Talk:Rayleigh scattering

Opalite Picture Dispute

The picture of opalite is used here to demonstrate Rayleigh scattering, and it is also used on the Tyndall effect page to demonstrate the Tyndall effect. If the Tyndall effect requires that the particles be about the same size as the wavelength, and Rayleigh requires that the particles be much smaller, then it can't be both. 24.113.123.50 (talk) 19:50, 31 January 2017 (UTC)

I agree - the opalescent glass is blue because of light bouncing off inclusions etc. that are LARGER than the wavelength of the light - that's the Tyndall Effect. If it was bouncing off the molecules in the glass, it would be Rayleigh Scattering. But it's not. Clear glass is blue green because of Rayleigh Scattering, Opalescent Glass is blue because of the Tyndall Effect. This image needs to be removed/changed. — Preceding unsigned comment added by 118.209.77.5 (talk) 05:26, 16 March 2019 (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. [1]). 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)

Given the strong consensus here. I've removed the image. Ergzay (talk) 22:32, 28 March 2015 (UTC)

I realize this is kind of an old discussion -- but since the change to the article just popped up in my watchlist: What makes everyone think that laser is illuminating dust? The source of scattering intensity could easily be some sort of aerosol. Other than a few intermittent spikes it scattering intensity, the beam looks pretty uniform. So, at the very least, dust particles shouldn't be the predominant species leading to scattering. Also, as was pointed out already in the above discussion, the beam is pointed away from the camera. Meaning that what we (the observer) are seeing is almost entirely backscattering, which would minimize the contribution from large particles. (+)H₃N-Protein\Chemist-CO₂(-) 13:41, 1 April 2015 (UTC)

Also -- if you look at the meta information in the photo it shows an exposure time of 2/1 seconds! So, the actual intensity of scattered light isn't nearly as high as people are assuming it to be. Might not even be visible by eye. (+)H₃N-Protein\Chemist-CO₂(-) 13:45, 1 April 2015 (UTC)

If it's an aerosol it's STILL not Rayleigh scattering. The act of being able to see the beam of a laser from strong intensity or from long exposures means its simply backscattering off of particles in the air, or moisture in the air.Ergzay (talk) 17:31, 17 April 2015 (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 equation the intensity of light is directly proportional to the sixth power of particle's diameter that cause particles of 40 nanometers diameter ( ${2\pi r}$ is approximately equivalent to the sphere diameter ) to be more than 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 ( below the wavelength of violet ) appear white in cold weather while storm clouds ( very far away over the wavelength of red ) appear blue sometimes, 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 that ozone layer reflects ocean's color — Preceding unsigned comment added by 41.218.181.44 (talk) 08:48, 18 October 2012 (UTC)

There is far more oxygen and nitrogen molecules than dust particles. More than 1000 trillion times as many, by my rough estimate. The article already has a graph showing the integrated scattering from air molecules alone amounts to about 20% of blue light. That's enough to explain why the sky appears blue without needing to invoke dust, ozone, or anything else. Dragons flight (talk) 21:11, 26 October 2012 (UTC)

Your assumption is not true Dragons flight, the average concentration of air pollution is 4 ppm (parts per million) which cause the share of gases in total intensity to be very negligible — Preceding unsigned comment added by 41.218.180.229 (talk) 07:43, 27 October 2012 (UTC)

You asked about particulate matter 40 nm and larger, which is rare globally. Ozone, nitrous oxide, carbon monoxide, etc., are similar in size to air molecules, so they don't matter for this discussion. Also, you can't use numbers for near surface pollutants if you want to measure the column integrated scattering. But the discussion about pollution is all irrelevant. As mentioned, clean air has enough scattering to appear blue, as you could calculate directly from the material in the article (about 20% of blue sunlight is scattered on its way through the atmosphere). Specifically, the Rayleigh scattering of sunlight into perfectly clean air will appear blue. Whether or not other pollutants (that may or may not be present depending on the environment) will add to or modify the appearance of the sky isn't really the issue. If you want to say the color of the sky isn't influenced by Rayleigh scattering you would need to show that the column integrated scattering due to the Rayleigh effect is somehow negligible, which is simply not true. Dragons flight (talk) 08:43, 27 October 2012 (UTC)

