From: "James F. Tims" <p00168@psilink.com>
Subject: Re: Why the sky is blue - again
Organization: Semper Excelsior
Date: Wed, 23 Dec 1992 03:54:44 GMT
X-Mailer: PSILink (3.2)

>DATE:   Tue, 22 Dec 1992 15:54:04 GMT
>FROM:   zellner@stsci.edu
>
> > OK.  Let's see if I'm getting anywhere.  All the light hits
> > the atmosphere and starts bouncing around, the odds of it bouncing
> > being some factor of how many molecules there are to hit, how big they
> > are, and so forth, and on the wavelength; the odds of bouncing get better 
> > the shorter the wavelength gets.  I assume the Rayleigh equations 
> > calculate the actual odds, given a molecular size, density, and a 
> > wavelength....
>
>That's essentially correct.  Actually you get a scattering efficiency - the
>ratio of the cross-section of light removed from the beam to the geometrical
>cross-section of the scatterer - which increases as the fourth power of the
>size/wavelength parameter.  You just multiply the scattering cross-section per
>particle by the number of particles in the beam. 
>
> > Blue takes enough extra bounces that a lot more of it bounces into the eye
> > than do the lower frequencies. The lower frequencies do in 
> > fact get back to the eye, but compared with the blue, they're small change.  
> > When looking through a lot of the atmosphere toward the rising horizon, 
> > more of the blue has scattered, leaving relatively more red, which has
> > miraculously managed to come through almost unscathed -- unscattered. 
>
>Generally there is only one "bounce"; the atmosphere is optically thin at
>visible wavelengths.  It's not the number of bounces that counts, but the
>probablility of bouncing.
>
> > Does this mean that UV-sky is "brighter" than the blue?  
>
>In the near UV the scattering efficiency as described above is higher.  But 
>the sun isn't very bright in the UV, and the higher optical depth starts to
>take a toll - more of the light is scattered back into space.  In the
>"vacuum UV", shortward of about 3000A, you start to get true atomic ABSORPTIONS
>by O3, O2, O, and nitrogen.  Recall to kick an atom into a higher electronic
>state or to ionize it you generally need 5 to 10 electron volts, which is in 
>the UV in terms of photon energy. 
>
> > Let me guess.  UV is short enough to where the molecule size starts to 
> > matter ...
>
>Nope.  Remember molecules are only a few Angstroms in size.  The wavelength 
>is still a thousand times larger.
>
> > O3 is more opaque than O2.
>
>The O3 absorption is a true absorption, not a pure-scattering phenomenon.  
>The photon doesn't just bounce, it's destroyed.  The atom that did the
>absorbing can re-radiate the energy at the same wavelength (resonant
>scattering), or it can radiate several photons at longer wavelengths as it
>cascades down through electronic energy levels (airglow), or the energy can 
>be thermalized by collisions between atoms.  But the process is entirely 
>different from Rayleigh scattering.
>
>Ben
>

I have profited from this thread, most definitely, and thanks to everybody.
I think I could almost pass the quick quiz.  My pick for the winning analogy
is the radio waves going through a cloud of baseballs.  Light can cover
a lot of ground in 1/10e7 seconds. I don't know -- .0456 miles -- 241 feet?
24 feet?  Now you know why I'm not an engineer or a physicist,  

  ,...,.,,
 /666;    ',    jim tims
////;    _~ -   p00168@psilink.com
(/@/----0-~-0
 ;'  . `` ~ \'
  , `    ' , >      I suffer the heartbreak of untidy thought processes.
;;|\..((   -C---->> Too much unfolded mental laundry.  "Hey!  Wanna hear
;;| >-  `.__),;;    why the sky's blue?" I was boring before, but now this! 
Newsgroups: sci.skeptic