Talk:Rayleigh scattering

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Mechanism?

I'm so sorry, but I find it mind boggling that someone takes the time to write up this semi-long article (it is well past the stub stage) and does not even mention the mechanism by which Rayleigh scattering works. I mean, sure, the formulas are nice to look at and so on, but a rough explanation of why that happens would, IMO, be one of the first things I'd want to add. I'll have a look in my text books later today and try to write something up.... 94.191.142.18 (talk) 11:08, 2 December 2009 (UTC)

Early comments

Would like to clean up all this blue sky stuff and archive most of this discussion any thoughts ? this page http://hyperphysics.phy-astr.gsu.edu/hbase/atmos/blusky.html#c2 has a good discussion to serve as a model Why is polarizability not included (is this the refractive index term  ?) for instance, most forms of the equation include a alpha^2 term (http://hyperphysics.phy-astr.gsu.edu/hbase/atmos/blusky.html#c2)

Although the lead paragraph mentions that rayleigh scattering is for things smaller then the wavelength, I think some units would help people (eg light is about ~500 nanometers, molecules are about 1 nm, any visible particle is much larger then 500 nm)

Rayleigh scattering (named after Lord Rayleigh) is the reason why the sky is blue. Rayleigh scattering of photons by Earth's atmosphere is dependant upon the size of the particles causing the scattering and the wavelength of the photon being scattered.

Neither of these first two sentences actually says what Rayleigh scattering is. The first says that it's the reason why the sky is blue. (So, now we have a name to a scientific explanation, but not the explanation.) The second says that the scattering is dependent upon something or other, but we still haven't been told what it is that is so dependent! --LMS

---

Have someone informations about Rayleigh's "blue sky law"? -- looxix 21:57 Mar 18, 2003 (UTC)

Why is Sky Blue? Answers

The question is not answered very well. Why not green or another shade of blue? The answer seems more dependent on the absorption of energy and the re-emission of a photon from the oxygen molecules, which is blue. The emission spectrum of the other gasses that make up the atmosphere are outside of our personal visual range. Another interesting question is, was the color of the sky always blue? I think not, as the content of the atmosphere has not always contained the 20% oxygen.

Although there is an absorption/emission line for doubly-ionized oxygen (O-III) at 500.7 nm, which is indeed in the blue-green, that is not responsible for the blue color of the sky, since other gases will also produce a blue sky, provided that the gases are present in sufficient quantities and the light source is bright enough (which the Sun surely is!). It is, as the article states, essentially a scattering effect.

Classically, atmospheric molecules can be treated as dipoles, which react to incident light much the same way that a spring reacts to an applied force. So long as the frequency of the stimulus is considerably below the resonant frequency, the amplitude of the response is proportional to the square of the frequency. The intensity is proportional to the square of the amplitude, so the total scattering is proportional to the fourth power of the frequency. This very rough treatment assumes the dipoles are of negligible size compared to the light waves, so (for instance) raindrop scattering does not obey the fourth-power law; rather, it follows Mie scattering, which is much less dependent on frequency and accounts for the generally grey or white color of clouds.

Note that the fourth-power law doesn't depend on the specific gas--it only requires that the resonant frequency be much higher than that of the incident light. That clearly is true for nitrogen, for instance--otherwise, we wouldn't be able to see sunlight through the predominantly nitrogen atmosphere. So a purely nitrogen atmosphere would also produce a blue sky.

As to the color of the early sky, in a reducing atmosphere containing hydrogen, water vapor, methane, ammonia--I don't really know. The Sun might have appeared bluish to our eyes, since hydrogen, methane, and ammonia all have absorption in the 600 to 900 nm range. (Uranus and Neptune appear bluish for a similar reason; their atmosphere absorbs the red light and reflects the blue light back to us.) The Sun was some 25 percent dimmer then, too, and emitted proportionally less blue and violet light. Possibly it would have been kind of dark (although our brains are very good at compensating for large changes in luminance and color balance). My guess is that the sky would have been blue back then, as it is now. BrianTung 23:17, 2 January 2006 (UTC)

The following paragraph (which was not written by me) I cut from the article, because its place is in the discussion.
The article is generally correect but light scattering depends on anosotropy. Anisotropy is the phenomenon where the speed of light in one direction is different in another direction usually perpendicular sources Nitrogen, the major source of gas in the atmosphere is isotropic. However there are small quantum fluctuations in the isotrpoy of nitrogen where it becomes anisotropic. This microscopic fluctuating is actually the source of the scattering on which the Rayleigh scattering and theory is based. Any comments?
Synapse 18:43, 10 January 2006 (UTC)

From the mailing list

>Dear Sir or Madam,
> 
> On that link to the rayleigh scattering coefficient ks is a mistake in
> the power of pi. Pi should be to the power of five and not to the power
> of six.
> http://en.wikipedia.org/wiki/Rayleigh_scattering
> 
> Please check this again. Thank you.
> 
> Best regards,
> Daniel Ploss

Can anyone find a reference to confirm this? -- Anon.

I changed it to to the 6th after some looking on Google. 69.142.2.68 10:51, 30 August 2005 (UTC)

The original reference for this article [1] has the "5" and "6" transposed between π and d. I've now re-written the maths from a solid reference, and I'm confident that it's now correct. The bottom two sections are still really chatty, though. -- DrBob 01:59, 31 August 2005 (UTC)

I found a different formula for Rayleigh scattering.

I found a different formula for the Rayleigh scattering coefficient. There the number of scatterers is in the denominator. It is from a lexicon of optics. I doubt that this is wrong. The formula is scattering coefficient b = (8*PI^3/3*N*lambda^4) * (n^2-1)^2, where N is the number of scatterers and n is the index of refraction.

