Rayleigh Scattering calculation

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  • Thread starter nmsurobert
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  • #1
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Main Question or Discussion Point

I've been trying to find a proof or a paper that goes into calculating why the sky is blue as opposed to the "the sky is blue because of scattering" explanation that all over the internet. I (think) understand the concept but I would like to see some numbers. I'm aware of the I = 1/lambda^4 formula. But I can't use that to explain the color of the sky. Is that a water down version of what is used to explain the color of the sky? I would think we have to take the incident wavelength into account as well as the size of the particle that causing the scattering.

Can anyone link me to something like this?

Thanks!
 

Answers and Replies

  • #2
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If you are aware of the formulas,
(for example see Wikipedia https://en.m.wikipedia.org/wiki/Rayleigh_scattering)
then I don’t think calculating is the issue. In particular equation six in the Wikipedia article shows the scattered intensity as a function of wavelength and angle and includes linked references if you’d like to see the derivation. That equation is not watered down, and it shows why the sky is blue.

Perhaps you would like some explanation as to what it means. When you look at the sky you are not looking at the sun. Light from the sun cannot go directly into your eye. Instead what you are seeing is sun light that was NOT traveling toward you. The light travelled directly from the sun to a point on a line along your line of sight. At that point it scattered from the atmosphere toward your eye. The line from the sun to the scatter point and the line from the scatter point to your eye make an angle. That is the scattering angle. Paths with multiple scatters are also happening, but the single scatter is much more probable. What the formula is telling you is that the probability of scattering (and therefore the scattered intensity) is much greater for shorter wavelengths (and smaller angles by the way) so the spectrum that reaches your eye is not white like sunlight, but weighted toward blue. For the very same reason reds are less likely to scatter and penetrate the atmosphere further making the light on the direct path a little yellow, or orange, or red depending on how much atmosphere it has to get through. That is why sunsets are red.

Bottom line, it is all right there in the scattering function.
 
  • #3
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Awesome. I was looking for a derivation for the equation but if thats what it is then thats what it is. Thanks.
 

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