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I've read at several places that a static magnetic (and electric for that matter) field can be thought of as made by virtual photons, at least that's what I understood.

Now, in Special Relativity we learn that an observer (in an inertial frame of reference) who sees only either an electric or a magnetic field will see both an $E$ and $B$ field if he starts moving. That is, he will see an electromagnetic field. And as far as I know an electromagnetic field is made of photons (not virtual photons).

Hence my question "is the property of being "virtual" for a particle (a photon in particular but any other particle is also welcome) dependent on the observer?"

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untreated_paramediensis_karnik
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  • You might also want to look at this: http://physics.stackexchange.com/q/221842/ – Lewis Miller Jan 17 '16 at 03:09
  • One talk about virtual photons in electric and magnetic field because the nature of these fields is not discovered. But somehow it is clear that photons are involved in electric and magnetic fields and in electromagnetic radiation too. Otherwise the (virtual) photons, which are inside the magnetic field, are not extractable, so how they give there properties to the EM radiation. Want to read more, see https://independent.academia.edu/HolgerFiedler and read my light-minded overview "Are photons composed particles" or the elaboration about "Complex one-dimensional structures of space" – HolgerFiedler Jan 17 '16 at 06:15

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"Virtual" is not a property of a particle at all. And it is not true that pure electric and magnetic fields are "made out of virtual photons" or that a combined electromagnetic field is "made out of real photons".

A "virtual particle" is not a particle. It is an internal line in a Feynman diagram, which is in turn a graphic notation for a certain integral. No particle states are associated to such internal lines in the formalism. You may tell stories about them, but in the end, whatever they are, they are not particles in any sense.

Note that the accepted answer to the question you link says

From a field theory point of view, all static fields, whether electric, magnetic, the weak nuclear force or the strong nuclear force can be thought of as being mediated by virtual particles. [emphasis mine]

where now "mediated" does not mean that there are actual particles that "make up" the field, it means that the Feynman diagram between two charges in which a virtual photon is present gives rise to the electromagnetic force in the classical limit, see this question.

The relationship between the electromagnetic field (and with that also pure electric and magnetic fields), is complicated. The energy of the electromagnetic field corresponds to the number of photons in it, see this question, but this is a rather useless statement if the field state is not a number eigenstate. It is electromagnetic radiation, i.e. electromagnetic waves, which corresponds best to "being made out of photons", although also there the link is not clear cut, see this question.

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ACuriousMind
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    I almost downvoted this for saying that a virtual particle is an internal line in a Feynman diagram. I really wish you guys would point out the simplest case in which a particular theoretic construct shows up instead of always jumping to QFT. Virtual particles show up in vanilla quantum mechanics. – DanielSank Jan 17 '16 at 03:07
  • @DanielSank: What is your simplest example? – ACuriousMind Jan 17 '16 at 03:09
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    Second order perturbation theory. The virtual transitions are clear as day: $\sum_{\text{virtual state }m} \langle \text{final} | m \rangle \langle m | \text{initial}\rangle / (\text{some denominator, whatever})$. – DanielSank Jan 17 '16 at 03:29
  • So out of what the electric and the magnetic field and the EM radiation is made? – HolgerFiedler Jan 17 '16 at 06:19
  • @DanielSank: The concept of "virtual particle" in QFT is actually a relic of an older form of theory, see this answer. You are correct that the "virtual particle" is technically analogous to virtual transitions, but I don't get what you want me to do in this answer: Do you think going on a tangent about second-order time-dependent perturbation theory in usual quantum mechanics would be useful? – ACuriousMind Jan 17 '16 at 15:34
  • @HolgerFiedler: That is not a question we would regard as answerable. It just doesn't make sense to ask it within the current framework. You can describe radiation as a classical wave, you can descibe it as a bunch of photons - those descriptions are all valid in some range, and useless in another. – ACuriousMind Jan 17 '16 at 15:36
  • As Lang as we are not r day to discover the inner structure of the electric and the magnetic fields and of the he EM radiation, we will "herumeiern" around with virtual photons. Since this fields as well as photons have dipole characteristics, they have to have an inner structure. In my paper I show, that it is possible to modelize such fields by the help of two quanta. It is a model and if one shows, that there are inconsistencies to the phenomenons of this fields, it will be easy to rip this model. It is your turn to do so. Be sure, I'm smart enough to know than the game is over -:) – HolgerFiedler Jan 17 '16 at 15:59
  • I mistyped (using the German keys), sorry. Here is, with what I wanted start: As long as we are not ready... – HolgerFiedler Jan 17 '16 at 17:53
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    @ACuriousMind I think it would be useful to explain new ideas in the simplest possible context whenever possible. Here is an example where someone asked about contour shifting in the context of relativity but I explained it using a damped harmonic oscillator. Should you write an answer that long? Obviously that's up to you, but again I do think keeping it simple is helpful and clarifying for the reader. As I said, I just ask that we not always jump to relativistic QFT to explain simple things. – DanielSank Jan 17 '16 at 18:20
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The answer is actually simple and definitive. No.

Indeed they have a mass (different from the respective "real" particle), but we know that $P^2=-m^2$, and since the square modulus of a 4-vector is observer independent, a virtual particle is observer independent. (Curved space effect aside, like Unruh effect)

Still, by definition you cannot observe it, so...

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