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Why does sunset light spread horizontally over the horizon, instead of being radially spread around the solar disk?

I understand physics about refraction, and wavelength, and gradient difference between blue and red colors. However I don't understand that shape in the picture, I don't understand why it is not point symmetrical around the sun.

Here is a picture below. As you may see there is a red circle over the oil rig marked with number 1, and this part is clear. However red color is stretched to the left and to the right following the line 2 like on second picture.

Also if we compare length of line 3 and line 2, line 3 obviously much shorter.

Also atmosphere refraction is not that big, to cause this effect, right?

So why it is happening?

enter image description here

enter image description here

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  • $\begingroup$ I know that blue light, w higher wave lengths, is easier to scatter (why the sky is blue). But that doesnt really say why that shape. I dont see a source of red light so far to the side, at the edges of the pic. Should just be more scattered blue light there, just as the horizon is when the sun is at, say, 30-deg above. $\endgroup$
    – Al Brown
    Commented Sep 2, 2021 at 9:13

2 Answers 2

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At sunset the sunlight that we can see passes through a large (many hundreds of kilometers) distance inside the atmosphere. As it propagates, part of it gets scattered away due to Rayleigh scattering. The smaller wavelengths are scattered more than the larger ones. This means that what remains is redder than what is emitted by the Sun. This makes the solar disk red.

Now, what you see around the solar disk is the in-scattered light — the light that passed through the atmosphere to some point of scattering, scattered towards the observer, and then passed again through the atmosphere to reach the observer. The act of scattering made the light bluer, but the two passes through the atmosphere scattered away a larger portion of blue, which resulted in overall reddening of the sky. Here we have two competing effects: out-scattering proportional to $\lambda^{-4}$ and extinction proportional to $\exp(-\ell/\ell_0)$ where $\ell$ is distance travelled. The exponential has larger effect than the constant $\lambda^{-4}$ when $\ell$ is large.

This happens over all the horizon at the azimuths towards the Sun $\pm90°$, as well as a bit above the horizon at the opposite azimuths. At higher elevations above the horizon the atmosphere passed is not as dense, so another effect dominates, leading to blue color.

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  • $\begingroup$ That makes a lot of sense. Thanks. Have been quite curious this one. Takes both effects to make it reasonable. So the intensity of the light at 2 could be pretty low, it’s just that we’ve also had atmosphere to take out even more of the blue. $\endgroup$
    – Al Brown
    Commented Sep 2, 2021 at 11:48
  • $\begingroup$ Thank you for explanation, having this knowledge, I think I can implement computer graphics shader to render nice sunset. Science is beautiful. $\endgroup$ Commented Sep 2, 2021 at 15:37
  • $\begingroup$ @DmytroBrazhnyk before you reinvent the wheel, see ebruneton.github.io/precomputed_atmospheric_scattering $\endgroup$
    – Ruslan
    Commented Sep 2, 2021 at 15:38
  • $\begingroup$ @Ruslan thank you for suggestion, will definitely check that. $\endgroup$ Commented Sep 2, 2021 at 16:39
  • $\begingroup$ @Ruslan Hi Ruslan, thank your for clarification, used your description and implemented that: youtu.be/A6YUiCqIAEo, and Beer-Lambert law to render night-evening afterglow works just fine. $\endgroup$ Commented Oct 18, 2021 at 14:33
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There is a really nice answer by @Ruslan, I would like to add a few things you might find interesting:

  1. There is something called afterglow, and this is (in part aside from Rayleigh scattering) responsible for the reddish (pinkish) horizon after sunset.

An afterglow is a broad arch of whitish or pinkish sunlight in the sky that is scattered by fine particulates, like dust, suspended in the atmosphere.

https://en.wikipedia.org/wiki/Afterglow

  1. Afterglow is caused by Mie (and Rayleigh) scattering, which is, in part (aside from Rayleigh scattering) the other dominant effect after sunset. Mie scattering is dominant when the (dust) particles are comparable size to the wavelength of (visible in your case) light. Mie scattering in this case causes (in part) the reddish (pinkish) dominance in the sunset horizon.

The backscattering of this light further turns it pinkish to reddish.

Why is the western sky yellow after sunset?

  1. Dust particles are (relative to the air molecules) heavy and can be found in the lower part of the atmosphere (closer to the surface), thus this effect is creating a (horizontally) elongenated shape. I believe this is what you are asking.
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  • $\begingroup$ The red horizon happens even without any aerosol, just with molecules. It's orange in this case, but this line is still present. So I don't think the question is about afterglow. $\endgroup$
    – Ruslan
    Commented Sep 3, 2021 at 8:13

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