On 11-Nov-98 Dennis R Redmond wrote:
>On Tue, 10 Nov 1998, Joseph E. Moryl wrote:
>
>> through the atmosphere it undergoes elastic (Rayleigh) scattering.
>>If all wavelengths were scattered equally then the sky would be simply
>>a bright white; instead the scattering goes as the frequency of the
>>light to the fourth power. What this means is that the blue end of
>>the visible spectrum is scattered more than the red end resulting in
>>a blue sky.
>
>Whoa. So, like, does this have to do with the chemical composition of
>the atmosphere, i.e. the proportion of nitrogen to oxygen, or is this a
>law of physics? If Neptune had clear skies, which apparently it doesn't
>(pretty cloudy in the outer planets) would they be blue-shifted, too?
>
>-- Dennis
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Since this seems like an invitation to freefall into complete bullshit, here is an ad hoc explanation. The scattering factor, the fourth power term itself is a crude average determined by an estimate of the particle surface area, its reflective quality (elastic collision properties), in addition to the density of these per unit volume. I think physics politely refers to such terms as 'empirical constants', i.e. look 'n see guess :). The chemical composition itself maybe incidental to the surface characteristics and density. In other words particulates with different chemistry, lead to the same power term, if their surface has the similar physical characteristics. So, Mars for example has blue skies with little or no oxygen or nitrogen(?), but mostly CO2(?). But the same idea of surface character goes to explain the color properties of things, with scattering effects, reflected wave lengths or elastic collisions and those frequencies absoped or inelastic collisions, which is why black gets warmer in the sunlight than white.
I don't suppose LBO is the proper forum to discuss the problems of dark
matter, mean density, and the Hubble sphere? Hmm, guess not.
Short form: don't open the air lock, just cause it looks pretty outside.
Chuck Grimes