Now per QED, electrical charges interactions are effected by photons. Suppose you are one of the two charges. How do you know to attract or repel the other charge? In other words, how do you know if the other charge is the same or opposite, since the critical interface you have with it is the zero-charged photon? It would seem that the photon would need to convey to the charges their relative polarity. How does the photon do that?
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$\begingroup$ "Now per QED, electrical charges interactions are effected by photons." That is a simplified just-so story told by taking Feynman diagrams too literally. Electrical charge interaction is through the gauge field, whose quanta are the photons. Please to not believe that the idea of exchange of virtual photons is a description of what we believe is "really happening" $\endgroup$– ACuriousMind ♦Nov 25, 2014 at 20:35
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1$\begingroup$ To the Curious Mind: Then, what YES to believe? You are so much convinced that "virtual particles" are only "on the paper". But, won't you say explicitly what you do REALLY believe? Of course, you have a certain picture in your mind. What it is? To deny is easy, but something has to be put in the place. If you exclude a real exchange of short-lived carriers ("virtual particles"), which other PLAUSIBLE option remains? $\endgroup$– SofiaNov 26, 2014 at 11:57
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Now per QED, electrical charges interactions are effected by photons. Suppose you are one of the two charges. How do you know to attract or repel the other charge?
You want something that does not exist - intuitive picture of physical process within a theory which is a demonstration of how far can one go with mathematisation of experience and ignoring intuitive pictures.
To study quantum electrodynamics you have to concentrate on its computational algorithms and neglect intuitive pictures, to study intuitive pictures you have to neglect QED.
Both are a good thing to study, just do not expect it is easy to make them consistent.
answered Nov 25, 2014 at 20:20
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$\begingroup$ Photons are not intuitive, they have been measured etc and are essential to QED since they alone convey forces among charges. So now we need to compute how the photon transmits the effect of the relative charges, most especially to account for the relative polarity of those charges. If your equations are assuming that relative polarity, then they are intuitive or rather assuming. We should not hand-wave with our equations. $\endgroup$– davAgNov 27, 2014 at 19:52
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1$\begingroup$ The photons have much to work with. They have both linear and angular momenta, phase, can be polarized, frequencies, spin; but they do not have charge. And somehow they must enable transmit the force to have two charged particles either attract or repel, since none of the other 3 forces (gravity, strong, weak) are suited for this job. $\endgroup$– davAgNov 27, 2014 at 20:00
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Since the field of each charged particle extends to infinity, the fields of two particles are "in contact" with each other (no "communication" is necessary). When the charges are not equal (+ & -), the fields "cancel" each other along the line connecting their centers. This causes the attraction of the particles. When the charges are the same (+ & + ; and - & -), the fields reinforce each other causing the particles to repel each other.
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The attraction of unlike charges and the repulsion of like charges is an experimental observation that has to be included in any model of electromagnetic reactions
When talking of photons one is in the quantum mechanical regime.
As in classical electrodynamics the sign of the charge defines the potential, attractive or repulsive, between the two charges, so in the quantum mechanical formulations the potential carries the charge , whether it is an attractive or repulsive potential. The solutions which are the state function of the system carry this information, and this is the way that the two particles "know" the charge of each other.
Carrying the information quantum mechanically means that if a scattering experiment is performed there will be a region in phase space where the attractive potential solutions will differ from the repulsive ones. One can get an intuition on this from simple quantum mechanical potentials , for example in these lecture notes in the figures on page 30 how there are parameter regions where the attraction and repulsion solutions diverge. Of course attractive situations also allow for bound states, which is another story.
When the distances are very large , the behavior is similar, so in this sense the two particles will "know" the charge of each other only close to each other.
For a quantum field theory description of virtual particle exchanges between positive and negative , negative and negative have a look at this entry, figures 1 and 2.
The bottom line is that the observed behavior of charges is accommodated by the mathematical models in all frameworks, classical and quantum.
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$\begingroup$ Anna, I feel that what the question is the same old question, are the virtual particles indeed virtual? See what the question-owner says: "It would seem that the photon would need to convey to the charges their relative polarity". I indeed saw an article (of some professor) saying that virtual particles EXIST, but are of short life. Why do you refute this possibility, can you explain? $\endgroup$– SofiaNov 26, 2014 at 12:05
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1$\begingroup$ @Sofia I am an experimental physicist. For me , exists means "can be measured by some proxy" , not that a mathematical formula exists. As there are other ways of measuring and fitting the crossections etc not needing the mathematics of virtual particles, I do not apply the verb "exists" to a virtual electron, as I apply it to the on mass shell electron that can leave a track in my chamber :). $\endgroup$– anna vNov 26, 2014 at 12:12
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$\begingroup$ The potential looks good, but it must be constructed of electromagnetic forces, i.e., photons. The electromagnetic field is responsible for that potential, and in turn the photon is directly involved. When electric charges interact, they accelerate, they radiate, they radiate photons to form electromagnetic fields. And per Maxwell, it's these fields that are responsible for voltages (potentials, magnetics, etc.). But your implication of potentials does have its merit, it may help clarify the role of the photon in the interaction in question. I am considering that. $\endgroup$– davAgNov 27, 2014 at 20:08