I've been able to confuse myself about the shape (i.e. surface ripple/bend/tilt) of wavefronts for transverse modes in an electromagnetic waveguide, like a rectangular cross section waveguide with perfectly conducting walls, or an optical fiber. Are they flat? Are they tilted? Are they rippled?

I understand so far, that the propagation constant $\beta$ will be influenced by both material and waveguide dispersion, as in $\beta = \sqrt{|\vec{k}|^2 - k_x^2 - k_y^2} = \sqrt{(\frac{\omega}{c}\cdot n(\omega))^2 - k_x^2 - k_y^2}$ .

So the wave vector has longitudinal and transverse components, and we have $|\vec{k}| \neq k_z$ and $|\vec{k}| = \frac{2 \pi n}{\lambda}$ . Thus the more transverse components the wave-vector of a mode has, the lower its phase propagation constant $\beta$, and the faster its phase velocity $v_\text{ph} = \frac{\omega}{\beta}$.

So a good way to experimentally excite transverse modes is, I suppose, shine in a beam at an angle at the input open/coupling end of the waveguide; but I guess this would in practice excite some funny superposition of transverse modes. Let's assume we somehow manage to just excite exactly one transverse mode. Typically the transverse modes have a field distribution with several nodes and antinodes across the cross section of the waveguide.

The mode field evolves down the wave-guide in a "self-consistent" fashion, such that its spatial distribution is conserved.

What specifically confuses me now is: If the field moves down the waveguide, and the wave vector can be defined as the vector standing perpendicular on the wavefront (I think it can), what shape does the wavefront have? I think it can't be flat, as that would mean there wouldn't be any transverse components of the wavevector, as far as I can see. Or is there some kind of symmetrical wavevector distribution? How can I formally find it? Thanks for your help in advance!

  • $\begingroup$ The wavefront of a given TE, TM or TEM mode is indeed flat, if what you mean by a wavefront is a surface of constant phase. The phase is constant on planes perpendicular to the direction of propagation, except for possible jumps of $\pm 180^\circ$ due to nodes in the wave "envelope". I know this is true for rectangular waveguides, I also think it holds for other infinitely long waveguides that are uniform in shape. $\endgroup$ – Puk Jul 22 at 20:42
  • $\begingroup$ Yes, by wavefront I understand a surface of constant phase. I didn't quite get the node-envelope jump thing. Maybe you could expand on this in an answer? What you haven't touched upon is: Where are the transverse wave-vector components that every transverse mode comes with, if the wavefronts are flat, as you say, and the wavevector is always perpendicular to the wavefront? Thanks. $\endgroup$ – Georg E. Jul 22 at 21:05
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    $\begingroup$ Don't worry about what I said about nodes and the phase jumps, I was just being overly careful not to say anything incorrect. I'm not sure how to answer your other question because I don't know what you mean by the wave vector $\vec{k}$. There isn't a single clear direction you can assign to it in a waveguide. It is instructive to think of the overall TE or TM wave as a superposition of TEM waves with different wave vectors $\vec{k}$, but that's about it. We can speak of a "propagation vector" $\vec{\beta}$ which always points along the waveguide, normal to the wavefronts. $\endgroup$ – Puk Jul 22 at 21:29
  • $\begingroup$ I was always (naively) thinking that a wave-vector can be associated with a periodic wave phenomenon, especially in a waveguide, where the concept even seems crucial to me to qualitatively understand waveguide dispersion. I think you just have a way deeper understanding of the physics than I... why for example do TEM modes have a definite $\vec{k}$ vector, while TE or TM modes don't? $\endgroup$ – Georg E. Jul 22 at 22:10
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    $\begingroup$ The "wave vector" you are after is $\vec{\beta}$. Since you know it points along the waveguide, you can treat it as a scalar $\beta$, called the propagation constant. For TEM, $\beta=k$, so it's the same as the wave vector you know and love. The fact that $\beta<k$ for TE and TM can be qualitatively understood from the fact that you can think of the wave as a superposition of TEM waves with $\vec{k}$ not strictly along the waveguide. The component of $\vec{k}$ along the waveguide gives you $\beta$ while the other components give you the transverse variation of the fields on a wavefront. $\endgroup$ – Puk Jul 22 at 22:23

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