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$q$ sets up an E-field, expanding outward at $c$, 'carried' by photons. The quantity of photons is finite - then, the longer the distance $r$ from $q$, the more 'diluted' the photons are about the spherical surface they span.

Eventually, $r$ is large enough that mutually closest photons are light years apart. Then, if E-field is, in fact, not mostly zero at that radius - what is transmitting the E-field to matter light-years between the distant photons? Whatever it is, it must be able to influence an arbitrary amount of matter in the 100 light years spanned between the two source carriers (see below) - simultaneously, even if the matter itself is separated by light-years. This implies superluminal speed of information transmission through space, violating relativity.

"It's a wave, not a particle" - the problem persists; can one wave simultaneously exert influence on quadrillions of particles (and separated by light years)? A wave's energy is finite, and finitely-reducible, and its speed is $c$, so this is impossible.

While the linked answer and others suggest photons are the 'carriers", I'm unsure whether there's a concensus - but whatever the carrier, as long as finite, above holds. All said; how is the electric field set up by a source charge not mostly zero very far away?


Diagram:


Update: I'll refine the inquiry to minimize any further ambiguity.

The root premise of the inquiry is localization. Whatever the 'thing' - a field, a particle, a wave - it isn't "omnipresent", and its interaction with matter is bound within a finite region of spacetime. This includes space itself; while possibly infinite, it isn't 'same' everywhere - its curvature isn't fixed, and matter behaves differently within its different subregions. Anything that isn't uniformly distributed over all of spacetime obeys this logic; e.g., in some models, electrons have an infinitely extending position probability distribution - but probability is unit-normed, and thus isn't uniform over all of spacetime (else $P = \infty$).

Whether as waves or as particles, photons (or whatever the carrier) must likewise obey the above. The "balls" in the diagrams represent localization, not particles.


Re: G. Smith's answer.

My inquiry concerns the E-field due to a charge. I particularly fail to make any sense of:

Finally, each charge does not have its own field. There is only only one quantum EM field pervading the universe, and all photons in the universe are quanta of this one field.

Whether true or not, somehow it's acceptable to casually introduce such a "non-mainstream" idea to a followup of a followup question. I digress; it seems more a matter of relevance than truth - again, I ask about the E-field due to a charge. If we agree that E-fields vary through space, and that one definite source of this variation is a net-charge, and that the transmission of this variation involves finitely many carriers, then all below "Update" applies.


The only scenario I see in which my arguments break down is if interpreting the quoted excerpt a certain way: that the EM-field is like the gravitational field - permeating all of spacetime, and affected by charges just like space is affected by matter. Then I've been heavily misled by both this and this answer to ever fail to bring this up (and to explicitly suggest photons are said carriers), likewise by my 'respected' university physics & engineering textbooks or professors. If this is the truth, I'll accept it, but there's sure a point to be made on not keeping such facts as "side ideas".

Even so, this puts in question the relevance of photons in context of E-fields (as it's been explained by the "answers"), but that'll be a separate question.

(Lastly, before there's again confusion, I clarify that the localization argument pertains to a field, set up by a localized entity, that's transmitted by finite carriers, which differs in context space & mass, or gravity)

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – David Z
    Commented Apr 19, 2020 at 1:59

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When you quantize Maxwell’s equations to get QED, the equations still hold. All that changes is the meaning of the potentials and fields; they are now quantum-mechanical operators. The field of a charge still extends arbitrarily far from the charge.

In fact, the quantum electromagnetic field exists even when there are no charges and no photons. For example, it has “quantum fluctuations” in the vacuum state. And a Big Bang does not require charges to produce photons; expanding spacetime is sufficient to produce them. (By the way, the quantum electron-positron field similarly exists when there are no electrons or positrons. Fields are more fundamental than particles, at least in many physicists’ ontology.)

Your conception of photons as being the field is incorrect. They are the quanta of the field. These quanta interact with matter as point particles, but the quantum field that they are quanta of exists everywhere, just like classical field in classical electromagnetism.

Finally, each charge does not have its own field. There is only only one quantum EM field pervading the universe, and all photons in the universe are quanta of this one field.

