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I am very frustrated with the fact that the E and B field of an EM wave are in phase in the far-field case.

Before I get into this question I am letting you know that I already read this link related to my question, but it didn't really help me so much.

First, I know that an EM wave propagates as shown in the GIF below.

enter image description here

But, how are E and B field in phase? According to Maxwell's equation #3 and 4, we know that the change of E field induces B field, and vice versa. I also attached these equations below. I deleted the J term since we are talking about the far-filed analysis which lets us ignore the B field induced by the current flowing, J, in the wire.

enter image description here

Let's assume that the wave propagates along the x-axis. According to Maxwell, when there is a maximum E field occurs (dE(t)/dt = 0) at a given point, x0, a minimum B field (B=0) should occur at the point. This should result in 90° phase difference between E and B field.

I know that we can calculate and prove that the EM wave propagates at a speed of light from Maxwell's equation #3 and 4 as shown below (source).

enter image description here

But the problem is that I don't know how to solve partial derivative equations (PDEs), and also I want some visualizations to understand this phenomenon better.

Is there any easy way to visualize why it is like this without solving PDEs?

Thank you.

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  • \$\begingroup\$ "Let's assume that the wave propagates along the x-axis. According to Maxwell, when there is a maximum E filed occurs (dE(t)/dt = 0) at a given point, x0, a minimum B field (B=0) should occur at the point. This should result in 90° phase difference between E and B field." A moment when both waveforms have a slope of zero at the same time doesn't sound like 90 degrees out of phase to me. If anything that's either in phase or 180 degrees out of phase. \$\endgroup\$
    – DKNguyen
    Commented Mar 23, 2020 at 19:44
  • \$\begingroup\$ physics.stackexchange.com/questions/181277/… \$\endgroup\$ Commented Mar 24, 2020 at 1:00
  • \$\begingroup\$ Hi, thank you for your answer to my question. In my example, when the slope of E is 0 and the magnitude of B is 0 (not the slope of it). The slope of B would be higher or lower than 0 which means that E and B are out of phase, but they have to be in phase according to Maxwell... \$\endgroup\$ Commented Mar 25, 2020 at 5:25
  • \$\begingroup\$ Sredni Vashtar, thank you for your link. I looked at it and it helped me understanding my issue a bit better. Though, I still don't understand fully. \$\endgroup\$ Commented Mar 25, 2020 at 5:36

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The impedance of free space is not a complex impedance quantity; it is resistive with a value of \$120\pi\$ or about 377 ohms. This means that the magnitude ratio of E to B is 377:1 and, each are linked in phase to one another due to that non-complex resistance. If you expect a full derivation of Maxwell's equations on this site, you are expecting too much - try stack exchange physics.

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  • \$\begingroup\$ I will go check that out. Thank you! \$\endgroup\$ Commented Mar 25, 2020 at 5:26
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You imagined that 90 degrees phase shift between E and B is a must. As a proof you used the time derivative in the basic equations.

It's true that d(sinX)/dt is cosX, but the differentiated B in the field equation is not claimed to be proportional with E, the equation gives the curl of E. And the same for the other equation.

University level field theory books and also this article https://en.wikipedia.org/wiki/Electromagnetic_wave_equation

show how the basic equations lead to the wave equation. The wave equation allows in-phase E and B.

Historically the case is extremely interesting. Maxwell did his groundbreaking work in his head. It was based on already known practically verified electricity laws and on some mathematical speculations of things which were then unmeasurable. The only speculated thing was the equation of the curl of B. Maxwell only thought that things could be in that way because it seemed to be useful and built a bridge over some unsolved problems which became contradictions with simpler assumptions. It took a long time before experiments (by Hertz) proved Maxwell's speculations were right.

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  • \$\begingroup\$ I'm not familiar with "rotor"; do you mean curl? \$\endgroup\$ Commented Dec 10, 2023 at 18:33
  • \$\begingroup\$ @TimWilliams Sorry, the English term is really curl! I'll fixed it to the answer. \$\endgroup\$
    – unawriter
    Commented Dec 10, 2023 at 21:51
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someone wise said "Whether Efield or Hfield, its all caused by electric charges"

When you have changing electric fields, the (very small time delays along a wire) might result in what philosophers have chosen to label magnetically-induced voltages and currents.

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