Maxwell Equations and Waveguide: why only TE or TM wave and not TEM?

1. Feb 22, 2005

chingkui

I have posted this in some other physics forum, however, I have not yet gotten any definite answer. Hope some of you can help.

I have been trying to understand why in a hollow metal tube, the Maxwell equation admit wave solution, and that they are always TE or TM. In all of the text books I read, they all say that the (E,B) solution for waveguide are:
E(x,y,z,t)=E(x,y)exp(j(kz-wt)); and
B(x,y,z,t)=B(x,y)exp(j(kz-wt))
(assuming the wave propagate in z-direction)
While I totally agree that TM and TE waves satisfy the Maxwell Equation and the boundary conditions for waveguide in hollow perfect conductor, I have a few questions:

1) Each one of the TE and TM mode is a solution, but are they THE solutions?Can there be other form of solution differing from TE or TM mode? So far, in all the derivation I have seen, the TE/TM wave solution is resulted because it is assumed that E and B has the form E(x,y,z,t)=E(x,y)exp(j(kz-wt)) and B(x,y,z,t)=B(x,y)exp(j(kz-wt)). Why is it that apart from the sinusoidal variation, E and B only depends on x and y, but not on z? (i.e. why isn't E(x,y,z,t)=E(x,y,z)exp(j(kz-wt)) with the z dependency also in E(x,y,z)?) What will happen if I send out a very complex EM wave composed of a continuous spectrum of frequency inside a hollow tube? Will the wave traveling inside the waveguide necessarily become one of the TE or TM mode?

2) Why is the wave always TE or TM? I have seen a proof (in Griffiths) that TEM wave solution is impossible in hollow tube waveguide, but sending out a very narrow beam of TEM wave (e.g. visible light) into the waveguide parallel to the tube (i.e. sending it in the z direction) doesn't seem to violate any physics, and it seems to constitute a valid solution to the Maxwell equation. So, what is the gap in my reasoning here? Or is there any gap (hidden assumptions) in the proof that TEM wave is impossible?

2. Feb 22, 2005

vincentchan

http://www.ece.msstate.edu/~donohoe/ece3324notes12.pdf#search='TE%20TM%20waveguide' [Broken]

goto page4 ~ 5 and you will see the answer, basically because the boundary condition.

a very complex wave is merely a superposition of infinitely many simple wave, and each simple wave will yeild a TM/TE wave in wave guide, therefore, you will not see a TEM wave....

Last edited by a moderator: May 1, 2017
3. Feb 22, 2005

chingkui

Thank you, but it still remains puzzle to me for my 2nd question. When I send a narrow beam of TEM wave down a hollow tube, parallel to it, the beam will not hit the wall of the tube. So, it is just like sending a beam in free space, the TEM wave should remain TEM all the time, shouldn't it?
I am thinking along this line: take a rectangular hollow metal tube, place it in front of your eye and look through it, you would see everything with absolutely no problem. It is just like you are looking without using the tube...
I can't tell what's wrong with my reasoning here....

4. Feb 22, 2005

kanato

so for 1, the idea is to write down a TE or TM solution, because that simplifies the problem IIRC into solving a 2-D laplace's equation. Once we have TE and TM solutions, we can make other solutions as a linear combination of them, so that we can get solutions that are neither TE or TM, but they can be expressed as having components which are TE or TM. TEM waves aren't possible because we can't meet the boundary conditions at the sides of the waveguide for both the electric field and magnetic field. As far as sending visible light thru a waveguide, we have to have some intuition as to what happens when we send polarized light thru a small waveguide like that, and I'm not so sure that it's plainly obvious what would happen in that case.

5. Feb 22, 2005

Claude Bile

This is only true for a plane wave. Real beams such as Gaussian Beams, or waves propagating in real (lossy) media, there will be some variation in z. This variation will usually be kept simple (i.e. seperable from E(x,y)) however, as most conventional methods for analysing waveguides require that propagation be unidirectional (or bidirectional). Analysing a curved waveguide for example is immensly tricky without using some coordinate transform.

It appears you are hinging alot of your arguments on your practical observations (Which should not be discouraged!). With regard to your practical observations I notice a couple of problems, firstly, microwaves are far more suitable for experimenting on hollow metal waveguides, and secondly Optical frequencies will not exhibit any modal behaviour in a metal waveguide due to the size of the waveguide (I assume you're talking about a macroscopic waveguide).