Since that formula doesn't contain a term for the size of the particles (d in the article), it's probably only valid for the limit of d<<λ, so it's valid in a different domain. I'll try to check. -- DrBob 14:45, 11 Oct 2004 (UTC)

Could you please give the source for the formula in the article. Thank you.

I also need the source for the intensity formula of Rayleigh scattering. Where was it taken from? 13:41, 4 Apr 2011 (UTC) —Preceding unsigned comment added by 87.77.200.134 (talk)

Hi all. The formula in the article comes from "Atmospheric Chemistry and Physics", 2nd Edition, by Seinfeld and Pandis on page 696 of the hardcover edition. The formula is written differently, though, but it maths out to the formula in the article. Hope this is useful. I'll update the references section of the article to reflect this. 134.197.0.22 (talk) 21:20, 9 January 2012 (UTC)

Why is the Sky Blue?

It should be mentioned that Raleigh Scattering is not physically a result of the size of the particles encountered by the light, but rather a result of interactions with the electron structure of the atoms (primarily a quantum mechanical interaction).

Simply put, the idea that size of particulate as being the primary physical reason for scattering in our atmosphere is no longer a modern view. This idea can't even be considered simplistic, because it is physically incorrect. While, Raleigh did observe the connection between scattering and size of the particle, this is not the reason for the scattering nor for why the sky is blue. Raleigh scattering is an easy to use formula and certainly important, however we shouldn't go too far - someone needs to explain that this effect does not consider light matter interaction as modernly viewed and that it posses Cretan inherent assumptions including the fact that spacing of particles is not considered. Application to atmospheric color should be amended to stress that the electronic structure of the particles, as governed by quantum mechanics (alternativly known as chemistry), describes the scattering of light in our atmosphere approximated by Raleigh Scattering.

Since no mention of where Raleigh's equations come from or when they are created are mentioned the reader is free to believe that his equations are modern and that the interpretation of them is the modern view. This is a serious oversight, because rule of thumb equations like this one often are often modified as theory develops. Lord Rayleigh's result was not fundamentally derived from more modern theory of quantum mechanics, which describe the interaction of light with matter.

Ultimately the question,"why is the sky blue?" is a matter of light - matter interaction, not light size interaction as this website asserts.

Furthermore, Raleigh's assumptions are not indicated here! How is the reader to know what circumstances his equation are valid for? I see no mention of pressure dependence which would effect spacing between molecules. Light when viewed as a wave must be considered to be able to diverge when passing through a gap on the order of a specific size, which would be effected by pressure.

The reason seems more clear in this link : Maybe we can incorporate it here ?? [2] Ap aravind (talk) 09:05, 1 February 2010 (UTC)

I have a PhD in physics and am under the impression that Rayliegh scattering does not explain the blueness of the sky. The sky is blue because of scattering from density fluctuations. Also, it was Albert Einstein who discovered this. I'v been keeping this knowledge in my head for 40 years, and I can't remember where I got it from. David Roemer (talk) 11:41, 2 May 2012 (UTC)

It is true that Rayleigh scattering by molecules in the air is modernly understood in terms of the interaction of photons with the molecular orbitals for the electrons, which is not how Rayleigh originally presented his argument. It is also true that Smoluchowski and Einstein worked out the importance of atmospheric density fluctuations in accounting for lateral scattering in the atmosphere. But Rayleigh's argument is not "physically incorrect." It is a good example of how the correct qualitative result can be obtained from simple dimensional analysis, without knowing the details of the dynamical interaction. To see this in the modern language of field theory, see Sec. 4 in this review: [3]. For an undergraduate-level review of the subject (including Rayleigh's argument and the contribution of Smoluchowski and Einstein), see Ch. 4 of Hecht's Optics, 4th ed. - Eb.hoop (talk) 18:25, 2 May 2012 (UTC)

I, too, vaguely remember learning that density fluctuations are the primary scatterers. Although, if the length scale of the fluctuations is small compared to the wavelength, then it's still Rayleigh scattering, isn't it? Spiel496 (talk) 18:47, 2 May 2012 (UTC)

Yes, it is still Rayleigh scattering. It's just that if the atmosphere were homogeneous, the scattered light would only propagate forwards. Smoluchowski and Einstein explained the role of density fluctuations in accounting for lateral scattering. (Again, see Hecht.) - Eb.hoop (talk) 18:58, 2 May 2012 (UTC)

Confusion of Terminology

You can't use terms from modern theory like "photn" in a document describing a theory that was developed prior for many reasons: 1. it creates ambiguity as to how modern the theory is 2. it suggests to the reader that the theory interprets light as photons, in fact the theory does not 3. it implies that the author at least heard the word photon before he died, this however is not possible

Simply put the use of the word photon in this context is confusing and erroneous. Consult any modern physics text book for confirmation. Suggestion (tippler and lewellen).

What's wrong?

The intensity I of light scattered by a single small particle from a beam of light of wavelength λ and intensity I₀ is given by:

$I=I_{0}{\frac {(1+\cos ^{2}\theta )}{2R^{2}}}\left({\frac {2\pi }{\lambda }}\right)^{4}\left({\frac {n^{2}-1}{n^{2}+2}}\right)^{2}\left({\frac {d}{2}}\right)^{6}$

where R is the distance to the particle, θ is the scattering angle, n is the refractive index of the particle, and d is the diameter of the particle.

So the smaller the diameter of the particle the less of the input beem goes out again. Particles of vanishing diameter would absorb all light, really big particles would emit more light than they receive.

