Fact 1: The cause of a sunset is from the angle of the sunlight, because the amount of scattering increases as the light passes through more and more of the atmosphere.

Fact 2: Higher frequency / lower wavelength light (towards the blue and violet side of the spectrum) scatters more than lower frequency / higher wavelength light (towards the orange and red side of the spectrum)

Fact 3: During a sunset, you can frequently still see blue skies, sometimes not even very observably far from where the sunset is occurring.

Combining all of these facts together, I wonder if it is possible to have a "black" sunset, where the light passes through so much atmosphere that light itself scatters too much to reach the observer directly. This would give the impression of an interesting blackness coming from the sunset, and the sun might even be visible without needing to use special glasses.

Would this kind of sunset be possible? Assume that the hypothetical planet in question can be changed however necessary to make this work - including size, atmosphere, axial tilt, distance to star, etc.

I'm talking about the possibility of a lack of light occurring only at the horizon due to atmospheric scattering.

Further clarification: Assume my question involves a clear sky, because cloud cover does not address the aspect of the most light scattering only at the horizon. Also, I'm not talking about a light that is black, I'm talking about a lack of light that is black.

Good answers will tell me yes or no, and will explain why or why not based on science.

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    $\begingroup$ Hi overlord, I'm a little confused by your phrasing (didn't downvote though!) If I understand you correctly, you're hoping for a blue sky directly above, but the sun itself is hidden behind blackness and the closer you get to the light source the darker it gets? But I'm also assuming things like clouds blocking the sun aren't what you're looking for? $\endgroup$ – Dubukay Sep 25 '19 at 16:33
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    $\begingroup$ I was under the impression Night only happened after sunset, not during. I specifically put "scattering" into my question and many people are glazing over this piece of information. $\endgroup$ – overlord Sep 25 '19 at 16:37
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    $\begingroup$ I don't think the snarky comments like "that's called a shadow" or "that's called night" are really helping here. They're not serious attempts at answering the question, and at the same time, they're not really asking for clarification. $\endgroup$ – F1Krazy Sep 25 '19 at 16:40
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    $\begingroup$ For information, you can safely look at the sun with no special glasses when is it very close to the horizon. It is a popular pass-time at the seaside for people to go on the beach early to see the sun rising (or late to see the sun setting, depending on the orientation of the beach). And I fail to see how the source of the light could possibly be darker than the rest of the sky; I would think that it would always have at least the luminosity of the sky lit by scattered light... $\endgroup$ – AlexP Sep 25 '19 at 16:49
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    $\begingroup$ Upvoted because this is an entirely valid question (which I'll post an answer to later). Don't beat yourself up over the bad responses, overlord. Downvotes tend to form a feedback loop where after the first, people who click on this question expect the worst and are in turn more likely to respond negatively. Maybe should have titled the question, "Can a sunset be entirely infrared" to stem the confusion, but other than that, you didn't do anything wrong. $\endgroup$ – Gilad M Sep 26 '19 at 8:11

The sun will be at least as luminous as the rest of the sky

Light is not scattered away. It is just scattered. Some of it will be scattered forward: this is why the sun won't ever be darker than the rest of the sky -- the luminosity of the sky lit by scattered light is the minimum.

And it is not hard to prove this experimentally. Think of a cloudy day; all the light coming from the sun is scattered by the small droplets of waters which form the clouds. You cannot guess where the sun is -- the sun is neither lighter nor darker than the rest of the cloudy sky. Light appears to come from all directions equally; there are no shadows.

