Re: Dissipation & Latent heat

[David Bowman <David_Bowman@GEORGETOWNCOLLEGE.EDU>]

Regarding John Denker's sneaky question:
> ...
> To generalize the idea...  One must decide what to say about the common and
> important case of _radiative_ cooling.  Is photon energy kinetic or
> potential or both or neither?

It's a good thing that John also included 'both' & 'neither' in his
multiple choices because the answer to the question depends strongly on
just what level of analysis one uses when considering the case of
photons/EM radiation.

For instance, suppose that the photons are considered merely as massless
particles.  Then in this case the total energy of a system of photons
(far from any interactions with any charges that may be lurking about) is
just the sum of the ultrarelativistic kinetic energies of the individual
photons.  The kinetic energy of a relativistic particle of mass m is
E_k = sqrt((m*c^2)^2 + (p*c)^2) - m*c^2 .  In the special case of
m = 0 this expression boils down to E_k = |p|*c.  The total energy
of a system of photons is just the sum of this expression over all the
photons present.  Thus, by this way of looking at things, all the photon
energy is kinetic.

OTOH, suppose we consider the system of photons as a region of space that
is to be treated as filled with (just) classical EM waves.  In such a
system the Hamiltonian function is (in a gauge such as the radiation
gauge, div(A)=0, where the gauge condition does not couple to the
electrostatic potential to the magnetic vector potential) essentially the
spatial integral of the square of the transverse part of the electric
field plus the square of the magnetic field (itself being always
transverse).  This Hamiltonian has *both* a kinetic energy contribution
from the square of the time derivative of the vector potential (from the
term involving transverse part of E) *and* a potential energy
contribution from the square of the curl of the vector potential (from
the term involving B).  Because *both* the kinetic energy and potential
energy terms are quadratic in the fields the system can be resolved as
a Fourier sum of non-interacting simple harmonic oscillator modes (one
such mode for each polarization state and allowed wave number).  Each of
these SHOs have their mean energy in thermal equilibrium partitioned with
exactly (1/2)*k*T being potential energy and (1/2)*k*T being kinetic
energy.  Unfortunately, this classical analysis is only valid for those
SHO (i.e. photon) modes whose frequency f is so low that the condition
h*f << k*T is valid.  If this analysis was to be applied to modes that
violate this condition the result would be the Jeans Ultraviolet
Catastrophe which comes about because boson modes whose frequency
obeys h*f ~= k*T do *not* obey the *classical* equipartition theorem,
but rather obey the quantum statistics of discrete bosons/SHOs.

If we do the above analysis correctly using the correct quantum
statistics for the discrete photon excitations of those SHO modes, we
find that instead of an ultraviolet catastrophe, we get the finite
Planck spectrum when including all the energies in all the SHO modes.
The spectrum is finite because the high frequency modes tend to be
'frozen out' because the energy, h*f, needed to create one excitation
(photon) of such a mode is much greater than the thermal energy, ~ k*T,
available to excite it.  In this case (violations of the condition,
h*f << k*T, for classical statistics) we cannot treat the field energy as
having separate potential energy- and kinetic energy-type degrees of
freedom because the potential energy (canonical coordinate-dependent)
terms do not commute with the corresponding kinetic energy (canonical
momentum-dependent) terms.  In this case both the canonical coordinate
*and* the canonical momentum for each mode need to be considered
*together* as a single quantum SHO mode/boson degree of freedom whose
thermal contribution to the thermodynamic properties of the EM field are
a single entity.  When the photon modes are considered as boson degrees
of freedom the Hamiltonian is a sum over normally-ordered products of
boson creation and destruction operators.  In this case the Hamiltonian
has *neither* kinetic energy- *nor* potential energy-type terms.
Instead, each term in the Hamiltonian is of the form: h*f*[a^dagger]*[a]
(where the quantities in the brackets are creation/destruction operators
on the Fock space of quantum states).

Thus, in answer to John's question, "Is photon energy kinetic or
potential or both or neither?" we see that the answer can be 'kinetic',
'both', and 'neither', depending on just which formal model one uses in
considering and analyzing the problem.

David Bowman
David_Bowman@georgetowncollege.edu