Talk:Diffuse sky radiation

Clarification please

The principle explained begs further clarification on 2 points:

"The sunlit sky appears blue because air scatters short-wavelength light more than longer wavelengths."

Question: Why is this?

"Near sunrise and sunset, most of the light we see comes in nearly tangent to the Earth's surface, so that the light's path through the atmosphere is so long that much of the blue and even green light is scattered out, leaving the sun rays and the clouds it illuminates red."

Question: Why then, does the sky not turn green and yellow first, before turning red in the evening?

Jakkeven (talk) 05:59, 7 October 2010 (UTC)

One item I'm confused about is that in the Color section the text says that the sky's color is equivalent to 474-476nm, but this wavelength substantially misses the peak of the spectrum in the first figure. Also, there seem to be several caveats and qualifiers from the author of the figure--might it be possible to find a better spectral curve? Mvsmith (talk) 10:00, 30 January 2011 (UTC)

The "equivalence" is not a spectroscopic one, but concerns the visual color appearance, as it is clear by the quoation reported. I'll try to clarify this in the article. Anyway, the spectrum reported is shifted towards shorter wavelengths, with respect to the spectrum of sunlight. --GianniG46 (talk) 16:12, 2 February 2011 (UTC)

Initial comments

This is taken from the excellent public domain NASA resource DICTIONARY OF TECHNICAL TERMS FOR AEROSPACE USE at http://roland.lerc.nasa.gov/~dglover/dictionary/content.html

Karada, are you sure this public domain material? -- looxix 00:04 Apr 18, 2003 (UTC)

It's USA government material. It's fine. HereToHelp 00:57, 28 September 2005 (UTC)

Merge with airglow

Can I suggest that instead of cleaning this article up, it is merged with the other sky background article (airglow)? Rnt20 06:48, 7 October 2005 (UTC)

Einstein and blue sky

The article claims that a relativistic correction is required to account for the blueness of the sky. I don't think this was shown anywhere in Einstein's work on scattering. Relativity has nothing to do with it. I am therefore removing the statement on the relativistic correction. DonSiano 18:15, 17 March 2006 (UTC)

Layman Question

If what is said is true, then why isn't the sky violet? This would seem to be correct, since the article says air scatters shorter-wavelength light in preference to longer-wavelengths. Matthew 00:17, 19 April 2006 (UTC)

Our perception of the color of the sky is based on how our brain processes the various short-wavelengths present in sunlight. There are multiple wavelengths present, but we only perceive a single color. The exact color of sky does not correspond to a specific wavelength. I am looking for a reference that can help explain the neuroscience of why the sky is blue. --Bjsamelsonjones 18:12, 29 June 2006 (UTC)

See 'the sky is purple?' question below for a continuation of this. ADNewsom (talk) 03:23, 11 June 2014 (UTC)

half-layman question

why isn't the sky ultraviolet, that is, transparent to us ? 86.104.188.234 17:39, 29 April 2006 (UTC) Stefan Udrea 17:45, 29 April 2006 (UTC)

seriously , I don't understand this:when the path of a light beam through the atmosphere is longer,the beams with higher energy (blue,violet,ultra-violet,etc.) should be scattered *less*, so they should get to our eyes, right ? while the beams of lower energy (red, orange) should be scattered more.It could be a stupid question, but nevertheless, I don't think anybody loses anything if the article gets more explanations. 86.104.188.234 17:43, 29 April 2006 (UTC) Stefan Udrea 17:44, 29 April 2006 (UTC)

That's due to the absorption of radiation by the atmosphere - which happens as well as scattering. Optical wavelengths aren't absorbed as much as x-ray/ultraviolet wavelengths. Also, the Sun doesn't emit equal amounts of em radiation at each wavelength (more visible than x-ray for instance), and all this together leaves us with a bluer sky. --MilleauRekiir (talk) 00:01, 9 July 2008 (UTC)

Figure what is highly suspicios

Ref from article:<<Approximately 23% of direct incident radiation of total sunlight is removed from the direct solar beam by scattering into the atmosphere; of this amount (of incident radiation) about two-thirds ultimately reaches the earth as photon diffused skylight radiation.[citation needed]>> Means 23%*75%=roughly 15% the light is coming in scattered form not a direct rays.

Two arguments against: I personally have measured the sun given flow from sun-size collimated spot where sun is, and compared it to identical size non-sun containing spot, both luminities. Extrapolating this small angle to whole semi-sphere gives figure that about 50% is scattered light output in to comparizon with direct sunlight, not any 15%. Measurement accuracy was about +/- 10% thus confident interval is 45-55% of direct sunlight. The geographical place where measurement was done, aprox 57 N and 21 E, summer. Second argument is (are) plethora of different Solar Panels producers texts where they swear the productivity of photovoltaics in scattered daylight is about 50% of the same at direct insolation case. And not the 15%.