What's wrong? 84.160.230.81 19:02, 5 October 2005 (UTC)

• Nothing is absorbed. Any light that isn't scattered is transmitted straight through. The equation gives the intensity of the scattered light, only. Vanishingly small particles will not scatter. Also, the equation is only valid for d < λ/10 (as stated in the article). Consider that R must be at least as big as d; therefore the scattering coefficient is always smaller than 1 in the valid range. -- Bob Mellish 19:18, 5 October 2005 (UTC)

More about S-Matrix

It would be very nice to have some "further reading", maybe some URLs or something like this about the S-Matrix.

Complex Index of Refraction?

I would like to see the extension to a complex index of refraction (ie, absorbing particles) included in the formula. My guess is that the only change is the add a absolute value inside the square.

192.249.47.9 (talk) 16:15, 11 December 2008 (UTC)

Rayleigh's Criterion

would this also be the page to add something about Rayleigh's Criterion? i just learned about it in class, and came here to see if there was more information. it has to do with the angular seperation of light sources and weathe ror not lenses can resolve the images of the two... heres a link to a page with more information on it ( http://www.fas.harvard.edu/~scdiroff/lds/LightOptics/RayleighsCriterion/RayleighsCriterion.html i wouldnt feel comftorble writing something myself on it seeing as i just learned about it

• We already have information about it at Rayleigh criterion. It's not anything to do with scattering, so it doesn't belong in this article. --Bob Mellish 21:31, 19 December 2005 (UTC)

RC CLARK's edit

The article is generally correct but light scattering depends on anosotropy. Anisotropy is the phenomenon where the speed of light in one direction is different in another direction usually perpendicular sources Nitrogen, the major source of gas in the atmosphere is isotropic. However there are small quantum fluctuations in the isotrpoy of nitrogen where it becomes anisotropic. This microscopic fluctuating is actually the source of the scattering on which the Rayleigh scattering and theory is based. Any comments?

This belongs in this section Daemon8666 20:23, 13 January 2006 (UTC)

I heard about the theory of Einstein and Smoluchowski, that describes scattering from density fluctuations on a local scale (with results similar to Rayleigh scattering) in media in which the scatterers are separated by distances small if compared with the wavelength of incident light. I didn't studied directly this E-S theory, but I found it in two texts about atmospheric optics. The first author uses E-S theory only for scattering in the middle region of atmosphere, which has higher density than the high atmosphere and reduces (by destructive interference) the effects of Rayleigh scattering. The second one states that E-S theory refers to matter that is taken to be continuous but with a refractive index which changes with position, and that Einstein didn't assume a discrete distribution of matter. For the moment I've no conclusion about that.--Ran.olo 07:59, 21 August 2006 (UTC)

Typical values for Earth's atmosphere

Does anybody know typical values for n and d in the atmosphere of Earth? Is the refractive index of the particle the same as that of air (approx. 1.0003 at sea level), or is a different value to be used for a single particle? In general, it would be nice if one could use the formula along with these values to make calculations with these. Getting information via Google about this seems to be very difficult (either the information is only about the principle but without numbers or it is drowning in details or you have to pay for it or Google's hit is about a completely different topic), so it would be nice if Wikipedia could provide such useful information. If I had such information I surely would add it to the article, but unfortunately I don't have any.--SiriusB 20:24, 26 January 2006 (UTC)

The "cleanup" tag

The section "An explanation of Rayleigh scattering using the S-matrix" appears to have been written very casually. I am in favor of deleting it and taking a fresh start to add more theory, if it is needed. I deleted one error, but I think there are too many others to fix easily. David R. Ingham 21:26, 14 March 2006 (UTC)

Oxygen is blue.

I understand oxygen is blue. Why is this omitted from the discussion?

The answer to the above question is that the gas oxygen is not blue, it is colorless. Liquid oxygen is a pale blue color, which I have shown to my students on many occasions. Colors are frequently present in substances having an unpaired electron, as is paramagnetism. Even in the upper reaches of the atmosphere, the temperature is still not cold enough to liquify oxygen, and even if it were, there would not be enough oxygen nmolecules at that altitude to make any difference in the color. Therefore, the color of liquid oxygen is irrelevant to the discussion at hand. - Dr. Art

Where are the experimental measurements associated with the support of the grand boggling mathematics? Some of us are wary of those who tend to "get lost in the mathematics".

"Logic is a systematic method for getting the wrong conclusion...

with confidence."

A scientist can never afford to discard data, however contradictory it may be to his ideas and intentions.

Edit: (Sometime around Wednesday, April 20, 2006) My apologies for talking. I presented no data. I do not have access to the necessary spectroscopy equipment.

I wonder if there is any argument that ozone is blue.

I also wonder if there is any debate over why a gas flame is blue.

I also wonder if there is any debate over why there is some blue gas just beneath a candle flame.

I also wonder if a piece of white paper held a distance of approximately one kilometer is blue (one kilometer having about as many oxygen molecules in a path as two inches of liquid oxygen, which is generally agreed to be blue.)

I am eager to see the experimental measurements that support the theory of Rayleigh scattering.Coucilonscienceorg

The color of liquid (and solid) oxygen is blue, with the solid being somewhat darker than the liquid. The same is true of ozone, with both liquid and solid forms being darker than they are for molecular oxygen. Solid ozone, in fact, is so dark as to be almost black.

This gives a clue as to why these facts, however true, are omitted from the discussion. They are irrelevant. If the sky were actually blue because of light absorption (which is why liquid oxygen is blue), the atmosphere would be opaque to the point that the stars could not be seen at night. In truth, though, they look nearly as bright as they do from above the Earth's atmosphere. So the sky must be blue for reasons other than the absorption of light by molecular oxygen.

