Talk:Rayleigh scattering

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)

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)