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A recent XKCD comic implies that the sky is blue as opposed to violet due to human physiology, and that animals more sensitive to shorter wavelengths will perceive the Earth's sky as the shortest wavelength that they can comfortably see.

Is this phenomenon also responsible for relativistic blue-shifting, as opposed to relativistic violet-shifting? In other words, might different creatures perceive bodies approaching them as violet or even shorter-wavelength colors that humans cannot see?

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Blue shifting is just a term which implies that the apparent wavelength of light that we receive is becoming shorter with time, or moving toward the blue end of the spectrum. What colour the light actually is depends on what the original frequency of emitted radiation was. –  Kitchi Dec 11 '12 at 13:31
@Kitchi: So I guess an ultraviolet source would be shifted away from the blue, yet be erroneously called blueshifted. –  dotancohen Dec 11 '12 at 14:53
It may be a good idea to treat the information from comics a bit less uncritically. We know that daylight is approximately "white" which means that all visible frequencies are comparably represented. The sky is blue "relatively to the solar white color" because the Rayleigh scattering prefers shorter frequencies. But it doesn't mean that the "really shortest ones" are the most represented ones. The Sun's emission of too-short wavelengths decreases and the atmosphere amplifies this suppression, e.g. it absorbs most of the ultraviolet (and the nearby violet) light. –  Luboš Motl Dec 11 '12 at 14:55
@dotancohen - Not erroneously. It is still being shifted toward the bluer side of the spectrum. That is if you assume the word "blue" to be generalized to "really short wavelength" and red to mean "really long wavelength". It is just for ease of notation, so to speak. Rather than a long sentence explaining that it's really ultraviolet but it's wavelength is becoming still shorter, we would just say "blueshifted". You'll see this terminology a lot especially in astrophysics/cosmology. –  Kitchi Dec 11 '12 at 15:50
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2 Answers

I tracked down a spectrum of the sky at an altitude somewhere below 51° and overlaid it on the colors of the spectrum:

      spectrum overlay

From this diagram, it appears that the intensity of the light admitted through the atmosphere diminishes significantly before reaching violet. Unless the perceiving retina was overpoweringly tuned to violet, it doesn't appear that the sky would appear violet. Since the violet part of the spectrum is so attenuated, it would not seem useful for an eye to be much more sensitive in the violet range.

My guess would be that the sky may appear a bit deeper blue, perhaps to the indigo, but probably not violet.

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Nice, thanks! This addresses the red herring, not the question, but it is a very concise way to address the colours of the sky. I wonder why we don't see the sky green, though, as green is no less dominant on the graph and our eyes are very sensitive to green. –  dotancohen Dec 11 '12 at 14:55
@Kitchi: Wouldn't the spectral response curve above include the effects of Rayleigh scattering? It is data from a ground-based spectrograph. Some details of the instrument are given in this paper. –  robjohn Dec 11 '12 at 16:31
No. The scattering affects only a small fraction of the light, so the spectrum of the unscattered light is essentially unchanged. The spectral dependence of the scattering primarily impacts the spectrum of the scattered light. –  Colin K Dec 11 '12 at 16:42
@Kitchi: actually, that plot is "Relative spectral distribution of the uneclipsed or partially eclipsed sky for solar elevation angle less than 51°" The blue responses look to be pretty similar for all plots. –  robjohn Dec 11 '12 at 17:00
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The blackbody spectrum of the sun is the following, given $T=5778 K$. I admit I'm just copying from Wikipedia.

$$I(\nu,T) =\frac{ 2 h\nu^{3}}{c^2}\frac{1}{ e^{\frac{h\nu}{kT}}-1}$$

The comic suggests that the reflection from scattering transforms the above spectrum by $1/\lambda^4$ (as in, it is multiplied by this). Light is a wave, so $\nu \lambda=c$. We need not worry about the total magnitude, so I will introduce the spectrum of scattered light (denoted $S(\nu)$) with the following shape.

$$ S(\nu) = \frac{\nu^{7}}{ e^{\frac{h\nu}{kT}}-1}$$

Next, we have a model for how the eye works. As described here, take the spectrum, multiply it by a basis (denoted x,y,z), which represents the function of the cones in our eyes. Then, the response of the cones is translated into colors in our brain by the matrix given in that reference.

[ R ]   [  3.240479 -1.537150 -0.498535 ]   [ X ]
[ G ] = [ -0.969256  1.875992  0.041556 ] * [ Y ]
[ B ]   [  0.055648 -0.204043  1.057311 ]   [ Z ].

I found a standard 1964 standard of the Colorimetric Tables from the CIE, which is some science organization that does this stuff. That gives me the data to find the x,y,z, so I plotted the original black body spectrum ($I$ above), the transformed spectrum after reflection ($S$ above), and the data from the 1964 standard here:


Take the reflected spectrum, multiply by the x,y,z functions, then sum (or "integrate") the result over the entire spectrum. I obtain the following values, the units are still arbitrary.

  • x: 628
  • y: 660
  • z: 980

Then multiply this by the matrix to get RGB values. Here are mine:

  • R: 532.9
  • G: 669.4
  • B: 936.2

Now, normalize these values so the highest (blue) is 256. Divide by 16. Take the floor, that's your first digit (values 9, 11, 16). Multiply that number by 16, subtract from the color values, take the floor again to get the 2nd digit (I get 1, 7, 0). Convert those two digits to their letter in Hexadecimal if needed to get "91B7F0". Go to Google, find a site to test it with. My result:

html test

Wow! That looks like sky blue. Exactly how it was supposed to look! What was wrong with the logic?

So why isn't the sky violet?

This is incorrectly applying a mental analogy. To the child, a rainbow has a clear ordering of colors, so if you shift by one over the wavelength squared, you're moving toward the violet. To some extent, this misses the stimulation of multiple cones (in our eye) by the same wavelength, to some extent it ignores the fact that the intensities of parts of the rainbow aren't constant (giving more mid-spectrum weight), and to some extent it ignores that the transformation still leaves a lot of intensity in the longer wavelength parts.

It turned out the model actually isn't insufficient in the slightest. This is in spite of the fact that we expect some imperfections. Think of multiple scattering events, Earth albedo, who knows what. But even the naive model explains what we see.

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fantastic answer Alan, +1 –  lurscher Dec 11 '12 at 17:59
Working in the printing industry, I know a few color scientist who would not be very happy about your color transformations. But it still makes a great answer for this site! +1 –  Jaime Dec 11 '12 at 18:08
Comment on my edit wars with myself: My "conventional wisdom" was that the sun's spectrum is peaked around green. This was wrong. I first argued that violet has a lower intensity than other parts of the rainbow due to this logic. The 5778 K blackbody spectrum does predict a lower intensity in the violet range, but not because the spectrum is peaked in the middle. It's peaked closer to red when plotted against wavelength, but I'm not sure if plotted against frequency. Either way, it seems to be a component of the answer so I now leave it. –  AlanSE Dec 11 '12 at 18:14
Thanks, Alan, this is a terrific answer. It doesn't address the question (relativistic blue-shifting), so I won't mark it as accepted, but it is extremely well-written and well-researched. This was fun to read! Thank you! –  dotancohen Dec 12 '12 at 22:53
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