# How electric field generates a sin/cos wave [closed]

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.

• 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). May 16 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 at 11:11
• 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 May 16 at 11:16

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
• @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. May 16 at 13:06