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Since virtual particles have charge and motion, their fields should interact with each other to produce Lorentz forces, even in a perfect vacuum.

Can these interactions propagate energy that is injected into their midst by, say, the acceleration of real charged particles?

James Bowery
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  • "Since virtual particles have charge and motion" - no, they don't. They're lines in a Feynman diagram, and nothing more in the modern formalism. For why people talk about them as if they were more, see this question. – ACuriousMind Dec 12 '15 at 18:48

1 Answers1

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Let us be clear on what a virtual particle is:It is a mathematical expression within an integral which when evaluated will give the probability or the lifetime of an event happening at (x,y,z,t), a crossection for the process. This is schematically represented with Feynman diagrams. Below are two diagrams that exchange a charged W meson which as a particle has a mass close to 100 GeV.

feynman diagrams

These are to first order in the expansion for calculating the crossections and the decay.

The crossection can be calculated from low input energies, much lower than the mass of the virtual W. The second diagram on the right gives the decay of the muon, which has a mass of 105 MeV, to compare with a real W which is around 100GeV. The line represents the quantum numbers of the particle that has to be exchanged, put the mass evidently has to be very off mass shell. The mathematical representation , the propagator, has the mass of the W in the denominator but the energy momentum variables of the interaction are way off when the integral is taken, as energy and momentum have to be conserved . A virtual particle is not the particle, it just carries the quantum labels.

Since virtual particles have charge and motion, their fields should interact with each other to produce Lorentz forces, even in a perfect vacuum.

A second lesson to take from these diagrams is that the interaction space, the vacuum you envisage in your question, is very small much less than a Fermi ( a unit appropriate for strong interactions). A virtual W- will not be found in the vacuum of space except within very very short distances under the integral for the interaction.

The virtual photon diagrams under special circumstances can be considered over macroscopic distances when talking of the repulsion of two electrons in vacuum, again a mathematical story but its effect is seen macroscopically in the electric field of the pair.

anna v
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  • When you say something reified by a physical theory "is a mathematical expression" and further qualify it with descriptors like "very very short" it gets rather confusing what your point is. Are you saying that virtual particles cannot interact with each other because they are merely mathematical expressions? Or are you saying that they can't interact with each other because electric potential, for virtual particles, doesn't fall off as 1/r as it does for real particles and that, therefore, there is no contradiction in claiming zero, as opposed to "very very small", interaction between them? – James Bowery Dec 12 '15 at 17:19
  • @JamesBowery the "very very small" is about the volume in space where the interaction is happening, the very short also. You are envisaging interactions long range in space. there is no long range. Everything that is happening is happening in a tiny volume. – anna v Dec 12 '15 at 17:24
  • Complicated Feynman diagrams allow interactions in virtual lines, the contribution of these diagrams is smaller than of the first order diagrams because of the expansion in the coupling constant. – anna v Dec 12 '15 at 17:27
  • Is one justified in claiming that the reason electric potential goes to 0 (as opposed to 1/r) over a very very short distance r is due to: t = the very very short time the HUP permits the charged particle to exist, 2) the electric potential propagating at c and, 3) the propagation of the electric potential at r=t/c ceases since its source charge "never existed"? – James Bowery Dec 12 '15 at 17:37
  • @JamesBowery I do not know that it goes to 0, why do you think so? Within the integral it will have the appropriate value to conserve whatever needs to be conserved. – anna v Dec 12 '15 at 17:48
  • I think it goes to 0 because if it does not go to 0 then energy propagation can occur through the following mechanism: If there is a "sea" of charged virtual particles between two charged real particles whose HUP-allowed existence overlaps in time, the real particles can have a Coulomb interaction that is, in part, mediated by a chain of charged virtual particle interactions connecting the two real particles. Indeed, even if it does go to 0 at some finite distance r, if the spatial density of virtual particles is less than r, such a chain can be constructed. – James Bowery Dec 12 '15 at 18:31
  • @JamesBowery That is not how it works with the real QED feynman diagram calculations. Virtual photons are exchanged between real charged particles. Charged virtual exchanges happen in diagrams as the one I showed above. – anna v Dec 12 '15 at 18:45