The white color of Stratosphere in picture simply answers you Mistake is not in Rayleigh equation, but in the assumption of color release without light separation caused by refraction — Preceding unsigned comment added by 41.218.180.229 (talk) 09:44, 27 October 2012 (UTC)

If the orange sun at sunset is predominantly caused by Rayleigh scattering at air molecules, why does it still appear white at sunset in the arctic? Must be dust scattering which is dominant and arctic air is clean?Superdoc1 (talk) 07:40, 16 August 2015 (UTC)

The orange sun at sunset is neither air molecules scattering nor dust scattering, actually it is superior mirage. Why there is no orange sun at sunrise? Because the lower end of the ozone temperature at sunrise and the great end of the ozone temperature at sunset. — Preceding unsigned comment added by 41.68.118.128 (talk) 13:56, 15 July 2019 (UTC)

Um, the sun IS yellow...

Okay, I KNOW! Before you get insane & start screaming to yourself about how the sun is white & I am a moron, you should really stop & think. According to NASA, the sun is 'technically' white due to the fact that it puts off an extremely broad spectrum of electromagnetic radiation (etc.). However,know this. Also according to NASA, spectral analysis of the 'white light' received from the Sun reveals that of all the colors in the spectrum, TWO are MOST PROMINENT. What are those two colors? You ask... YELLOW & GREEN This is quite ironic when you investigate stellar classifications because they move in the common spectral pattern, starting from blue (o) & ending up in red (k)... what is ironic is that once you get to the classification of the sun, it jumps from blue to white... Umm... one is left to wonder why this is when for all we know, every star could appear white from a relative distance to that of Earth's from our own.. Furthermore, yes, actually... our star is mostly yellow-green... What is the opposite of white noise? SILENCE! In the physics based part of the equation, white & black are not attributes of chromaticity... they are purely the root functions of luminosity. there are numerous colors that don't actually have a frequency & wavelength & this puts them close to the same category. Colors like Magenta are not technically 'natural notes' on the color wheel... This is also true for white and black... Lastly, anything regarding the suns color is by and large purely ignorant. If you wanted to judge the suns color, you would first be required to turn off it's own light & then inspect it with your own light source. Preferably a source of light you have predetermined to exhibit a particular spectral pattern so that you can further justify your observations... As it stands, one would be wise to realize the fundamental difference between light & color. Oh... & go look at the sun, because it doesn't just shine, it ripples, it flashes, it exhibits no qualities that would indicate that one would have any justifiable reason to assume it is a single color at all... the claim that it is white is just... well, wrong. Those are photons! Lawstubes (talk) 18:06, 10 January 2013 (UTC)

There's a discussion of the solar specrum here: Sunlight#Composition_and_power. Compare that to diffuse sky radiation and consider also the luminosity function of the human eye and color temperature. --Kkmurray (talk) 01:53, 11 January 2013 (UTC)

Actually, it depends on how you measure color. In frequency space, sunlight at the surface of the sun peaks in the infrared-red. In wavelength space, it peaks in the blue. 129.63.129.196 (talk) 21:37, 28 March 2013 (UTC)

Color of Sky Explanation Inaccurate, Omits Violet

The graph fades to black at the left side where it needs to show violet. There is about as much violet as there is green (the graph in the diffuse sky radiation article is inaccurate). The reason we don't see violet in the blue is explained here http://patarnott.com/atms749/pdf/blueSkyHumanResponse.pdf - humans cannot distinguish blue-violet from blue-white. You can find a more in-depth explanation of metamers here http://blog.asmartbear.com/color-wheels.html - Wiki's own article (http://en.wikipedia.org/wiki/Metamerism_%28color%29) is rather inadequate. I think "why the sky is blue" should have its own article since a full, accurate explanation is outside the scope of this one. At the very least the current landing spot for "why is the sky blue" (http://en.wikipedia.org/wiki/Why_is_the_sky_blue) should mention metamers. — Preceding unsigned comment added by 173.14.140.253 (talk) 23:52, 27 January 2013 (UTC)

