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If electromagnetic waves cause disturbances in the Electric Field… what “is” in this E Field which photons Interact with?

I ask because in Vacuum, there are no electrons to excite. So what is “it” that's adding up in the E Field as a disturbance in wave propagation?

Rain
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1 Answers1

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You are confused about electromagnetism. The thing is, there generaly (if we leave quantum mechanics out of it) is only electromagnetism, which can manifest itself as magnetic field, electric field, or usualy electromagnetic field depending on the reference frame you're observing it from.

What you need to know is that moving electric charges create a magnetic field that is propagating away from the charges at the speed of light.

Static electric charges create an electric field that is propagating away from them at the speed of light.

A changing magnetic field (accelerating charges) create an electric field around them.

Photons don't interact with electric field (under normal circumstances). They interact with charged particles like protons or electrons by exerting force on them.

You can Picture empty space as a space where there are many different fields which are 0 on average. When a photon, which is just an excitation in the electromagnetic field travels trough that space it is a wave of electrical and magnetic potential that is traveling trough the empty space. When it's at a certain point it increases a value of electric field in that point. Which you can actually measure. Now if you would to put an electron in that same location, the electric field (caused by photons) would interact with it, pushing it in some direction.

MaDrung
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  • So you're saying a Field is "nothing" but a representation of what we observe. Because if there are no interactions with any matter (emptiness), then it is basically only a traveling photon. I'm confused about the E Field because it increases and decreases in magnitude due to disturbances, and I'm not seeing what is "in" in that magnitude. No explanation. – Rain Feb 10 '17 at 11:28
  • Another question - In air, at radio/microwave frequencies, I was told by a physicist in university that they do not interact at all with the electrons in the air because they are bonded and the Photons at this regime are too low in energy. So, basically, a wave propagates with no electron interaction or any other interaction. Just nothingness and travelling photons? – Rain Feb 10 '17 at 11:31
  • Quantum mechanics explains why this is so. Bassicaly electrons are only stable at certain energy levels. So if a photon doesn't have a higher energy than that, it doesn't interact with it. But a more accurate Picture would probably be that instead of thinking about a photon, you think of a moving electromagnetic field. The photon interacts with every atom, accelerating them. The atoms shake with their own frequency depending on the medium they're in. This creates it's own electromagnetic field that is then traveling on. But because it lags behind the original field....... – MaDrung Feb 10 '17 at 11:38
  • it adds up with the original wave in such a way that it partialy reduces it's value. When this happens, we could observe that the front of the wave is actualy being reduced to 0. Which we could then interpret as light slowing down, when it's actualy always traveling at the speed of light, just that the interaction with the medium act's to counter it's fron wave. We get the illusion of light traveling slower. – MaDrung Feb 10 '17 at 11:40
  • Thank you. It's become clearer in regards to electrons. I have other photon interaction questions, but will post as a different question to not confuse the readers. You were very helpful. – Rain Feb 10 '17 at 12:02
  • Well in general. You don't need electrons to have electric fields. But charges are the ones whos movements create outwards propagating electric and with it magnetic fields. It doesn't need a medium to propagate trough. But if there is a medium with charges there it interacts with it in different ways. You just don't always notice it if you don't go deep enough. – MaDrung Feb 10 '17 at 12:04
  • So if no electrons are required to propagate a wave, is it 'correct' to say that the disturbance in the EM field is driven by only Photons? So without photons, no EM Field would allow for propagation? – Rain Feb 10 '17 at 12:17
  • It's not that they are driven by photons, they are photons. :) But the problem is because of wave-particle duality and complexity of the problem that is causing the confusion. Even when we say that photons are nothing else than electromagnetic waves, it's actualy wrong, because it depends on the experiment. It's a difficult subject. Every electrical exchange of energy is driven by virtual photons. – MaDrung Feb 10 '17 at 12:20
  • But if EM fields 'are' Photons, are we saying the universe is a universe of photons in every point in the fields... ? This is indeed a very different way to explain it which no book I read described it that way. It's usually a "it is what it is".. and therefore... we don't "really" understand 'photons'... but physics just quantifies its observational behaviors – Rain Feb 10 '17 at 12:24
  • Photons is a disturbance in the em field that is moving in the space. But why you are confused is quite logical, because when a field exerts some force in a particle, it is said that a virtual photon is used to transmit that energy from the object who created the electric field. The virtual photons travel in every direction around an electron, but if they do not hit any particle for them to transmit the energy to, those photons get returned to the source particle. It gets much more confusing than that, which is why I believe there is no easy answer to this question. – MaDrung Feb 10 '17 at 12:30
  • That a photon existed at all that was used to transfer the energy is known only after the fact. It's a mathematicaly sound predictive model of many things, but is not really intuitive just as most of quantum mechanics isn't. The reason being it probably doesn't describe the actual mechanics of the universe, but is good at predicting a lot of it mathematicaly and statisticaly. – MaDrung Feb 10 '17 at 12:35
  • I should not get started on virtual photons as well. But it further explains my confusion of the term photon. Because if they start calling it virtual. It really does mean that photons are just a name given to a "force" in EM fields. And further explains your answer, that photons are the disturbances themselves. It's the force. (As far as I see understand it anyways) – Rain Feb 10 '17 at 12:41
  • Couldn't thank you enough for your patience in explaining what I didn't know. I can't vote because I'm still new to this forum unfortunately. But I have voted anyways. – Rain Feb 10 '17 at 12:42