How does a photon carry force?

In summary, the conversation discusses the question of how a photon carries or transmits the force of the electromagnetic field. The participants bring up theories and concepts such as Feynman's QED, interference between cases of photon exchange, and the direction of ripples in the EM field. Ultimately, it is suggested that the force from a photon can exist in any direction, even if the EM perturbation is in a different direction.
  • #36
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?
 
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  • #37
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.
 
  • #38
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 ?

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:
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)

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?]
 
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  • #39
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.

All of that is beyond current physics however, which is where the mathematics comes in. I could be wrong, but the thread linked to by Tiny Tim was pretty definitive as I remember. Hans De Vries had some interesting points to make as well.
 
  • #40
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.

I was looking for that link, but couldn't find it.

You don't have to go into momentum space to understand this, though. You can do it in ordinary position space. Consider an electron some distance from a proton. The probability amplitude to absorb a photon is larger on the side of the electron's wavepacket closest to the proton. And that probability amplitude is imaginary. So when you add the contribution to the new wavefunction from the electron absorbing a photon to the contribution from the electron not absorbing a photon, the effect is to multiply the electron's wavefunction by a phase, which the phase depending on how close you are to the proton. This sets the electron in motion towards the proton.
 
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  • #41
The simplest intuitive answer is: virtual photons that mediate attraction forces have negatie energy. Virtual particles violate conservation of energy, right? They can violate it both ways - too much or too little.

When you throw a ball forward, you are pushed backward. But when you throw a ball of negative weight forward, you will be pulled forward, too.

And now slightly more formally:
Virtual particles are physically real, but they are not particles. There are no esoteric photons floating here and there. It's the electromagnetic field that carries forces. Just as photons are excitations of electromagnetic field, the electromagnetic field is a logical anti-excitation of virtual photons.

But we can not think of virtual particles as actual particles. They are not.
 
  • #42
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?]


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.
 
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  • #43
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.

Wow, those photon detectors have something coming to them! :wink: Yes, photons are real, detected, and have appreciable physical properties.
 
  • #45
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.

:eek: :bugeye: :confused: :cry:

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.

Wow! Now that is pretty cool. I am not clear on the details yet (I can only access the arXiv paper at the moment), but it seems this would definitely support the "reality" of virtual phonons. One thing that is a bit worrisome is that the arXiv paper seems to say that the evanescent waves cannot be measured or interact with things ... this would contradict my own experimental experience. I have used total internal reflection techniques in spectroscopy, and we can observe coupling of the evanescent waves to sample material, and thus record an infrared spectrum. A colleague down the hall uses near-field techniques to precisely measure the displacement of beads in a magnetic tweezers experiment. In both cases we rely on the interaction of the evanescent waves with our samples. I guess I'll have to read that other paper at work tomorrow.
 
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  • #46
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.

Günter Nimtz is the same guy who claimed to have observed violations of special relativity in superluminal tunneling, where it turned out that all he has is some clever form of pulse shaping mixed with a bad understanding of what group and phase velocity mean.

One should be pretty careful about the conclusions drawn in his papers.
 
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  • #47
SpectraCat said:
:eek: :bugeye: :confused: :cry:

:smile:

Very well 'said' Cat!
 
  • #48
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.

Have you see http://math.ucr.edu/home/baez/physics/Quantum/virtual_particles.html" ?
 
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  • #49
JDługosz said:
Have you see http://math.ucr.edu/home/baez/physics/Quantum/virtual_particles.html" ?

Just a detail …

that's not "the" Physics FAQ, it's John Baez's Physics FAQ. :wink:
 
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  • #50
tiny-tim said:
Just a detail …
that's not "the" Physics FAQ, it's John Baez's Physics FAQ. :wink:

Yes... where the devil lies. :-p
 
  • #51
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?
You are not misreading me, that's a fact.
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?
 
  • #52
Hi FireBones! :smile:
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?

"not obeying this law" is usually called "off mass-shell" … see http://en.wikipedia.org/wiki/On_shell_and_off_shell" :wink:
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.

Neither real nor virtual photons "carry" force … the force comes from the field, and the virtual photons are mathematically connected with the field.
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?

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! :rolleyes:) 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"). :smile:
 
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  • #53
tiny-tim said:
Hi FireBones! :smile:



"not obeying this law" is usually called "off mass-shell" … see http://en.wikipedia.org/wiki/On_shell_and_off_shell" :wink:


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! :rolleyes:) 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"). :smile:

Wow, I understood that completely... which probably explains the sudden nose-bleed... and blindness... and incontinence... and headache. :wink: Ahh... virtual particles... *gulps aspirin*
 
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  • #54
Force = change in momentum

Photon carries momentum !
 
  • #55
presario said:
Force = change in momentum

Photon carries momentum !

:rolleyes:
 
  • #56
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.

Hmm...what about the neutron star? i think electrons and protons collide with neutrons and release neutrinos by the effect of the weak force.
 
  • #57
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.

That is a matter of degeneracy... a unique case.
 
  • #58
tiny-tim said:
Just a detail …

that's not "the" Physics FAQ, it's John Baez's Physics FAQ. :wink:

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?
 
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  • #59
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?

He meant to clarify that it was the physics FAQ of a different site, not this one. I think.
 
  • #60
Usenet Physics FAQ

Frame Dragger said:
He meant to clarify that it was the physics FAQ of a different site, not this one. I think.

Thanks, Frame Dragger! :smile:

Yes, PF has its own FAQs on various topics, which are normally referred to first. :wink:
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.

Yes, I assumed that since it was on the well-known John Baez website that it was organised by him.

I now realize that he's just one of eleven hosts in seven different countries …

From http://math.ucr.edu/home/baez/physics/index.html" …
This is the web version of the Usenet Physics FAQ (Frequently Asked Questions). Its purpose is to provide good answers to questions that have been discussed often in sci.physics and related usenet newsgroups. The articles in this FAQ are based on those discussions and on information from good reference sources. That does not mean that they are always perfect and complete.​
…
Thanks go to John Baez, Ronen Ben-Hai, Jay Brown, Jon Butterworth, Alan Cairns, Dave Edsall, Harald Falkenberg, Philip Gibbs, Amarendra Godbole, Chris Hillman, Chung-rui Kao, Matt McIrvin, Joe Mirando, Matthew Parry, Han-Tzong Su, Nathan Urban, Johan Wevers, Sam Wormley, and the various organisations for hosting us! If any other non-commercial sites would like to mirror this FAQ, please contact the editor.​
 
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  • #61
Frame Dragger said:
:rolleyes:

Yes, that is the whole point, with QED picture there is no "action at a distance" since I said that force is change in momentum, anything that carries momentum can exert a force by passing on the momentum.
Every photon can only be specified by its momentum and its polarization
 
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