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I've been reading up on the internet as to why the sky is blue. The answer usually cites Rayleigh scattering that I've checked on wikipedia: https://en.wikipedia.org/wiki/Rayleigh_scattering:

$$ I=I_0 \frac{1+\cos^2\theta}{2R^2}\left(\frac{2\pi}{\lambda}\right)^4\left(\frac{n^2-1}{n^2+2}\right)^2\left(\frac{d}{2}\right)^6 $$

This answer raises more questions in my mind, that I hope some people can help to answer.

First of all, I can't understand the $\lambda^{-4}$ dependence in that equation. It means that the scattered intensity goes to infinity as $\lambda\to 0$. It also means that the observed intensity $I$ can be greater than the incident intensity $I_0$.

On several web pages, the $\lambda^{-4}$ dependence has been cited to explain why the sky is blue, but that doesn't make sense either. According to this reasoning the sky should be purple or indigo which has a higher frequency than blue. I've seen another explanation online that says that the sunlight impinging on our atmosphere has less indigo frequency than blue. However I can see the indigo part of a rainbow; it doesn't seem significantly dimmer than the blue part, so sunlight must have a decent indigo frequency content and as mentioned, the 4th power is a very strong dependency. This argument says that the sky ought to be indigo.

A second question I have is concerning the angle dependency. $1+\cos^2 \theta$ has a maximum at zero and at 180, and a minimum at 90 degrees. According to this dependency the sky should look brightest when looking at 0 degrees (towards the sun), and 180 degrees (with the sun to your back), but it should have half the intensity at 90 degrees. This doesn't match with our experience of the sky. Given the hand-wavy nature of the explanations, I wonder if Rayleigh scattering truly is the explanation for why the sky is blue.

AccidentalFourierTransform
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Chah
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3 Answers3

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  • $\lambda^{-4}$ can't go to infinity because Rayleigh scattering is valid only for scatterers smallers than the wavelengh (so $\lambda$ can't go to zero).

  • Not violet sky: this is addressed in many pages, see links above. The main difference with rainbow is that blue sky is very whiteish, so your tri-stimulus vision does not interpret the same way having 1 (highest) only vs 3 band-captors triggered.

  • $\cos(\theta)$ : observing $180°$ requires being in a plane. Plus you confuse 1 scattering with the sum of scatterings along the line of sight. Seen from space (thus with the same optical depth that from the floor) the atmosphere is very bright blue as well while reflecting the Sun (but take care not looking a color-treated satellite images, as most are).

Look at this Computer Graphics paper integrating and simulating the sky based on these equations: https://hal.inria.fr/inria-00288758 and it's youtube video: https://www.youtube.com/watch?v=0I7Af2Ev5iQ

Fabrice NEYRET
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  • Thanks for your response. Regarding the lamba^(-4), yes I see what you mean. Regarding the violet sky, if I summarize correctly, you (and others) are saying it is because of the imperfect response of the 3 colored cones in our eyes? If this is so, then if we use a spectrometer to measure the frequency of light coming from the sky, will we see the wavelengths more in the violet region than in the blue region? Regarding the angular dependence. Am I correct that the equation predicts half intensity at 90deg, relative to 0 and 180 deg? Thanks also for your links. I'll go and take a look! – Chah Mar 03 '16 at 09:13
  • I wouldn't say it's imperfection. It's the way physicals colors goes to perceptual colors. the fact that we share the same name for some is a trap ! The simplex of perception loops between blue and red (while physical colors don't). Violet is uper blue and thus already in the strange part. Purple is even a perceptive color that does not exist in the rainbow ! Spectrometer: I don't think so, because the Sun (+filtered by atmosphere) is also weaker in violet (UV filters have tails). Half: yes. But for 1 isolated scatter only, not a column. BTW, for more precise, consider also polarisation. – Fabrice NEYRET Mar 03 '16 at 10:38
  • Spectrometer: according to the graph below, spectrometer peaks at blue (450 nm) then decrease . more or solar/sky spectrum (above/below atmosphere) here : https://en.wikipedia.org/wiki/Sunlight – Fabrice NEYRET Mar 04 '16 at 01:05
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You have not understood that the observed spectrum depends on $I_0 \lambda^{-4}$, where $I_0$ is the incident spectrum of light - i.e. it is also wavelength dependent. A pure $\lambda^{-4}$ scattered light spectrum would only be recovered if you shone pure white light at the atmosphere.

