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After reading the really interesting question and discussion Terrestrial Exoplanet Skies – I've Built a Visual Sky Chart. Is it Accurate? I was reminded that in this video at 40:20 Elon Musk mentions that at dawn and dusk Mars' sky (would appear) blue, and red during the middle of the day, the reverse of Earth's sky.

Is this true? If so, why?

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  • $\begingroup$ I don't want to copy text from other web pages, but I want to present some links: 1, 2, 3, 4, 5, 6. $\endgroup$
    – Uwe
    Dec 20, 2017 at 15:42
  • $\begingroup$ @Uwe properly credited block-quotes from other sites or sources are a standard part of stackexchange answers. Those are a lot of good sources. Hopefully someone will combine that information and distill it down to a simple, clear answer that explains the reversal of red/blue in daytime vs sunrise/sunset. $\endgroup$
    – uhoh
    Dec 24, 2017 at 7:10
  • $\begingroup$ I'm not sure if this deserves a new question or not, but: how would the famous sailing rhyme work on Mars? IE, does stormy/windy weather make the crepuscular light more red, more blue, no effect, or something else? $\endgroup$
    – Roger
    Feb 27, 2019 at 15:11

3 Answers 3

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The picture thumbnailed below from here, originally from NASA, does indeed show a reversed effect from what we see on Earth. Note that the hue of the Sun is not reversed; instead of the orange or red Sun on Earth turning blue or green on Mars, the Sun remains white.

When the Sun is rising or setting on Earth, blue light is scattered over a broader range than red, but the narrow range over which the red light is scattered has more intense red light. Hence the switch from blue to red as one's line of sight approaches the Sun. Mars presents a similar effect, but with the colors in reverse order.

The Sun itself appears to violate this rule on Mars, appearing white instead of blue or green. This is because the direct light of the Sun locally overpowers the scattered light around it; in the thin Martian atmosphere only a little light of any color is scattered away. On Earth, enough light is scattered to color the Sun but it's still much whiter than the surrounding, reddened sky. We see that difference when contrasting components of the direct sunlight's spectrum are highlighted by optical conditions.

Mars Sunset Source click image for full size.

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    $\begingroup$ I made a small edit to include a thumbnail that shows the effect clearly, feel free to roll back or edit further. $\endgroup$
    – uhoh
    Feb 28, 2019 at 12:52
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I came to this interesting question while discussing a related one: Do the gases in the Earth's atmosphere affect the color of a lunar eclipse?.

After thinking on the possible cause for this,"reversed" as compare to Earth, phenomenon I came to identify the directionality of Mie scattering as a possibility.

Digging a bit more to find corroborative material I have found this link that nicely explain the situation on Mars, and includes a truly beautiful comparison of photos of a red sunset on Earth and blue-ish sunset on Mars:

https://blog.briankoberlein.com/two-worlds-one-sun/

It is especially nice as it reviews the Rayleigh diffusion that is dominant in our atmosphere first, making it easier to understand the comparison to the optics of Mars' atmosphere. Moreover, after explaining the current answer it makes a point related to the question I mentioned above. So it even complements my answer to that.

Not to limit myself to a link-only answer, I would like to stress few points that will make the understanding even easier:

• while the intensity of Rayleigh scattering is very sensitive to wavelength, Mie scattering is normally relative agnostic to that.

https://en.m.wikipedia.org/wiki/Mie_scattering

However, it has a directional character, and particularly so when the particle have magnetic properties such those of the magnetite dust present in the atmosphere of Mars. In this case a remarkably forward vs backward scattering asymmetry can occurs, resulting in shorter wavelengths propagate straight as compared to longer ones.

https://www.osapublishing.org/josa/abstract.cfm?uri=josa-73-6-765

...unfortunately though, after a paywall.

This is what causes the blue sunset on Mars.

For sake of precision, martian midday has a more iyellow sky dome, then sky is progressively reddening til sunset arrives.

The point here is to correctly see the forward radiation as a scattered too, which semantically is not straightforward. The meaning is that the photons do indeed interact with the particle and are then diffuse along various directions. It just happens that in this particular Mie case the predominant one coincide or is a prolongation of the incident one.

As a curiosity, this link show a martian sky but in the Alps. While the NASA explains the bluish tint of sun on Mars by the forward scattering I mentioned above, it does not say if the mechanism is exactly the same in the alps case. It just mention the hard physics cause, i.e. Sahara's sand. https://apod.nasa.gov/apod/ap210218.html

• always dealing with colour is important to recall that we are not dealing always with monochromatic light. In discussions such a this one we must keep in mind that the colours we perceived are always due to a particular spectral distribution and even a combination of other factors as brightness and contrast. In terms of photons number the solar spectrum just out of our atmosphere, whatever we named its colour, is this:

Photon flux vs Wavelength, AM0

solar flux vs wavelength

Whatever colour we perceive, the spectrum impinging to our retina is always what results after the source spectrum has been modified by the various phenomena occurring in between. A modified ratio at different wavelength can suffices for us to perceive different colours. Whose more interested can look for the CIE colour space, for instance

https://en.m.wikipedia.org/wiki/CIE_1931_color_space

I do apologize but I have no pictures editing tools aboard the little phone I am using right now

Edit: the end of my answer above covers the aspect of brightness and contrast. However it does so in a very general way as it applies to all circumstances. A nice answer to the question would be combining the one above with that by Oscar Lanzi below. To which amount the various effects contribute to the observe facts remain opens, but we got an idea of the sunset colour reversal.

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    $\begingroup$ I've cropped the image a bit. I can find a similar but better one if you like, or you can just add it later. I think there won't be any problem leaving it like this for a while. $\endgroup$
    – uhoh
    Feb 3, 2018 at 11:21
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    $\begingroup$ @uhoh thanks a lot . I have to make a little editing than is fine. It went when already while checking the preview. Thanks again for your assistance $\endgroup$
    – Alchimista
    Feb 3, 2018 at 11:42
  • $\begingroup$ This is a great answer, thank you! I can't yet say I know completely why the differences in the angular distribution between Mie and Raleigh scattering leads to this hue reversal, but I'm on my way now thanks to your discussion! I'll check out the paywalled paper next week. $\endgroup$
    – uhoh
    Feb 4, 2018 at 10:32
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The Martian atmosphere is too thin. When the sun is high in the Martian sky the thin atmosphere and suspended dust particles make the sky red. As the sun 'sets'the light must penetrate more atmosphere and this is when all colors but the color blue are absorbed. Hence a blue sunset / sunrise.

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  • $\begingroup$ You should mention the color of the dust particles. What is the influence of the thin atmosphere without the dust? $\endgroup$
    – Uwe
    Dec 26, 2017 at 20:50
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    $\begingroup$ You might also take a look at @Uwe's comment below the question and the six helpful links there. Just for example Mie scattering vs Rayleigh Scattering. $\endgroup$
    – uhoh
    Dec 26, 2017 at 21:23
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    $\begingroup$ For the "Is this true?" part of the question, please add some supporting references to your answer. There should be some way for readers to verify this is true. For the "If so, why?" part of the question please also make sure your explanation is supported with external references. This might be good as an answer to a question on an exam, but it would not be a good way to explain to the class what is happening so that they understand. Thanks! $\endgroup$
    – uhoh
    Dec 27, 2017 at 20:40

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