  • $\begingroup$ But what if the oranges and reds got scattered as well? $\endgroup$ – overlord Sep 25 '19 at 17:18
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    $\begingroup$ During the process of scattering, the light essentially spreads out. A sunset works that way because the orange and red wavelengths don't get scattered as much - causing their light to pass through the atmosphere directly. $\endgroup$ – overlord Sep 25 '19 at 17:20
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    $\begingroup$ Have you never in your life been out on a cloudy day? Red light is scattered less than blue light, so at sunset the sun appears red. But if all sunlight light is scattered then the sun is simply invisible in the sky -- all the sky appears to be the same. It is the same process which makes a lamp covered by a lampshade appear as if the lampshade was the source of the light. $\endgroup$ – AlexP Sep 25 '19 at 17:37
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    $\begingroup$ I give up. What does it matter that it is clouds or anything else that scatters the light? Once all the light is scattered the entire sky appears to be uniformly lit, light comes from all directions equally. The entire point is that scattering does not have a preferred direction. $\endgroup$ – AlexP Sep 25 '19 at 18:21
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    $\begingroup$ Don't give up, I finally get it now $\endgroup$ – overlord Sep 25 '19 at 18:32

You have a fundamental misunderstanding of light, color, and visibility (which is probably driving the downvotes).

From Wikipedia:

In the visible spectrum, black is the absorption of all colors. Black can be defined as the visual impression experienced when no visible light reaches the eye.

According to Google, light is:

the natural agent that stimulates sight and makes things visible.

Thus, light cannot be black. Objects appear black because they absorb light.

A sun that is black is actually a black hole.

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    $\begingroup$ I'm not under the impression that black lights exist. But if all of the light near the horizon gets scattered away, you would have black. $\endgroup$ – overlord Sep 25 '19 at 16:58
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    $\begingroup$ So you're telling me the downvotes are because I didn't entirely understand an entire field of science? If everyone knew what they were asking about - well, they wouldn't need to ask, would they? $\endgroup$ – overlord Sep 25 '19 at 17:26
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    $\begingroup$ @overlord If all the light entering an atmosphere is scattered away, then every ray of light is interacting with an object. We typically call an event of this magnitude in an atmosphere a "cloud." $\endgroup$ – Frostfyre Sep 25 '19 at 18:10
  • $\begingroup$ @overlord - You wouldn't have "black", you'd have darkness, which is the absence of light. If there's sunlight, the medium would have to be opaque, which even in any conceivable circumstance our atmosphere wouldn't be. $\endgroup$ – Mazura Oct 4 '19 at 2:42

I am interpreting OP's question as "What conditions would allow for all visible light to be scattered away, making the sun look black?". Let's get cracking with some hard-science.

Disclaimer : I am not going into the why of the phenomena at all, just throwing together equations. Please research on your own.

Given the fact that the scattering cross-section for a given gas is (under the Rayleigh scattering model) : $$ \sigma_s = \frac{2\pi^{5}}{3}\frac{d^{6}}{\lambda^{4}}\left({\frac{n^{2}-1}{n^2+2}}\right)^{2} $$ $\left(\sigma_s=\text{scattering cross-section},d=\text{particle size},\lambda = \text{wavelength}, n = \text{refractive index of light}\right)$
And that the fraction of light scattered for a particular wavelength is : $$ df_s = -Cf_s\sigma_sdl \implies f = e^{-c\sigma_s\cdot l} \left(f(0) = 1\right) $$ $\left(C=\text{numerical density of particles},l=\text{Distance traveled}\right)$
Let consider only the longest, reddest light our eyes can see, as it would suffer the least scattering. Actually, since the accuracy of this model depends on the particle size being about a tenth or less of the wavelength, and we would like the largest possible size to maximize the scattering, let $d = \frac{\lambda}{10}$, also let's combine the two equations : $$ \ln(f) = -Cl\frac{2\pi^{5}}{3}\frac{\lambda^2}{10^6}\left({\frac{n^{2}-1}{n^2+2}}\right)^{2} $$ Note : By law of atmospheres, numerical density, or density of any kind, depends on the height from the surface, assuming a constant temperature of course, but that is a deep deep rabbit hole I am not getting into.