I concur. If Rayleigh scattering were the only mechanism involved, then the sky would be violet. You have to include Rayleigh scattering, the fact that the incident sunlight is a blackbody spectrum and not a constant spectrum, bulk attenuation, and the human eye response in order to show mathematically that the sky is blue. The article should mention all of this. If you only take Rayleigh scattering into account, mathematically you end up with a violet sky. 129.63.129.196 (talk) 21:34, 28 March 2013 (UTC)

Still no mechanism!

I am buffled by the experts that run into lengthy discussions in this talk age, but do not contribute this crucial part to the article. C'mon guys. 213.8.52.148 (talk) 05:27, 20 February 2013 (UTC)

There's nothing to disagree about while we stick to the title subject, Rayleigh scattering. It's the result of scattering from small objects, typically molecules and atoms; anything much smaller than the wavelength will serve but they must also be randomly distributed in space. That's the mechanism - see e.g. "Subtle is the Lord", a biography of Einstein by Abraham Pais, pp.102-3. Of course, light scattering in the atmosphere and the resultant colouration is more complicated, but that is not the subject of the article.TSRL (talk) 20:21, 18 March 2014 (UTC)

to techie for encylopdia

"The size of a scattering particle is parameterized by the ratio x" a word like parameterized should not appear Instead of the first equation, thre should be something more simple along the lines of Is = Io lambda^-4 d^6, emphasizing that when the medium is not perturbed, at low concentraions, for a given experimental setup - the usual things that apply in teh real world - the most importantt things are the steep dependence on wavelenght and particle size — Preceding unsigned comment added by 50.195.10.169 (talk) 21:32, 27 August 2013 (UTC)

Compton scattering is an INELASTIC scattering. First sentence is wrong! — Preceding unsigned comment added by 2001:718:1401:58:0:0:2:A214 (talk) 16:57, 5 September 2013 (UTC)

This article doesn't mention Compton scattering. — HHHIPPO 16:52, 16 October 2013 (UTC)

No, but our article on Compton scattering does. Perhaps the unsigned IPv6 above got directed to the wrong talk page? (+)H₃N-Protein\Chemist-CO₂(-) 11:20, 17 October 2013 (UTC)

Interesting, at the time of their comment Compton scattering was indeed wrong. I wonder how they ended up here. — HHHIPPO 15:46, 17 October 2013 (UTC)

No, it was right, assuming it said elastic. Both energy and momentum are conserved, the same before the collision as after.TSRL (talk) 16:52, 17 October 2013 (UTC)

Seems I was wrong - it is an elastic collision, producing inelastic scattering (WP). As it happens, I may be lunching tomorrow with a career long neutron scatterer (if he doesn't get scattered away by other interactions) and I'll try to find out usage in his area.TSRL (talk) 19:16, 17 October 2013 (UTC) Yup, I was wrong!TSRL (talk) 20:22, 18 March 2014 (UTC)

Date?

The article is missing vital information: When and in what publication did Rayleigh propose his theory? It should also be noted that Johann Wolfgang von Goethe was first to study and theorize on its effects as well as those of Mie scattering in his Theory of Colours in 1810. Goethe's conclusions were basically that Rayleigh scattering (resulting in a violet-cyan spectrum) was due to light interacting with black objects (such as the blackness of space), that Mie scattering (resulting in a yellow-magenta spectrum) was due to light interacting with turbid objects (such as earth's atmosphere), and the larger the angle of the sunlight reaching us (such as during sunrise and sundown), the more it is shifted towards the Y-M spectrum because of having to cross a much larger mass of turbid atmosphere than when reaching us from above, where it has to cross a much smaller amount of turbid atmosphere. --2.240.198.214 (talk) 20:33, 16 March 2014 (UTC)

