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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.

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  • $\begingroup$ 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). $\endgroup$
    – AXensen
    May 16 at 11:06
  • $\begingroup$ Hi. Do you have some recommendation for the book ? the smaller the better and the more focus on my questions, the better. Thanks. $\endgroup$
    – Matt
    May 16 at 11:11
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    $\begingroup$ 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 $\endgroup$
    – AXensen
    May 16 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

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  • $\begingroup$ 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. $\endgroup$
    – Nadav Har'El
    May 16 at 11:54
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    $\begingroup$ 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. $\endgroup$
    – Nadav Har'El
    May 16 at 12:38
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    $\begingroup$ @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. $\endgroup$
    – AXensen
    May 16 at 13:06
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    $\begingroup$ @Matt 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. $\endgroup$
    – AXensen
    May 16 at 13:22
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    $\begingroup$ 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. $\endgroup$
    – naturallyInconsistent
    May 16 at 14:54

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