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Re: Sunsets



At 04:50 PM 8/29/00 -0700, Daniel Schroeder wrote in part:
The scattered blue-rich light is the blue sky, and the unscattered
red-rich light is the red sunset.

Right.

Note that this explanation is strongly supported by observations, e.g. the
polarization of the skylight. Light coming from sky locations 90 degrees
from the sun is very highly polarized.

Also note that if it weren't for the atmosphere, the sun would appear white
not yellow.

Imagine that you're an air molecule and an EM wave comes by, with a
frequency that's too low to excite your electrons into excited
states. ... your electrons (and nuclei), being
accelerated charged particles, are going to emit their own EM
wave (dipole radiation). The rate at which they emit energy is
proportional to their acceleration squared
<snip>

That argument is narrowly true up to the snip, but it is not the correct
explanation (and does not directly lead to the correct explanation) of the
red sunset / blue sky.

That argument would be fine for light scattered by a single air molecule,
but in fact there are quite a few air molecules. If we assume
(temporarily) that the air is a homogeneous fluid, then the total effect of
all the molecular scatterings collectively would be nothing but coherent
forward scattering. Huyghen's principle and all that. We would wind up
with an estimate of the index of refraction, nothing more. We expect no
scattering by clean homogeneous air.

Since clean are does scatter, we must retract the assumption of homogeneity.

A fellow named Einstein had something to say about this.

Now that you know the outline of the correct argument, it is an easy bit of
thermodynamics homework to model the atmosphere as a fluid with an index of
refraction _and_ density fluctuations.
1) Calculate the amount of light Rayleigh scattered by the fluctuations.
2) Calculate how high you have to fly before you see the stars come out
in the daytime.

Of course dust particles contribute additional Rayleigh scattering.