Rich L. is exactly right: The real issue here is motivation. Let me
suggest one motivation:
Planck's Law (https://en.wikipedia.org/wiki/Planck%27s_law
) applies to a
perfect blackbody which does not occur in nature, but is only an
idealization for what occurs in the natural world. Physical observations
of blackbody radiation are at best, close approximations. So, let's work
with those close approximations.
What Hawking discovered (https://en.wikipedia.org/wiki/Hawking_radiation
based on Bekenstein
that perfect black holes emit the same Planck radiation spectrum as
perfect blackbodies. But if you listen to Susskind's lecture
at about 32 minutes (or refer to another source which makes similar
points), it is clear that photons above a certain cutoff energy, near a
black hole, will be captured by the black hole and unable to escape to
be seen by a distant observer, while others below that energy will
bounce off and will be able to escape.
If you want to find this energy boundary, you can use a particle in a
) approach, whereby
a photon particle with a wavelength smaller than the Schwarzschild
diameter of the black hole box will become trapped, while a photon with
a larger wavelength will escape and so can be observed from afar. If you
do the calculation for a black hole, you will find that this boundary
occurs at slightly longer than 1/8 of the Wien peak wavelength. So, for
black holes, there is an ultraviolet (UV) cutoff in the Planck spectrum,
which we can ascribe physically to the black hole gravitational field
holding back the highest energy photons. In fact, Susskind as referenced
above uses this approach to derive the temperature of Hawking radiation
from Bekenstein's black hole relation.
The question then arises whether the same cutoff exists for an *ordinary
blackbody*, which is *not* a black hole (so far as we know based on
present theory). There are two possible answers: yes or no.
If no, then perfect black holes and perfect blackbodies emit the same
spectrum above ~1/8 of the Wien peak wavelength, but do NOT emit the
same spectrum at shorter wavelengths. Rather, blackbodies still have a
spectrum over this domain, while black holes do not. Black holes revert
over this high-UV domain to being truly black. This now breaks the
spectral identity between blackbodies and black holes at very short
wavelengths / high (UV) energies.
If yes, then the spectral identity between black holes and ordinary
blackbodies remains intact over the entire spectrum domain. But, if yes,
then we have to explain how the statistical thermodynamics underlying an
ordinary blackbody spectrum can give rise to such a UV cutoff without
the *apparent* involvement of black holes to trap photons with
wavelengths shorter than the black hole Schwarzschild diameter.
Jos Bergervoet makes the very astute observation that eventually we *do*
expect deviations. The black-body curve at least might have deviations
at the Planck-energy, but why not earlier? Let's flip this a bit: We
know that the fluctuations at the Planck energy are so dense (Wheeler
1957, 1962), that the Planck vacuum will be filled with a sea of
ultra-tiny black holes. Accordingly, these will emit Hawking blackbody
radiation and there *will* be a UV cutoff because the entire vacuum
across the whole sea of fluctuations acts as one omnipresent
photon-trapping box. The question is whether we can ever observe this
from where we sit in the natural order.
To this question, when we observe an ordinary blackbody, what we are
really doing is emptying out a cavity as best we can, then heating that
empty space using the experimental tool of a physical housing
surrounding the cavity, and observing the spectrum coming out of a hole
in the wall of the housing. So, we are really observing the Planck vacuum
by probing it with thermal energy, but at temperatures many orders of
magnitude removed relative to what would be the intrinsic temperature of
that vacuum. This remoteness of our observation simply damps the spectral
curve of the vacuum down to lower temperatures, because of redshifting
and screening effects.
So, if we were to observe this cutoff in an ordinary blackbody, there
would appear to be *no other explanation* for this but that we are
observing the Planck vacuum from a relativistically very remote frame of
reference. And, for all we do not know about quantum gravity, what we
*do* know is that the Planck vacuum is a place where quantum gravity
*does* come into play. So, by observing such a cutoff, we would for the
first time be observing a phenomenon directly rooted in and attributable
to quantum gravity.
Next, over to Douglas Eagleson's suggestion to try the NIST optical
group, which I suspect would not be averse to bagging the first
experimental observation of a quantum gravitational effect.
PS:I will also note without elaboration unless someone asks for it, that
one can combine the Bekenstein bound
) with the Wien
deductively arrive at the above black hole-based cutoff near 1/8 of the
Wien peak. This is a second motivation which independently supports the
first motivation detailed above.