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Re: blackbody radiation



At 06:22 PM 4/26/2004, you wrote:
///
Imagine two boxes, one a "true"blackbody radiator (such as the box with
aluminum foil referred to by John D.) and the other a collection of
tunable lasers. These lasers are tuned and have intensities in such a way
that when someone plots the intensity verus wavelength of the light coming
out of the box the resulting curve matches that of a blackbody.

Now if a thermometer is used to measure the temperature inside the two
boxes the temperature of the first box, the one with the "true" blackbody
in it, will correspond to the peak wavelength according to Wien's law,
whereas the second will not.

This is an apparent paradox to me. What is wrong?


Justin Parke

Lacking some other power source in the black box, it will emit radiation
corresponding
to room temperature. The other sources will be lasing (or should it be
masing?)
at appropriately low microwave frequencies.
A question of interest is the power flux at the two apertures.
One could reasonably suppose, that if the power flux and spectrum were
identical
from the two apertures, a bolometer set to intercept the radiant energy
would read
equally at both emissions.

But now, suppose that one quadrupled the intensity of each of the masers.
If the synthetic spectrum were unaltered, then the spectrum would not
now represent a black body, but rather a pink body: a contribution at higher
frequencies would be lacking.

To extend this parable: if now we replaced the maser array with an intense
narrow beam radar signal at a specific frequency, the temperature at an
absorbant surface would not be well-predicted by the radar frequency,
but rather by its power density.



Brian Whatcott Altus OK Eureka!