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This question received answers I deem completely irrelevant to its true intent, so I'll ask it differently: how do we know? As OP noted, field lines are a theoretical concept - but what empirical evidence do we have that electric field lines do in fact have limitless range? We need not even go that far; answering the original requires answering the following:

How quickly does the electric field propagate? That is, if we (a) added or removed a charge from the Universe, or (b) moved the charge, when would other charges "feel" its effect?

  1. If instantly, then we transmitted information through space instantly; this violates relativity

  2. If not instantly, then the field propagates with finite speed; this necessitates the concept of a field "carrier", whatever it might be (i.e. "some physical thing that affects other physical things")

In case 2, the carrier cannot be infinite in quantity, as there isn't anything infinite within a finite closed system (e.g. a volume spanned by field after some initial time). However; infinite range demands an infinite carrier - else, the larger the radius, the less of the surface area spanned actually experiences any field. So, for a very long range, only tiny patches of regions will experience anything from the charge; the vast majority of matter at the radius will experience exactly zero E-field. Thus: infinite range is impossible.

Am I wrong?

David Z
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OverLordGoldDragon
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2 Answers2

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Even though your question is phrased a bit aggressively, it does demonstrate an important point. Yes, you are correct that the photons emitted from a source, which make up an electromagnetic wave, are discrete, and there are a finite amount of them. So over enormous distances, the photons get diluted, until you get to the point that any detector sensitive enough to see anything at all is just seeing the individual photons as they come in.

This is not only known, it's an essential consideration when doing long-distance optical astronomy, where the detectors don't measure the electric field, but rather see "clicks" for individual photons coming in. This results in "shot noise" due to the discreteness of the photons, just like how shot noise arises in electronics due to the discreteness of electrons. You can even enhance the sensitivity of telescopes by correlating their click times, using the Hanbury-Brown--Twiss effect.

Now, I guess your specific question is what happens to the electric field at such distances. It is not correct to say that the electric field is zero in most places and nonzero at others. Instead, classical electromagnetism breaks down in this limit: we have to remember that electromagnetic fields are fundamentally quantum objects, like everything else. Instead of having a definite field value, we have a superposition of field values at each point. The typical field values fall off as $1/r$ just as advertised (otherwise astronomy would not work), but there's a spread. Measuring the field at a point just collapses this superposition.

The point is that we cannot say that field lines either do or do not go on forever -- that is a overly casual application of words. What's really going on is that the entire idea of a field line eventually stops making physical sense.

This isn't just some weird theorist's fantasy, either. The fact that the electromagnetic field is quantum plays an important role in particle physics (under the name of "quantum electrodynamics"), in astronomy (as mentioned above), in atomic physics (as "quantum optics"), and even in electrical engineering (as "cavity QED" and "circuit QED"). We don't talk about this in introductory E&M courses because the mathematical complications are great, but it is relevant when you make precision devices and measurements.

