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As pointed out in [1] the fields at an interface that is not a perfect conductor on either side are related by

\begin{equation}\label{eqn:fieldsAtInterface:20}

\begin{aligned}

\ncap \cdot \lr{ \BD_2 – \BD_1 } &= \rho_{es} \\

\ncap \cross \lr{ \BE_2 – \BE_1 } &= -\BM_s \\

\ncap \cdot \lr{ \BB_2 – \BB_1 } &= \rho_{ms} \\

\ncap \cross \lr{ \BH_2 – \BH_1 } &= \BJ_s.

\end{aligned}

\end{equation}

Given the fields in medium 1, assuming that boths sets of media are linear, we can use these relationships to determine the fields in the other medium.

\begin{equation}\label{eqn:fieldsAtInterface:40}

\begin{aligned}

\ncap \cdot \BE_2 &= \inv{\epsilon_2} \lr{ \epsilon_1 \ncap \cdot \BE_1 + \rho_{es} } \\

\ncap \wedge \BE_2 &= \ncap \wedge \BE_1 -I \BM_s \\

\ncap \cdot \BB_2 &= \ncap \cdot \BB_1 + \rho_{ms} \\

\ncap \wedge \BB_2 &= \mu_2 \lr{ \inv{\mu_1} \ncap \wedge \BB_1 + I \BJ_s}.

\end{aligned}

\end{equation}

Now the fields in interface 2 can be obtained by adding the normal and tangential projections. For the electric field

\begin{equation}\label{eqn:fieldsAtInterface:60}

\begin{aligned}

\BE_2

&=

\ncap (\ncap \cdot \BE_2 )

+ \ncap \cdot (\ncap \wedge \BE_2) \\

&=

\inv{\epsilon_2} \ncap \lr{ \epsilon_1 \ncap \cdot \BE_1 + \rho_{es} }

+

\ncap \cdot (\ncap \wedge \BE_1 -I \BM_s).

\end{aligned}

\end{equation}

Note that this manipulation can also be done without Geometric Algebra by writing \( \BE_2 = \ncap (\ncap \cdot \BE_2 ) – \ncap \cross (\ncap \cross \BE_2) \)).

Expanding \( \ncap \cdot (\ncap \wedge \BE_1) = \BE_1 – \ncap (\ncap \cdot \BE_1) \), and \( \ncap \cdot (I \BM_s) = -\ncap \cross \BM_s \), that is

\begin{equation}\label{eqn:fieldsAtInterface:80}

\boxed{

\BE_2

=

\BE_1

+ \ncap (\ncap \cdot \BE_1) \lr{ \frac{\epsilon_1}{\epsilon_2} – 1 }

+ \frac{\rho_{es}}{\epsilon_2}

+ \ncap \cross \BM_s.

}

\end{equation}

For the magnetic field

\begin{equation}\label{eqn:fieldsAtInterface:100}

\begin{aligned}

\BB_2

&=

\ncap (\ncap \cdot \BB_2 )

+

\ncap \cdot (\ncap \wedge \BB_2) \\

&=

\ncap \lr{ \ncap \cdot \BB_1 + \rho_{ms} }

+

\mu_2 \ncap \cdot \lr{ \lr{ \inv{\mu_1} \ncap \wedge \BB_1 + I \BJ_s} },

\end{aligned}

\end{equation}

which is

\begin{equation}\label{eqn:fieldsAtInterface:120}

\boxed{

\BB_2

=

\frac{\mu_2}{\mu_1} \BB_1

+

\ncap (\ncap \cdot \BB_1) \lr{ 1 – \frac{\mu_2}{\mu_1} }

+ \ncap \rho_{ms}

– \ncap \cross \BJ_s.

}

\end{equation}

These are kind of pretty results, having none of the explicit angle dependence that we see in the Fresnel relationships. In this analysis, it is assumed there is only a transmitted component of the ray in question, and no reflected component. Can we do a purely vectoral treatment of the Fresnel equations along these same lines?

# References

[1] Constantine A Balanis. *Advanced engineering electromagnetics*. Wiley New York, 1989.