Sunlight is much weaker in the violet part of the spectrum. So, even though violet light is more efficiently scattered, the mixture of wavelengths simply shifts from appearing white (to our eyes) for pure, unfiltered sunlight, to blue for scattered sunlight. The plot below (from the wikipedia page on sunlight shows what claims to be spectra of direct sunlight versus that scattered from a blue sky. You can indeed see that the ratio of light at ~450 nm (blue) to that at 360-400 nm (violet) is bigger in scattered sunlight than in direct sunlight, but the still broad spread of wavelengths results in a blue appearance.

In detail: taking the ratio of the (scaled) blue-sky curve and direct sunlight curves. The ratio is 2 at 400 nm, and 1 at 465 nm and 0.5 at 575 nm. This compares to $(465/400)^4 = 1.83$ and $(465/575)^4= 0.43$. Given the limitations of reading the plot, that seems quite close to a $\lambda^{-4}$ dependence.

Rayleigh scattered sunlight

You cannot just apply the Rayleigh scattering cross-section for arbitrarily small wavelengths. If the photons become energetic enough then the cross-section turns over and becomes the constant Thomson scattering cross-section.

Finally, the eye has a logarithmic response to brightness. I have not made the measurements myself, but I could easily believe that the brightness of the sky fell by a factor of two between close to the Sun and at right angles to the Sun. Further issues that complicate a straightforward interpretation are that any Mie scattering caused by larger particles in the atmosphere is much stronger at small angles and that the air mass of scatterers will also vary as a function of zenith distance - your quoted relationship has to be multiplied by number of scatterers. There are also issues around the range of angles that scattered light comes from.

