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It is known that a generic harmonic electric field which propagages in free space has an elliptical polarization (if you want to read the proof, look at this, slides 72 - 99).

My question is: what about the polarization of E and H fields in a waveguide or a transmission line? I have always read about elliptical, circular and linear polarization for electromagnetic waves in free space, but I have never heard/found something referred to guided waves. So, I'll say that waves in waveguides are simply linearly polarized. But:

  • I do not know if this is true. I have never read also the sentence "Waves in waveguides have linear polarization".

  • I do not understand why we cannot have elliptical polarization in waveguides. As you may see from the slides is the consequence of the harmonic behaviour of E and H fields, which is described by the wave equations. In theory, it is true also for waveguides...

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Let's directly answer your questions, first, then address your misconception:

So, I'll say that waves in waveguides are simply linearly polarized

That can't be the case. How would you, in a round waveguide, like a circular tube waveguide or, very prominently, a coax cable, satisfy any boundary conditions with a linearly polarized wave (in cartesian coordinates, at least).

You can very well have a waveguide with e.g. a circular polarization; for example, when you open a satellite dish feed (satellite TV is circularly polarized), there's parts of the waveguide system that are linearly polarized (because that's easy to pick up with a monopole) and parts that are circularly polarized, and there's the transition between the two.

So, nothing to see here: wrong claim, can't be justified.


Polarization is something that you define from the direction of the E-field.

In free-space propagation, that's easy: at any given time and point, a plane wavefront will have exactly one direction of the E-field.

It doesn't work if you have a mode propagating in your waveguide that has more than one half-wave width. Higher-order modes usually can't be assigned any polarization at all; not even an elliptical one. The concept polarization doesn't apply!

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  • \$\begingroup\$ Thank for your answer. It is more clear now, but I do not understand what you mean with the sentence "In free-space propagation, that's easy: at any given time and point, a plane wavefront will have exactly one direction of the E-field". In case of Elliptical polarization, the direction of E rotates in time, so it changes, so what do you mean with that statement? Moreover, I do not understand why this concept does not work for modes woth more than one half-wave width. \$\endgroup\$ – Kinka-Byo Apr 13 at 12:12
  • \$\begingroup\$ read the sentence twice: "at any given time and point". \$\endgroup\$ – Marcus Müller Apr 13 at 12:13
  • \$\begingroup\$ For "modes with more than one half-wave width" do you mean that the waveguide dimensions is greater than lambda/2? \$\endgroup\$ – Kinka-Byo Apr 13 at 12:39
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    \$\begingroup\$ Maybe I have got it: it is basically due to the fact that E field is not uniform in each plane wavefront, but it has a wave behaviour which changes its direction along that waveform. So, in few words, can we say it is due to the fact that in waveguides we have non uniform plane waves? \$\endgroup\$ – Kinka-Byo Apr 14 at 2:48

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