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I've been reading up on photons, and find myself puzzled by an element of them...

How can photons have an electric field without having a charge? Correct me if I am wrong but I believe electric fields can only be created by charged particles, which photons aren't. When I did some research most other sources also told me this, so what is going on here?

hft
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Gamaray
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    Photons don't create an electric field, they substitute the electric field, that is, they are part of a model which describes the electric field. – Photon Aug 30 '22 at 19:41
  • maybe my answer here to a different question may help https://physics.stackexchange.com/questions/718926/how-do-we-know-em-fields-are-created-by-particles-and-not-vice-versa/718932#718932 – anna v Aug 31 '22 at 03:41
  • 1)Maxwell's equation, specifically faraday's law: $E=\int \int_s \frac{d \phi}{dt}$, implies that a time-dependent magnetic field that is changing induces an electric field. So really, there is no need for a charge. 2) The electric field that is described by Faraday's law(as above) is NOT conservative; the field due to a charge IS. – F.N. Sep 02 '22 at 04:50
  • Insert obligatory reference to ICP here – Carl Witthoft Sep 02 '22 at 12:32

7 Answers7

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I believe electric fields can only be created by charged particles

There are two things that can produce a (disturbance of the) electric field:

  • A charged particle

  • A changing magnetic field

Since an electromagnetic wave has a changing magnetic field component, it can produce a (disturbance of the) electric field without a charged particle being present.

Initially, there was a charged particle involved at the source of the electromagnetic wave, but this particle doesn't travel with the wave.

This is similar to how if you drop a rock in a pond and produce a water wave, the rock doesn't have to travel along the wave to sustain the wave.

The Photon
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    The OP must be feeling blessed to get an answer by the photon itself – ɪdɪət strəʊlə Aug 30 '22 at 19:46
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    @ɪdɪətstrəʊlə I thought they were going to start the answer with: "Let me tell you a little about myself..." – Lawnmower Man Aug 31 '22 at 16:08
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    @LawnmowerMan, you'll notice the answer doesn't actually mention photons at all. I think OP has two main misconceptions: Does an EM wave carry charge, and how does the photon relate to the EM field. I really only addressed one of these in my answer. The one about the photon would take a bit more space to answer. – The Photon Aug 31 '22 at 16:49
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    That the charges don't have to accompany the photons is ... a good thing (for the observer). – Peter - Reinstate Monica Sep 01 '22 at 12:14
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    I think "co-exists" with a changing magnetic field, rather than "produced" by , is more accurate. – jensen paull Sep 01 '22 at 16:49
  • This comment section keeps on giving, @LawnmowerMan. Nice name for those of us old enough! :) – AnoE Sep 02 '22 at 11:31
  • @ThePhoton - Fundamentally, I don't think a charged particle or a changing magnetic field "produce" and electric field... The EM field exists everywhere and the things you listed create purturbations in that field. I think it's important to distinguish that – ScottishTapWater Sep 02 '22 at 14:58
  • Thanks for shedding some light on this. – Florian F Sep 02 '22 at 15:53
  • @ScottishTapWater, everywhere I've ever worked or studied, "produce an electric field" has been used as everyday shorthand for "produce a disturbance in the electric field". I agree it's an important distinction to make when teaching new learners. But it usually isn't because we learn Maxwell's equations before we learn the proper definition of a field. – The Photon Sep 02 '22 at 15:57
  • @ThePhoton - Sure, I agree with you there, but given the level of the question, I think it would improve your answer if you acknowledged the caveat/distinction and then proceeded with your current explanation – ScottishTapWater Sep 02 '22 at 16:06
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Other answers have made a connection between sources and fields at the level of Classical Electrodynamics. However, photons are quantum entities, and it is difficult to ignore Quantum Electrodynamics (QED).

Although photons are introduced as excited states of a quantized field intended to describe electromagnetic fields, they do not have an electric (or magnetic) field. The electric (and magnetic) field of an electromagnetic wave emerges as a special state of many photons (a coherent state). Additional useful information about the relation between photons and electromagnetic field can be found here.

