We know that in free space, a propagating electromagnetic wave is always transverse. However, along a dielectric waveguide, the propagating wave can have longitudinal components. The exist of the longitudinal component makes the group index of refraction not equal to the phase group of refraction of the waveguide.

My questions is can one by using some combinations of propagating waves from any direction (even from the perpendicular direction to the waveguide) to make the field purely transverse or longitudinal in presence of a cylindrical fiber waveguide?

I am highlighting these two words here, and define the waveguide/optical fiber axis direction as the longitudinal direction and the plane perpendicular to the waveguide axis is the transverse plane. So, if a combined field oscillating along the fiber axis, I treat this as a longitudinal local field. Without this definition, one would get confused if we allow the light incident from any directions.

To be clear, meanwhile, I am talking about the electromagnetic field at least on one/periodical crossing section(s) of the propagating field or at least on a line, not just one or two specific/trivial points in space or the entire space. Thanks.

  • $\begingroup$ There are all sorts of longitudinal waves in space. Are you meaning that the only waves that do not interact with the background plasma must be transverse "free" modes of the system? $\endgroup$ Commented Oct 7, 2015 at 15:15
  • $\begingroup$ Thanks for asking. I just defined the longitudinal direction in the question. For example, if you shine a light perpendicular to the waveguide axis with a linear polarization oscillating parallel to the fiber axis, I would treat the local field as longitudinal for my purpose. Let me know if there is any further confusions. $\endgroup$ Commented Oct 7, 2015 at 16:27
  • $\begingroup$ I would look up articles on magnetosonic or fast mode waves (or kinetic Alfvén waves) in the auroral acceleration region. Bob Lysak wrote some papers on how that region of space can act like a wave guide. I do not recall off hand whether there were longitudinal waves involved, but it is worth a look. By the way, it is relatively easy to generate a longitudinal electrostatic wave without a wave guide in a plasma (e.g., ion-acoustic waves, lower hybrid waves, etc.). $\endgroup$ Commented Oct 8, 2015 at 10:39

2 Answers 2


Pure longitudinal modes, where either the electric or the magnetic field alone are wholly longitudinal (not needfully both vectors at once) are ruled out by Poynting's theorem. Such a wave could needfully propagate power only transversely to the guiding direction (by Poynting's theorem).

Pure TEM modes (i.e. no longitudinal component) with compact support can only exist on waveguides with two closed Jordan curves, one needfully inside the other, at each cross section. This means we have to have conductors for the electric field lines to end on, which means that they are practically ruled out for optical waveguides (although you could probably come close at longer wavelengths using metal coated waveguide technology). TEM modes of course exist in co-axial cables and microwave striplines at microwave frequencies.

The words with compact support are important, for a plane EM wave is of course a purely TEM mode.

To understand briefly why the TEM situation arises, it can be shown that TEM modes exist where the transverse field configurations of the electric and magnetic fields are exactly the same at each cross section as for electro- and magneto-static fields; see, for example, my answer here. Conversely, one can show a TEM field needfully has this analogy with static fields. But this means that if the field has compact support, then there is a zero potential equipotential formed by some closed Jordan curve that bounds the support region, i.e. a hollow conductor. The only solution to Laplace's equation, unless there are singularities within the boundary, with this behavior is a constant potential within the boundary, i.e. no electric of magnetic fields. So this means that there must be a singularity within or, equivalently, a second equipotential Jordan curve within. So we are talking about a strip line, with two different equipotential plates, or a generalized co-axial cable.