Time to plug stuff in. The lowest frequency perceived by our eyes has the wavelength $750 nm = 7.5 \cdot 10^{-7} m$. The minimum intensity our eyes can perceive is about $10^{-10}Wm^{-2}$, and in the $740-760 nm$ band, the sun emits $747100 Wm^{-2}steradian^{-1}$, assuming it to be a perfect black body at $5800 K$. Earth recieves $50.73Wm^{-2}$ of that, hence the fraction we must scatter away is $(1- 3.86\cdot10^{-12})$, and so $ln(f) = -26.28$. The distance $l$ can be deduced to be $\sqrt{2*height_{atmosphere} \cdot radius_{Earth} - height_{atmosphere}^2}$ by the Pythagorean Theorem, so about $1125km$. And the refractive index for air is $1.0003$.
Hence, the deduced concentration must be (in almost Earth-like concentrations): $$ C = 5.09 * 10^{18}m^-3 $$ Which corresponds to a pressure of $0.021 Pa$. Well I guess this is what you get for assuming constant density for the atmosphere.

Conclusion : The procedure for increasing accuracy would be the same, just with better models and fewer assumptions. Of course I attempted to calculate for relatively Earth-like conditions, but the values may be switched out for any other gases, planetary sizes, etc. I am guessing you will have better luck with gas giants, but it is time I took my leave.

  • $\begingroup$ Would this cause a constant state of night or darkness through the entire sky? $\endgroup$ – overlord Sep 25 '19 at 19:41
  • $\begingroup$ Note that you find 0.021 Pa of air pressure at altitudes in excess of 20 km on Earth. $\endgroup$ – Frostfyre Sep 25 '19 at 20:16
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    $\begingroup$ It doesn't have to be on Earth. I'm just frustrated because I've done tons of resesrch into light scattering and I'm having a hard time understanding it and everyone punished me by downvoting just because they didn't like my question. $\endgroup$ – overlord Sep 25 '19 at 20:41
  • $\begingroup$ @overlord As AlexP responded in his question, the sun's image must be atleast as luminous as the rest of the sky. It would mean that the sky is a blanketing black haze. However it is not necessary that if visible light in the star's spectrum then the visible light must still be present in the "sky", as it can always be lost to space. I am hoping that is what is happening in my model. $\endgroup$ – Varad Mahashabde Sep 27 '19 at 13:33
  • $\begingroup$ @Frostfyre Inaccurate models give inaccurate results ¯_(ツ)_/¯ $\endgroup$ – Varad Mahashabde Sep 27 '19 at 13:42

Short answer: no, it cannot happen.

Long answer:

What is scattering?

Atoms or molecules which are exposed to light absorb light energy and re-emit light in different directions with different intensity. This phenomenon of light is called scattering of light. Scattering is a general physical process where some forms of radiation, such as light, sound, or moving particles, are forced to deviate from a straight trajectory by one or more paths due to localized non-uniformities in the medium through which they pass.

By definition scattering happens in any direction, and as such it won't create an area of complete darkness. The very reason you can see within your shadow on Earth is scattering. If you are on the Moon, or whenever there is no atmosphere, your shadow is pitch black, except from direct reflection from surrounding surfaces.

  • $\begingroup$ But if you think of a sunset as "the blue and green" being scattered around the area of the sunset, couldn't the same thing happen to the yellow, orange, and red colors? $\endgroup$ – overlord Sep 25 '19 at 17:16
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    $\begingroup$ @overlord: You could indeed. The problem is that light is re-emitted at a lower frequency than light absorbed, which is why you get red and orange sunsets--green, blue and UV gets re-emitted in yellows and reds. Some of the actual red and yellow gets re-emitted in IR, and is invisible, but the extra red and yellow coming from re-emitted higher frequency light more than makes up for it. $\endgroup$ – nzaman Sep 25 '19 at 17:20

Offering the trivial solution: in any environment with no atmosphere, you will get a black sunset. With no atmosphere to scatter, you get a black sky all the time.

