0
$\begingroup$

There are several answers on this site and elsewhere about why the sky is blue and why sunsets are reddish. But I could not find anything that discusses the relationship between the blackbody spectrum and the spectrum of colors seen above the horizon after sunset.

enter image description here

enter image description here

In particular, the "physical spectrum" (meaning the set of colors seen in the sky at increasing altitude above the horizon) after sunset looks identical to the blackbody spectrum. Perhaps this is related to the fact that sunlight is approximately blackbody to begin with, but I can't find any source that discusses this. For example, consider the following questions:

  1. If sunlight had an "even" spectrum in the visible range, instead of blackbody, would sunsets look considerably different? Would they stop looking like the blackbody spectrum?

  2. Is the color spectrum at each fixed location in the sky actually a blackbody spectrum (whose color temperature varies with altitude), or only approximately a blackbody spectrum? (Assume even that the sun is a perfect blackbody.)

  3. Since sunlight is blackbody, is it inevitable that any type of scattering would create a blackbody-like gradation of colors in the sky, i.e., not necessarily Rayleigh scattering?

In short, why does the physical spectrum of colors seen in the sky after sunset mimic the blackbody spectrum?

$\endgroup$
11
  • $\begingroup$ The "blackbody spectrum" you showed first is just a representation of perceived colour vs temperature of the black body. Not an actual spectrum. This is all just a coincidence.... (for example, the sun's colour in the first image is between 5500 and 6000...actually close to our perception of white) $\endgroup$ Mar 4 at 23:49
  • $\begingroup$ I agree with the first statement—perhaps I should have avoided the term "blackbody spectrum" to refer to a set of colors. But the statement that this is a coincidence doesn't seem to follow (and I find it hard to believe). $\endgroup$
    – WillG
    Mar 4 at 23:53
  • $\begingroup$ why not? Sky is blue due to scattering and so is the red sunset. A sunset in mars looks blueish due to different atmospheric contents. The first image is about squeezing a whole spectrum to one single color, each individual layer of the sky's colour is that colour. $\endgroup$ Mar 4 at 23:55
  • $\begingroup$ But why? I could imagine Rayleigh scattering shifting the overall color of each portion of the sky in such a way that the result looks nothing like the blackbody colors. But this is not the case. $\endgroup$
    – WillG
    Mar 4 at 23:58
  • $\begingroup$ Or let me add a different take on your questions: the sun's output lies in the 5500 to 6000K of the first image. 1. An even spectrum in the VIS would look green, sunset would probably still look red. 2.No. 3. no. $\endgroup$ Mar 4 at 23:59
2
$\begingroup$

Red/orange, white and blue are the colors you see when the intensity of light as a function of frequency (or wavelength) is very roughly linear across the visible range. To get other hues, you need narrow peaks or troughs in the spectrum, and Rayleigh scattering and blackbodies don't produce such spectra for different reasons.

Spectra from Rayleigh scattering are monotonic in the frequency. Blackbody spectra of different temperatures are all translated copies of a single shape on a log chart, as seen here:

At around 6000 K the visible part is roughly flat (white), at lower temperatures it's brighter at the red end, and at higher temperatures it's brigher at the violet end (but not sharply peaked there, so it appears blue, not violet).

So it's coincidence in the sense that Rayleigh scattering and blackbody emission are totally different processes, but not coincidence in the sense that you can expect to get similar colors from any physical process that doesn't produce narrow peaks or troughs in the spectrum.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.