under very clear conditions that should induce a dark-blue sky (Angstrom's beta = 0.005) on 5 Jan 2009. Second, for Kuwait Airport (sea level) under hazy conditions, conducive of a whitish sky (beta = 0.667) on 1 May 2009. Both are for an average sun position (Z=45°). From the plot below, it is obvious that there is relatively more violet at Mauna Loa than in Kuwait under such conditions. However, to grasp the complete picture, I think a more elaborate analysis using the CIE chromaticity system would be in order.

Dear Chris

Thank you for your lucid comment. According to the experiments with flat response detectors, it is obvious that Mie scattering in hazy atmosphere changes the sky color. However, in clear sky, you mean that the sky really contains more violet content than it looks like by our eyes? and a flat response detector reveals this fact. In fact, it approves that the atmosphere does not absorb violet stronger than blue.

In this case, our eyes may be an illusive detector to deceive us! However we detect violet and we see the violet color in the rainbow. According to spectra above, perhaps we need further investigation on the spectal respose of human eye. I believe that some facts are still missing.

I went a step further by convolving the diffuse irradiances in the plot above with the CIE spectral response of the eye. This results in the effective eye response to the previous scenes, albeit now in relative units only. It is clear than now the peak eye response (at 555 nm) becomes the main factor The Kuwait scene peaks at that wavelength, while the Mauna Loa scene peaks at 540 nm, just barely closer to the blue. Since we don't see the sky as yellow, something else is at play here. That's why I think we need to use a full-fledged colorimetric analysis, because all colors subjectively affect the eye's perception.

In summary, I am not convinced yet. The sky should look like violet based on sensitive spectrometry and in agreement to Rayleigh scattering. If it resembles light(sky) blue,as it is , then it may be more related to the eye sensitivity curve. However, the above comments above indicate peak eye response at 540 nm closer to blue. Since we don't see the sky as yellow, something is missing and this question is still live requiring deep quantitive investigation.

A mix of yellow with blue will produce green, and a mix of blue-violet with orange+red will produce purple or pink. I think if we convolve and integrate all factors and functions together, the colorimetric (photopic curve-based) analysis will like ly show what we already see. The closest laser colour is probably 473 nm. Oddly though, the "sky blue" region of the visible light spectrum is itself not so wide in nanometers and not so much of the solar energy is contained within that region. It's probably that mix of magenta (violet + orange + red) and blue-green (yellow + blue) that gives the perceived sky-blue. Inside rainbows, the violet is seen clearly because of the spatial separation of the violet rays, despite the drop in eye sensitivity.

This is how I understand it. There are four factors: (i) the emitted solar spectrum (more blue than violet); (ii) Rayleigh scattering (more violet than blue); (iii) the spectral sensitivity of the eye (more sensitive to blue than to violet), and (iv) the perseption of the color. The last point refers to the already pointed out fact that the apparent color depends on the relative contribution of different wavelengths detected by the eye, not just where the peak value is.

As a second example, Solar spectrum peaks at green, but we see Sun as yellow, because the combined effect of solar spectrum, Rayleigh scattering and the spectral sensitivity of the eye gives more move weight to wavelengths on the longer size of the peak (yellow, orange, red). This tilts our impression of the color to yellow.

By the way, you can see an approximate curve for the spectral eye sensitivity, e.g., in Wikipedia (http://en.wikipedia.org/wiki/Spectral_sensitivity). As you can see, eye is very insensitive to violet.

Dear Timo

According to the approximate spectral sensitivity of eye and photographic emulsion given by:

(http://en.wikipedia.org/wiki/Spectral_sensitivity),

for blue (450-490 nm) and violet ( 380-450 nm), eye may be able to detect part of violet comparing blue( cut off wavelength ~ 400nm), and the sky photo does not show to be violet but looks like blue (even with cut-off wavelength ~ 350nm), eventhough the corresponding sensitivity extends to violet. On the other hand, we can see blue and violet colors easily whereas according to that curve, the eye sensitivity is very poor over that spectral range relatively. Therefore, I do not assure of the definition and measuring methods of spectral sensitivity of the sensors: eye, photographic film, CCD, etc. If those sensitivity curves are true, then

sky should be seen deep blue or violet. Hence, I agree with Mikhail that the colorimetric analysis may help to answer the question.