The first footnote lists publications and dates. The remaning content of your post is difficult to follow, and I don't see a way to incorporate into the article. Perhaps you could concentrate on one particular fact that you feel should be addressed. Spiel496 (talk) 17:43, 17 March 2014 (UTC)

I think that any article on a scientific theory should have the date when it was first proposed in the lead, not just in a footnote, just as we're putting the name of its discoverer Rayleigh in the lead as well. And well, I just meant that there should also be a short mention of Goethe's Theory of Colours as part of an introductory history or introduction or overview section of scientific theories on why the sky is blue (and sunrises are red/magenta) before Rayleigh, and I think Goethe was pretty close to Rayleigh and Mie (just as there are one or two close late 19th century forerunners to Einstein's theory of special relativity).

Goethe said that the blue sky is due to light interacting more or less directly with the blackness of space, and Rayleigh said that in order to see the blue effect of the scattering, you require a black backdrop. Goethe said that the magenta sunrises and sundowns are because light interacting with a turbid object or medium results in a Y-M spectrum and that the sunlight has to cross much more turbid atmosphere at such an extreme angle, and Mie pretty much repeated that, only dwelling on the molecular structure of gasses in detail where Goethe simply talks of turbidity. Of course, Goethe was no mathematician, but it's notable how close his Theory of Colours came to the underlying principles of Rayleigh scattering and Mie scattering. --2.241.26.109 (talk) 23:46, 17 March 2014 (UTC)

The idea of "light interacting more or less directly with the blackness of space" sounds nice, but it has nothing to do with Rayleigh scattering. Spiel496 (talk) 18:35, 18 March 2014 (UTC).

Agree with Spiel1496. Does Goethe say scattering off small, randomly placed entities and/or mention the (wavelength)^-4 dependence? If not, why include him? My own feeling just now is that the subject of this article should be its main topic and "the colour of the sky in the presence of an atmosphere" another. BTW, the sunset colours are primarily the result of a longer path in the atmosphere, scattering away the shorter wavelengths, rather any change in the properties of the atmosphere. Goethe's poetry is fine, and sings well when set by Schubert, but his well intentioned forays into science are premature and not generally successful.TSRL (talk) 20:41, 18 March 2014 (UTC). We do have such an article, Diffuse sky radiation which I missed, and much of the discussion on this talk page would perhaps be better there. There's quite a lot of overlap between the two talk pages.TSRL (talk) 08:53, 19 March 2014 (UTC)

But that's just what Goethe says: That it's to do with the amount of atmosphere (aka "longer path" through the atmosphere), just as you're saying it TSLR, not with any sudden "change in the properties of the atmosphere" that you're making up now. And if Goethe's bringing the blackness of space into it really doesn't have anything to do with Rayleigh, then why haven't I been the first by far on this talkpage to point out that Rayleigh says that you require a black backdrop in order for the blue effect to be seen?

Ah! This shows why physics requires mathematical, rather than purely verbal description. I'd read "cross much more turbid atmosphere" to mean a region of much greater turbidity, rather than a much greater region (longer path) of atmosphere of unchanged turbidity. Language is often rather ambiguous; a bonus for poets but not scientists. On the blackness of space (a region of few scatterers): of course atmospheric scattering will only dominate the view in the absence of brighter sources. There's no interaction involved.TSRL (talk) 14:48, 2 April 2014 (UTC)

Again, this is not about mathematical details or the detailed molecular structure of certain gasses, but about the basic principles of a.) the behavior of light in earth's turbid atmosphere (i. e. resulting in an increasing Y-M shift the longer the path), and b.) the fact that the blue color has to do with the blackness of space. The rest are but details which facilitate us to calculate the exact amount of scattering in detail, not the underlying aforementioned principles.