The only case when indeed something near the 15% is possible, is when strong lower atmospheric layer clods are abundant so much the meteorologists says "strongly cloudy day". (Dont miss with heavy rain-clouds, when 1000x weaker light even is possible). — Preceding unsigned comment added by 5.179.10.194 (talk) 13:32, 22 May 2023 (UTC)

Section moved from article

I moved the paragraph below here from the article. It doesn't seem to fit with the more common explanation of the blueness of the sky in terms of the wavelength dependence of Rayleigh scattering, as given in the article. If someone can adjust the text so that it's clear how these ideas fit together, this might be fine, but as it is it's just confusing.--Srleffler 15:27, 2 September 2006 (UTC)

Individual gas molecules are too small to scatter light effectively. However, in a gas, the molecules move more or less independently of each other, unlike in liquids and solids where the density is determined by the molecule's sizes. So the densities of gases, such as pure air, are subject to statistical fluctuations. Significant fluctuations are much more common on a small scale. It is mainly these density fluctuations on a small (tens of nanometers) scale that cause the sky to be blue.

I don't have a reference handy, but it is my understanding that this paragraph is the correct explanation. It could be better worded. Here's how it fits in: The sky is blue because of the wavelength dependence of Rayleigh scattering. Rayleigh scattering is the scattering of light off objects smaller than the wavelength. Those "objects" are the density fluctuations talked about in this paragraph. The individual atoms scatter light as well, but not enough to explain why the sky is so bright. Spiel496 16:37, 11 February 2007 (UTC)

I don't usually consider it essential to cite references for basic physics, because others can check my work. I this case I had other things to do and so, didn't explain my derivations and sources when questioned. Sorry to have caused inconvenience. I will provide convincing arguments if I have time.

The sky is purple?

Yeah ok. Before I believe this, someone ought to add some explanation for why a picture of the sky doesn't look purple. It looks like the exact same blue I perceive when I look at the real sky. If it is really a flaw with the human eye, then why in the world would a camera reproduce that flaw. I don't think that it does, which is why I dispute this until somebody clears this up. 8 November 2006 User:69.174.226.90

I agree, the physiological section does not benefit the article. It comes dangerously close to that meaningless sort of statement "it's really violet, but it looks blue". For an explanation to the "why not violet?" question, just look at the sky spectrum shown at the top of the article. It's a very broad peak centered at 500-550nm -- bluish-green. Perhaps the shorter violet wavelengths are absorbed, I don't know, but they aren't being lost in the eye. Spiel496 16:28, 11 February 2007 (UTC)

No one seems to have any defense of this section, so I removed it for the reasons above. Spiel496 06:28, 2 March 2007 (UTC)

Another idea: I don't think the violet rays are being lost in th eye. They are not emitted by the sun. The sun has it's peak emission at 500nm (Yellow) wavelengths, so these are naturally more scattered than the violet rays since there are more of them. Don't have any sources though, just simple physical intuition. User:132.229.227.66 10:45, 2 March 2007 (UTC)

I have to disagree that violet rays are not being emitted by the sun. They can clearly be distinguished in rainbows. ADNewsom (talk) 03:22, 11 June 2014 (UTC)

Incidentally, The eye has its peak sensitivity to light around green, colors to either side, like red or purple, would be less bright even if they were equally intense. Darryl from Mars (talk) 06:24, 10 July 2012 (UTC)

The spectrum in this article is inaccurate - if you view the picture details, it states it was taken through glass. It should really be cut off somewhere around 500nm to hide the inaccurate portion, or be replaced by an accurate spectrum. If you look at other spectra of the sky, they continue increasing in intensity at shorter wavelengths, with a peak around 400-420nm, for example figure 1 on page 7 of the external article link Atmospheric Optics (.pdf), Dr. Craig Bohren or page 2 of the referenced paper "Human color vision and the unsaturated blue color of the daytime sky" .

Therefore a physiological section is needed, though I haven't looked at the one that was removed LeBleu (talk) 21:43, 9 October 2017 (UTC)

References

Risk of skin cancer in the shade??

The included spectrum seems misleading, from the lengthy caveat: It was taken behind four panes of glass, so anything below 400nm is not shown.

My question is: if scattering is stronger for smaller wavelengths, wouldn't UV light be scattered most? If this is the case, I would think that it is important to present here if only to warn people of the danger of skin cancer being outside in a shady area. 121.44.119.136 21:32, 3 December 2007 (UTC)

That would be why you can get sunburned on a cloudy day. :) 98.246.109.37 (talk) 10:33, 1 May 2009 (UTC)

Ocean emitting light?

