TE and TM modes in optical fibers

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In step-index optical fibers, TE and TM modes can be derived from Bessel functions of order ν=0, with TE modes involving components Eφ, Hr, and Hz, while TM modes involve Hφ, Er, and Ez. Although TE and TM modes are conceptually valid, practical propagation typically occurs in the form of linearly polarized (LP) modes, which are approximations of the exact modes. Each mode can propagate independently due to the linear nature of the governing equations, although some modes, like evanescent modes, do not propagate in the traditional sense. The discussion highlights confusion about whether TE and TM modes can propagate alone, clarifying that they can exist conceptually but are not commonly observed in practice. Overall, while TE and TM modes are theoretically significant, LP modes dominate in actual optical fiber applications.
EmilyRuck
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In a step-index optical fiber, considering Bessel functions of order ##\nu = 0## and no ##\phi## dependence, it is possible to obtain two separate sets of components, which generate respectively TE and TM modes. In the former case, only ##E_{\phi}##, ##H_r##, ##H_z## are involved; in the latter, only ##H_{\phi}##, ##E_r##, ##E_z##.

The exact modes in optical fibers are: TE, TM, EH, HE. Also, some approximated modes can be identified, known as LP: each of them may be regarded as the composition of two or more exact modes.

This gives me some confusion: are the exact modes able to propagate alone, or not? Is a single TE mode available for propagation in a step index fiber? A single HE11 mode?

And, if not, how do they propagate, then?

Thank you anyway
 
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Modes are a set of solutions of linear equations (in this case Maxwell's equations in the fiber), out of which you can compose all solutions. If you deal with self-adjoint differential operators you get a complete orthonormal set of eigenfunctions of these operators, and all solutions can be written in terms of these functions, which is a generalized Fourier series (or transformation if you have continuous "eigenvalues").
 
EmilyRuck said:
Summary:: Do they propagate alone, or are they only used to compose other modes?

In a step-index optical fiber, considering Bessel functions of order ##\nu = 0## and no ##\phi## dependence, it is possible to obtain two separate sets of components, which generate respectively TE and TM modes. In the former case, only ##E_{\phi}##, ##H_r##, ##H_z## are involved; in the latter, only ##H_{\phi}##, ##E_r##, ##E_z##.

The exact modes in optical fibers are: TE, TM, EH, HE. Also, some approximated modes can be identified, known as LP: each of them may be regarded as the composition of two or more exact modes.

This gives me some confusion: are the exact modes able to propagate alone, or not? Is a single TE mode available for propagation in a step index fiber? A single HE11 mode?

And, if not, how do they propagate, then?

Thank you anyway

Yes- you can launch a single-mode into a fiber and that mode can propagate unchanged. It's also possible to launch a linear superposition of modes into the fiber, and each mode will propagate independently from each other.

Lastly, it's possible to use a fiber as a sensor by examining how the mode amplitudes change to measure external mechanical strain, external electromagnetic fields, etc. In fancy terms, the external potentials act to perturb the mode structure of the fiber- transferring energy from one mode to another. Polarization-maintaining fibers are excellent for this application.
 
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vanhees71 said:
Modes are a set of solutions of linear equations (in this case Maxwell's equations in the fiber), out of which you can compose all solutions.

Yes, I get it.

vanhees71 said:
If you deal with self-adjoint differential operators

Sorry, I don't know them.

vanhees71 said:
you get a complete orthonormal set of eigenfunctions of these operators, and all solutions can be written in terms of these functions, which is a generalized Fourier series (or transformation if you have continuous "eigenvalues").

If I correctly understood, yes, modes somehow represent the spectrum of the optical fiber. Any real field propagating in this structure can be described as a composition of modes.
But my post was about a slightly different scope: are all these modes individually able to propagate? In particular when dealing with optical fibers, very often, only LP modes are considered; but they are not eigenfunctions, and they are not even real modes, just approximations.
I did not find notes or texts explicitly talking about a single TE or TM mode which is able to propagate alone, hence my doubt.
 
Andy Resnick said:
Yes- you can launch a single-mode into a fiber and that mode can propagate unchanged. It's also possible to launch a linear superposition of modes into the fiber, and each mode will propagate independently from each other.

Ok, thank you so much!