Hmmmm. Solids have absorption bands, gases have lines. Also, you're ignoring the amount of absorption, since a weak line might only contribute a visible color if miles of dense gas were present. So the real question is simple: what is the absorption spectrum of the gas mix in our atmosphere, and what color would human eyes perceive when observing a material having this spectrum? Second question: does this absorption spectrum contribute any significant color, or is it insignificant when compared to the frequency spectrum created by Rayleigh scattering? --Wjbeaty 01:24, 25 August 2006 (UTC)

I've found that the only significant absorption contribute to the blue skylight should be the one from ozone, but it is appreciable only at twilight. --Ran.olo 19:33, 26 August 2006 (UTC)

The sky on Mars, when it is relatively free from dust, is blue-violet, and its atmosphere contains no oxygen to speak of; it is thin and consists mostly of the colorless gas carbon dioxide. When carbon dioxide condenses, it goes through no liquid phase at Martian pressures and freezes directly into colorless dry ice. So the color in the skies of Mars also cannot be due to the color of oxygen.

As for "getting lost in the mathematics," while it is true that some scientists do over-formalize, it is important to remember that they are caught at that by people who specifically understand that mathematics. And when they do so, they point out either what is wrong with the mathematics, or why it is inapplicable. It is an error to say, "Wow, that's a lot of math," and simply by virtue of that impeach the reliability of that math. You have to show why it is wrong or inapplicable. Not being able to understand it is no excuse, since by this point, there are lots of people who do understand the math.

In short, just because one doesn't understand an explanation, does not mean the explanation is wrong. It is quite conceivable that one's understanding is insufficient. And when an explanation is commonly accepted by people who are trained in experimental science, it becomes more than conceivable--it is almost inevitable. That insufficiency can be repaired, of course, but you can't expect it to happen in a single afternoon, or without effort. BrianTung 23:35, 25 April 2006 (UTC)

I apologize for being stupid. The question I am asking is where the experimental evidence for the particular intensity of the particular blue color we see in the sky. In this entire discussion, noone has mentioned a single reference to experimental verification that Rayleigh or any other type of scattering could produce the particular intensity of the particular blue color we see in the sky.

Furthermore, what is the predicted intensity of color we would perceive from this scattering? How can it be explained that the blue of the sky matches the 500.7 nanometers of the strong emission line of doubly ionized oxygen? I see that there are references to evidence that the sky on Mars is blue. might that evidence can be shared with us?Coucilonscienceorg 08:50, 28 April 2006 (UTC)

Oxygen is not blue

I can not even begin to address the fundamental misconceptions presented in the previous comment. Anyway, now that that is out of the way, I cleaned up the s-matrix bit a fair amount, still needs a bit of proper prodding, but should be bordering on acceptable. --Meawoppl 05:41, 18 April 2006 (UTC)

Someone should just include an absorption spectrum, rather than argue about this. Oxygen and water vapor have some absorption bands in the red end of the spectrum, but they are not a major contribution to the sky color. DonPMitchell (talk) 18:29, 26 June 2009 (UTC)

Emission spectra

If the atmosphere of a planet contains atoms or molecules with particular emission spectra, will the sky of the planet be the color of the emission spectra, or the color of the scattering? Perhaps some of the excellent astronomers present can present some data in this regard.Coucilonscienceorg 09:48, 28 April 2006 (UTC)

The variaty of the universe is infinite, and one can think of many different planets and circumstances. The color of the star(s) is also of importance. How the sky would look under different circumstances can be calculated. If for example you would have a very red star, the sky would still not be red as you might think, but rather the same blue in zenith but fading to a rather reddish brownish color near the horizon.

Again, this is not relevant on earth where the sun is has a nice white-yellowish color. Albester 13:04, 10 May 2006 (UTC)

The statement, "If for example you would have a very red star, the sky would still not be red as you might think, but rather the same blue in zenith but fading to a rather reddish brownish color near the horizon." Must be verified by experimental evidence or else it is inadmissible in a scientific discussion.

Many who have posted in this discussion have mentioned the idea that spectroscopy is predicated on the idea that the change in energy state of electrons is the cause of photoelectric emmission and absorption. This idea is extremely strongly supported by experimental evidence. Considering the lack of reference to any experimental evidence by the scattering hypothesizers, and the prevalence of accusation, off topic remarks, and other fallacious gestures, this whole discussion needs serious editing if it is to be considered a scientific discussion. In other words, the majority of people seem to like to believe that the sky is blue in honor of Mr. Rayleigh. At least the majority of people no longer think the world is flat. Too bad majority of people have no bearing in a scientific discussion where evidence is presented to support theory. When Edwin Hubble presented the idea that Universe is expanding, he had Marvin Humason's evidence presented immediately before hand. Only then did Einstein start to think that the cosmological constant was the greatest blunder of his life.Coucilonscienceorg 18:10, 17 June 2006 (UTC)

I think that the "red star" example is pertaining to a scientific discussion. One of the step of scientific method is to make predictions based on the physical laws of a certain model (rayleigh's law of scattering in our case). Experimental verification is only the next step: if it doesn't confirm expectations, we have to change the model on which we built our prediction. --Ran.olo 11:30, 20 August 2006 (UTC)

The main reason light from the sky is blue

If Rayleigh scattering is the "main reason" for the blue of the sky, what are the other reasons? Why aren't these other reasons mentioned anywhere?Coucilonscienceorg 01:55, 1 May 2006 (UTC)

It is the main reason,and in the case of earth the only reason. What you mentioned earlier about oxygen being blue in liquid state etc simply have no bearing on the question of why the sky is blue. I don't mean to be rude, but I find your comments about logic and mathematics insightfull as to how you think, but that opinion is not shared by many, has no acceptance among serious persons and as such they should not be put here on wikipedia. Perhaps you can start your own webpage explaining your own theories? Albester 12:59, 10 May 2006 (UTC)

