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I am trying to follow a derivation for the reflectance and transmittance of a plane wave from an anisotropic material ($\varepsilon_a,\varepsilon_b,\varepsilon_c$). The derivation uses a $4 \times 4$ tranfser matrix method. Isotropic homogeneous materials occupy $z\leq 0$ ($n_1$) and $z \geq L$ ($n_2$), and the anisotropic material occupies $0 < z < L$, extending to infinity in the $x$ and $y$ directions. The book states that the field phasor in the material is best represented as

$$\textbf{E}(\textbf{r}) = \textbf{e}(z) \exp[i q (x \cos \psi + y \sin \psi)]$$

where $\textbf{e}(z) = e_x(z) \hat{\textbf{x}} + e_y(z) \hat{\textbf{y}} + e_z(z) \hat{\textbf{z}}$, $\textbf{h}(z) = h_x(z) \hat{\textbf{x}} + h_y(z) \hat{\textbf{y}} + h_z(z) \hat{\textbf{z}}$, $q$ is the complex-valued wavenumber, and $\psi$ is the angle from the $+x$-axis to the $+y$-axis.

The fields for the regions $0 \leq z$ and $z \geq L$ are written as:

$$ \textbf{e}_{inc}(z) = (a_s \hat{\textbf{s}} + a_p \hat{\textbf{p}}_{inc}) \exp[i k_0 n_1 ( x \cos \psi + y \sin \psi) + k_0 n_1 z \cos \theta_{inc}]$$

$$ \textbf{e}_{ref}(z) = (r_s \hat{\textbf{s}} + r_p \hat{\textbf{p}}_{ref}) \exp[i k_0 n_1 ( x \cos \psi + y \sin \psi) - k_0 n_1 z \cos \theta_{inc}]$$

$$ \textbf{e}_{tr}(z) = (t_s \hat{\textbf{s}} + t_p \hat{\textbf{p}}_{tr}) \exp[i k_0 n_2 ( x \cos \psi + y \sin \psi) + k_0 n_2 (z-L) \cos \theta_{tr}]$$

where

$$ \hat{\textbf{s}} = - \hat{\textbf{x}} \sin \psi + \hat{\textbf{y}} \cos \psi$$

$$ \hat{\textbf{p}}_{inc} = -(\hat{\textbf{x}} \cos \psi + \hat{\textbf{y}} \sin \psi) \cos \theta_{inc} + \hat{\textbf{z}} \sin \theta_{inc}$$

$$ \hat{\textbf{p}}_{ref} = (\hat{\textbf{x}} \cos \psi + \hat{\textbf{y}} \sin \psi) \cos \theta_{inc} + \hat{\textbf{z}} \sin \theta_{inc}$$

$$ \hat{\textbf{p}}_{tr} = -(\hat{\textbf{x}} \cos \psi + \hat{\textbf{y}} \sin \psi) \cos \theta_{tr} + \hat{\textbf{z}} \sin \theta_{tr}$$

$$ \sin \theta_{tr} = \frac{n_1}{n_2} \sin \theta_{inc}$$

Here are my questions:

  1. The problem is solving for the $z$-dependence of the electric field. But why only have $\textbf{e}(z)$ and not $\textbf{e}(x,y,z)$? If we can assume an $\exp[i q (x \cos \psi + y \sin \psi)]$ form for $x$ and $y$, why not for $z$? Generalizing, could we have assumed the form for any two pairs of variables and then solved the problem for the third variable?

  2. If we have a plane wave entering a complex material, how do we know that the wave propagating inside the material will also be a plane wave (which is what has been assumed for the $x$ and $y$ direction)?

  3. Given that the material is anisotropic, how would we know that the transmitted wave will obey Snell's law? In addition, as given, $\psi$ is the same inside and outside of the anisotropic material. But, how do we know that $\psi$ won't change inside the anisotropic material?