Andy Resnick said:
Lastly, it's possible to use a fiber as a sensor by examining how the mode amplitudes change to measure external mechanical strain, external electromagnetic fields, etc. In fancy terms, the external potentials act to perturb the mode structure of the fiber- transferring energy from one mode to another. Polarization-maintaining fibers are excellent for this application.

Oh, I don't deal with this, but it is great.
 
EmilyRuck said:
Summary:: Do they propagate alone, or are they only used to compose other modes?

In a step-index optical fiber, considering Bessel functions of order ν=0ν=0\nu = 0 and no ϕϕ\phi dependence, it is possible to obtain two separate sets of components, which generate respectively TE and TM modes. In the former case, only EϕEϕE_{\phi}, HrHrH_r, HzHzH_z are involved; in the latter, only HϕHϕH_{\phi}, ErErE_r, EzEzE_z.
It strikes me that a bit of basic reading about conventional waveguides could help you get into this subject. Waveguides are very often rectangular section (hard to make optical fibres this shape) and the analysis is much more straightforward. Piling in with the necessary Bessel functions for circular guides can be avoided if you start off with rectangular modes first.
 
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EmilyRuck said:
Yes, I get it.
Sorry, I don't know them.
If I correctly understood, yes, modes somehow represent the spectrum of the optical fiber. Any real field propagating in this structure can be described as a composition of modes.
But my post was about a slightly different scope: are all these modes individually able to propagate? In particular when dealing with optical fibers, very often, only LP modes are considered; but they are not eigenfunctions, and they are not even real modes, just approximations.
I did not find notes or texts explicitly talking about a single TE or TM mode which is able to propagate alone, hence my doubt.
Yes they are. Since the equations used to describe the waves are linear each mode propgates independently of any other, i.e., if at some initial time the solution is a particular superposition of modes it stays this superposition at any time.

Note however, that "propagate" must be taken with a grain of salt since not all modes really "propagate" in the literal sense, because there are also "evanescent modes", which die out exponentially.
 
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vanhees71 said:
Yes they are. Since the equations used to describe the waves are linear each mode propgates independently of any other, i.e., if at some initial time the solution is a particular superposition of modes it stays this superposition at any time.

Ok! My doubt arouse because of sentences like:

In almost all practical optical fibers one would not find light propagating in the form of TE, TM or HE modes. Light propagates in the form of linearly polarized modes which are briefly called as LP modes. The TE, TM and HE modes thus remain only to academic investigations, because in practice light inside an optical fiber is almost linearly polarized.

from this document, page 6. However, if I correctly understood what you state, TE and TM modes are, at least conceptually, valid and existing modes, and they are able to propagate by their own.

vanhees71 said:
Note however, that "propagate" must be taken with a grain of salt since not all modes really "propagate" in the literal sense, because there are also "evanescent modes", which die out exponentially.

Yes, of course. This depends on the frequency of the input signal. I was implicitly assuming that it is above the cutoff for at least one TE and one TM mode.
 
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sophiecentaur said:
It strikes me that a bit of basic reading about conventional waveguides could help you get into this subject. Waveguides are very often rectangular section (hard to make optical fibres this shape) and the analysis is much more straightforward. Piling in with the necessary Bessel functions for circular guides can be avoided if you start off with rectangular modes first.

Yes, it is not a good starting point. I have read about the dielectric slab, for example, where modes are much simpler and immediate. My problem is not about modes themselves, but about some unclear information on what modes actually propagate in optical fibers (refer to the quote I just posted in the previous message). I wasn't sure if TE and TM modes exist at all, or are able to propagate at all.
 
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EmilyRuck said:
from this document, page 6. However, if I correctly understood what you state, TE and TM modes are, at least conceptually, valid and existing modes, and they are able to propagate by their own.

that's not what it says on pg6 of that doc

the cluster TE01, TM01 and HE21 degenerate to form one single linearly polarized mode. The dominant mode HE11 is already a linearly polarized mode as can be seen from figure 7.1. In almost all practical optical fibres one would not find light propagating in the form of TE, TM or HE modes. Light propagates in the form of linearly polarized modes which are briefly called as LP modes. The TE, TM and HE modes thus remain only to academic investigations, because in practice light inside an optical fibre is almost linearly polarized.
 
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