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Is the word "vacuum" equal to "nothing"? Imagine two charges of opposite sign, far from one another in vacuum, and moving with constant velocity - s.t. none radiates energy, none emits photons. At some time though, the course of their movement brings them closer. They begin to accelerate toward one another. How do they feel the presence of one another? During their constant movement no photons are emitted, no "information" spreads in the universe about their presence.

Sofia
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    The vacuum is far from "nothing" - check out "vacuum polarization" http://en.wikipedia.org/wiki/Vacuum_polarization –  Nov 09 '14 at 11:03
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    They feel and respond to each others electrical field. – Steeven Nov 09 '14 at 11:23
  • Also, read up on 'fields'. The particle does not act on the other particle. It act based on the local field. – Hennes Nov 09 '14 at 11:24
  • @DirkBruere Wow, that article is abysmally written. I'm in pain from trying to parse the run-on sentences, comma abuses, and just general lack of flow and readability. The second "paragraph"--which is really a single sentence--is perhaps the most egregious. – zibadawa timmy Nov 09 '14 at 12:27
  • @Hennes: What is the field here? I know the general concept of field, but what is filed in this particular case. What is it constituted of? It can't be virtual photons field, because virtual photons are only mathematics, and maths cannot make things move. – bright magus Nov 09 '14 at 12:27
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    "maths cannot make things move" - priceless. – ProfRob Nov 23 '14 at 14:11
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    @brightmagus, I'm afraid that attitude towards mathematics might leave you generally unsatisfied with the whole field of physics :) – lionelbrits Nov 23 '14 at 14:29
  • @lionelbrits: No need to worry. I'm only dissatisfied with unphysical physics :) – bright magus Nov 23 '14 at 14:42
  • Is there any other kind? – lionelbrits Nov 23 '14 at 14:50
  • ... than unphysical physics? Sure - the physical one. – bright magus Nov 23 '14 at 15:24
  • Possible duplicates: http://physics.stackexchange.com/q/197/2451 , http://physics.stackexchange.com/q/2244/2451 and links therein. – Qmechanic Mar 04 '15 at 06:39

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"Vacuum" actually means the ground state of space. When there are real particles in space, "it" is in an excited state, and no longer the vacuum. But we usually think of a small region of empty space as approximating a vacuum, for reasons of locality, etc. If you have two real particles in space, and talk about their mutual interaction, then you necessarily have photons. This is because the force between the two particles is mediated by the electromagnetic field, which, when quantized, leads to photons. This is because the $\vec{E}$ and $\vec{B}$ fields, like any physical thing, needs a quantum mechanical treatment. For example, if you wanted to "derive" the Coulomb force between two electrons from quantum electrodynamics (this is done as an exercise in Peskin and Schroeder), then you need to consider scattering events like the following:

Scattering of two electrons in lowest order perturbation theory

So it's not that the space between the two charged particles is "empty", but that there is an exchange of virtual photons between them (the $\gamma$ in the diagram).

However, while electron number is fairly good label for a state, photon number isn't (being massless). The state of the EM field with a single (macroscopic) point charge in it is already a horribly complicated wavefunction. But we can say that this state has maximum amplitude around a field configuration for which $\vec{E} = \frac{1}{4\pi\varepsilon_0 r^2} \hat r$. On the other hand, $\vec{E}$ and $\vec{B}$ can roughly be thought of as conjugate pairs, like $x$ and $p$, so there is also an uncertainty relation at play, so our field configuration is not exact. The best we can do is a coherent state. This is all perturbation theory, so it doesn't give the complete picture, but it's a good undergraduate level of understanding.

Maybe someone has a good link for coherent states in quantum optics. This one might be a start.

lionelbrits
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  • Why I have NECESSARILY photons? Also, I don't ask about the state of the EM field with a single electron, I ask how does an electron, distant from a proton, "feels" the presence of the proton. Which "carriers of information" pass between the two particles, given that they are IN VACUUM? You try to answer my question, but I don't see how the things that you say are connected with it. Please be more "close to the question". – Sofia Nov 23 '14 at 13:09
  • I have edited my answer hopefully to your satisfaction. Btw, some people read capital letters as SHOUTING :) – lionelbrits Nov 23 '14 at 14:27
  • the number of photons is not a good quantum number because they are bosons, not because they are massless. At one point neutrinos were thought to be massless, but we counted the lepton numbers that were conserved. – anna v Nov 23 '14 at 14:34
  • Are you suggestion that particle number is not conserved in cryogenic Helium-4? Also, lepton number isn't the same as "number of that particular particle", which is what I was trying to get at. Also, lepton number conservation is related to charge conservation of a certain U(1).

    You can also talk about coherent states for fermions, e.g. in doing path integrals. There, of course, you have a superposition of having either 0 fermions or 1.

     – lionelbrits Nov 23 '14 at 14:49