Physicspunks

[Jim Choate]

One of the ways to model a 'black body' radiator is using this technique.
The point being that the energy of the photons emitted through the hole
have a statistical energy distribution that matches (at least close enough
for experiments) the radiation emission of a black body at a
characteristic temperature.

On Fri, 17 Aug 2001, Anonymous wrote:

> The photon discussion a few weeks back got me reading about cavity
> radiation.  I'm puzzled.  Perhaps somebody can point me in the right
> direction.
> 
> For those who don't already know, cavity radiation is a surprising
> phenomenon which required quantum theory to model.

Actually statistical mechanics, not quite the same thing as quantum
theory. But you can't talk about QT w/o SM.

> Metals radiate energy in the form of light.  Each metal has a 
> characteristic "radiance" for each temperature, the amount of energy it 
> radiates.

That should probably be the amount of energy it radiates at particular
frequencies.
 
> If a block of a metal is hollowed out and a small port is drilled to
> see in, the radiance of the cavity is substantially higher than that
> of the surface of the metal.  As if that weren't shocking enough, it
> turns out that the radiance of cavities is the same no matter what
> kind of metal is used.  (This is so counterintuitive that I almost
> don't believe it!)
> 
> My book suggests that this is intuitively sound because if cavity
> radiation had different values, a violation of the second
> law of thermodynamics would occur.
> 
> Set up two cavity radiators of a metal X and a different metal Y like
> so:
> 
> XXXXXXXXXXXXXXXXXXXX YYYYYYYYYYYYYYYYYYYY
> XXXXXXXXXXXXXXXXXXXX YYYYYYYYYYYYYYYYYYYY		    	 
> XXXXXXXXXXXXXX 	 XXX YYY   YYYYYYYYYYYYYY  	       	       	 
> XXXXXXXXXXXX   	  XX YY	     YYYYYYYYYYYY  	       	       	 
> XXXXXXXXXXX    	       	      YYYYYYYYYYY  	       	       	 
> XXXXXXXXXXXX   	  XX YY	     YYYYYYYYYYYY  	       	       	 
> XXXXXXXXXXXXXX 	 XXX YYY   YYYYYYYYYYYYYY  	       	       	 
> XXXXXXXXXXXXXXXXXXXX YYYYYYYYYYYYYYYYYYYY  	       	       	 
> XXXXXXXXXXXXXXXXXXXX YYYYYYYYYYYYYYYYYYYY
> 
> If X and Y have the same temperature and Y has a higher level of
> cavity radiation, then X will get hotter and Y will get cooler.  In
> other words, you would have a perpetual motion machine.  So the
> radiance of each cavity must be the same.

Actually not, as Y cools its emissions drop (looking at your example in
reverse). At some point X becomes hotter than Y and the net travel of
photons goes the other way. It's worth remembering that what is actually
happening is that both bodies are emitting radiation continouly, only the
statistical 'average' of the results model this 'one way' flow. Eventually
they'll reach a point where the average energy emitted by one, and
absorbed by the other, equals. From that point both objects will emit
radiation into their surroundings and both will cool until that exchange
balances.

> So far so good, but this argument should also apply to surface
> radiance, which we know to be different:
> 
> XXXXXXXXXXXXXXXXXXXX YYYYYYYYYYYYYYYYYYYY
> XXXXXXXXXXXXXXXXXXXX YYYYYYYYYYYYYYYYYYYY
> XXXXXXXXXXXXXXXXXXXX YYYYYYYYYYYYYYYYYYYY
> XXXXXXXXXXXXXXXXXXXX YYYYYYYYYYYYYYYYYYYY
> XXXXXXXXXXXXXXXXXXXX YYYYYYYYYYYYYYYYYYYY
> XXXXXXXXXXXXXXXXXXXX YYYYYYYYYYYYYYYYYYYY
> XXXXXXXXXXXXXXXXXXXX YYYYYYYYYYYYYYYYYYYY
> XXXXXXXXXXXXXXXXXXXX YYYYYYYYYYYYYYYYYYYY
> XXXXXXXXXXXXXXXXXXXX YYYYYYYYYYYYYYYYYYYY
> 
> If Y has a higher level of surface radiance than X, then one would
> expect X to grow hotter, also making possible a perpetual motion
> machine.
> 
> Clearly, something went wrong somewhere.  Can anybody clue me in?

The flat radiation of the surface isn't the same as with the hollow
spheres. For the hollow spheres the greatest part of the radiation is
re-absorbed by the relevant body (and re-emitted, and re-absorbed, ...).
This re-adsorption is what leads to the black body behaviour. It acts as
an averaging mechanism. The plates just emit into space and the majority
of radiation is lost, not reabsorbed.


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