- #36
Frame Dragger
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I'm suggesting that "virtual photons" describe an intermediate period which current theory cannot. Actually, I'm suggesting no such thing, did anyone even BOTHER to pay attention to what Tiny Tim posted?
lightarrow said:But which is (I am just asking) the physical meaning of a particle (I'm talking of virtual particles of course) which energy E and momentum p don't obey E2 = (cp)2 + (mc2)2 ?
lightarrow said:How much or to which extent can we believe in the physical existence of a matematical tool which is very useful in physics but cannot be directly measured? (Difficult question, I know)
SpectraCat said:Hmmm ... I have heard it said a lot that virtual particles were added just to "make the math work out", but that does not gibe with my understanding. My understanding was that they were added to QED to make the *physics* work out. Specifically, unless there is some stream of particles that are subject to special relativity being exchanged by the electron and proton, then they are undergoing "spooky action at a distance", right? Isn't this what is illustrated by the thought experiment of what would happen if the proton suddenly blinked out of existence? According to SR, there must be a delay before the electron "knows" that the proton is gone. This can only happen if the proton and electron are somehow exchanging "information" about each other's presence, and unless I misunderstand, that is the function of the virtual photons.
FireBones said:Jeblack,
I found a discussion that appears to explain what is going on by passing to momentum space, and I think it is saying the same thing you are.
http://math.ucr.edu/home/baez/physics/Quantum/virtual_particles.html"
I need to reread this discussion, but it is along the lines of what I was looking for. An honest effort to explain how photons actually can transmit force.
FireBones said:Can you explain why virtual photons do not obey this law? I'm not challenging that you are correct, but I hadn't seen before that virtual photons fail this equation. Or am I misreading you?
But this gets back to what I had asked earlier. What does it mean to measure something "directly." When we say we "detect" a real photon what we mean is that we see an excitation occur and we say this is a detection of energy being transferred from some other source to the atom/electron/phonon/whatever that absorbed the photon. We did not "see" the photon midflight, we saw its effect and we call that a "detection" of the photon.
But isn't the same thing happening with, say, an electron being accelerated to a proton? We see a change in momentum of the electron and deduce that momentum has been transferred..but by what agent? Can't the measurement of a change in momentum count as just as much a "detection" of a "virtual" photon as the the change in energy of an electron/atom/phonon/whatever be a 'detection" of a "real" photon?
I do have a further question though... photons are the mediator of the EM force. [There have been a few threads that start out with the question 'What is a photon?' and occasionally someone will answer "the force carrier of the EM force." as the only answer.]
But this means that all photons have to be force carriers, not just these virtual ones. While I alluded to virtual ones in the Feynman book, it seems that this is unnecessary. If "the carrier of the EM force" is a suitable definition for "photon," than any photon should be transmitting EM force...not just the virtual ones.
And so now I'm wondering what in the world that means for just everyday, ordinary photons. For example, we say something like "An electron drops from n=3 to n=1 orbital and a photon is released." What does this electron moving orbitals have to do with a conveyance of EM force?
Is it possible for this force to be attractive rather than repulsive? If only virtual photons can carry repulsive force, can someone tell me how the argument in http://math.ucr.edu/home/baez/physics/Quantum/virtual_particles.html" changes when the photon in question is 'real' rather than virtual?
[Perhaps the answer to that last question is that "real" photons are detected, hence collapsing their wave function, hence you no longer have the interference between the two momenta amplitudes?]
seniornegro said:A photon isn't real. Its a quanta of EMR. Its like a gallon or a pint...
A photon being a quanta of energy and energy being the amount of work that can be performed by a force, the photon carries the gas needed / transferred or expended by a force.
seniornegro said:A photon isn't real. Its a quanta of EMR. Its like a gallon or a pint...
A photon being a quanta of energy and energy being the amount of work that can be performed by a force, the photon carries the gas needed / transferred or expended by a force.
acentauri said:For what it's worth, classical http://iopscience.iop.org/0295-5075/76/2/189?ejredirect=migration".
http://arxiv.org/abs/0907.1611" a related paper.
acentauri said:For what it's worth, classical http://iopscience.iop.org/0295-5075/76/2/189?ejredirect=migration".
http://arxiv.org/abs/0907.1611" a related paper.
