To get cutoff frequencies and eigenmode field distributions for a waveguide, one uses following equation: $$1/\epsilon ∇ \times 1/\mu ∇ \times E = \omega^2 E$$ With $ \omega^2 $ as eigenvalues. This works fine. But what if I would like to calculate the wave number $k_z$ ? If the waveguide is not changing in $z$ direction, then the E-Field can be described as: $E(x,y,z) = E(x,y)e^{jk_zz}$. Following that, the upper equation becomes: $$1/\epsilon ∇ \times 1/\mu ∇ \times E(x,y)e^{jk_zz} = \omega^2 E(x,y)e^{jk_zz}$$

But I need an eigenvalue equation whre $k_z$ is my eigenvalue. And in this case it is in the exponential, which is not good for me.

  • $\begingroup$ I'm voting to close this question as off-topic because it is a physics/mathematics question and not a computational science one. $\endgroup$ – nicoguaro Aug 31 '16 at 2:09
  • $\begingroup$ yes please could the admin migrate it to physics stackexchange? $\endgroup$ – Kosha Misa Aug 31 '16 at 14:46
  • $\begingroup$ I've flagged your question for migration; but to be honest I disagree that it's off-topic here -- it's clearly related to computation, even if not directly, and just because it's on-topic somewhere else doesn't necessarily mean it's off-topic here. But your wishes are of course respected. $\endgroup$ – Christian Clason Aug 31 '16 at 14:49
  • $\begingroup$ I see - my question concerns mathematics and physics. It doesn't even have to be brought into the numerical domain. I was searching for an analytical solution. But on the other hand, without an analytical solution, there can be no computational science :) $\endgroup$ – Kosha Misa Aug 31 '16 at 14:54
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    $\begingroup$ @nicoguaro That's ok; I thought it was a reasonable question to ask if someone wanted to do numerical simulation of waveguides (and you gave a good answer), hence I voted to leave open. In the end, the community (or moderators) decide. But I stand by my comment that the scope of this site (as documented in the help center and on meta) is not restricted to purely computing questions (although, for historical reasons, the line tends to be much more closely drawn for pure software questions than for pure math questions). We're such a small site, we should be as inclusive as possible. $\endgroup$ – Christian Clason Aug 31 '16 at 21:15

Assuming that you have scalar permitivity and permeability, you can relate your wavenumber with your angular frequency by $$\omega = kc$$ where $c = 1/\sqrt{\epsilon_0 \mu_0}$ is the speed of the wave in vacuum. Then your equation would read $$1/\epsilon_r\nabla \times 1/\mu_r\nabla \times E_n(x,y) = \alpha_n^2 E_n(x,y)$$ where $\epsilon_r$ and $\mu_r$ are the relative permitivity and permeability, and the wavenumber for your propagating wave is given by $$k_n^2 = k^2 - \alpha_n^2 \, .$$ Then, when $k^2 > \alpha^2$ you have propagating waves. And you need to find the eigenvalues for the transverse section.

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  • $\begingroup$ I just expanded the question, since I just realized I made a mistake adding a 2. I erased my other comments since now everything is included in the answer. $\endgroup$ – nicoguaro Aug 31 '16 at 14:18
  • $\begingroup$ great, thanks! So correcting my earlier comment: I compute the eigenvalue equation $Ae(x,y)=α^2 e(x,y)$. My eigenvalues are $α^2$ and my $k_n=\sqrt{ω^2/c^2−α^2}$, where $ c^2 = 1/(μ_0 ϵ_0) $. Right? $\endgroup$ – Kosha Misa Aug 31 '16 at 14:32
  • $\begingroup$ Yes. I suggest that you erase your old comments. And if you find that my answer solve your question you can pick it as answer. One more thing, as I mentioned before this question is probably better suited for Physic StackExchange, so you can ask for migration. $\endgroup$ – nicoguaro Aug 31 '16 at 14:34
  • $\begingroup$ I have reviewed your answer and it seems to me, that there is a problem. The $rot$ operator is not defined in $2d$. And here we have a $2d$ E-Field vector. in that case the $E(x,y)$ must be something like $E(x,y,z), z=?$. The $\delta E_z/\delta_z$ derivative should be $jk_z$ which again is part of the searched eigenvalue. Do you understand my point? $\endgroup$ – Kosha Misa Aug 31 '16 at 16:46
  • $\begingroup$ That the function depends on two variables does not mean that it lies in $R^2$. Indeed, if you want your field to propagate in the $z$ direction then it should not have any component in that direction, since the waves are transversal waves. $\endgroup$ – nicoguaro Aug 31 '16 at 18:13

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