What you mentioned about oxygen undergoing a chemical change in its transition from liquid to gas needs experimental evidence, otherwise it is hot air (pun intended). The comments relating to logic lead into one of the fundamental rules of science: any conjecture or hypothesis MUST be supported by experimental evidence. I did some time with chemists, and spent some time with spectrometers. The color of liquid oxygen, solid oxygen, or gaseous oxygen is considered by chemists to be a result of electron energies, not scattering. There seems to be a great deal of disagreement regarding what exactly is causing the scattering in the Rayleigh conjecture. Given that the equation involves the sixth power of the size of the scattering object, perhaps that should be addressed conclusively. If the oxygen atom is found to be the right size, then some experimental evidence that the color is not due to change of electron energies, as chemists say, is in order. The contention that the sky is blue instead of violet or some other color because of the human eye limitations can be easily verified with a spectrometer. Until it is, chemists won't agree.Coucilonscienceorg 17:55, 17 June 2006 (UTC)

The spectrum of the sun has been measured. So has the spectrum of sunlight at Earth's surface. The difference is the spectrum caused by atmospheric absorption. To look for coloration contributed by air, look at the visible band between 400nM and 700nM in atmosphere absorption spectrum below. --Wjbeaty 01:39, 25 August 2006 (UTC)

Come on guys. Rayleigh Scattering is fairly well proven theory and does explain the color of the sky. Maybe this whole section (i.e. "Why is the sky blue") should be removed to a separate article so that Rayleigh Scattering can be described without controversy (if we even should call it that!) It may also help to add some links to data.

There is no sky...

...there is only air above us. It may sound obvious to the point of stupidity, but it bears mentioning that there is no solid surface up there. When children look upwards on a bright sunny day, they see an apparent solid surface, and they hear it called by the name "sky." They wonder why this solid-looking "sky" thing is colored blue. Unfortunately most of us grow up without really grasping the idea that the "sky" is just a cloud of air being lit by sunlight from the side, with the black of space in the background. Or in other words, since there is no "sky," it's the air which has a blue color. Does my statement seem crazy? If so, this shows that at some level you believe in a "sky" which has color, while air does not. Rayleigh scattering explains why a miles-thick layer of air has a blue color. But Rayleigh scattering cannot explain the color of the sky, since "the sky" is an illusion. The bright blue stuff up there is called air, not sky. --Wjbeaty 08:25, 15 June 2006 (UTC)

Okay smartass, but nobody asks the question "Why is the thick cloud of air above the earth incorrectly referred to as a sky (despite the fact that "sky" is a ludicrous and childish concept) tinted blue and not simply transparent?" —Preceding unsigned comment added by 164.92.175.76 (talk) 16:13, 25 February 2008 (UTC)

there is a sky above us

when you look at a very distant dark object, a mountain for example, you see it blue. So, are you looking at the air or at a mountain made blue by air?

That's not honest: a false dichotomy. There is no "or," instead it is "and."

If we look at a distant blue mountain, then we are observing a genuine substance: the air. AND we are looking at a distant mountain behind the air layer. The mountain only appears blue to us because a blue substance is in front of it. The mountain has not been made blue, as observers close to the mountain can attest.

Above is how we describe small transparent objects we can hold in our hands (for example, viewing a distant mountain through a piece of blue glass.) To use a different set of rules to describe air is to be inconsistent; introducing mystery where there is none. --Wjbeaty 00:54, 25 August 2006 (UTC)

In the same way, when you look at the blue stuff above you, are you seen air or the dark interstellar empty space seen through the air?

Again a false dichotomy. We see blue air, AND we see a black background. Space has not been made blue, as anyone observing space from above the atmosphere can attest. Instead there is a bright blue transparent substance in front of the dark background.

If instead we combine blue air with black space and call it "sky," then we have erased a simple and understandable explanation, and replaced it with a mysterious "sky" which cannot be described because it does not exist.

Sorry, but I can't see at the same time the blue air AND the black background, because air changes the black color of the background. So, when I'm far from mountains, I SEE blue mountains but I KNOW that the mountains are black, with blue air between me and them. When I'm close to mountains I see them black, but I don't see any blue, even if I know that there is still air. The same for air and black empty space. And I can call "sky" the system made by these two objects, even if sky itself is not a solid object. in the same way, when I see the light spectrum caused by sun that shines onto water droplets, I call it rainbow.--Ran.olo 20:45, 26 August 2006 (UTC)

You cannot see the blue of the air if you don't have a dark background... And you can call "sky" the space made blue by the atmosphere. it's only matter of point of wiew. --Ran.olo 12:52, 20 June 2006 (UTC)

Ah, but your point of view is wrong because this is an encyclopedia, not an everyday conversation. Scientific explanations should always be brutally honest and as simple as possible. They should state the simple truth in simple words. They should carefully point out all illusory objects or substances such as "sky," and never substitute them for real substances such as air. Or preferably, they should avoid introducing illusory objects in the first place.

Therefore, if someone asks why the sky is blue, the honest answer is, first, that there is no substance or object called "sky" above us. The concept known as "sky" derives from the Medieval concept of "heavenly firmament," a solid blue-colored dome above the Earth. Instead of "sky," there is only a thick layer of air which is illuminated from the side by sunlight, and which is backed by the blackness of space. The layer of air has a blue color. This color is a "structural color," not a pigment, and is explained by the frequency selective mechanism which is part of Rayleigh scattering. And anyone who is lifted many tens of KM upwards will prove to themselves that "sky" is an illusion: they will discover that this bright blue layer of gas is now below them, allowing them to observe interstellar space with no colored substance in the way. --Wjbeaty 00:54, 25 August 2006 (UTC)

