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The answer to this question probably lies in QFT, which I know just enough about to appreciate my current lack of understanding of the subject, if you follow me.

About a year ago I asked our Professor and was told:

a. "It's a property of photons."

b. To think in terms of inverse momentum, which sort of makes sense, provided you have the background.

So, in other words, is there an intuitive picture available to help understand the observed fact that two like electrical/magnetic charges repel and two opposite charges attract and a photon seems to "know" this and alter momentum values accordingly?

I do own a few QFT books, so if it's in page X of of say, QFT in a Nutshell, or Srednicki, that would be a great help.

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  • $\begingroup$ Zee, chapter "Coulomb and newton: Repulsion and Attraction" $\endgroup$ – innisfree Jun 15 '15 at 15:37
  • $\begingroup$ @innisfree thanks very much for that, afraid you were going say Weinberg:) $\endgroup$ – user81619 Jun 15 '15 at 15:55
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For the forces between elementary particles we have Feynman diagrams, where there exists a mediating particle for the interaction. In the simplest diagrams: for the strong it is the gluon, for the weak it is Zs and Ws and for the electromagnetic it is the photon.

Look at these diagrams some of the tree level Feynman diagrams of compton scattering , photon on a proton .

The mediator of the interaction is not a gauge boson, thus electromagnetic interactions happen even though the mediator ( the exchanged particle) is not the photon.

compton scattering

Here is Bhabha scattering, where the electron and the positron ( attractive force) are of low energy:

bhabha1bhabha2

      annihilation                                  scattering

So the question should be how can there be attractive and repulsive forces. To really answer one would have to do the mathematics that the Feynman diagrams dictate and the result will tell us the the force is attractive.

I have found useful for intuitive understanding the analogue with boats throwing balls to each other, and transferring momentum for repulsion, and boomerangs for attraction.

repulsive

repulsive analogue

attractive

Momentum conservation directly for the repulsive, angular momentum conservation in the attractive. As all analogues it should not be stressed too much. Here we have a ball and a boomerang, but in the Feynman diagrams the e+e- has an extra diagram to add to the calculation, which the e-e- or e+e+ does not have.

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  • $\begingroup$ As always, thanks very much anna. I would not be asking these questions if I did not feel I could understand the math, I just want to finish off my GR study first. I just can't resist asking a (good) bit ahead of my current background. $\endgroup$ – user81619 Jun 15 '15 at 14:51
  • $\begingroup$ how can i see that boomerang stuff in QFT? $\endgroup$ – innisfree Jun 15 '15 at 15:40
  • $\begingroup$ @innisfree Feynman diagrams are the calculational part of QFT itp3.uni-stuttgart.de/lehre/Archiv/ss13/quantenfeldtheorie/… . It is what the creation and annihilation operators of QED end up in . $\endgroup$ – anna v Jun 15 '15 at 15:58
  • $\begingroup$ it's not clear to me that two people on boats throwing boomerangs has anything at all to do with QFT or EM. is it anything more than a nice picture to show if something asks this question? has it actually got anything to do with reality? $\endgroup$ – innisfree Jun 15 '15 at 16:08
  • $\begingroup$ @innisfree I offered it as an analogue of how, when common sense would tell you that throwing something from a boat will repulse the other boat, there is a consistent way of attracting the other boat. $\endgroup$ – anna v Jun 15 '15 at 16:49
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Virtual photons are not actual particles, they are just a mathematical abstraction used for perturbation expansion.

As such, they are off-shell : their physical properties are not the same as those of real particles, and they do not have to obey classical equations. Most famously, they are off the mass shell, that is, their energy and momentum do not follow the law

$E^2 - p^2 = m^2$

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  • $\begingroup$ Doesn't answer the question $\endgroup$ – innisfree Jun 15 '15 at 15:30
  • $\begingroup$ @innisfree: It does not, but it gives the OP a very important piece of information that is missing in the way Feynman diagrams are being interpreted by the novice who hasn't been properly warned. I don't think one should vote down a stop sign just because it doesn't answer the question who should go first at an intersection. $\endgroup$ – CuriousOne Jun 16 '15 at 3:33
  • $\begingroup$ @innisfree A warning for you: " missing in the way Feynman diagrams are being interpreted by the novice " sounds very condescending in a non instructive manner. If photons are just mathematical abstraction, it will be hard to explain photoelectric phenomenon. $\endgroup$ – user121963 Jun 25 '16 at 21:02
  • $\begingroup$ If you share the same master and the same source of funding, then yes you will get a stop sign in an intersection because the master of yours wants to decide which one gets funding. However, if you don't even use their funding and their resources, it will not be any stop sign. What you have is a multiple lane high way. $\endgroup$ – user121963 Jun 25 '16 at 21:08

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