Blue-shifting as opposed to violet-shifting

Question

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?

Answer 1

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

      

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.

Answer 2

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:

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.