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We know that electromagnetic waves fly with the speed of light, but my question is not about waves. Consider a very strong electromagnet that creates a substantial field 3 meters away. Then we send a proton accelerated to near the speed of light to fly by. The proton interacts with the field of the magnet and is deflected.

The field is generated by the electrons in the wire of the electromagnet. The interaction of the proton with the field is a quantum exchange of energy, momentum, etc. between the proton and these electrons. Whether we label these exchanges as "mediating virtual photons" or not is not really important here. My question is about the speed of these exchanges.

What are the current views on the timing of these exchanges? Are they instantaneous or limited by the speed of light? If instantaneous, would they not violate casualty by transferring information and energy faster than light?

If the exchanges happen at the speed of light, a number of problems arise. First, the exchanges must be directional. Say, if an electron in the wire emits "a virtual photon" toward the proton, then the proton would be 3 meters away from that position by the time "the virtual photon" arrives. So the electrons would have to aim at the future position of the proton to hit it. This makes no sense and probably is one of the reasons why the "virtual particle" model is not in favor. Secondly, a virtual photon would have to exist for a nanosecond that would severily limit its energy per the uncertainty principle.

Can someone please clarify the actual physics behind the electromagnetic interactions from the timing standpoint? By physics here I mean physical observables, something we can measure. Using quantum fields of mathematical probabilities if fine as long as they are linked to observable values.

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  • $\begingroup$ Re: feel free to replace the electromagnet there with a static charge But charge and magnetic dipole are entirely different things. In Coulomb gauge we have equation $\Delta \phi= - 4\pi \rho$ without any time derivatives, which means instantaneous propagation of $\phi$. But there is no causality violation since charge is a conserved quantity and there would be the same charge at any moment of time (no information is transferred superluminally). Magnetic dipole is not a conserved quantity, you could turn it on and off, so information about it would propagate at a speed of light… $\endgroup$
    – A.V.S.
    Commented Feb 22, 2020 at 17:41
  • $\begingroup$ … Incidentally, the no-hair theorems of black hole physics is a good indicator for the speed of propagation: the fields carrying information on mass (grav. monopole), angular momentum (gravimagnetic dipole) and charge propagate instantaneously, so they have no trouble escaping the black hole. Information on magnetic dipole moment however propagates at the speed of light and so black hole has no independent magnetic dipole moment (it does have induced by frame dragging magnetic dipole proportional to charge and spin). This take on no-hair thm., IIRC, belongs to J. Bekenstein. $\endgroup$
    – A.V.S.
    Commented Feb 22, 2020 at 18:08
  • $\begingroup$ So the recoil of the static charge is delayed by the speed of light relative to the deflection. And if we change the frame to make the other charge static, the situation reverses per relativity of simultaneity. Got it, thanks! An interesting insight on black holes. I get it perfectly well and it is quite elegant. Intuitively it also should be equivalent outside to all matter and charges located at the horizon. $\endgroup$
    – safesphere
    Commented Feb 22, 2020 at 18:34
  • $\begingroup$ Here is a relevant paper on the subject: arxiv.org/abs/gr-qc/9909087 . Though the title speaks about “speed of gravity” it does have a section on EM as a warm-up. $\endgroup$
    – A.V.S.
    Commented Feb 22, 2020 at 19:25
  • $\begingroup$ @A.V.S. Thanks! I had a question based on this very paper :) Also with no good answer: physics.stackexchange.com/questions/492870/… $\endgroup$
    – safesphere
    Commented Feb 22, 2020 at 21:23

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Virtual photons are primarily a mathematical concept used to make some particle interaction calculations easier. They are not physically real--hence the name--but sometimes they can assist in thinking about an interaction.

As for your electromagnet, if the magnet was already turned on before the proton arrived, then the field already existed and is static, so it doesn't matter when the proton arrives. To use the virtual photon picture, a static field is created by an object constantly throwing out virtual photons in all directions, whether or not there is another object to intercept them. So, when the proton passes by the magnet, there are already virtual photons there to push it to the side.

If the field is initially off, then it has to be turned on at least $d/c$ seconds before the proton arrives, where $d$ is the distance from the electromagnet to the proton's path through the field and $c$ is the speed of light. When the electromagnet is turned on, the field it creates will be established at a point after a delay due to the speed of light.

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  • $\begingroup$ Thanks for the answer, but it doesn't answer my timing question in terms of physical observables. I'm not buying the idea of a charge constantly "throwing out virtual photons". It just makes no sense, especially considering they don't exist. Not buying the "field" idea either, as it is not observable at the quantum level. When the interaction happens, there is an exchange of energy and momentum, which are observable. What is the timing of this exchange between two charged particles? Simultaneous? Instantaneous? Delayed at the speed of light? Directional? It would be great if you can clarify :) $\endgroup$
    – safesphere
    Commented Oct 22, 2017 at 19:46
  • $\begingroup$ @safesphere I'm not sure what you mean by "fields aren't observable at a quantum level." In Quantum Field Theory, fields are all that exist. Individual particles are stable ripples in the fields. For example. there is a single electron field that spans the universe. Every electron is a ripple/wave in this field. The electron field can interact with the electromagnetic field (the ripples of which are photons), and so electrons have an electric charge that creates a detectable electric field around them. $\endgroup$
    – Mark H
    Commented Oct 23, 2017 at 6:05
  • $\begingroup$ "There exists" means different things in math and reality. Quantum fields "exist" only in the imagination. They don't exist in the physical sense, because they are fields of probability, which is a mathematical abstraction. You can use them to predict the distribution of electrons to be detected, but the observables here are electrons, not the fields. The "fields only" viewpoint seems to be an emerging occupational hazard of losing physical reality behind formulas :) See this perfect answer: physics.stackexchange.com/questions/360902/… $\endgroup$
    – safesphere
    Commented Oct 23, 2017 at 7:09
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    $\begingroup$ @safesphere Given how weird nature is turning out to be and how successfully the mathematical formulas are at predicting what reality does, I'm hesitant to call waves and fields imaginary. It may be the case that mathematical abstraction is the best way to get at reality that is so far from our everyday experience. If the formulas that let us predict the distribution of detected electrons say they move as waves prior to detection, then the accurate prediction of the final distribution of detected electrons lends weight to the reality of waves. $\endgroup$
    – Mark H
    Commented Oct 23, 2017 at 8:00
  • $\begingroup$ @safesphere In anna's answer that you linked to, the tracks in the bubble chamber are made by high energy particles, which are actually predicted to act like particles due to their very short wavelength (large momentum). $\endgroup$
    – Mark H
    Commented Oct 23, 2017 at 8:00

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