Wjbeaty, I would say you are wrong because in the modern world we call "sky" as the atmosphere or gaseous molecules ABOVE us. If you refered back to history, you simply contradicted yourself. Okay back to your statement: "The concept known as "sky" derives from the Medieval concept of "heavenly firmament," a solid blue-colored dome above the Earth." I believe we all knew what the people were talking about rite? Well a solid blue-colored dome above the Earth to them is our gaseous atmosphere above us, so simply that means f(x)=y. A different concept might be = to another concept using different terms. If you're going to say math is not related to science well, it is related and also, as I said another concept can sometimes equal that concept. --InternationalEducation 7:03P.M., 31 December 2006 (UTC)

Simple question

If the sun's emission peaks at yellow, then why does it say that "there are more violet photons to scatter than there are blue ones"? Is this an error, or am I just misunderstanding the article? -- kier07 9:32, 30 July 2006 (UTC)

The error was the assertion that the sun's emission peaks in the yellow. It does not. It peaks in the blue. See Solar radiation. There's a nice graph there.--Srleffler 05:36, 5 September 2006 (UTC)

Sounds like an error to me. The most fundamental reasons for a non-violet sky are color perception, as well as the fact that there are far more blue photons than violet ones.

Also, I have no idea what a particular person further up on this page is babbling about. There is absolutely no fixed wavelength of blue light in the sky. The shade of blue varies with the amount of scattering taking place, and is not a constant 500.7 nm. It is a proven fact that the sky over highly populated areas is more of a light blue or even grey, due to the presence of more scattering particles in the air. So there's no exact match to emission lines of doubly ionized oxygen.

I'm baffled that anyone could believe the sky is blue because of the emission spectrum of oxygen (which implies that the sky is actually glowing due to photon emission as an electron drops to a lower energy level). This really has nothing to do with oxygen at all, as is evidenced by the blue color of the Martian sky when dust-free. 17:00, 31 July 2006 (UTC)

Sounds like an error also to me: visible spectrum of sunlight outside the earth's atmosphere has a maximum between 450 and 500 nm, i.e. in blue, so blue is the dominant sunlight wavelength scattered according to Rayleigh. But if we "add" Rayleigh's scattering law to sunlight spectrum we'll obtain a spectrum of scattered light that is not peaked in blue. Skylight contains light of all visible wavelenghts, but blue dominates in our color perception. --Ran.olo 11:02, 20 August 2006 (UTC)

Is the sky red?

I find in this article that sky is reddened by sunset, but I don't think it really happens. I mean, sunlight is red at sunset (or sunrise) because it scatters a lot in the low atmosphere, so every object (e.g. clouds but also big molecules in the atmosphere, like dust) illuminated by (or seen through) this direct light will be red. But if I look up at the sky, I still see it blue, because it is coloured by the blue scattering that still dominates in the high parts of the atmosphere. So at sunset we have two different sources of light: direct, red light from the sun and scattered, blue light from the sky, and it explains why clouds at sunset are often pink (=red+blue).--Ran.olo 22:04, 18 August 2006 (UTC)

If our entire atmosphere was sitting on a desktop before you, so that you could observe it without being immersed within it, you would discover that the air behaves much like smoke, or aerogel, or like water stained with milk. From your perspective, when illuminated from the side, the air behaves as a blue fluid. When illuminated from behind, the air is red. The color varies with the angle of illumination and observation, and for several different observers these materials will have different colors. Angle-dependant colors are no mystery, and are quite common in structural colors such as bluejay feathers, butterfly wings, etc. --Wjbeaty 01:04, 25 August 2006 (UTC)

ok, maybe it's another misunderstanding caused by word 'sky'... --Ran.olo 19:48, 26 August 2006 (UTC) Ran.olo did you do a lab with food coloring in school because red and blue makes purple, maybe you could say red and white makes pink. InternationalEducation 7:11P.M., 31 December 2006

Red + blue DOES make "pink", actually magenta. This is additive color synthesis (superposition), which corresponds to experimental setup1: project 2 light beams, one blue, and one red, onto the same white surface. "food coloring", just like mixing paints, is substractive synthesis (removal by absorption), that is experimental setup2: project one single white light beam on a white surface after it has passed through 2 filters in a row, one blue one red. The only part of the light that was not absorbed by either one or the other filter will be purple. (Mystero 80 15:21, 19 July 2007 (UTC))

Glaciers etc.

I don't know why the bit on glaciers being blue keeps getting added to this article but the idea that glaciers are blue due to rayleigh scattering off of bubbles is totally dead wrong. Why is ice without bubbles just as blue as bubbly ice? The reason that ice is blue is the SAME REASON THAT WATER IS BLUE and that is the slight absorption of water in the far red region of the spectrum due to an overtone of the OH group stretch in the infrared region. Very simple, no bubbles needed. The bubble/rayleigh scattering thing is just a gross misconception. [4] --Deglr6328 22:46, 23 August 2006 (UTC)

Why blue not violet

The section on why the sky is blue instead of violet seems to have some problems. In particular, it claims there is a rapid falloff in sensitivity below 450 nm, but this is clearly not supported by the graph, which shows that the S receptor peaks at 420 nm. The article formerly claimed that receptor peaked at 450 nm, but I have adjusted it to agree with the graph. It's clear from the graph that the eye's sensitivity to monochromatic violet light is not much less than its response to monochromatic blue light. The more likely explanation involves how the brain interprets the mixture of wavlengths present in skylight. An alternative version of this section is developing at Diffuse sky radiation. Perhaps that should be copied here?--Srleffler 05:41, 5 September 2006 (UTC)

I was told it was because the violet light gets scattered again on its way from the sky down to you leaving the blue dominant. I make no guarantee that explanation is correct. Man with two legs 13:34, 9 October 2006 (UTC)

Are you saying the violet light is scattered back out into space? If so, that doesn't make sense to me. It's not as if the blue light scatters exactly once. It will scatter off many particles before reaching your eye. You can think of the blue light bouncing around all over the air before hitting your eye, so that it's more-or-less equally likely to approach you from one part of the sky as another (as opposed to the red light, which bounces only "a few" times, and so hits your eye on more-or-less a direct trajectory from the Sun.) This is all very approximate, of course, but the point is if it's already approaching you from a basically random direction, then even after more scattering the direction of the light will still be random. So the violet light wouldn't be any more likely to be aimed back up into space as the blue light.

