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If we got an oscilating electron up and down, it will definately produce changing electric field. I won't go into magnetic field topic. Let's only restrict the question to electric field.

Since we got a changing electric field, we know that it produces sin or cos wave. My question relates to how it produces the sin or cos wave. The reason I have doubts is the way charge A produces a field for charge B is A sends virtual photons to B. at t=1, A sent off photons. at t=1.1, it again sent off photons from new position. The key is: "new position". but if you look at the sin wave of changing electric field, at some t values, the graph shows 0 for E field magnitude. Definately, it shouldn't be 0. I might be asking it in a wrong way, but I will update the question if something is not clear.

Matt
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  • I think the only decent answer to this question is that you should just use an electrodynamics textbook. You should read sections on electromagnetic waves, and Larmor radiation or dipole radiation (the simplest examples of how electromagnetic waves are initially produced). A lot of your questions have really simple answers that are given in E+M textbooks (esp your issue with zero, why the solution is sines and cosines). – AXensen May 16 '23 at 11:06
  • Hi. Do you have some recommendation for the book ? the smaller the better and the more focus on my questions, the better. Thanks. – Matt May 16 '23 at 11:11
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    Griffiths "Introduction to electrodynamics" (easy to find as a PDF online) sections 9.1 and 9.2 explain electromagnetic waves. Sections 11.1 and 11.2 explain the two ways of producing them that I mentioned. But if you don't like the descriptions there I can really assure you that this will be described with very similar chapter titles in every e+m textbook – AXensen May 16 '23 at 11:16

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I think your question mixes the classical view of electromagnetism (electric field, etc.), with quantum field theory (virtual photons, etc.).

If you want to know why an accelerating electron moving back and forth generates a moving wave, the answer is best understood from classical electromagnetism, and by the way, involves the magnetic field which you tried to ignore - the accelerating electron generates a magnetic field, a changing magnetic field, which in turn generates a changing electric field which generates a changing magnetic field which... ends up a moving wave in both electric and magnetic fields - this is the electromagnetic wave. You can find a gazillion books and online resource on this topic, a random one I found in Google is: http://labman.phys.utk.edu/phys222core/modules/m6/production_of_em_waves.html

But the question asks how virtual photons, which is the realm of quantum field theory. I guess it hinges on why in the classical limit (high enough intensity - many photons, not just one), you can measure a zero of the the electric field. This zero doesn't come from a single virtual photon, but from the superposition of many virtual photons. I'm not familiar enough with the theory to give a good answer, but it has been asked here before - e.g., see Virtual photon description of $B$ and $E$ fields