Most physicists would argue that it's only through the agreement with observation of mathematical (quantitative) predictions can you start to think a qualitative model may be right. The more you can quantify the model, the more rigorous the tests. Rayleigh scattering does not ask much of the scatterers: they need to be much smaller than wavelengths in the relevant spectral range and not absorb much of the light in that range, as well as being randomly distributed in space. Cheers,TSRL (talk) 15:43, 2 April 2014 (UTC)

And, well, diffuse sky radiation is made up of Rayleigh scattering (for the V-C spectrum) and Mie scattering (for the Y-M spectrum). It's not like Mie and Rayleigh would have nothing to do at all with diffuse sky radiation, they're just the two different types of it. --2.240.228.185 (talk) 10:37, 2 April 2014 (UTC)

Atmospheric filter

Is it coincidence that the sky away from the sun looks light blue and the difference between the [spectrum above & below] the atmosphere is somewhat greater in blue? On a possibly related issue, is some 'average' of the yellow sun and the blue sky (maybe plant green too), a basis for our evolved perception of white?
--Wikidity (talk) 03:32, 24 March 2014 (UTC)

My view is that this, together with many of the comments about the sky, belongs in Diffuse sky radiation, not here. The mechanism of Rayleigh scattering is the subject of this article and the colour of the sky is just one example of this at work. There is more to the colour of the sky than just Rayleigh scattering, including star type, the structure and composition of the atmosphere and the evolution of the eye, so are these not better handled separately?TSRL (talk) 07:56, 24 March 2014 (UTC)

BTW, the link just brings up a Bad title error.TSRL (talk) 08:06, 24 March 2014 (UTC) This does work: File:Solar_Spectrum.png. These spectra are of direct, rather than scattered sunlight so scattering will reduce the ground level intensity at the shortest wavelengths but, though I don't have the numbers to hand, by much less than the ozone absorption.TSRL (talk) 08:41, 24 March 2014 (UTC)

Small size parameter section opening is a bit confusing

TSRL, in your recent addition to the "Small size parameter" section, you wrote the phrase "The wavelength dependence is the consequence of the dominant dipole-induced-dipole mechanism". There's a lot of hard words in there for the typical encyclopedia reader. I would like to reword it in a more accessible way, but, even with a degree in physics, I'm not sure what it means myself. Can you (or anyone who's up to it) expand on it here. What is a dipole-induced-dipole mechanism? Does that mean a dipole induced by the electric field of the light? Spiel496 (talk) 01:08, 3 April 2014 (UTC)

I'd hoped to find an accessible internal or external link for this and didn't want to dwell too long on it. It's important, though, as the source of the λ^-4 dependence. Yes, you're right; the field - think DC for a moment - induces a dipole, like that on the dielectric in a capacitor. If we now use an AC field, this dipole will oscillate, one end first positive then negative This oscillating dipole will then re-radiate (scatter) the initial field, just like a current driven dipole aerial in any radio transmitter, because the charges are accelerating. I'll see what I can find, othrwise include something like the above. Cheers,TSRL (talk) 08:06, 3 April 2014 (UTC)

Looking at it again, there is a description of this in the opening paragraph or lead. Maybe we should shorten the lead and put some of is into the article proper.TSRL (talk) 08:14, 3 April 2014 (UTC)

Leaving the lead for now, I've dropped the techie dipole-induced dipole phrase and replaced it with a link to dipole radiation, which also refers to Rayleigh scattering. See what you think.TSRL (talk) 08:31, 3 April 2014 (UTC)

It looks better. It is a good goal to explain the λ^-4 dependence. Spiel496 (talk) 13:53, 4 April 2014 (UTC)

Bold text

Error of Omission: Coherent Scattering

Hi everyone:

I have learned, the hard way, that the blue sky argument is much trickier than it first appears !

In particular, this otherwise diligent and well-written article propagates a common mistake. The mistake (oversight) concerns the statement:

"The fraction of light scattered by a group of scattering particles is the number of particles per unit volume N times the cross-section."

This is true only in the case of incoherent scattering--when the scattering medium is not dense and i.e. the scattering particles are far apart compared to the characteristic wavelength of the incident beam.