Dear Colleagues,

perhaps thing is that impurities, contaminants and water droplets in the atmosphere scatter not according Rayleigh law, but by the laws geometrical optics, which shifts violet to blue.

What Vladimir talks about is explained by Mie scattering on large particles, such as aerosols. However, for most of them, scattering is also more intense at shorter than longer wavelengths. The power law is only weaker in their case, compared to molecules: wavelength^-4 for molecules, vs. wavelength^-1.3 for rural aerosols, for instance. Therefore, we cannot expect any shift in color from that process. However, it makes the sky appear with a less intense blue than if no particles existed. If the concentration of these particles--particularly large humid particles--becomes extremely high, the sky becomes hazy and whitish or yellowish, depending on the type of aerosols.

Dear Chris,

your are right. But one little note. Mie scattering theory does not account that layers of large particles or aerosols may be dislocated beneath the wide layer that produces violet color. Therefore the aerosols reflect backward and scatter outside the part of violet radiation more intense than longer wavelengths! This is the same that makes the sky yellow in the close to the sun area, especially in dirty atmosphera - you just pointed to this fact too.

If elastic and inelastic scattering events were the only sources of spectral shift, then sky should appear violet. The power density of sun shows strong violet in the spectral range. The eye detects violet, the same as blue according to its sensitivity with minor discrepancy. The absorption rate at violet is not much greater than blue.

Hence, this question remains live for further profound study.

I have climbed up to ~3750 m over sea level where the sky color resembles dense blue, however it is not violet yet !

Rayleigh scattering may not be the single dominant mechannism of scattering around blue-violet specral range, otherwise the sky might have seen deep blue!

What is the competitive mechanism to prevent violet scattering being dominant?

Parviz, it really isn't only about scattering. You also need to account for the solar spectrum (not so much power at violet), the sensitivity of the human eye (not so sensitive to violet), and the fact that the color perception is a result of how the three difference cone cells in the eye are stimulated, and these all are sensitive to different ranges of wavelengths.

So, it is rather complicated, but still, I see no obvious obstacles for modeling this, if all the functional dependencies are available. I would imagine that somebody has actually done this already.

Dear Timo

Thank you for your comment. Regarding eye sensitivity, we can discern violet color of rainbow, hence I believe that we can see violet color in sky. Furthermore, violet spectral range of the solar radiation is relatively significant, to my knowledge. This question remains unsolved to me!!

There are two parts in solving this issue. The first task is to establish the spectrum of the downwelling diffuse radiation at the surface. This is a radiative transfer problem, and this is where the Rayleigh scattering by molecules, absorption by gases (not much at visible), as well as scattering and absorption by aerosol comes in. As described by the Rayleigh approximation, shorter wavelengths get scattered more efficiently, but because of this, the optical depth of the atmosphere is also higher at shorter wavelengths. As the optical depth is significantly below unity, multiple scattering is not dominating. Nevertheless, some multiple scattering takes place, and because of that, radiative transfer computations are needed to derive the spectrum at the surface. A quick googling reveals that the diffuse spectral irradiance peaks in the range of 450-500 nm wavelength, which is blue, and drops very quickly below 450 nm, or at violet. Check Figure 3-2 at

http://cpl.usu.edu/files/publications/publication/pub__3324930.pdf

The next step is then to deal with eye sensitivity and how the color impression is formed. I am no expert on this, but there probably are existing tools to compute also this. It is obvious, however, that it is not as straightforward as to where the maximum power is. We see Sun as yellow, although its spectrum peaks at shorter wavelengths.

Finally, note that being able to see something says nothing about the sensitivity. As a simple example, let's assume that at wavelength X the cone cells detect every tenth photon, and the wavelength Y they detect every other. In this case, you would need five times more photons at the wavelength X compared to Y to have the same sensory response. Yet, obviously both wavelengths are seen, because there is a sensory response at both wavelengths.

Solar irradiation spectrum

Solar irradiation spectrum and Rayleigh scattering due to  gas molecules in sky

  Ref: Wikipedia

The strong wavelength dependence of the scattering (~λ−4) means that shorter (blue) wavelengths are scattered more strongly than longer (red) wavelengths. This results in the indirect blue light coming from all regions of the sky. Rayleigh scattering is a good approximation of the manner in which light scattering occurs within various media for which scattering particles have a small size parameter.