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  • $\begingroup$ Question updated $\endgroup$ Commented Apr 18, 2020 at 15:35
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    $\begingroup$ the EM-field is like the gravitational field - permeating all of spacetime, and affected by charges just like space is affected by matter Yes, that’s correct. BTW, the strength of this EM field (in addition to matter) at every point helps determine spacetime curvature there. $\endgroup$
    – G. Smith
    Commented Apr 18, 2020 at 16:19
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    $\begingroup$ The current Standard Model of particle physics is that there are seventeen quantum fields pervading the universe. They explain all elementary particles except for gravitons. These fields interact with each other, and these field interactions explain all non-gravitational forces between the particles. The gravitational field is a classical field which determines the metric of spacetime, so it is not yet properly unified with the seventeen quantum fields. $\endgroup$
    – G. Smith
    Commented Apr 18, 2020 at 16:32
  • $\begingroup$ By the way, your disparagement of my answer as “non-mainstream” is incorrect. It is exactly what is taught in any course on quantum field theory. It is not some personal theory of mine. $\endgroup$
    – G. Smith
    Commented Apr 18, 2020 at 16:43
  • $\begingroup$ Well, that's some enlightening information right there. My "non-mainstream" comment is more of a criticism of the education system, as despite lengthy searches in both textbooks and online sources (not explicitly on QFT), this was not mentioned anywhere. Yet, this info defines the entire framework of thought one has about electric fields. $\endgroup$ Commented Apr 19, 2020 at 0:11
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In QED we understand the electromagnetic force as being transmitted by photons, and we understand the speed for transmission of the electromagnetic force as being equal to the speed of light. If there is a force between charges light years apart, then it would indeed have to result from photons which were transmitted years ago.

The formulae of quantum theory are probabilistic. According to the formulae, the greater the distance, the less the chance of a photon being transmitted, and the fewer photons are exchanged. The matter is complicated because we cannot actually say how many photons are exchanged (there is no photon number operator for this exchange). There is no bound on the number of photons with vanishingly small energies. Also the photons involved in the exchange are emitted by one charge and absorbed by the other, and are not detectable in any way except through the force which they transmit. I.e. we cannot directly observe the underlying process, we can only calculate its consequences and find that they match observation.

The classical force is calculated from the expectation of the photon exchange. For near charges, we can think that many photons are exchanged and the expectation gives a close match to the observed force. For distant charges we may like to think that very few photons are exchanged (or that those exchanged have vanishingly small energy). In this case we should expect that the expectation is a less good approximation to the force --- the force should become quantised, meaning that if we were able to detect it, it would only exist when a photon is exchanged. It would be lovely if we could do such an experiment to show force becoming quantised at large distances, but unfortunately I see no prospect that we could. The force becomes very small, and will be drowned out by noise long before it actually vanishes.

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  • $\begingroup$ This answer yet again appears to contradict G. Smith's by implicating finite carriers, and I seem to be the only one to think it's a contradiction - am unsure how else to formulate my question to yield consistency. If your answer is true, then the localization dilemma seems to persist, so the question's yet again unanswered. Guess I'll be thinking of a followup question. $\endgroup$ Commented Apr 19, 2020 at 15:27
  • $\begingroup$ @HolgerFiedler Interesting topic, thanks for linking $\endgroup$ Commented Apr 19, 2020 at 15:27
  • $\begingroup$ @HolgerFiedler, You may also like ejtp.com/articles/ejtpv10i28p27.pdf in which qed is constructed as a theory of particles. $\endgroup$ Commented Apr 19, 2020 at 15:38
  • $\begingroup$ @OverLordGoldDragon, the equations of both classical and quantum electrodynamics are both clear that there is no instantaneous transmission, whether one thinks of the quanta or the field as fundamental (I cannot think the field fundamental). $\endgroup$ Commented Apr 19, 2020 at 15:42
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The same is true for any spherically expanding wave function: the probabilty of detecting a particle per steradian is constant but of course the probability per surface unit decreases with $r^2$. Secondly, if you don't detect a photon it does not mean that the field is zero. It does not mean that the probability to detect one was zero.

What exactly is your problem with this fact ?

My answer agrees fully with @G.Smith 's, I am just using other words.

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    $\begingroup$ Downvote unjustified +1 $\endgroup$
    – safesphere
    Commented Apr 18, 2020 at 17:13

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