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Polarisation and k vector are not vertical to each other

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Hi

I try to simulate a waveguide and I see that the polarisation vector and the k vector are not orthogonal.

I did it by the attachment mph file and taking the data into a MATLAB script to take a dot product between the vectors to check if they are perpendicular.

what I have found is that these vectors are not orthogonal excluding in the plane z=0 but in every other place it isn't. which is not expected according to theory.

Is anyone encountered in such similar case?

thanks in advance



3 Replies Last Post 2019年1月3日 GMT-5 16:31
Robert Koslover Certified Consultant

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Posted: 6 years ago 2019年1月2日 GMT-5 14:28

I haven't downloaded your files, but... You said you are modeling a waveguide. Waves in a waveguide are not the same as free-space plane waves. Rather, each waveguide mode can be expressed mathematically as a superposition of such plane waves, reflecting off the walls at a characteristic angle that is dependent on the frequency relative to cutoff. But anyway, within a waveguide oriented along the z-axis, the effective k vector is along z (with k sub g = 2 x pi / lambda sub g) , with the "sub g" referring to the guide wavelength (as opposed to the free space wavelength). E will be orthogonal to z for all TE modes and H will be orthogonal to z for TM modes. But E will have a non-zero z component for TM modes and H will have a non-zero z component for TE modes. It is only for TEM modes (which can exist in free space or in multi-conductor (i.e., 2 or more conductors) transmission lines) that both E and H will be perpendicular to z.

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Scientific Applications & Research Associates (SARA) Inc.
www.comsol.com/partners-consultants/certified-consultants/sara
I haven't downloaded your files, but... You said you are modeling a *waveguide*. Waves in a waveguide are *not* the same as free-space plane waves. Rather, each waveguide mode can be expressed mathematically as a superposition of such plane waves, reflecting off the walls at a characteristic angle that is dependent on the frequency relative to cutoff. But anyway, within a waveguide oriented along the z-axis, the effective k vector is along z (with k sub g = 2 x pi / lambda sub g) , with the "sub g" referring to the *guide* wavelength (as opposed to the free space wavelength). E will be orthogonal to z for all TE modes and H will be orthogonal to z for TM modes. But E will have a non-zero z component for TM modes and H will have a non-zero z component for TE modes. It is only for TEM modes (which can exist in free space or in multi-conductor (i.e., 2 or more conductors) transmission lines) that *both* E and H will be perpendicular to z.

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Posted: 6 years ago 2019年1月3日 GMT-5 07:40

Thank you for your detailed reply

You are right about a waveguide where there is indeed an effective K vector but there is not expectation for a big change so they will not be vertical at all.

But in a simple case like the file I was uploading would expect that at least there will be symmetry around the center of the box (it isn't even a waveguide, just a box) and in the plane Z = 0 when the light progress direction is X indeed they are vertical and even on the sides of the box, but in the rest of the box they are not vertical (the dot product values are close to 1) that seems a little strange.

thanks again

Thank you for your detailed reply You are right about a waveguide where there is indeed an effective K vector but there is not expectation for a big change so they will not be vertical at all. But in a simple case like the file I was uploading would expect that at least there will be symmetry around the center of the box (it isn't even a waveguide, just a box) and in the plane Z = 0 when the light progress direction is X indeed they are vertical and even on the sides of the box, but in the rest of the box they are not vertical (the dot product values are close to 1) that seems a little strange. thanks again

Robert Koslover Certified Consultant

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Posted: 6 years ago 2019年1月3日 GMT-5 16:31

OK, I just downloaded your file. I don't have the wave-optics module, but I have the RF module.

I noticed the following:

  1. You have a driving boundary condition that I don't understand: E0z = E0(w0/w)exp(-(r/w)^2)exp(-i(kxx+kzz+(kzz^2)/(2R)-phi)) I presume you trying to launch some kind of shaped-beam or something. In any case, that doesn't look like a waveguide mode to me.

  2. Only one of the walls of your box is apparently a conductor. That is to say, that is what I see when I open the file. You have one driven wall, 4 scattering-condition walls, and one conducting wall opposite the driven wall. That's not a waveguide. It's a conducting plate illuminated by a wave with a seemingly-strange field distribution.

Unless what I saw was caused by me not having the right module, then the above issues are big contributors to your problem. Check and re-check your boundary conditions carefully. Oh, and for the field-driven surface, you also might want to consider using a waveguide port boundary condition with a specified mode. Good luck.

-------------------
Scientific Applications & Research Associates (SARA) Inc.
www.comsol.com/partners-consultants/certified-consultants/sara
OK, I just downloaded your file. I don't have the wave-optics module, but I have the RF module. I noticed the following: 1. You have a driving boundary condition that I don't understand: E0z = E0*(w0/w)*exp(-(r/w)^2)*exp(-i*(kx*x+kz*z+(kz*z^2)/(2*R)-phi)) I presume you trying to launch some kind of shaped-beam or something. In any case, that doesn't look like a waveguide mode to me. 2. Only one of the walls of your box is apparently a conductor. That is to say, that is what I see when I open the file. You have one driven wall, 4 scattering-condition walls, and one conducting wall opposite the driven wall. That's not a waveguide. It's a conducting plate illuminated by a wave with a seemingly-strange field distribution. Unless what I saw was caused by me not having the right module, then the above issues are big contributors to your problem. Check and re-check your boundary conditions carefully. Oh, and for the field-driven surface, you also might want to consider using a waveguide port boundary condition with a specified mode. Good luck.

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