# Electromagnetic boundary conditions for a 'current' interface

Hello,

I am at present analyzing the electromagnetic interaction of a layer of paritcles in air when illuminated by an electromagnetic wave.

This can be done by considering the layer of particles as an interface with surface current (as opposed to a 'normal' interface with Fresnel coefficients) and finding the reflection and transmission coefficients of a plane wave from such an interface.

The source of the current is the polarization of the particles by the illuminating wave.
So the current is proportional to: $\vec{J} \propto \omega\rho \overset{\leftrightarrow}{\alpha}. \vec{E}$ where, $\rho$ is the density of paticles and $\alpha$ is the polarizability of the particles (which is a tensor) and $\vec{E}$ is the total field (incident + reflected).

The orientation of the current would depend on the polarization of the EM wave. For 's' polarization with the direction of electric field, say, only along 'y' axis, the orientation of the current will also be only along 'y' axis. For 'p' polarization the current will be in the x-z plane but different from the orientation of the electric field ( which will also be in the x-z plane) due to the effect of the polarizability tensor. I'll first explain the case of 's' polarization where I am not having any problems applying the boundary conditions and then come to the 'p' polarization.

[I have attached a file with the same explanation of the problem but with figures regarding the orientation of the fields]

When I apply the boundary conditions for such an interface for an impinging 's' polarized wave; I easily get the reflection and transmission coefficients. Considering that the tangential electric field should be the same on both the sides I get:
$$E_{i} + E_{r} = E_{t}$$
and for the magnetic field, applying the circulation law for a tiny loop at the interface, I get:
$$( H_{i} \cos \theta_i - H_{r} \cos \theta_r) - H_t \cos \theta_t = J$$
where the subscripts, i, r and t denote the incident reflected and transmitted waves.
For air on both sides of the particles, we have $\cos \theta_i = \cos \theta_r =\cos \theta_t = \dfrac{k_{z}}{k_0}$ with $k_z$ being the z-component of $k_0$

From these two equations and considering that $E/H = \mu_0 c_0$ and that the incident wave has unit amplitude at the interface we can get the reflection coefficient $$R = E_r/ E_i = \frac{-J \omega \mu_0}{2 k_z}$$

and similarly I can get the transmission coefficient too. These have been verified to satisfy the energy conservation statement as well :
$S_{1z} - S_{2z} = -\vec{J}. \vec{E}$ with $S_1$ and $S_2$ being the Poynting vector on either side of the interface .

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The problem occurs when I try to do a similar analysis for a TM wave for a particular case of polarizability. Consider the case when we have polarizability such that even when the electric field is in the x-z plane; the current has only a component along z-axis.

If I try to apply the boundary conditions for such an interface when a TM plane wave of unit amplitude impinges on it, I get for the magnetic field:

$$H_i + H_r = H_t$$

For the electric field, instead of applying the continuity of tangential fields (which doesn't help too)
I apply the Gauss's law:

$$E_{tz} - (E_{iz} + E_{rz}) = \sigma/\epsilon_0$$ with $\sigma$ being the surface charge density.
Dividing by $E_i$ ; expressing the magnetic fields in terms of the electric fields and since we have air on both sides we get the following two equations:

$$\frac{k_\rho}{k_0} \big[ (1 + R) - T \big] = - \sigma/\epsilon_0$$
$$(1 + R) - T = 0$$
with $k_\rho$ being the inplane wave vector.

As you can see these two equations are not solvable! I have not been able to understand what is going wrong and why we can apply the boundary conditions to find the reflection and transmission coefficients for such a simple problem!

Could it be that these boundary conditions do not take into account the anisotropic behavior ? How will I be able to overcome it in such a case?

Any thoughts/suggestions will be greatly appreciated.

Thanks

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