On Tue, 24 Jul 2001, Tim May wrote:
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).
<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'. The incident photons strike the mirror. A current is induced. That current is electrons moving in a resistor. Making heat, losing energy. Note, we are NOT talking about photons here but J/C. That current re-emits photons that retain both frequency and temporal/time related coherence (see Maxwell's Equations for more detail). However, the total number of photons MUST be reduced from the incident beam. This also means the incident photons can not be the same as the emitted photons. The photons (as opposed to 'a photon') lose energy. Blipverts strike again! </groan> Physics J. Orear ISBN 0-02-389460-1 In numbers, z_o is the area of the incident beam j_o is the surface current per unit distance I_o is the induced current I_o = (z_o)(j_o) j_o = (c * E_o)/(2 * 3.14159... * k_o) Go look those up yourself. -- ____________________________________________________________________ Nature and Nature's laws lay hid in night: God said, "Let Tesla be", and all was light. B.A. Behrend The Armadillo Group ,::////;::-. James Choate Austin, Tx /:'///// ``::>/|/ ravage@ssz.com www.ssz.com .', |||| `/( e\ 512-451-7087 -====~~mm-'`-```-mm --'- --------------------------------------------------------------------