In the case where the medium is dense i.e. there are many particles within the characteristic wavelength of the incident beam, then coherent scattering must be considered. In this case one must add scattering amplitudes arising from the dense particles. One must subsequently square these amplitudes to derive the scattered intensity. This boosts the scattered intensity with an additional factor N, resulting in an N² dependency of the scattered intensity. In other contexts this is called 'superradiance'

https://en.wikipedia.org/wiki/Superradiance

A fulsome analysis of the propagation of light in dense medium adds a lot of complexity. It involves so-called 'extinction theorems' that explain, for example, the index of refraction of the medium as well as the reason the observed backscattered intensity is weak even though the Rayleigh backscattering amplitude is strong.

There is a simple, grosso modo explanation that I believe sheds some light :-) on the situation:

- In dense media, one must define appropriate scattering centres.

- In some cases, impurities are the 'true' scattering centres (not the molecules that make up the 'matrix' of the medium).

- iIn the atmosphere, the scattering centres really should be considered to be the natural density fluctuations. These fluctuations occur in all gases.

- These fluctuations satisfy Poisson statistics which implies that they are proportional to the square root of N (where N is the average particle number in a given volume).

- In other words, we can see that the two new factors cancel each other: (a) coherent scattering demands a square (N²) while (b) the analysis of density fluctuations introduces a square root.

Conclusion: the above gross argument suggests that the contentious statement is correct but only due to a fortuitous cancellation of 2 errors (a) and (b)!

...What to do?

While I believe that the above gross analysis has merit, it is not rigorous. Furthermore, it would unduly complicate the article. Therefore I don't recommend incorporating it.

One solution would be to add qualifiers to the analysis as currently presented : "behaves as..." "produces a scattering intensity analogous to...."

Another option would be to reduce the depth of the analysis of the Blue Sky portion of this article. After all, the topic is "Rayleigh Scattering" which is a general and fundamental phenomenon that applies to many situations.

I await feedback from the wisdom of the collective :-).

Riccbdr (talk) 17:06, 23 August 2015 (UTC)

PS I know what to suggest:

- The current narrative very quickly jumps to the blue sky phenomenon. This may give readers the impression that Rayleigh scattering is exclusively related to blue sky.

- To prevent this, and to broaden the perspective, we should first mention that Rayleigh scattering is a very fundamental, atomic-level process whereby one incident photon encounters a small particle, the photon and particle interact, and this interaction generates one emitted photon with the same energy as the incident photon.

- This 'one-photon-in-and-one-photon-out' is the 'building block' so to speak, of many important processes, including light propagation within transparent media, as well as light reflection from mirrors.

- When applied to light propagation in the atmosphere, Rayleigh scattering is the fundamental process responsible for the index of refraction of the air as well as the blueness of the sky.

- I will suggest polished text soon, as well as a new diagram. Not much will need to be changed in the current article.

Riccbdr (talk) 11:37, 26 August 2015 (UTC)

I haven't heard refractive index being attributed to Rayleigh scattering before. Can you expand on what you said? Are you saying that the coherent scattering leads to the refractive index, but the scattering off density fluctuations leads to the blue color? Also, I'm having trouble wrapping my head around "coherent scattering" being applied to randomly-positioned gas molecules illuminated by white light. Nothing about that seems coherent. Does coherent refer to the fact that neighboring molecules are experiencing approximately identical electric fields? I think Riccbdr has made some good points, and I've heard this claim elsewhere that the scattered light is from density fluctuations. The important next step, however, is to identify a secondary source (text book or web publication) that explains it well, to use as a reference. Spiel496 (talk) 20:23, 27 August 2015 (UTC)

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Thanks for your patience Spiel496

Yes you got to the core of my understanding: the index of refraction is essentially a coherent scattering phenomenon while the Blue colour of sky is basically incoherent scattering from density fluctuations. Both are based on Rayleigh scattering as the fundamental atomic process involved. This is not a totally rigorous description. It's a simplified picture of what's happening. But it's much better compared to our usual description which breaks the rules of scattering theory from the get-go.