A portion of the beam of light coming from the sun scatters off molecules of gas and other small particles in the atmosphere. Here, Rayleigh scattering primarily occurs through sunlight's interaction with randomly located air molecules. It is this scattered light that gives the surrounding sky its brightness and its color. As previously stated, Rayleigh scattering is inversely proportional to the fourth power of wavelength, so that shorter wavelength violet and blue light will scatter more than the longer wavelengths (yellow  and especially red light).

In terms of the molecular polarizability α, proportional to the dipole moment induced by the electric field of the light. In this case, the Rayleigh scattering intensity for a single particle is written as below:

I ~ I0  α λ -4 (1 +cos 2teta )

However, the Sun, like any star, has its own spectrum and so I0 in the scattering formula above is not constant but falls away in the violet.

In addition the oxygen in the Earth's atmosphere absorbs wavelengths at the edge of the ultra-violet region of the spectrum. The resulting color, which appears like a pale blue, actually is a mixture of all the scattered colors, mainly blue and green. Conversely, glancing toward the sun, the colors that were not scattered away — the longer wavelengths such as red and yellow light — are directly visible, giving the sun itself a slightly yellowish hue. Viewed from space, however, the sky is black and the sun is white.

What is your point, Parviz? Is this a counter argument to my latest explanation as a whole, or to my point that we don't see Sun as green, even though the radiation peaks there? Or is this somehow your view why sky should be violet? I must say that I am at a loss as what is there left to explain.

Dear Timo

Sky should look like violet instead of light blue in the case of Rayleigh scattering because the solar radiation at violet range is still strong (as seen above and more precisely here), quite opposite of what is written in Wikkipedia!! and is prevalent in general Physics periodicals.

Dear Parviz,

The diffuse downward solar flux at the surface is not just about Rayleigh scattering and which wavelength is scattered the most. It is a radiative transfer problem, which introduces additional factors. Chris's plots above illustrate this also. See how the spectrum peaks at blue in Kuwait, but at violet in Mauna Loa. The latter has only about 70% of the atmosphere above it, and much less aerosol, so there is less multiple scattering.

The starting point of any investigation is the spectrum of diffuse radiation at the surface, solved through radiative transfer computations, because this is the spectrum of the radiation that enters the eye (assuming we are talking about the color of the sky at the surface). Here the Rayleigh scattering is already accounted for, as are all the other factors that contribute to it. Just to give you one example of a process that you are possibly ignoring, the optical depth of the atmosphere is larger at violet than it is at blue, because of the Rayleigh scattering is stronger at violet. Whenever Rayleigh scattering occurs, halve of the radiation goes up and half goes down. At violet, this happens more, because violet light is scattered more effectively, so less violet light is actually penetrating all the way to surface.

The second step is to look at what the human eye makes of it. I have no expertise on this, so I leave it for those who have.

In any case, I do not have anything more to contribute to this discussion.

I am not convinced because the solar spectrum over visible light may be assumed nearly  uniform while the Rayleigh scattering emphasizes  the violet scattering is dominant, however the real color of sky is not violet,  instead looking like light blue and there is no concrete explanation for this obvious spectral shift otherwise we postulate that  Rayleigh scattering fails or there is some mysterious factor. Moreover, even we assume that  the eye is not a  reliable detector, the sensitive optical detection verifies bluish sky instead of violet !

I have not found a perfect answer to my question:

Why is not sky violet?

Why does the sky appear blue in color photographs and digital cameras ? While the human eye may have specific sensitivities should Silver halides of film or photo detectors be accurate ?

It's interesting to quantify the Rayleigh Blue sky color perception by using the CIE color matching functions to calculate the chromaticity of the Rayleigh Blue sky (as @Chris Gueymard alluded to earlier). This works out to about x,y = 0.23,0.23. However when we do this and render the blue sky on a laptop computer it doesn't quite seem to match what I (and at least one other researcher) perceives. The rendering looks more aqua colored than the reality on a day free of haze. The question becomes whether the CIE approach is accurate for this color spectrum that has lots of blue and violet light. I am planning to recalculate this with a more realistic solar spectrum to see if it helps.

Cameras seem to vary in how they show the blue sky so it depends on the sensors and post-processing algorithms.