At 10:45 PM -0500 7/24/01, Jim Choate wrote:
>On Tue, 24 Jul 2001 [EMAIL PROTECTED] wrote:
>
>> You stated that every photon interacts, loses energy and is re-emitted.
>
>Sure, it has it's momentum changed. Think about it. The photon comes in
>from one direction and is absorbed/interacts with the atoms. As a result
>they get re-emitted (reflected) in the exact opposite direction. The point
>is the photons that get re-emitted ARE NOT THE SAME PHOTONS THAT WERE
>ABSORBED.
>
>You can't do that without losing something. photons only have one thing,
>energy as represented in their wavelength. The beam that gets re-emitted
>is less energetic than the beam that came in. Even if it does have the
>same phase and time coherence as the incident one. 2nd law of
>thermodynamics.
>
>You're confusing the intermediate vector boson as the carrier of
>information with the information itself.
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).
This is the essence of Planck's and Einstein's work in the first
decade of the 20th century. Photons don't lose a "little" bit of
their energy. They either get completely absorbed, aka the
photoelectric effect, or they hold _all_ of their energy. Ironically,
the key experiment was done with photons of varying energies striking
(scattering) off of a metal plate. The photoelectric effect
established that a photon only gives up its energy when it is
energetic enough to knock an electric out of a orbital. Photons do
not "give up some energy" except in this all or nothing way.
The only way photons change their (apparent) energy is through
Doppler shift, which is really a frame of reference situation. Red
shift, for example. (Ditto the Mossbauer Effect, where gamma photons
alter energies slightly.)
Whether a photon moving through a medium is the "same" photon or a
series of absorbed/emitted photons is an interesting topic to
discuss. But the one thing we _know_ is that such photons do not
"lose energy" in the way you describe. If they did, then blue light
would turn into red light. It doesn't. A blue photon is a blue photon
is a blue photon.
It's not your ignorance of high school physics (or high school math,
or high school history, etc.) that's annoying, it's your oracular
pronouncements of flawed theory. This is why people call you a crank.
--Tim May
--
Timothy C. May [EMAIL PROTECTED] Corralitos, California
Political: Co-founder Cypherpunks/crypto anarchy/Cyphernomicon
Technical: physics/soft errors/Smalltalk/Squeak/agents/games/Go
Personal: b.1951/UCSB/Intel '74-'86/retired/investor/motorcycles/guns