As to the question above, assuming the graph is accurate and the short wavelength receptor peaks at the boundary between blue and violet, then my guess is that the small amount of longer wavelength light coming from all directions mixes with the blue-violet light to make something that we perceive as being closer to blue. Whereas the light on the ultraviolet side can't contribute because our eyes can't detect it. --Tim314 21:27, 9 October 2006 (UTC)

It is not correct to say the direction is random. All the light I see from a blue sky is going down, not up. Clearly a blue photon once scattered downwards direction has little chance of being scattered up again or the sun would appear red at noon. If this did not apply to violet light which therefore was scattered into space, that would shift the colour of the light coming downwards towards blue. So it still looks plausible to me. Man with two legs 11:09, 10 October 2006 (UTC)

The "physiological explanation" section is seriously flawed. First, it seems to be saying that the sky is really violet, but it appears blue. That is a meaningless assertion. If something appears blue, then it is blue. That is what it means to be blue. The fact (?) that the eye is less sensitive to violet is irrelevant, because certainly the brain would compensate for a static disparity like that. Second, it contains the phrase "although the scattered light contains more violet photons than blue ones..." This statement is unsourced and it contradicts the spectrum[5] of the sky shown in Diffuse sky radiation, which clearly shows more light in the blue (470 nm) than in the violet (410 nm). Finally, a long (apparently flawed) discussion on a physiological effect doesn't add to the understanding of Rayleigh scattering. Finding nothing redeeming about this content, I'm am removing it now. Spiel496 03:39, 12 February 2007 (UTC)

I agree. This thing about violet appearing blue is flawed, as Spiel496 clearly states. the physiological detection of light is interesting, but i can't see that it's useful to exlpain it in this way. Is it really much differnet to saying, "hot objects may appear to be another colour, but they're really Infrared, it's just our eyes can't see Infrared"?Spute 21:07, 12 February 2007 (UTC)

I don't follow you, Man With Two Legs. Some of the light *is* scattered back up, but of course you can't see this from the ground because it never gets to your eyes. Even in a plane at 30 thousand feet, there's still probably not enough air below you to scatter significant blue light back up to you (or it's drowned out by the light reflecting off the ground and off clouds), but at 60 thousand feet you can see a blue glow from the atmosphere below (see these photos:[6]). I'm not sure what you mean about the sun being red at noon -- you're saying even the yellow light would be scattered back up when it's directly overhead? Actually, there's more atmosphere between you and the sun when it's on the horizon, which is why the sun looks red at sunrise and sunset. -- Tim314 14:02, 12 March 2007 (UTC)

I am suggesting that blue and violet light are both scattered in all directions including down (potentially seen) but the violet light is more likely to be further scattered so that much of it goes up again and is not seen. That is the explanation someone told me a long time ago, and it still sounds plausible to me. I am not willing to bet actual money on it though unless someone models it formally. The point about the sun not being red at noon is to show that the scattering of blue light is a small effect, but the scattering of violet light will be much greater. It has just occurred to me that violet light scattered up could be scattered again down killing off the effect I have just described. Perhaps the fact that air gets denser as one approaches the ground is why that does not happen. However these speculations are OR and so not suitable for inclusion in the article, but since the theory I have described is at least 25 years old, there should be someone out there who knows for sure. Man with two legs 20:20, 12 March 2007 (UTC)

Violation of Conservation of Momentum

I'm TAing for a graduate chemistry class, and this is something that's always bugged me: Doesn't rayleigh scattering violate the conservation of momentum? I can't see how you can absorb a photon and reemit at the same frequency in another direction without transferring some momentum to another particle (thereby incurring an energy loss and frequency shift). Or is a molecule already in motion absorbing? But this would suggest that anisotropic materials exhibit less rayleigh scattering (and thus less absorbance), than materials with tons of molecular motion, which doesn't make sense to me.

Strictly speaking, in Rayleigh scattering the photon is not absorbed and reemitted, but regardless I think the answer to your question is that the momentum of the photon is very small compared to the momentum of the scattering particle. It's sort of like bouncing a billard ball off of a bowling ball. The momentum of the bowling ball is so large that any impulse (change in momentum) due to the interaction with the billard ball is insignificant. Optiksguy 19:37, 26 October 2006 (UTC)

Einstein and Smoluchowski

In 1908 the Polish physicist Marian Smoluchowski became the first scientist to ascribe the phenomenon of critical opalescence to large density fluctuations. In 1920 Albert Einstein showed that the link between critical opalescence and Rayleigh scattering is quantitative. DFH 14:26, 17 November 2006 (UTC)

Why does greater than blue wavelengths aren't scattered?

From the article: "...by particles much smaller than the wavelength of the light". Those particles are also inevitability smaller than green, yellow and red wavelengths, so why aren't them scattered as well?

Sorry for my English, I hope you understood me. Brass 08:27, 21 December 2006 (UTC)

They are scattered, but the angle is different.86.138.101.53 01:56, 8 January 2007 (UTC)

Not correct. Shorter wavelength light is scattered more than longer, as is described in the article. Man with two legs 10:28, 8 January 2007 (UTC)

Why does the sky turn green during a tornado?