I removed this line from the article because it doesn't make much sense. The ocean can appear blue because it's reflecting the sky. It almost makes it sound like the ocean itself emits light regularly. "This is partialy also because of the ocean and how deep it is making it appear blue on the surface whilst blue light is filtered into the atmosphere." Ghiles (talk) 06:02, 28 February 2008 (UTC)

Image of the overcast sky

There is an image at the bottom of the article showing an overcast sky. However, it has a strong blue tinge and I don't think this is accurate. It would be quite easy to get a photo of a blue overcast sky through either incorrect white balance settings on the camera, or through image manipulation, but I don't think it is common for an overcast sky to be blue. The only time I can remember seeing a similar sky is after sunset, where (this is just my guess) the weak blue scattering of the upper atmosphere is then filtered through the clouds, but there is no direct sunlight on the clouds as the sun is now below the horizon. If this is indeed the case with the image in question, then it should be explained as such, but it is misleading to just caption it 'overcast sky' because it probably leads readers to wonder why it is blue when overcast skies are usually neutral. Please do let me know if I've misunderstood the phenomenon. Diliff | (Talk) (Contribs) 14:59, 26 August 2008 (UTC)

I agree. It's probably just poor white balance. Maybe a better picture would include some other object in the foreground, as a reference. Spiel496 (talk) 17:43, 26 August 2008 (UTC)

Huh?

"Scattering varies as a function of the ratio of the particle diameter to the wavelength of the radiation. When this ratio is less than about one-tenth, Rayleigh scattering occurs in which the scattering coefficient varies inversely as the fourth power of the wavelength. At larger values of the ratio of particle diameter to wavelength, the scattering varies in a complex fashion described, for spherical particles, by the Mie theory; at a ratio of the order of 10, the laws of geometric optics begin to apply."

That's some fancy word speak but it doesn't help the average reader understand anything. —Preceding unsigned comment added by 65.23.116.46 (talk) 06:38, 7 February 2009 (UTC)

I agree

This is a very true comment. CamrynRocks! (talk) 17:54, 19 April 2010 (UTC)

Elastic scattering

The last paragraph of the introduction was originally

The important processes in the atmosphere (Rayleigh scattering and Mie scattering) are elastic. No energy transformation results, only a change in the spatial distribution of the radiation.

The second sentence isn't quite right, so I removed it. Elastic scattering does not mean no energy transfer, it means no energy is transformed from kinetic energy to some other form (such as bind energy), or vice versa. In particular, air particles may very slightly decrease the energy of the scattered photon, although I believe this will be imperceptible. I welcome anyone who would like to rewrite the second sentence. Njerseyguy (talk) 23:54, 16 November 2010 (UTC)

Sky is Blue because that is its colour

I read somewhere that the sky is blue because that is what colour it really is. The article claimed that air and water look clear at small volumes and the effect of it's inherent colour is not noticeable at this scale. I still think that light scattering is involved because that would explain a red sunset. Maybe both phenomena are important. what do you think?

I'm sorry I can't find the article at the moment. I'll have a look later

I also had a thought about cyan coloured alpine lakes and streams.. is this caused by some sort of chemical composition in the water? — Preceding unsigned comment added by 193.206.74.232 (talk) 11:39, 2 December 2011 (UTC)

I don't understand the phrase "that is what colour it really is".

It sounds like you are saying "the sky is blue because it is blue".

A thing is blue either because (1) it predominantly scatters blue light,

allowing red light to transmit; or (2) because it absorbs red light and then

scatters the leftover blue light. This article says #1 explains the blue sky. Spiel496 (talk) 14:43, 2 December 2011 (UTC)

Ah, yes I think you are right. The two must be one in the same. Do you have any thoughts about different coloured lakes or rivers? — Preceding unsigned comment added by 193.206.74.232 (talk) 21:19, 16 December 2011 (UTC)

It's true that water is, to an extremely small degree, blue in the 'absorption' way, and not just scattering. I don't think the same can be said for air on the scale of our atmosphere. Darryl from Mars (talk) 06:17, 10 July 2012 (UTC)

P.S. To be clear though, most of the color variation in bodies of water come from organisms and sediments like algae and silt. Darryl from Mars (talk) 06:20, 10 July 2012 (UTC)

Not Earth Specific

I understand that this article is talking about Earth because it's a great example of this, but surely the process isn't Earth specific, which is the strong implication of how the article is currently worded.

There is currently no indication that this process applies to any other body with an atmosphere or that the Earth's details are merely an example of it.

Should the article be updated to account for this? -- Shhac ^talk_contribs 18:10, 4 July 2015 (UTC)

Issues with paragraph "As a part of total radiation"

The section "As a part of total radiation" appears to me to be relatively unconnected to / interwoven with the context. Furthermore I cannot see a (clear) explanation and definition of certain quantities and terms, e.g.:

• "total solar radiation" here is (apparently) meant to be limited to radiation hitting a plane that is at an angle with the horizontal;
• "β is an angle from the horizontal" is not meant to be some viewing angle indicating a position in the sky but appears to be related to the tilt of a radiation receiving plane.

My first impression on reading the paragraph was that it was going to give (an approximation for) the radiant emittance of diffuse sky radiation as a function of zenith angle and azimuth angle.

Only after consulting the source mentioned, I have reached the conclusion that the formulae are actually meant to give an approximation for the amount of radiant energy hitting a plane surface whose normal is at an angle β with the local upward direction, and then probably only when this tilt is in a north-south direction.

While this information in itself is interesting to me, I am not at all convinced it belongs here (in its current form).Redav (talk) 13:10, 23 December 2020 (UTC)