To summarize, here are the 2 main problems that arise from the usual explanation:

(A) Using numbers from the article, we can estimate there to be about a million Nitrogen gas molecules per cubic wavelength of bluish light. In this case we are simply not allowed to use the straight scattering cross section. As you insightfully pointed out, very many of these molecules are 'inside' the photon field, experiencing the same EM fields--hence they are all re-emitting in phase with each other. As you said, it has nothing to do with the light's optical coherence length which would govern a length scale of many wavelengths (unless it were strange light found only in laser physics labs :-).

(B) As the current article points out, only a small amount of light is scattered (exp-5 for every meter of travel). So what happens to the rest? To be consistent with scattering theory, the answer would need to be "absolutely nothing." The majority of the light sails right through the 'target' (air in this case), totally unaffected and therefore travelling at c. We are thus left with nothing to explain the index of refraction. So now what are you gonna do to explain n?...Invent some new, ad-hoc, forward-scattering cross section outta thin air (sic)?  :-)

I totally agree with you about the need for good references. I will try to find. Some years ago, I spent a an embarrassingly long time on this problem. Back then I found lots of repetitions of the usual errors. Unfortunately, I since threw out most of my notes!

...Pleasure doing 'business' with you! Riccbdr (talk) 15:32, 2 September 2015 (UTC)

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I almost forgot to mention: The coherent scattering picture suggests that Rayleigh scattering in directions perpendicular to light travel as well as in rearward direction (i.e. all off-forward-axis scattering) would self-cancel. To see this, simply, sketch one wavelength of light and pepper it with a million particles inside each 1/2 wave (you know what I mean :-). Taking as example, the scattering in the perpendicular direction (90° from light propagation), you can see that the amplitudes arising from particles within first 1/2 of the wave are exactly out of phase with the amplitudes from the particles in 2nd half of the wave. Similar geometrical factors essentially kill all off-forward-axis scattering. Amplitudes scattered in forward direction, however, always interfere constructively.

In other words, the simple coherent scattering picture suggests that forward Rayleigh scattering is heavily favoured and this is consistent with what we observe about light propagation. This model also suggests that in absence of density fluctuations or rarified impurities, there would be essentially no scattering off-axis i.e. no blue sky. Riccbdr (talk) 16:54, 2 September 2015 (UTC)

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Some resources:

1) Here is an inspiring summary (see approx 1/2 way down section on "Electromagnetic Scattering"): https://en.wikipedia.org/wiki/Scattering

2) Here is an important part of the story which I never pursued (and subsequently forgot that it even existed):

There is an Einstein-Smoluchowski analysis for blue sky based on density fluctuations. A high-quality introductory reference: http://www.osti.gov/accomplishments/nuggets/einstein/daytimea.html Riccbdr (talk) 18:27, 2 September 2015 (UTC)

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Comprehensive references were requested above. Here are 2 + 1:

- ( 15p ).

- ( 7 tri-columns pages ).

- note that a short version was published in 1981 as ( 2 pages. pdf: available in your preferred science papers free repository )

Fabrice.Neyret (talk) 17:35, 21 March 2019 (UTC)

Small-size parameter approximation

I am confused about this section. It cites, as an example, nitrogen. But the result it supplies, the cross-section of 5.1 x 10(-31), does not appear reachable from the equation provided. For one thing, the value d in the equation isn't clearly the diameter of one nitrogen atom, or of a nitrogen molecule. Another problem is that it isn't clear what value to use for n, the index of refraction. Is it the index of refraction of the atmosphere at atmospheric pressure? The section includes the text, "... In detail, the intensity I of light scattered by any one of the small spheres of diameter d and refractive index n". Does a nitrogen atom (or molecule) have a definite refractive index? And what about the index of refraction of the rest of the space? I get the impression that the resulting cross-section wasn't derived using this equation, alone. (the 4th equation in this section.) 67.5.237.141 (talk) 07:36, 20 October 2015 (UTC)

New picture

I just uploaded a new self-made picture, which may help to explain the subject: RayleighScattering.gif. As i'm not active on the English wiki, feel free to use this picture. Regards, Pingel (talk) 19:46, 28 March 2017 (UTC)

Image for Tyndall scattering?