GiorgioP-DoomsdayClockIsAt-90
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  • Here you say photons are excitations of the electromagnetic field but then imply that photons do not have electric or magnetic fields. I don’t get it. What’s the difference? – Jagerber48 Sep 01 '22 at 01:55
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    @Jagerber48 A quantized field is an operator-valued function of space-time. When we speak about the observable electromagnetic field, we speak about a vector real-valued function of space-time. The two things are not the same. Notice that the electromagnetic field operators and number operators do not commute. Therefore in an eigenstate of the number operator, we cannot assign a precise value to the fields. Coherent states are special superpositions of an infinite number of states with a varying number of photons that have a well definite value of the em field. – GiorgioP-DoomsdayClockIsAt-90 Sep 01 '22 at 05:02
  • In quantum mechanics classical variables become operators. This statement applies to classical variables that are functions of space-time (i.e. classical fields). That is, we would expect a classical field like the EM field to become "an operator-valued function of space-time". – Jagerber48 Sep 01 '22 at 21:56
  • Next, consider a quantum harmonic oscillator. I would say the electromagnetic field is analogous to the position operator $\hat{x}$ because $\hat{x} \propto \hat{a}^{\dagger} + \hat{a}$ and $\hat{E} \propto \hat{a}^{\dagger} + \hat{a}$. Suppose the quantum harmonic oscillator is in the state $|n\rangle$. We don't stop saying $\hat{x}$ is the position operator just because we can't assign a "precise value" to it. Likewise, just because we can't assign a "precise value" to $\hat{E}$ when the photon field is in state $|n\rangle$ doesn't mean we stop saying $\hat{E}$ is the electric field.. – Jagerber48 Sep 01 '22 at 21:58
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To understand the nature of light has been a challenge. It is true to assert that light is an electromagnetic wave in which electric fields and magnetic fields oscillate in phase with each other, dependent on time. To produce such a wave, initially there should be a source (possibly a charged particle that oscillates very fast) that creates a time dependent electric field or magnetic field. Now, a time dependent electric field induces magnetic field and a time dependent magnetic field induces an electric field. So, a self-sustaining process begins and as these fields oscillate, an electromagnetic wave propagates. Using Maxwell's equations, we can derive a wave equation ( a partial differential equation that depends on time and position) for the electromagnetic wave, and we can see that the wave velocity we obtain from this wave equation is exactly the speed of the light. So overall, light is a transverse electromagnetic wave, that has oscillating electric and magnetic fields.

The electromagnetic energy of light comes in discrete packages, energy is quantized, unlike other waves we know of (sound). Discoveries like the Ultraviolet Catastrophe or the Photoelectric effect experiment demonstrates that light shows particle behavior. So, we have come to the understanding light is constituted of packages or quanta of photons. We may say that the photon is the fundamental particle of light. The photon is a wave-particle meaning that it acts both as a wave and a particle.

The reason I talked about electromagnetic energy being discrete to clarify the concept of photon. I hope my answer was helpful.

Yigit Yazgan
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Photons are electromagnetic fields. That’s why they have an electric field.

Jagerber48
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  • This answer is misleading. In the presence of a single photon, the expectation value of the electric field is exactly zero. – GiorgioP-DoomsdayClockIsAt-90 Aug 31 '22 at 17:33
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    @GiorgioP Sure the expectation value is zero but the probability distribution has support away from zero. In other words $\langle E^2 \rangle$ is non zero. Quantum Photon states are 1:1 with quantum states of the electromagnetic field. Photons are electromagnetic fields. – Jagerber48 Aug 31 '22 at 19:26
  • Photons are not electromagnetic fields more than water molecules are water waves. The electromagnetic field is a macroscopic phenomenon emerging from microscopic photons. Typical properties of an electromagnetic field (amplitude, phase) are not properties of single photons. – GiorgioP-DoomsdayClockIsAt-90 Aug 31 '22 at 21:05
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    @GiorgioP Disagree. The photon state determines probability distributions for amplitude and phase of the EM field. – Jagerber48 Aug 31 '22 at 22:06
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    The water molecule : wave analogy is NOT valid for the relationship between photons and EM waves and it leads to many misconceptions. – Jagerber48 Aug 31 '22 at 22:07
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In my youth, we had a game involving a long stretch (maybe 12m) of heavy hemp rope (probably 2.5cm diameter) and two combatants. Whoever dropped their end of the rope, lost. Due to the inertia of the long rope, most actions arrived diluted enough at the other end to be harmless. The main strategy involved creating a kink or loop at your side that would travel the length of the rope comparatively undisturbed and then knock the rope out of the other's hand upon arrival.

