Recent content by smyri

Interestingly, Einstein's original paper of 1905 (The most widely known and least cited paper) does not mention the Michelson Morely expt. Einstein sometimes used the constancy of light as a starting point simply because was, to him, "the thing we knew most about". (The Meaning of Relativity...

Great answer...perfecto! Thanks a million Ich. Yes, I see it now...the smaller dm, when moving, is more massive!

The reason I started this line of questioning is I was trying to understand where all the radiation energy from the big bang had gone simply because the radiation field has cooled from an unbeleivably high temperature to only 2.7 degrees Kelvin...but as there are many photons involved in this...

Ich.. I see your analogy with the windscreen and the stone. I guess if the windscreen is moving relative to the stone, it supplies some of the energy that gets dissapated in the smashing of the windscreen...i.e. the energy of the collision is not that of the stone alone. Is it possible this...

Thank you Ich...I can see now that I need to see the photon frequency (or energy) in the rest frame of the absorber to determine the effect. So, if a UV photon was emitted from the emitter and this photon was absorbed by the absorber as an IR photon; I now realize there'l be no photo effect...

Sorry, that was a very confusing description...I mean... Suppose an emitter sends out a UV photon and it hits an absorber at rest relative to the emitter. In all frames of reference we see an ejected photo-electron. Suppose now only an IR photon is emitted under otherwise identical...

Thanks for all your thoughts; esp. Ich...the only thing stil puzzling me is the question; what is the photon energy with respect to the absorber? I guess it's determined by the relative intertial speeds between the emitter and absorber and the frequency of the photon in the emitter's frame of...

And then there's also the question I have with a single photon either having enough energy to cause the photo electric effect or not. If the "absorber" is approaching the red photon fast enough, suddenly the photon will have enough energy to eject an electron but before it didn't? I don't...

jcsd...are you saying one photon would not conserve together the momentum-energy? What about a pair of photons? Can they do it? Do these two photons then need to be viewed from a frame in which the newly created pair would have a zero COM? That would overcome my problems as together they...

I guess I still need to know that if this photon in one frame seems to have enough energy to create a proton-antiproton pair; where did this energy come from if the original photon was emitted by a mere red torch?

..or more simply stated.. If the photon's energy depends on it's frequency; then we can turn a red photon into a UV, gamma, IR or radio wave photon merely by changing our frame of reference to a long departed photon source which can be considered, to all intents and purposes, out of the...

I don't know how to understand doppler shifts and yet conserve energy. Consider a red photon is emitted by a torch. This photon is let free to travel through space. Any observer at rest relative to the torch (which by now may be millions of miles away) will see it as a red photon...

So the electron isn't orbiting; therefore it isn't accelerating and so it won't emit photons. The "wave" is truly a stationary state.

The electrons are not orbiting...that's a classical picture being imposed on a quantum state. The electron "wave" is only telling you where you are most liekly or most unlikely to find the electron with that angular momentum and energy if you place a detector "in the way" and try to detect it...

When symmetry is spontaneously broken, the vacuum state is not the lowest energy state. You "shift" the fields using a gauge transformation so that the new fields do have the vacuum state the lowest. This involves incorporating the Higgs field to achieve this. The new equations have a mass...