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