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I have read that TM (transverse magnetic) and TE (transverse electric) modes of the electromagnetic field of light travelling in a glass optic fiber cable coexist. I want to know the precise mathematical justification for this (not physical intuition). Is this a banal observation or a deeper fact? Does it follow immediately from the refractive indices and Snell's law? I just do not see how at the moment. Can someone explain it?

EDIT: People asked for clarifications. I talk about light propagating in a multi-mode optical fiber with total internal reflection, assuming perfect rotational symmetry.

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  • $\begingroup$ Are you talking about electromagnetic waves? Am I missing something? Every electromagnetic wave has to have nonzero electric field and magnetic field as well - they are pumping the energy to the other just like the harmonic oscillator pumps the energy from the kinetic energy to the potential and back. There can't exist any periodic solutions that would have $B=0$ or that would have $E=0$ everywhere. It's trivial to see from Maxwell's equations. Nothing qualitative changes about these matters in a dielectric such as glass. $\endgroup$
    – Luboš Motl
    Commented May 26, 2016 at 17:57
  • $\begingroup$ Did you try to find a book that explains the em modes in fibers? $\endgroup$
    – CuriousOne
    Commented May 26, 2016 at 18:04
  • $\begingroup$ @LubošMotl If you mentioned that they (the electric and the magnetic field component of light)are pumping the energy to the other, which of this or this sketch are illustrating your words? $\endgroup$
    – HolgerFiedler
    Commented May 26, 2016 at 19:32
  • $\begingroup$ Yes, this is how an electromagnetic wave always basically looks. $\endgroup$
    – Luboš Motl
    Commented May 26, 2016 at 19:33
  • $\begingroup$ Chris, there are two kinds of fibers: monomodal and multimodal, you should specify which one you're talking about. Also a precise reference would be better than "I have read". $\endgroup$
    – L. Levrel
    Commented May 26, 2016 at 19:37

2 Answers 2

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Let's consider an optical fiber as a rotationally symmetric structure consisting of a core with a slightly higher refractive index as the surrounding cladding. We want to find the solutions of Maxwell's equations for the electromagnetic field that can propagate along this structure. First one would express Maxwell's equations in cyllindrical coordinates and then solve the resulting set of differential equations with the separation of variables technique. The solutions would then consist of a product of Bessel functions for the radial coordinate, harmonic functions for the azimuthal coordinate and an exponential function for the propagation direction. Since all the field components are related to each other via Maxwell's equations, one usually considers only one of these components for the electric and magnetic fields, respectively, namely the $z$-component, which point in the direction of propagation.

Next, one would impose the boundary conditions between the core and the cladding to ensure that all the tangential components of the fields are continuous across this boundary. This fixes all but one parameter (apart from the modal index): the propagation constant. To find the propagation constant one composes a charateristic equation, which needs to be solved numerically for all the different values of the modal index, because it is rather complicated.

When one sets the modal index to zero one finds that the charateristic equation factors into two parts. So now either one of the two factors can become zero independent of the other. This is a specially situation that only applies for the zero modal index. In this case it turns out that the two solutions, respectively, represents the TM and TE solutions. In other words, these solutions either do not have an electric field component in the $z$-direction (it is purely transverse, hence TE) or they do not have a magnetic field component in the $z$-direction (hence TM). For all other values of the mode index both the electric and the magnetic fields have $z$-components. They are therefore called hybrid modes.

It is important to know that the TE and TM modes are not the modes that propagate in single mode fiber. The mode that propragates in single mode fibre is a hybrid mode HE$_{11}$.

This is an explanation in words. The mathematics is quite involved and for that I'd suggest you consult a proper text book on the topic.

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Maxwell's equations are linear. As a consequence, if some TE modes and some TM modes can propagate in a given guide (not necessarily a fiber), then their sum can as well.

Note that optic fibers, unlike rectangular waveguides but similarly to two-conductors guides (coaxial or not), can propagate TEM modes. But a TEM mode is not a superposition of TE and TM modes.

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  • $\begingroup$ Could you expand on the first part? Can the sum of a TE mode and a TM mode still propagate in the fiber because of the rotational symmetry? $\endgroup$
    – Chris
    Commented May 26, 2016 at 20:50
  • $\begingroup$ Chris - there is not much to expand here. One only needs to understand what linearity means to agree that Chris' sentence is complete and self-explanatory. Otherwise: The linearity of equations implies that for solutions $TE$ and $TM$, $TE+TM$ is also a solution. But if you decompose a solution $S$ to $S=TE+TM$, linearity does not imply that $TE$ and $TM$ will be solutions separately. $\endgroup$
    – Luboš Motl
    Commented May 27, 2016 at 3:46
  • $\begingroup$ Yes I do get the reasoning. But then this would hold for any "waveguide" (this term was new to me by the way). And that puzzles me, because I was of the impression that this is a property specific to the optical fiber. $\endgroup$
    – Chris
    Commented May 27, 2016 at 4:45
  • $\begingroup$ Yes, it does hold for any waveguide, even more, for any system whatsoever (the only limitation is nonlinear materials, which are quite rare or require very high intensities as far as I know). $\endgroup$
    – L. Levrel
    Commented May 27, 2016 at 11:25
  • $\begingroup$ How do you define "waveguide"? How do you define a "mode"? $\endgroup$
    – Chris
    Commented May 27, 2016 at 12:27

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