  • $\begingroup$ And so, is your answer to the OP's question "Yes" or "No"? I'd rather an answer helps me understand durably, rather than understand briefly. Also I think constructs like "Conversely, one can show a TEM field needfully has this analogy with static fields." are just plain hard to understand and possibly gratuitous. What are you saying? Is there not some way that this can be written in plain English without the frills and flowery allocutions? How can an electromagnetic field be needful? A good SE answer should be accessible to many/most, not simply demonstrate that you know the answer. $\endgroup$
    – uhoh
    Commented Aug 14, 2017 at 8:28
  • $\begingroup$ Could you add some clarifications, perhaps something like for each cross section; Jordan curve $\rightarrow$ simple closed curve, and compact support $\rightarrow$ goes to zero outside of some finite boundary? This reads more like a personal blog than a helpful SE answer for the OP and future readers. $\endgroup$
    – uhoh
    Commented Aug 14, 2017 at 8:38
  • $\begingroup$ @unoh: the converse holds, i.e. the static field condition is necessary and sufficient. Feel free to edit as you see fit, but be careful of the topological sense. I think "support" is widely enough understood and readily looked up; it replaces a whole clunky phrase and is commonplace in EM theory. $\endgroup$ Commented Aug 14, 2017 at 10:22
  • $\begingroup$ I think there is something really important to learn here, my frustration comes from feeling that it's behind a few meters of glass and hard to get a sufficiently good view of it to understand. Plain speaking and simplification to get the point across in informal settings has never been a taboo in physics, nor in mathematics, but I understand also not wanting to say something that's "slightly wrong" when the correct language is available. Still, should the OP take away from this a "Yes" or "No" to the question as asked? $\endgroup$
    – uhoh
    Commented Aug 14, 2017 at 10:32
  • $\begingroup$ OK, I've just asked Need help understanding another answer on TEM coaxial cable modes. $\endgroup$
    – uhoh
    Commented Aug 14, 2017 at 11:03

If your question is whether waveguides can support TEM (with transverse electric and magnetic field) waves, then the answer is generally no. There is a fundamental theorem that for a TEM wave the waveguide must be open. An example waveguide which supports TEM waves is the parallel plate waveguide. Closed waveguides such a circular or rectangular do not support TEM waves. [see Jin Au Kong "Electromagnetic Wave Theory", 1986, p. 196].

Note also since the waveguide modes are orthogonal, you cannot expect full cancellation of the transverse components (in the full cross section) by superposition of modes unless the modes have no transverse components in the first place.

This is generally not true for cancellation only in a part of the cross section. There probably one can achieve cancellation of the transverse components (I am mot fully sure there), but one generally would need infinitely many modes with excitation of a certain phase and amplitude.

  • 1
    $\begingroup$ AFAIK, a waveguide having two layers of boundaries seem to support TEM modes as well. And indeed, I am asking if a part of the field can be purely transverse or longitudinal, not the whole space of course. I feel there may be some way to do it, but not sure... Thanks for responding! $\endgroup$ Commented Oct 6, 2015 at 16:30
  • $\begingroup$ "There is a fundamental theorem that for a TEM wave the waveguide must be open." Does this theorem have a name? Is it possible to refer to it somehow with a link? I'm not challenging your conclusion, it's just that if answers started popping up everywhere invoking unnamed fundamental theorems, the site could quickly descend into chaos. Answers of the form "There exists a theorem such that..." are just not really useful. I would like to read about this as-yet unspecified fundamental theorem in order to be able to verify your answer and vote on it. Thanks! $\endgroup$
    – uhoh
    Commented Aug 14, 2017 at 6:50
  • $\begingroup$ Can you reconcile your answer with the sentence in Wikipedia Coaxial Cable article: In radio-frequency applications up to a few gigahertz, the wave propagates primarily in the transverse electric magnetic (TEM) mode, which means that the electric and magnetic fields are both perpendicular to the direction of propagation. This seems to be describing TEM in a closed waveguide, am I mis-understand ing "for a TEM wave the waveguide must be open" Maybe you mean two conductors versus one? Coax is closed; fields outside are zero. $\endgroup$
    – uhoh
    Commented Aug 14, 2017 at 6:59
  • $\begingroup$ @uhoh: I added a reference to the theorem. As opposed to the circular waveguide, the coaxial cable seems to support TEM waves. Perhaps the term "closed waveguide" is too imprecise and needs a proper definition. Probably you are right with the two conductors vs. one, but unfortunately I cannot look the reference up anymore to check what was the precise argument in the theorem was (I took the reference from my notes but do not have the book anymore). I hope it will help you and clarify the issue. $\endgroup$
    – Andreas H.
    Commented Aug 14, 2017 at 8:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.