knzhou
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  • Finally, a substantive answer. The core seems to be here: "instead of having a definite field value, we have a superposition of field values at each point" - though, the picture remains incomplete. What 'generates' said superposition at each point - the wave nature of the photons? If so, shouldn't the position probability distribution drop off about some mean (e.g. line from charge to original direction of emission)? Since # of photons is finite, isn't it correct to say: if we set up enough sensors at a radius, some will receive no photons, thus measuring zero E-field? – OverLordGoldDragon Apr 15 '20 at 21:28
  • This followup can admittedly be a question of its own, which I can post separately if necessary - but so far only a part of the picture's been revealed. – OverLordGoldDragon Apr 15 '20 at 21:29
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    @OverLordGoldDragon I was intentionally vague there because, if you really want to know what the superposition looks like, you need to learn a lot more about how the field is generated in the first place, and it gets into a lot of nitty gritty details. However, an important point is that the $E$-field and the number of photons are "incompatible observables", the math of quantum mechanics tells us they can't have definite values simultaneously, just like the uncertainty principle constrains momentum and position values. – knzhou Apr 15 '20 at 21:32
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    To take it to an extreme, even if you got a box and carefully removed all the photons, a perfect $E$-field meter inside would still see a nonzero spread of values! That is, the $E$-field is not precisely zero even when you definitely have zero photons. (This is the "quantum vacuum" people talk about.) That's why in this limit we usually focus on detectors sensitive to photon number rather than $E$-field. Conversely, a state with exactly zero $E$-field everywhere actually necessarily contains a wide superposition of $A$-field by the uncertainty principle, and hence many photons. – knzhou Apr 15 '20 at 21:34
  • The reason we can get away with ignoring these issues in everyday life is just the same reason that we can ignore the Heisenberg uncertainty principle -- absolutely everything has a spread of values but in everyday life all of the spreads are small. – knzhou Apr 15 '20 at 21:35
  • Thanks for the response. On quantum vacuum - sure, but the question concerns the E-field due to some charge $q$; if we 'absorb' all the photons originally emitted by $q$, will there still be an E-field due to $q$? If yes, that begs even more questions. If no, we're partly back to square one. – OverLordGoldDragon Apr 15 '20 at 22:58
  • On "incompatible inobservables" - unsure how this addresses photon spreading; if E-field is fixed at $r$ from source, but probability of observing a photon must decay w.r.t. some field line, then the photon 'exerts' an E-field uniformly over varying distances. ... without dropoff, it's thus infinite, again violating relativity. -- I think these ideas warrant their own question, unless you're fine just continuing here. Whether you decide to proceed with this or not, thanks for the info. – OverLordGoldDragon Apr 15 '20 at 22:59
  • @OverLordGoldDragon Sorry I didn't respond to this yesterday, but these are each very large questions (which, even asked separately, many people would probably completely justifiably respond to by saying, e.g. "read a whole circuit QED / quantum optics / cavity QED textbook") which should be asked separately! – knzhou Apr 17 '20 at 07:03
  • Fair enough; I plan on accepting your answer, after posting a followup question to mention in this original, as parts remain unanswered. – OverLordGoldDragon Apr 17 '20 at 15:35
  • Followup posted – OverLordGoldDragon Apr 17 '20 at 23:40
  • Btw, you have $1/r$ instead of $1/r^2$ there - edit too short to submit – OverLordGoldDragon Apr 17 '20 at 23:49
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    That’s intentional: radiation fields fall off as $1/r$, you’re thinking of static fields, which fall off as $1/r^2$. – knzhou Apr 18 '20 at 00:22
  • Ah, "field values" as in wave amplitude - alright – OverLordGoldDragon Apr 18 '20 at 00:27
  • Since this site enforces "meta-commentary" rules being edited into questions, I ask that you edit in a clarification that the answer isn't 'complete' for the question's intent, and preferably link the followup; it's an important caveat to inform the reader about. With this I'll re-accept your answer. – OverLordGoldDragon May 19 '20 at 23:37
  • @OverLordGoldDragon The site rules are against meta-commentary in answers too. If you don't think my answer fully answered your question, then you just don't have to accept it, and that's fine! – knzhou May 20 '20 at 00:55
  • This bookkeeping is stifling 'exchange' of knowledge. – OverLordGoldDragon May 20 '20 at 01:02
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Yes, you are wrong.

The gauge boson (force-carrying particle) for electromagnetism is the photon. It is massless, therefore its range is infinite. Photons travel at the speed of light (c), which is therefore the speed with which electromagnetic disturbances propagate through a vacuum. No material medium is required to "carry" photons; the vacuum constants for permeability and permittivity yield the numerical value of c.

niels nielsen
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  • So, I'm right; "for a very long range, only tiny patches of regions will experience anything from the charge; the vast majority of matter at the radius will experience exactly zero E-field." – OverLordGoldDragon Apr 15 '20 at 20:22
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    No you are not correct. The vast majority of matter will not experience ZERO E-field based on this answer. Photons can travel in all directions. You might be confusion the state of feeling an E-field with feeling a change in the E-field. –  Apr 15 '20 at 20:57