Rayleigh equation as explanation for sky being blue

Question

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.

Answer 1

• $\lambda^{-4}$ 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

Answer 2

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.

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.

Answer 3

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.