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Re: Sunsets



My model predicts the correct 1/lambda^4 dependence of the scattered
intensity.

Do you have any evidence indicating that's actually the correct dependence?
If one unphysical calculation agrees with another unphysical calculation,
that doesn't prove that either one is correct. Have you got any real data?

I could not immediately lay my hands on quantitative data for the spectrum
of blue sky-light, but I can tell you that photographers model it using a
color temperature of about 9000 K,
http://www.fact42.com/light/articles/wave/wave.html
and if you look at the Planck spectrum, in the optical band the
alleged lambda^4 dependence doesn't look anything like the 9000 K
spectrum; it looks very much hotter than 9000 K. To say it another way,
the blue of the sky isn't as deep blue as you might think; it is not
nearly as deep as, say, the blue phosphor on your CRT.

John is certainly right in saying that the solar spectrum, multiplied
by 1/lambda^4, gives a spectrum that approximates a blackbody (over
visible wavelengths) much hotter than 9000~K. To match the peak in
the spectrum, you'd need a blackbody of about 14500 K, and to best
match the optical range, you'd need a blackbody hotter than 20,000 K.

However, we would expect this blue-dominated spectrum to be washed
out (by a variable amount) due to scattering from particulates and
aerosols that are larger than optical wavelengths. With enough such
stuff in the air, the sky can even appear white, as has been pointed
out. My guess is that the photographers' number, 9000 K, is an
approximate figure for conditions at certain locations. If those
locations tend to be humid (producing lots of aerosols), then I'm
not surprised that the effective temperature would be only 50% hotter
than the sun's surface.

How are you going to explain to your students about optical fibers? It is
well known that you can see through hundreds of kilometers of glass, with
negligible scattering. The optical path in the glass contains orders of
magnitude more molecules than the optical path through the sky. If the
"molecule by molecule" theory is correct, either the sky must be
transparent (i.e. stars visible in the daytime) or glass fibers must be opaque.

In a solid or liquid, the spacing between adjacent molecules is pretty
uniform, so the interference of scattered light from different
molecules will be much more complete than in a gas, where the molecules
are more or less randomly distributed.

Dan