Nadav Har'El
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  • Oh, and I forgot to mention, that E=0 doesn't mean the electromagnetic field is zero - where E is zero, B is non-zero. Again, you can't ignore the B, it's part of the electromagnetic field. – Nadav Har'El May 16 '23 at 11:54
  • Well, on the graph, an accelerating charge shows EM wave. but are you saying that on the graph, the Electric field only itself is not a wave ? If yes, thats what I'm asking why. It's true it might not be EM wave, but it is still a wave. I ignored Magnetic field to only focus on Electric field graph. – Matt May 16 '23 at 11:54
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    The electric field is a vector field, not a scalar field, so I'm not sure what you mean by a "graph" of it, I guess a graph of the magnitude of the field. The electric field can definitely be 0 at a certain position in space at a certain time. But at the same position and time, B will not be 0. Anyway, if you want to understand how EM works, I think that looking at QED (quantum electrodynamic) and its virtual photons will only confuse you more. – Nadav Har'El May 16 '23 at 12:38
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    @NadavHar'El "at the same position and time, B will not be 0" - this is exactly incorrect. In free space the zeros for B and E are at the same place $E=E_0\cos(\omega t-kz)\hat{x}$, $B=(E_0/c)\cos(\omega t-kz)\hat{y}$ (wave propagating in the z direction). The wave "persists despite this zero" not because $B$ is nonzero, but because $dB_y/dz$, $dB_y/dt$, $dE_x/dt$, and $dE_x/dz$ are not zero. – AXensen May 16 '23 at 13:06
  • what I'm trying to understand is the way photons move and travel when E changes, do the photon movement have a waveform ? or we say it's a wave because magnitude of E itself is a wave?. I believe Magnitude of E is a wave, not the photon movement. there's no way photons would move from A to B with a sinx function movement. – Matt May 16 '23 at 13:12
  • Note that I think my above reply like that because in a rope, when you oscilate, you actually see its movement as a sinx function, but for EM, where photons move, no chance they move as a wave. Even when E changes and new photons are emitted from a new position, it shouldn't produce anything like wave. Is my logic correct ? – Matt May 16 '23 at 13:12
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    @Matt https://bohr.physics.berkeley.edu/classes/221/2021/221.html notes 40 and 41 on this website explain what a photon is and how it relates to the E and B field. In theory this should perfectly answer your questions, but this is an advanced topic. These notes are meant as the end of the second semester of quantum mechanics for PhD students. I think it might be understandable on its own, but it will take some effort. Unfortunately it might be a case of - you need a more solid foundation of E+M and quantum mechanics before this can be properly understood. – AXensen May 16 '23 at 13:22
  • Thanks. I'm not asking for deep understanding. Just need to clarify what wave means in EM wave. Here I repeat my logic. - "in my opinion, the movement(whatever propagates - whether virtual photons or whatever) in the vacuum because of E and B changing, those don't have a waveform. What we mean that EM is wave is its magnitudes that change have waveform, but not the movement of virtual photons or whatever." Could this be correct ? – Matt May 16 '23 at 13:28
  • @Matt The "waveform" is seen in $E_x$ and $B_y$. If you want to know why they behave like that, read those textbook sections and come to an understanding of what classical EM waves are. If you want to discuss photons instead, you should sit down and read what photons are by reading a derivation of quantum electrodynamics, and no amount of qualitative statements via text will help with this. And finally, there are other kinds of waves than scalar quantities defined on a line. There are vectors in 3d, tensors in 4d (gravitational waves), etc. Not every wave has a clearly identifiable "waveform" – AXensen May 16 '23 at 14:21
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    Actually, you are all just caught up in bad pedagogy. When textbooks cover the electromagnetic wave, they tend to seek the simplicity of linearly polarised light, and in that case, you have sines and cosines, and the infinite plane waves of E and B fields will have periodic planes of zeroes, giving the illusion that we could have isolated each photon between the planes. This is an illusion, because if you just let the electrons to oscillate circularly, making circularly polarised light, there will be no plane of zeroes, and the photon has to be completely delocalised throughout the whole box. – naturallyInconsistent May 16 '23 at 14:54
  • @naturallyInconsistent what would be the right way to understand this ? I asked this question to make my get started from something, but seems nobody gave an yes or no to it, but I will repeat one more time. "in my opinion, the movement(whatever propagates - whether virtual photons or whatever) in the vacuum because of E and B changing, those don't have a waveform. What we mean that EM is wave is its magnitudes that change have waveform, but not the movement of virtual photons or whatever." Could this be correct ? – Matt May 16 '23 at 18:31
  • I mean, the way google explains it is on the x-direction, E moves as a sin wave, but if you oscilate electron up and down, I'm not sure how E itself would move through space. Can't believe classical physics can't explain this. My explanation is that it's not E that moves through space. The graph we see http://electron9.phys.utk.edu/optics421/modules/m1/emwaves.htm just should be showing that E changes its magnitudes in the form of waves., but it's not moving as a wave. – Matt May 16 '23 at 18:36
  • I already gave the viewpoint that is least misleading. Simply consider electrons moving in circles. That makes E and B fields that are moving in circles, both of them having both of sine and cosine parts. Then the E and B fields have constant magnitude throughout the volume you are studying, and are perfectly nice waves. Quantising them gives you photons that are completely delocalised. – naturallyInconsistent May 17 '23 at 01:48