Choate Prime Physics

[drs]

 >> 12:39 AM 7/25/2001 -0500, Jim Choate wrote:
 >On Tue, 24 Jul 2001, Tim May wrote:
 
 
	IAAP, so let me make a suggestion. Don't argue about physics. The
winner comes down to the person or persons least wrong since every
description given so far is an attempt to extend some part of an 
incomplete picture in a way that is _WRONG_, leading to arguments
about misstatements.

 >
 > > You're gibbering about things you have no clue about. Babbling about
 > > "the intermediate vector boson" when you clearly don't even
 > > understand high school physics is especially bizarre.
 > >
 > > Photons are _quanta_, as in quantum theory. Their energy is given by
 > > the usual E = hv (v is nu, frequency). They aren't "less energetic"
 > > when they scatter (i.e., are reflected). A photon fired at a surface
 > > will scatter/reflect with precisely the energy it had when it hit the 
 > > surface, unless it is absorbed (in which case it knocks electrons out
 > > of atoms...the photoelectric effect in a vacuum, thermalized in
 > > ordinary solids).

	Scattering is not reflection. In fact, photons do lose energy
scattering and the first experiments that demonstrated this is where
scattering gets its name: Compton scattering. The compton scattering
formula is easily derived from conservation of 4-momentum. Reflection
from a mirror is easily described by maxwell's equations, but is more
difficult in terms of photons. While the description of the photoelectric
effect is more or less ok, the term "thermalize" means applies to 
contimuum scattering of electrons in the conduction band through collisions
with other electrons, not to discrete transitions. Describing a reflection
as scattering can be done, but not in the length of this response.

 > ><groan>
 > >
 > >Here is what actually happens. It's called "The Radiated Electric Field".
 > >
 > >Some 1st year engineering physics books will have it listed in the index
 > >under 'mirror'.
 
	Let me suggest that everyone defer to "Classical Electrodynamics" 
by Jackson, as a definitive reference. It's the canonical physics text on
the subject and is practically universal as the text for a first year
graduate course in any physics program. If you know anyone that's
a physicist, they should have a copy.

 > >The incident photons strike the mirror.
 > >
 > >A current is induced.
 
	This mixes terminology in a bad way. First of all, any _SINGLE_ photon
scattering from a surface will be scattered in an arbitrary direction with
the appropriate change in energy. A _reflection_ is the collective behaviour
of photons with particular energies that particular interact with the
conduction electrons in a way which is easy to describe with maxwell's
equatiions, but not very easy to describe photon by photon, since from a
miscroscopic view, reflection is a statisitical phenomena. For example,
gold which is fairly thin, will be transparent in the UV, not reflective.

 >
 > >That current is electrons moving in a resistor. Making heat, losing
 > >energy. Note, we are NOT talking about photons here but J/C.
 
	The energy radiated from a resistor is no different than the
energy radiated from a blackbody, except perhaps in how ideal it
is.

 > >That current re-emits photons 

	Currents don't emit photons. Accelerated charges do. 
 
 > >that retain both frequency and temporal/time

	If free charges behaved this way (i.e., had memories of what changed
their momenta), it would violate both relativity and quantum mechanics.
A charge is at rest in its own rest frame before the absorption and
after the absorption. How would it retain any knowledge of its previous
state without (hint) communicating with neighboring charges?
 
 > >related coherence (see Maxwell's Equations for more detail). However, the

	Maxwell's equations do not describe photons.
 
 > >total number of photons MUST be reduced from the incident beam. This also
 
	The number of photons is irrelavent, since you cannot count them
by virtue of the uncertainty principle between the number of photons
and the phases. The ones you count are the properly symmetrized linear
combination that produces the signal in the detector and that is always
equivalent to a single photon. So the above statement is meaningless.

 > >means the incident photons can not be the same as the emitted photons.

	They are identical. Or not. Depends on how you write the wavefunctions
in any of an infinite number of equivalent ways. See bose-einstein
statistics.

 > >The photons (as opposed to 'a photon') lose energy.

 >The photons don't lose energy: the beam or flux is diminished in 
 >intensity.  Your improper choice of terms is what's getting creating the 
 >misunderstandings.
 
	This is closer to correct, but still doesn't contain the idea that
it's meaningless to talk in terms of photon numbers except in a very
specific context where one recognizes the limitations and applicability
of the picture.

-- 
drs