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Suppose a waveguide with 2 Perfect electric conductor at both boundaries. The waveguide is filled with a lossy media modelled with a conductivity $\sigma$.

Solving for the following Maxwell's equations (TE mode) : \begin{eqnarray} \mu \partial_t\tilde H_x &=& - \partial_y\tilde E_z \\ \mu \partial_t\tilde H_y &=& + \partial_x \tilde E_z \\ \epsilon \partial_t\tilde E_z + \sigma \tilde E_z &=& + \partial_x \tilde H_y - \partial_y\tilde H_x \end{eqnarray}

one can find that a valid solution adopting a complex notation is: \begin{eqnarray}\label{eq:ShapeComplexSol} H_x &=& +jE_0\frac{k_y}{\omega\mu} \cos(k_yy) e^{j(\omega t - k_x x)}\\ H_y &=& - E_0\frac{k_x}{\omega\mu} \sin(k_yy) e^{j(\omega t - k_x x)}\\ E_z &=& + E_0 \sin(k_yy) e^{j(\omega t - k_x x)} \end{eqnarray}

provided the dispersion relation : \begin{equation}\label{eq:PECdispersion} k_0^2n^2 = k_x^2 + k_y^2 ~~\mbox{with}~~ k_y =\frac{m\pi}{a} ~~\mbox{and}~~ n^2 = \mu\epsilon\left(1 - j\frac{\sigma}{\omega\epsilon}\right) ~~\mbox{and}~~ \end{equation}

where $k_y$ is imposed by the PEC boundary conditions.

Note that since the refractive index n is a complex number, $k_x=\beta-j\alpha=|k_x|e^{j\phi}$ is a complex number too.Consequently, the real part of the solution is :

\begin{eqnarray}\label{eq:RealSol} \Re(H_x) &=& -E_0\frac{ k_y }{\omega\mu}\cos(k_yy)e^{-\alpha x}sin(\omega t - \beta x) \\ \Re(H_y) &=& -E_0\frac{|k_x|}{\omega\mu}\sin(k_yy)e^{-\alpha x}cos(\omega t - \beta x + \phi) \\ \Re(E_z) &=& +E_0 sin(k_yy)e^{-\alpha x}\cos(\omega t - \beta x) \end{eqnarray}

Which does not satisfy the 3rd equation of Maxwell's equations. Indeed, \begin{eqnarray} \partial_x H_y -\partial_y H_x &=& \frac{E_0}{\omega\mu}sin(k_yy)e^{-\alpha x}(|k_x|(\alpha cos(\omega t - \beta x + \phi) - \beta sin(\omega t - \beta x + \phi)) +k_y^2sin(\omega t - \beta x)) \\ &= &\frac{E_0}{\omega\mu}sin(k_yy)e^{-\alpha x}sin(\omega t - \beta x) (|k_x|^2 + k_y^2) \\ &\neq&E_0sin(k_yy)e^{-\alpha x}(-\omega\epsilon sin(\omega t - \beta x) + \sigma cos(\omega t - \beta x)) \label{eq:modulus}\\ &= &\epsilon \partial_t \tilde E_z + \sigma \tilde E_z \end{eqnarray}
My question is the following :

Since taking the real part of the fields is not a valid solution of the presented Maxwell's equations anymore, how can one find the corresponding real modes that would actually be found if one were to carry an experiment?

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  • $\begingroup$ You may wish to explain what waveguide you are considering and what coordinates you are using. If this is a rectangular waveguide, why does $E_z$ only depend on 2 coordinates? $\endgroup$ – akhmeteli Nov 24 '16 at 13:46
  • $\begingroup$ It is a planar waveguide so no dependence on z. Indeed, the Maxwell's equations are written in 2D. $k_x$ is the propagation axis and is the transverse axis hence the quantization of the component $k_y$. $\endgroup$ – Ronan Tarik Drevon Nov 24 '16 at 16:59
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I believe the real part of the fields is still a valid solution of the Maxwell equations. The modes will be attenuated in the direction of propagation $x$.

I don't quite understand though why the component of the magnetic field in the direction of propagation $H_x$ does not vanish for a TM mode.

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  • $\begingroup$ that's not correct! take the real part. Replace it into the original Maxwell's equations and you will see that it does not satisfy the system. As for the Field component along the direction of propagation, this is a common results that one of the field components is along the direction of propagation. The denomination TE or TM is a matter of convention. I agree here it should be TE mode hence only the electric field is transverse to the direction of propagation. $\endgroup$ – Ronan Tarik Drevon Nov 24 '16 at 17:34
  • $\begingroup$ @RonanTarikDrevon: I don't see why my answer is incorrect. The Maxwell equations have real coefficients. Therefore, if they have a complex solution, the real part of the solution is also a solution. If you disagree, show me your calculations. With all due respect, I am not under any obligation to "replace" anything anywhere - we are discussing your problem, not mine. $\endgroup$ – akhmeteli Nov 24 '16 at 17:44
  • $\begingroup$ @RonanTarikDrevon: In particular, be careful when you calculate the real part of $e^{j(\omega t- k_x x)}$ and $j e^{j(\omega t- k_x x)}$, where $k_x$ is not real. $\endgroup$ – akhmeteli Nov 24 '16 at 17:48
  • $\begingroup$ @ akhmeteli: I have edited the post to show why I think the real part is not solution. I do not understand your argument when you say "The Maxwell equations have real coefficients. Therefore, if they have a complex solution, the real part of the solution is also a solution". Do you have further details ? Indeed, this is common to say the real part of the complex solution is also solution but it does not seem to be true in this case. $\endgroup$ – Ronan Tarik Drevon Nov 24 '16 at 17:55
  • $\begingroup$ @RonanTarikDrevon: 1. Did you check that the complex solution satisfies the Maxwell equations? 2. With all due respect, I don't like your calculations: you should choose one form for complex $k_x$, rather than two forms: one using $\alpha$ and $\beta$ and the other using $|k_x|$ and $\phi$. As of now, I am not sure the Maxwell equations are not satisfied for the real part. $\endgroup$ – akhmeteli Nov 24 '16 at 18:08

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