ProfRob
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  • directional intensity: Note that it becomes more obvious when seen through a polarizer: $90^o$ is highly polarized (and thus can be exinguished by turning the polarizer), and this has a close relation to the fact that the average intensity varies with angle. – Fabrice NEYRET Mar 03 '16 at 22:18
  • Thanks for the plot. The blue curves on that plot, I believe we would interpret as a blue sky. For the yellow/orange curves, I do not think our brains would see them as blue. The spectrum is almost flat, peaking in the red for the "late afternoon" curve. Those curves don't seem to show a lambda^(-4) dependency either. Regarding x2 intensity. That's equivalent to 1 stop of a camera lens. That is detectable by our eyes. I don't quite get Fabrice's explanation why if a single particle has the angular dependency, but the column doesn't have the same dependency. I'm also not clear on the polarize – Chah Mar 04 '16 at 01:44
  • Ok I just went to the wiki page where the legend is a little bit bigger on my screen. The blue curves are for the sky and the yellow curves are for the sun. So that makes more sense. Except the fact that the sky spectrum doesn't seem to be showing lambda^(-4) dependency. (doubling of wavelength -> 16x reduction in amplitude) – Chah Mar 04 '16 at 01:56
  • @Chah The curves seem very close to an $I_0 \lambda^{-4}$ dependence to me. Maybe you forgot the $I_0$ bit, which seems to be key to your misunderstanding. As for the brightness bit - measure it then. You can't look close to the Sun and you can't look at two widely separated patches of sky at once, plus your eyes are accepting light from a large range of angles. – ProfRob Mar 04 '16 at 07:46
  • We would expect a 16-fold reduction from about 400nm to 800nm if it follows a lambda^(-4) trend. (I0 is common to both and can be divided out in the ratio of the two values of lambda) We're only seeing something like a 3-fold reduction from ~450 to ~800. That doesn't sound like a lambda^(-4) dependency. It's off by a factor of about 5. – Chah Mar 04 '16 at 08:13
  • If you've ever taken a photo spanning a large area of the sky, I think you'll see that there isn't a factor of 2 intensity difference between the 90 degree angle and the 180 degree angle. Incidentally, don't take the photo into the sun. Take the photo with the sun to your back. The equation tells us that the scattering should be maximum at 0 and at 180. At 0 we're looking into the sun. So use 180. – Chah Mar 04 '16 at 08:16
  • I just read your explanation of why you think it's a lambda^(-4) trend, but I don't think that's correct. Scattered intensity proportional to lambda^(-4) means doubling lambda, gives 16x reduction in intensity. You can't take the ratio of the direct sunlight curve to the blue-sky curves. They are not normalized to the same value. There is no way that the blue sky has anywhere near comparable energy or intensity as the sun. The legend of the figure says (x18) and (x3.8) etc.. that should be the scaling factor for each curve. You can't take the ratio of the 2 curves. They aren't the same scale. – Chah Mar 04 '16 at 08:41
  • @Chah Doubling lambda does not lead to a 16 time reduction and you still do not appear to get it. The proportionality is $I_0 \lambda^{-4}$, not $\lambda^{-4}$. $I_0$ is not "common to both", it is a function of wavelength. The relative scalings of the curve has no bearing on the issue whatsoever it is simply a constant of proportionality. It is clear that your misunderstandings are either in mathematics or that you do not understand the equation that you have taken from wikipedia. If the Sun is "at your back", then you are looking below the horizon. – ProfRob Mar 04 '16 at 08:55
  • @Rob. See this page: http://hyperphysics.phy-astr.gsu.edu/hbase/atmos/blusky.html and this one:http://math.ucr.edu/home/baez/physics/General/BlueSky/blue_sky.html I being proportional to lambda^(-4) does indeed mean that a double of lambda results in a 16x reduction in I. In the 2 pages cited, they use (700/400)^4 or a 10x reduction. similarly (800/400)^4 should give a 16x reduction. – Chah Mar 04 '16 at 09:01
  • @Chah and you still don't get it! The scattering formula is proportional to $I_0 \lambda^{-4}$. The $I_0$ is the spectrum of the Sun. Your 16 times reduction due to $\lambda^{-4}$ is accompanied by a 3 times increase in $I_o$. – ProfRob Mar 04 '16 at 09:11
  • Where are you getting it from that I0 is a function of lambda? I've never seen that statement made, and if it's true, then I is in fact no a function of lambda^(-4) at all. Give a citation for your assertion that I0 is a function of lambda. – Chah Mar 04 '16 at 09:12
  • Rob, then why do those web pages I cited state that scattered energy at 700nm is 10 times less than that at 400 nm writing (700/400)^4? If what you say is true? – Chah Mar 04 '16 at 09:14
  • Firstly: "where am I getting it from"? What did you think $I_0$ was? It is the incident intensity of light, which is obviously a function of the wavelength considered. Secondly: not everything you read on the web is true, however you have not read the web page correctly. It adds "for equal incident intensity". But the incident intensity is not equal, it is a function of wavelength. The second web page says "blue light is scattered more than red light... by a factor of 10". Yes of course that is correct, but there is less blue light to scatter in the first place. – ProfRob Mar 04 '16 at 09:22
  • How do you know I0 is not the total incident energy over all frequencies? Isn't it just your assumption that I0 depends on lambda? If I0 were a function of lambda, they would write it I0(lambda). Like I said, give a citation to show your assumption is correct. I've given you 2 pages showing that I(lambda) is proportional to lambda^(-4) with no other hidden dependencies on lambda. Meaning I0 is not a function of lambda. – Chah Mar 04 '16 at 09:34
  • My final comment. I teach a course on electromagnetism featuring Rayleigh scattering and yes - the first web page you have pointed me to is very sloppy. It is the scattering cross-section that depends on $\lambda^{-4}$ not the scattered intensity. The scattered intensity also (of course) depends on the incident intensity, and that depends on wavelength too. Good luck in your Physics studies. – ProfRob Mar 04 '16 at 09:42
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Rayleigh Scattering was calculated using classical theories of EM radiation.

Perhaps you should also look at Quantum Spectroscopy explanations. Scattering is different from absorption, at dawn and dusk water vapour has a greater effect, hence "Red sky at Morning/Night'.

Also consider the sky colour on Mars, where very little water vapour, CO².

Arif Burhan
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