The closest I have gotten to obtaining the first equation is by making use of the continuity of the tangential component at the first interface. So we have the following:

$$ \hat{\textbf{z}} \times (\textbf{E}_1 - \textbf{E}) = 0$$

where $\textbf{E} = \textbf{e}_{inc} + \textbf{e}_{ref}$ and $\textbf{E}$ is the field inside the anisotropic material. Then we can solve for the $x$ and $y$ component of $\textbf{E}$ at $z = 0$ to obtain

$$ \textbf{e}_{x} = (-a_s \sin \psi - r_s \sin \psi - a_p \cos \theta_{inc} \cos \psi + r_p \cos \theta_{inc} \cos \psi) \exp[i k_0 n_1 ( x \cos \psi + y \sin \psi)]$$

$$ \textbf{e}_{y} = (a_s \cos \psi + r_s \cos \psi - a_p \cos \theta_{inc} \sin \psi + r_p \cos \theta_{inc} \sin \psi) \exp[i k_0 n_1 ( x \cos \psi + y \sin \psi)]$$

But now there is all this junk at the beginning. Not to mention these contain $k_0 n_1$ while the former has $q$ and this doesn't address the $\textbf{e}_{x}$ component.

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  • $\begingroup$ Hi, which book are you referring to and where in it is this derivation? $\endgroup$
    – Amey Joshi
    Commented Mar 20, 2022 at 19:36
  • $\begingroup$ @AmeyJoshi The Transfer-Matrix Method in Electromagnetics and Optics $\endgroup$
    – ngc1300
    Commented Mar 20, 2022 at 19:37
  • $\begingroup$ @AmeyJoshi Were you able to find it? $\endgroup$
    – ngc1300
    Commented Mar 20, 2022 at 21:49
  • $\begingroup$ I'm afraid I couldn't yet get a copy of the book. $\endgroup$
    – Amey Joshi
    Commented Mar 21, 2022 at 7:16
  • $\begingroup$ @AmeyJoshi I have it. Is there some way I could share it with you? Google drive maybe? $\endgroup$
    – ngc1300
    Commented Mar 21, 2022 at 15:16

1 Answer 1

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  1. The geometry of the problem is such that all materials extend to infinity in $x$ and $y$ directions. It means that their properties do not change along those directions. The medium, considered as a whole, varies only in the $z$ direction. It has one set of properties from $z = -\infty$ to $z = 0$, another from $z > 0$ to $z = L$ and yet another from $z = L$ to $z = \infty$. That is why we consider the electric field's amplitude as a function of $z$ alone.
  2. The phase of the fields is written in the form $\exp(iq(x\cos\psi + y\sin\psi))$, which suggests that the wave is travelling in the $xy$ plane. Its wave vector does not have a $z$-component. There is no $z$ in the phase because $\vec{k}$, the wave vector, has no $z$ component. Recall that the phase is $\exp(i\vec{k}\cdot\vec{r})$.
  3. We are considering a plane, monochromatic wave. An arbitrary wave can be considered as a Fourier sum of these waves. An arbitrary wave can change its form as it gets into another medium because of several reasons. (a) Sometimes the medium is dispersive so that 'plane waves' of different frequencies travel at different speeds. But this situation is not relevant when we assume that the incident wave is monochromatic. (b) If the medium is nonlinear then a plane wave may generate harmonics which may travel at different speeds. It appears that the authors may have ruled out this possibility.
  4. Snell's law is really a kinematic property. It follows from the continuity of the tangential components of $\vec{E}$ and $\vec{H}$ across the boundaries of two media. It does not depend on the permittivity or the permeability of the media. (That is also the reason why Snell's law is valid for sound waves as well.) Not just Snell's law but also the equality of the incident and the reflected angles is also a consequence of the same equations. [Refer to Jackson's Classical Electrodynamics, 3rd edition, section 7.3 for a detailed and more illuminating explanation.]
  5. Please note that $\psi$ is just the angle between $\vec{k}$ and $\vec{r}$. It is not related to the angle of incidence, angle of reflection or angle of refraction. It is just a variable that will have certain values in a medium. Eventually, you may get relation between phases of the three waves but not between $\psi$'s alone.

Please let me know if any of these points are not very clear. I will try my best to explain them.

I would also recommend Marion and Heald's 'Classical Electromagnetic Radiation' as a helpful reference. It is not as formidable as Jackson's book.

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  • $\begingroup$ Thanks for your answer. Regarding 2. If it doesn't have a z component then how does it travel to the other side? It has to have a z componenet. Why we can't we assume $\exp[i q z cos \theta]$ for the z-direction and then also solve for any other possible z-dependence for the amplitudes? If I were given this problem this is the way I would have naturally approached it. $\endgroup$
    – ngc1300
    Commented Mar 25, 2022 at 21:47
  • $\begingroup$ I agree. I would have considered the wave vector to have a nonzero z component. $\endgroup$
    – Amey Joshi
    Commented Mar 25, 2022 at 22:34

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