Why does the sky turn green during a tornado? -Hamster2.0 02:11, 31 October 2007 (UTC)

Laser picture dispute

I believe that the primary source of scattering for this is off of dust rather than rayleigh scattering, therefore the picture may not be entirely appropriate, discuss and remove if agreed 193.60.83.241 (talk) 20:56, 15 May 2008 (UTC)

I agree entirey. The laser is visible because its light is scattered at large dust/mist particles. So, the underlying effect is mere reflection rather than Rayleigh scattering. If a third person agrees, please remove the picture. – Torsten Bronger (talk) 06:41, 30 May 2008 (UTC)

I agree, the laser beam is mainly observable due to reflection of the beam on larger than wavelength dust particles. Rayleigh scattering is restricted to particles smaller than the lights wavelength. If only air was present or very fine dust the beam would still scatter, but to a significantly lower degree, so much so that it would be almost invisible. Astro.scope (talk) 14:46, 14 December 2009 (UTC)

What you're saying contradicts the rest of the article. When sunlight shines on the atmosphere, Rayleigh scattering is a primary source diffuse radiation. Why shouldn't the same be true of some air being lit by a laser? I suppose one could argue that the dust particles near the ground are larger, but that isn't obvious. Spiel496 (talk) 17:42, 30 May 2008 (UTC)

Do you know anything abou lasers ? They give monochromatic light ! Take e.g. a green laser. Rayleigh scattering will scatter away a bit of the green light. But the beam, the scattered and reflected light will stay green. —Preceding unsigned comment added by 217.233.128.229 (talk) 07:46, 17 June 2008 (UTC)

Yes. Spiel496 (talk) 13:45, 17 June 2008 (UTC)

Scattering of dust is not Rayleigh scattering. This is in direct contradiction with the first paragraph of this article and the article on the Tyndall effect. The most economic course of action is to delete the picture and its subscript. 129.241.172.204 (talk) 09:53, 29 September 2011 (UTC)

Scattering by small dust is Rayleigh scattering. Rayleigh just means d << λ. For all I know, most of the light is coming from sub-100nm particles. Spiel496 (talk) 18:50, 29 September 2011 (UTC)

The "proof" is what I wrote in the motivation of my edit. A beam is much more visible when it is directed toward you than away from you, which is typical of Mie/Tyndall scattering, whereas Rayleigh would be symmetrical: think to a thin sunbeam entering a dark room. In this case, moreover, you can distinguish a lot of dust, whose radius I think improbable to be smaller than λ/2π (about 80 nm), which is the boundary between Rayleigh and Mie (see the first figure in Mie scattering). And, even if the dust were small, the figure would be out of place, in the paragraph "From molecules". Finally, perhaps one should prove that the laser beam represents Rayleigh scattering, to keep the figure, rather than disprove it to remove it. Maybe that changing the figure position could be a reasonable compromise. --87.7.187.76 (talk) 08:50, 18 March 2012 (UTC)

I agree, the photo does not belong in a section titled "molecules". And I would concede that the burden of proof really should be on the one arguing to keep the figure, which I suppose is me. I don't have numbers to back it up. Yes, there are probably some big particles contributing. But, as you point out, their scattered light would be directed primarily forward, away from the camera, so it seems plausible that most of the light we see is coming from small stuff. I'll just repeat what I said above: The rest of the article implies that Rayleigh scattering is responsible for much of scattered light we see when sunlight passes through the air. I don't see why the situation should be any different for a laser beam. Spiel496 (talk) 20:30, 19 March 2012 (UTC)

I think that at sea level there are much more (and much larger) dust particles than in higher atmosphere. Remember that the optical properties of high-altitude dry atmosphere agree fully with pure molecular Rayleigh scattering, so that they can be used to estimate Avogadro's constant (see, e.g. [7]). Anyway, since you agree that the photo is out of place there, the better thing to do is simply to change its position.--87.11.215.15 (talk) 13:41, 20 March 2012 (UTC)

Run-ons and dense math do not mix!

This is confusing! Someone should fix this:

in particular, the scattering coefficient, and hence the intensity of the scattered light, varies for small size parameter inversely with the fourth power of the wavelength.

I would do it but I'm not certain what bits are important. Craig Pemberton (talk) 03:51, 28 July 2009 (UTC)

I took a shot at it. Spiel496 (talk) 16:37, 28 July 2009 (UTC)

regarging 'updating info'

hello it would be nice if you could update certain of your pages and information...."the sun does not rise or fall" —Preceding unsigned comment added by 78.151.154.143 (talk) 09:36, 23 August 2010 (UTC)

Pronounciation

How is "Rayleigh" pronounced? Ray-lee? Rah-lee? — Preceding unsigned comment added by 71.201.125.93 (talk) 00:40, 13 January 2012 (UTC)

It's pronounced "RAY-lee". You can hear it here. I also have it on the authority of physicists trained in Britain. - Eb.hoop (talk) 18:32, 2 May 2012 (UTC)

Why sky blue

Sunlight scattered by gases with very negligible intensity because according to Rayleigh scattering particles range between 30 and 40 nanometers are trillion times more intense than gases ( nitrogen & oxygen ) and much less intense than haze ( about 200 nanometers ). On the other side tiny particles and haze appear white in cold weather, which means that atmospheric blue color is Rayleigh scattering independent.

It is more easy to explain in my mother language as shown in this link

http://www.kabel.0fees.net/Sky/Sky-color-is-unknown.htm

that ozone layer reflects ocean's color — Preceding unsigned comment added by 41.218.181.44 (talk) 08:48, 18 October 2012 (UTC)

Atmospheric white color of Mars ( right click and copy shortcut )

Why storm clouds are blue ( right click and copy shortcut )

Why sky is not blue at 29 km hight ( right click and copy shortcut ) — Preceding unsigned comment added by 41.218.181.44 (talk) 11:12, 18 October 2012 (UTC)