My question is, what does the assumption of standing waves in a
blackbody cavity in the Rayleigh-Jeans derivation have to do with
reality?  The Rayleigh-Jeans derivation works for long wavelengths
and, if you throw in Plank's discretization of energy, it then works
for short wavelenghts and you no longer have the ultraviolet
catastrophe problem anymore.  That suggests that the initial
assumptions of the Rayleigh-Jeans derivation match reality (the
discretization-of-energy assumption aside).  But I can't make sense of
why only photons with zero amplitude at the wall (at both the sending
and receiving end) should necessarily be a requirement inside a
blackbody cavity.  What I've read is that the assumption was based on
the surfaces' being at thermal equilibrium, and therefore having no
electric field.  This really doesn't clarify it for me, though.

I assume that you do not want  standard reference, or textbook,
 but rather  direct response to  your specific question.

 I will do that, but, plese, do ask for clarification, if you feel I did
not answer what you are asking. This is the that type of question
which may require an iteration.

I agree 
"that the assumption (of photon having zero amplitude) because
the surfaces' being at thermal equilibrium .." 
makes little sense.

More exactly: photon of a given energy corresponds to a an electromagnetic
wave with a given wavelength. There is an electric vector and magnetic vector
and actually only one of them has zero amplitude at the wall.

Which one, that depends on the material of the wall. This is usually not
discussed.  For a similar reason, shape of the cavity is not dicussed.
The reason is simple: It has no effect on the final result. So, implicitly,
derivation assumes a rectangular box, and waves are than plane waves,
independent in x y z direction. It is more simple than to assume let's say
an elipsoid. You would get more complex resonant modes, but we only care about
the count at a given energy. So, similarly,  if you care which vector
will be zero at the wall, (You have choices Ex, Ey ..  Hz or even
mixes of those)
 you would make some additional assumptions. Let's say wall being
conductive or not. Either way, the result will be the same.

So, as a specific example let's look at:
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/rayj.html

There is an equation for Ex 
and author says 'the usual':
.." The solution to the wave equation must give zero amplitude at the
walls, since a non-zero value would dissipate energy and violate our
supposition of equilibrium..."

He says Ex (which is better then saying photon) BUT 

you can still question that, since the wall is black, and absorbing,
and equilibrium only means that 'on the average' as much of light energy goes
to heat the wall, as much of light is 'emitted' by that wall at that T.

So, you may say (let's say the opposite) the partial d Ex / dx  is zero,
rather than Ex, and you will get a different solution than the one shown
 (but just the phase will differ) but the number of them, number of solutions,
 will be the same.

 To use a mechanical analogy: wall of a pipe can be slippery or sticky,
 and if you are solving flow of a liquid, you need to select the 'right'
 boundary condition, but if you are just counting resonant modes, it will
come out the same. (because number of the degrees of freedom is the same)
as dicussed a bit more here (or Hilbert-Courant textbook etc).
http://people.ccmr.cornell.edu/~muchomas/P214/Notes/IntroWaves/node1.html

So, if you a student of physics, you are getting extra A+ for critical
thinking, and would suggest you look next at Debay's derivation of specific
heat - since it deals with the same issue - counting the degrees of freedom
(in this case of a string) and (forthe same reason as here) it does not matter
if the string ends are fixed, or tied to a elastic rubber band. What
matters is that we need a 'cutoff frequency' - as is apparent apriori,
and as was already recognised by Kelvin, and mentioned in his famous
lecture on the
'two clouds' ...

 and that is the true history. If you are interested in history of the
discovery, look at Stephan Brush - books and review articles...

hedgie

Thank you very much!  I'm glad you understood my question, and I
slowly was able to understand yours.  Thank you to the others for your
contributions as well.  (And no, I get no A+, no longer being a
physics student... but sometimes some things that I was supposed to
learn still haunt me and demand resolution.)  Thanks again.

Hi, I found a few websites that might help out: 

http://www.colorado.edu/physics/phys2170/phys2170_spring96/notes/2170_notes1_5.html

http://physuna.phs.uc.edu/suranyi/Modern_physics/Lecture_Notes/modern_physics2.html#b

http://www.vigyanprasar.com/dream/june2001/article-1.htm

A fairly complete derivation of the cavity radiation is at:

http://www.phys.virginia.edu/classes/252/black_body_radiation.html

Here it is stated...

"The waves are contained in the oven, so the electric field intensity
drops rapidly to zero on approaching and going into the walls, because
inside the walls the electric energy will be rapidly dissipated by
currents or polarization. In fact, then, the boundary condition at the
walls is much like that for waves on a string fixed at both ends,
where the wave amplitude goes to zero at the ends."

As far as I can tell this is the classical answer and what Rayleigh
Jeans must have been basing their thinking on. It is of course an
extremely idealised problem which gives the wrong answer anyway. If
you want to think beyond an idealised flat wall and think what is
really happening with the absorption and emission of photons then you
have to do it with quantum mechanics and that of course gives you the
right answer.

Interestingly later in the above derivation when encountering
problematic solutions it is stated...

"This problem can be fixed by thinking more carefully about the
boundary conditions at the walls. It is true that any component of the
electric field parallel to the wall will be attenuated rapidly by
currents or polarization in the wall. However, there could be an
electric field perpendicular to the wall, because there could be
surface charge on the wall."