In reply to Stephen A. Lawrence's message of Sat, 06 Sep 2008 08:12:25 -0400: Hi,
Thanks, that helped. However it raises another question. What about circularly polarized radiation? [snip] >> This makes me wonder how an ordinary photon manages to go through umpteen >> cycles >> between source and destination with a "stopped clock". :) > >It doesn't. A photon is the same no matter when you sample it. > >The wave function associated with it "goes through multiple cycles" >(which are distributed in space) but the photon itself does not >oscillate in any sense of the word. > >Remember, the photon is traveling with the wave front, and ON THE WAVE >FRONT the E and B fields are "stationary". If, at the crest of the >wave, E points up, then it's that up-pointing E vector which is >traveling through space; at the crest it always points up, but the crest >is moving at C. Any observer in any inertial frame will see an >oscillating E field as the photon passes, of course, because the >up-pointing E field at the crest is preceded and followed by >down-pointing E fields -- but they're all moving along through space in >tandem. > >If you could travel at C, and you flew along with a radio wave (which is >easier to measure than a light wave), and you sampled the E and B >fields, you would find that they didn't seem to be changing. This is >one of the problems with traveling at C: In a frame of reference moving >at C the traveling wave no longer looks like a solution to Maxwell's >equations, because @E/@t = @B/@t = 0. The way out of this box chosen in >special relativity is to let @t -> 0 when you travel at C. > >A "traveling wave" is exactly that. It is not a "changing wave"; rather >it's a fixed pattern which travels through space. > > > >> >> [snip] >> Regards, >> >> Robin van Spaandonk <[EMAIL PROTECTED]> >> >> Regards, Robin van Spaandonk <[EMAIL PROTECTED]>