How could it knock the rope out of the other's hand without having a hand? What was traveling along the rope was not a hand. It was a self-sustaining package of energy in the form of a combination of movement and displacement that traversed across the rope using its mechanical properties as a medium.

In a similar vein, a photon is not a charged particle but a self-sustaining package of energy in the form of a combination of electrical and magnetic energy that traverses vacuum using its quantum mechanical properties as a medium.

It can be created by movement of charged particles, but like a kink arriving in the rope, you don't get much of a clue regarding the origin other than its energy.

Spiritman
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Other anwers here are correct, but there is another perspective that might be helpful, which is to consider the electromagnetic field in terms of retarded potentials. One way of solving the electromagnetic equations is to use Jefimenko's Equations, which say that the electromagnetic field at a given point in spacetime can be calculated from the charge distribution (and its rates of change) on the past lightcone of that point.

The past lightcone of a point is all the points in space and time from which a light ray would arrive there at exactly that moment. If an observer is located there, it describes all the events they can see (with lightspeed signals) at that moment.

So if you consider some point just as a radio wave passes through it, you can work out the field there by drawing a lightcone from that point back to the beginning of the universe. All the charged particles whose paths cross that lightcone contribute. We simply add up all the contributions to find the field. For our radio wave, when you trace the lightcone back into the past, we find it passes through the antenna that emitted the radio wave - the moving charges in the wire are the source of the electromagnetic field.

So, although there are no charged particles close (in conventional terms) to the point where the radio wave is passing, the charged particles are there on the past light cone. Since in relativity the distances shrink to zero as you approach the speed of light, there is a sense in which the charges are in fact extremely close!

Charge on past lightcone

This next bit goes well beyond your question, but I should probably mention that there is a spot of controversy around the idea retarded potentials, which is to do with where the time asymmetry comes from. Maxwell's equations are time-reversible, so if we can derive the potential at a point from the past lightcone, then we ought to be able to derive it from the future lightcone using the advanced potentials in exactly the same way. So the question is, why do we not see the future? Why does the effect go only one way?

The answer seems to be to do with a point I skipped over earlier, which is that Jefimenko's equations also have a 'boundary term', describing the field that comes from 'infinity'. You have to do the integration over a defined region, and the result is the sum of what is going on inside the region (moving charges and so on), plus what enters the region from outside (the boundary conditions). Our observations of the universe indicate that the boundary term contributed by the start of the universe (the past) is zero, and the boundary term contributed by the end of the universe (the future) is whatever it needs to be to cancel all the stuff going on on the future lightcone. It's still a tiny bit mysterious. Richard Feynman and John Wheeler did some work trying to figure out why, but I don't think Feynman in particular was ever really satisfied with the explanation.

Miyase
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Nullius in Verba
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A wave is harmonic motion moving in space, and harmonic motion have potential and kinrtic energy. These potential and kinetic energy of a wave comes from medium through it moving. There is no space void of matter but that matter is comsidered as no-matter because it is distribited through out space.

Imagine an electric field or a magnetic field stored without matter. A photon is disturbance of electric and magnetic field of medium and a disturbance doesn't carry matter so it doesn't need its own charge.

Neil Libertine
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