This image is used on several pages including this page and Tyndall effect and others. On some it's cited as an example of Tyndall scattering, on others Rayleigh scattering. We need to be consistent if we're explaining the difference to people! --RProgrammer (talk) 16:19, 28 April 2020 (UTC)

One or two mechanisms in amorphous solids

In glass, Is it the "microscopic variations of density and refractive index" or variations in the dielectric constant, or are these different ways to view a single mechanism ? - Rod57 (talk) 12:24, 13 December 2021 (UTC)

Equation Terms

"R is the distance to the particle" - distance to what? Closest approach? Observer? Baltimore?

What is 'd'? Diameter of the particle? It isn't specified.

"Averaging this over all angles gives the Rayleigh scattering cross-section" - shouldn't that be 'Integrating'?

2600:1700:C280:3FD0:D9D6:C44E:B21A:B3D8 (talk) 23:02, 23 March 2022 (UTC)

Flag this whole article for being far too technical

C’mon fellas and lasses, this is Wikipedia… particle physics needs to be simplified for the masses here… 2600:387:C:6D15:0:0:0:8 (talk) 07:21, 17 April 2022 (UTC)

Similar concept applies to gamma rays

Is it worth mentioning in THIS article that the same concept applies to scattering of gamma rays by nuclei? 2001:8003:E41C:1C01:8D09:B458:4705:8FB3 (talk) 11:16, 22 August 2022 (UTC)

Polarisation

There is no discussion of the polarisation properties of Rayleigh scattering, and how it causes polarisation of scattered sunlight (though there is an image). It might be worth adding a more detailed account of the effect of scattering on the Stokes parameters, perhaps by showing the phase matrix. Greatanarch (talk) 20:42, 11 January 2023 (UTC)

Why is the sky blue?

I don't think this article says why Rayleigh scattering is not refraction, reflection, or diffraction.

Also, I don't think it explains what Rayleigh scattering (or scattering) is.

..

The article currently says:

Rayleigh scattering results from the electric polarizability of the particles. The oscillating electric field of a light wave acts on the charges within a particle, causing them to move at the same frequency. The particle, therefore, becomes a small radiating dipole whose radiation we see as scattered light. The particles may be individual atoms or molecules; it can occur when light travels through transparent solids and liquids, but is most prominently seen in gases.

I think that paragraph says: “the particle (e.g. an oxygen atom.. or the electron?) absorbs (accepts?) and re-emits (the blue)”.

The article currently also says:

A portion of the beam of light coming from the Sun scatters off molecules of gas and other small particles in the atmosphere.

The images at [2] [3] [4] look (to me) like absorption (of the light wave) and re-emission. Those words match [5].

But this one [6] says “no absorption”. The illustration looks like there is… and the re-emission can be “backwards”.

..

It seems (to me) the light (or EM wave) doesn't get “diffracted” on its pass through the atom. It seems to be absorbed by something (not an electron or the nucleus) and re-emitted… the new direction is generally forwards, but for light at a higher amplitude (which is a field strength not a distance) and a shorter wavelength, that direction is more greatly changed.

Note: blue light is (around) 475 nm. An oxygen or nitrogen molecule is (around) 0.3 nm… about 1/1600th… a lot lot less than the 1/10th cited as required for Rayleigh scatter.

MBG02 (talk) 04:55, 23 April 2024 (UTC)

Thanks. I rewrote that section with a reliable source.

Rayleigh scattering is classical elastic scattering from inactive particles. The wave or ray schematics for elastic scattering are indistinguishable from absorption and reemission. The elastic scattering model is much simpler and adequately describes the blue sky effect. Johnjbarton (talk) 15:52, 23 April 2024 (UTC)

How is “scatter” not composed of any of (reflection, refraction, or diffraction)?

How is the light “bent”… with some of it going backwards?

Is the light absorbed and re-emitted… if so, is it by the electrons, the nucleus… or what?

MBG02 (talk) 04:34, 25 April 2024 (UTC)