Re: blackbody radiation

[John Denker <jsd@AV8N.COM>]

James Mackey asked:
> ... to explain exactly why there are black body radiators, and how
> they radiate.  After giving a typical txtbook answer, I tried to
> think about what it is fundamentally. Blackbody radiation is EM
> radiation, but what mechanism produces it, especially in objects a
> low temperatures.

A crucial part of the story is that the result (i.e.
the radiation) is _independent_ of the mechanism that
produces it.  That's why the typical textbook doesn't
dwell on the mechanism.

It is independent of the mechanism because it represents
thermal equilibrium.  By way of analogy, if you've seen
one reversible heat engine you've seen 'em all ... they
all have the same properties, independent of mechanism.

>  What about a gas of neutral neutrons?

  (I won't ask what a non-neutral neutron is. :-)

Two answers:
 -- If the question meant to ask about real neutrons,
they can radiate, because they have a magnetic moment.
But they aren't very strong radiators, so you might
need to have a huge number of them and/or wait a long
time if you want to bring the EM field into thermal
equilibrium with them.

 -- If the question meant to refer to some hypothetical
particle that did not interact at all with the EM
field, then we agree that no, that would not be an
acceptable way to produce blackbody radiation.

> I am used to
> thinking of EM waves being generated by accelerating charges, or at
> least charges changing states, but this doesn't seem to work in
> general.

The Maxwell equations are what they are.  Electric
charges are the only source term (unless you want
to throw in magnetic monopoles, and there's no
evidence that such are relevant to this discussion).

In the case of real neutrons, it is conventional to
imagine them as having zero _total_ charge but having
a nontrivial internal distribution of + and - charge.
Its magnetic dipole is attributed to ultramicroscopic
Amperian currents.  Again the result is independent
of the details;  if you flip a particle with a magnetic
moment, usually all you need to know to calculate the '
radiation is the value of the magnetic moment, because
that will usually swamp the contributions from
quadrupole and higher moments.

===============

The usual classroom demonstration is to pass around two
shoeboxes.
 -- One has a small hole.  Immediately behind the hole
is a piece of black cardboard.
 -- The other has a same-sized hole.  It is a wide-open
opening into the interior of the box.  The inside is
lined with the aluminum foil, which is one of the
most unblack things you can think of.

Yet if you look at the two boxes, the black cardboard
is nowhere near as black as the wide-open hole.  The
point is that aluminum absorbs a little bit on every
bounce, and light that goes into the hole would
have to bounce hundreds of times before it got out,
and it's not likely to make it.

If aluminum can be a black body, then almost anything
else you can think of will work, too, given enough
of it and a suitable geometry.