Other than that, as others have pointed out, it can't happen. There's a logical reason for this. Consider that light is a spectrum. If you can see the sun at all, then that light is not being scattered all that much. If you are seeing a sunset, then there is at least some set of visible wavelengths that are sufficiently unscattered to get a meaningful image of the sun. The wavelenghts that scatter slightly more than this will always be visible at sunset.

You can test this out yourself. On a very clear day, with no moutains nearby, look opposite a sunset. The sky is darker out there than it is closer to the sun. If you look at this a little bit earlier, what you'll notice is that it's more blue than the sky closer to the sun.

The other trivial solution, as others have pointed out, is to scatter the light so sufficiently that you can't see the sunset at all. We experience this on Earth as an overcast day.

(Incidentally, if you want something really trippy, go out on a clear day after sunset and look away from the sunset. You'll see a red/brown region from regions of the atmosphere that are only lit by the redder wavelengths -- the blues already scattered out. There's a time where you can actually see a red/brown band all the way around you, 360 degrees. Light scattering is trippy!)


Yes! ...Definitely not on Earth though.

As stated in previous answers, the light from a setting sun has to pass through a lot more atmosphere to reach your eyes than that from a midday sun, so shorter wavelengths are more likely to be scattered before they reach you due to Rayleigh scattering. This is why a setting sun is red.

So what happens if we further increase this atmospheric distance? A simple application of the Pythagorean Theorem tells us that at sunset, light travels through a total distance of $\sqrt{2hR-h^2} \approx \sqrt{2hR}$, where $h$ is the height of the atmosphere and $R$ is the planet's radius. This means that either a much larger planet or a much thicker atmosphere will do the trick - I think the first option is more visually interesting, since it means that the sky could still be blue at midday (ignoring the effects that the correspondingly large mass would have on the atmosphere), but it's up to you. At any rate, none of this is new information so far.

What the other answers have neglected is that Rayleigh scattering is angle-dependent. The intensity goes as $I(\lambda, \theta) \sim \frac{1+\cos^2\theta}{\lambda^4}$ - It's as much as half as weak at large angles as it is at small ones. This is why only the part of the sky near the setting sun is red; the light further out needs to scatter at a larger angle to reach your eyes, so Rayleigh scattering is weaker, and thus more of the blue light reaches you.

With this in mind, consider a setting sun in an atmosphere of a giant planet, such that Rayleigh scattering at an angle of 0 scatters away all visible-spectrum light. At large angles, some of the visible (red) light might yet remain. Thus, you could have a truly bizarre sunset where the sun and the sky around it looks black, as though the sun were obscured by a thick black fog before it had even set below the horizon. Meanwhile, further from the sun, its reddish light would still be visible.

  • $\begingroup$ What exactly is preventing light which has been scattered on the direct path from being perceived on an indirect path? After light has been scattered once it does not go on strike and refuse to be scattered again. If all the light has been scattered away from the direct path what you get is a diffuse illumination of the entire sky. Hint: the sky is not black at the antisolar point. $\endgroup$ – AlexP Sep 26 '19 at 18:15
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    $\begingroup$ Light can scatter in the atmosphere many times. But it's more likely to stay close to a straight line (leading to that square-cosine dependence I mentioned). If the light is already heading straight toward you, it's a lot more likely to reach you than light that is heading away or off at an angle. Also, you get diffuse illumination in the large-scattering limit, but for lower frequencies that are less easily scattered this does not occur. That's why the sky is blue and not white. The sun is white, because the red light mostly stays along a straight path from the sun to your eyes. $\endgroup$ – Gilad M Sep 26 '19 at 19:33
  • $\begingroup$ @AlexP The more of an indirect path light takes, it's intensity decreases. The sun peaks in green, but never have I ever seen a resemblance between a sunset and a lime :-) . $\endgroup$ – Varad Mahashabde Sep 27 '19 at 13:39

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