Claude.

6. Feb 22, 2005

vincentchan

the gap (hidden assumption) you mention about is the size of the waveguide must be much much less than the wavelength....

7. Feb 23, 2005

chingkui

Then my next questions are:
1) If I send a spherical wave of a single frequency f (above the cutoff frequency), how would I calculate the resulting E and B far far away from the source? In all the EM books I read, this is not mentioned... what about if the source is more complicated, involving a spectrum of different frequency? Is the E and B field expressible as a linear combination of the TE/TM modes? Is there a general way like matching coefficients in Fourier series in computing the solution?
2) If I want to excite a mode, say TE11, what kind of EM source do I need?
3) Can waveguide be used to transmit information? When I measure the E and B field a long way from the source, how can I reconstruct the E, B waveform of the source?
Thanks.

8. Feb 23, 2005

kanato

1) Different frequencies are always separable.. Maxwell's equations are all linear, so once we find a solution for one frequency, you can add a solution for another frequency and still have a solution.. that's why all the problems are always done with a single frequency of radiation.. once you have one solution, you just write it down with $$\omega_1$$ and write it down again with $$\omega_2$$ and add them, and you have a wave with multiple frequencies. If you want to have a really complex waveform, you can just find its fourier series, and do the same thing.

If you have a spherical wave, as you get more and more away from the source, it begins to look more like a plane wave. With waveguides, the idea is that they're really long, so planewave solutions are really what you want to look for once the wave is any distance inside the waveguide from the source.

9. Feb 23, 2005

Claude Bile

1) Most analytic solutions assume a plane wave source, but the answer will always be a linear combination of the possible modes within the waveguide PROVIDED the coupling length (usually denoted z_c) is much smaller than the length of the waveguide.

Because the modes are a function of wavelength, if you use a source with a continuous spectrum, I would say you would find the solution as a function of wavelength and then integrate across the spectrum of the source (as an educated guess).

2) Experimental Physicists go to great pains to excite single modes, so your question is not trivial from an experimental viewpoint. In other words, I don't know of the top of my head, I'm not even sure if it's possible.

3) Absolutely (Think optic fibres). For long haul communications, single-mode fibre is preferred as they have a higher bandwidth, but multimode waveguide (such as fibre microscopes they use in surgery) can be used to transmit images. The easiest way to reconstruct a source would be to form an image of the source using some sort of lens configuration.

Claude.

10. Feb 25, 2005

chingkui

In the use of waveguide and transmission line, it seems that we avoid attenuation by perfectly satisfying the boundary condition (i.e. no E field parallel to the boundary surface is allowed). However, in practical use of, for example, optical fiber and coaxial cable, the cables are not perfectly straight and boundary condition cannot be satisfied exactly. Shouldn't it cause rapid attenuation and communication become impossible over long range? Is there a way to estimate how rapid the attenuation is (from first principle)? Any probabilistic/statistical analysis of the situation? Or is that too complicate to analyze and that we can only rely on experiment?

11. Feb 27, 2005

Claude Bile

The primary loss mechanism in optic fibres is from Rayleigh scattering and absorption. These losses is unavoidable, but can be reduced using

- High purity glass.
- Uniform refractive index profie.

Optic fibre communications systems operate typically around a wavelength of 1.5 microns, which is where attenuation is minimum for silica. Commercial grade fibres are also doped to minimise dispersion at this wavelength as well.

The typical loss of a commercial grade optic fibre is < 0.1 dB/km. Repeaters are used roughly every 500 km or so to compress and amplify the pulse. EDFAs (Erbium Doped Fibre Amplifier) are used to do this, as they are purely optical devices and don't limit the bandwidth of the communication channel.

Analysing the loss relies on applying the theory of Rayleigh Scattering and Absorption, though any estimates do rely on experimental values (e.g. absorption of silica at a particular wavelength). Maxwell's Equations can quantify loss for certain waveguides (leaky modes, coupled mode theory etc), but not for optic fibres as the loss mechanisms are different.

A final note. The boundary conditions must ALWAYS be satisfied, as the boundary conditions are defined by Maxwell's equations. If they weren't satisfied, then Maxwell's equations would be violated.

Claude.