Talk:Rayleigh scattering

message: 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)
replies: text: 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)
thread_title:  Opalite Picture Dispute 

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

text: 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)
replies: text: 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. ). 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)

text: 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)
replies: text: 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)
replies: text: 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)
thread_title: Laser picture dispute

message: 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)
replies: text: I took a shot at it. Spiel496 (talk) 16:37, 28 July 2009 (UTC)
thread_title:  Run-ons and dense math do not mix! 

message: 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)
thread_title:  regarging 'updating info' 

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

message: 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)
replies: text: 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)

text: 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)
replies: text: 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)

text: 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)

text: 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)
thread_title:  Why sky blue 

message: 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)
replies: text: 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)

text: 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)
thread_title:  Um, the sun IS yellow... 

message: 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)
replies: text: 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)
thread_title:  Color of Sky Explanation Inaccurate, Omits Violet 

message: 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)
replies: text: 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)
thread_title: Still no mechanism!

message: "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)
replies: text: This article doesn't mention Compton scattering. — HHHIPPO 16:52, 16 October 2013 (UTC)
replies: text: 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)
replies: text: 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)
replies: text: 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)
replies: text: 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)
thread_title: to techie for encylopdia

message: 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?<BR>
--Wikidity (talk) 03:32, 24 March 2014 (UTC)
replies: text: 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)
replies: text: BTW, the link just brings up a Bad title error.TSRL (talk) 08:06, 24 March 2014 (UTC)
thread_title:  Atmospheric filter 

message: 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)
replies: text: 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)
replies: text: 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)
replies: text: 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)
replies: text: It looks better. It is a good goal to explain the λ-4 dependence. Spiel496 (talk) 13:53, 4 April 2014 (UTC)
thread_title:  Small size parameter section opening is a bit confusing 

message: 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)
replies: text: 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)
thread_title:  Error of Omission: Coherent Scattering  

message: Comprehensive references were requested above.
Here are 2 + 1:
- "Light Scattering on Molecules in the Atmosphere",  Benjamin Witschas, chapter 5 
in book "Atmospheric Physics, Research Topics in Aerospace". Springer-Verlag, 2012, ISBN 978-3-642-30182-7.  DOI: https://doi.org/10.1007/978-3-642-30183-4_5  pdf: https://link.springer.com/content/pdf/10.1007%2F978-3-642-30183-4_5.pdf  ( 15p ).
- "Rayleigh scattering:
Questions of terminology are resolved during a historical excursion through the physics of light‐scattering by gas molecules", Andrew T. Young, Physics Today 35, 1, 42 (1982) https://doi.org/10.1063/1.2890003 pdf: https://www.researchgate.net/profile/Hans_Moosmuller/post/Does_Rayleigh_Scattering_really_explain_blueness_of_sky/attachment/59d6338dc49f478072ea23ef/AS%3A273644943544335%401442253446960/download/Young1982PhysicsToday.pdf  ( 7 tri-columns pages ).
- note that a short version was published in 1981 as "Rayleigh scattering", Andrew T. Young. Appl. Opt. 20. (1981) https://doi.org/10.1364/AO.20.000533  ( 2 pages. pdf: available in your preferred science papers free repository )
Fabrice.Neyret (talk) 17:35, 21 March 2019 (UTC)
thread_title: ========

message: 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)
thread_title:  Small-size parameter approximation 

message: 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)
thread_title:  New picture 

message: 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)
thread_title:  Image for Tyndall scattering? 

message: 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)
thread_title:  One or two mechanisms in amorphous solids 

message: "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)
thread_title:  Equation Terms 

message: 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)
thread_title:  Flag this whole article for being far too technical 

message: 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)
thread_title:  Similar concept applies to gamma rays 

message: 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)
thread_title:  Polarisation