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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 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)
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)
Given the strong consensus here. I've removed the image. Ergzay (talk) 22:32, 28 March 2015 (UTC)
This is confusing! Someone should fix this:
I would do it but I'm not certain what bits are important. Craig Pemberton (talk) 03:51, 28 July 2009 (UTC)
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)
How is "Rayleigh" pronounced? Ray-lee? Rah-lee? — Preceding unsigned comment added by 71.201.125.93 (talk) 00:40, 13 January 2012 (UTC)
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 ( 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)
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)
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)
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 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)
"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)
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)
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)
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)
Bold text
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)
=-=-=-=-=-=-=-=-=-=-=-
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)
=-=-=-=-=-=-=-=-
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)
=-=-=-=-=-=-=-=-
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)
Comprehensive references were requested above. Here are 2 + 1:
- [1] ( 15p ).
- [2] ( 7 tri-columns pages ).
- note that a short version was published in 1981 as [3] ( 2 pages. pdf: available in your preferred science papers free repository )
Fabrice.Neyret (talk) 17:35, 21 March 2019 (UTC)
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)
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)
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)
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)
"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)
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)
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)
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)
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:
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:
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)