SpectraCat said:
FireBones said:In Feynman's QED, he talks about how a nucleus keeps an electron in orbit by exchange of photons, but I don't see how a photon can provide a push, much less a pull.
tiny-tim said:Just a detail …
that's not "the" Physics FAQ, it's John Baez's Physics FAQ.
You are not misreading me, that's a fact.FireBones said:Can you explain why virtual photons do not obey this law? I'm not challenging that you are correct, but I hadn't seen before that virtual photons fail this equation. Or am I misreading you?
lightarrow said:But which is (I am just asking) the physical meaning of a particle (I'm talking of virtual particles of course) which energy E and momentum p don't obey E2 = (cp)2 + (mc2)2 ?
How much or to which extent can we believe in the physical existence of a matematical tool which is very useful in physics but cannot be directly measured? (Difficult question, I know)
FireBones said:Can you explain why virtual photons do not obey this law? I'm not challenging that you are correct, but I hadn't seen before that virtual photons fail this equation. Or am I misreading you?
But this means that all photons have to be force carriers, not just these virtual ones. While I alluded to virtual ones in the Feynman book, it seems that this is unnecessary. If "the carrier of the EM force" is a suitable definition for "photon," than any photon should be transmitting EM force...not just the virtual ones.
lightarrow said:About the reasons for which they don't obey the law, instead, I have less knowledge.
Let's try with an example, if the conclusions are wrong I hope someone helps correcting me.
Two electrons collide head-on along the x-axis with equal and opposite speeds. The virtual photon image describes this process with the exchange of a virtual photon between the two electrons. If every electrons, because of the velocity change in the process, acquires a momentum p, then the virtual photon carries a momentum p. Now let's see the energy. Every electron, after the collision, has exactly the same energy as before, so if energy conservation still have to hold, the virtual photon should have zero energy. How's possible that a real particle has momentum but no energy?
tiny-tim said:Hi FireBones!
"not obeying this law" is usually called "off mass-shell" … see http://en.wikipedia.org/wiki/On_shell_and_off_shell"
Neither real nor virtual photons "carry" force … the force comes from the field, and the virtual photons are mathematically connected with the field.
That's a valid proof of why, if a single virtual photon was exchanged, it would have to be off mass-shell.
But the correct mathematical description of a collision between two electrons is that an infinite number of virtual photons (and virtual electrons! ) are involved, of all possible momentums (moreover, if they are on mass-shell, they also appear at all possible locations in space-time, but if they are off mass-shell, they have no location, and exist in an imaginary "momentum-space").
presario said:Force = change in momentum
Photon carries momentum !
LostConjugate said:The photon provides kinetic energy, the direction does not matter. The reason why is because there is no physical state that exists where the electron crashes into the nucleus. This means that if an electron is given kinetic energy it will always be beneficial to keeping the electron in orbit.
IceMan815 said:Hmm...what about the neutron star? i think electrons and protons collide with neutrons and release neutrinos by the effect of the weak force.
tiny-tim said:
JDługosz said:The first sentence is "This is the web version of the Usenet Physics FAQ." The Title is "Usenet Physics FAQ". The Copyright statement doesn't mention him. The main text mentions some history of original creation and maintenance, but his name is absent. The "Thanks" at the bottom does mention him, along with many others. The individual page cited is by Matt McIrvin.
Am I missing somewhere where proper citation format is given?
Frame Dragger said:He meant to clarify that it was the physics FAQ of a different site, not this one. I think.
JDługosz said:The first sentence is "This is the web version of the Usenet Physics FAQ." The Title is "Usenet Physics FAQ". The Copyright statement doesn't mention him. The main text mentions some history of original creation and maintenance, but his name is absent. The "Thanks" at the bottom does mention him, along with many others. The individual page cited is by Matt McIrvin.
Frame Dragger said: