# Help understanding the electromagnetic boundary conditions between 2 linear media

I don't quite understand this set of conditions from Griffiths, Introduction to Electrodynamics (equation 9.74):

\begin{align} \epsilon_1 E_1^{\bot} = \epsilon_2 E_2^{\bot},\quad &\mathbf{E}_1^{\parallel} = \mathbf{E}_2^{\parallel}\\ B_1^{\bot} = B_2^{\bot},\quad &\frac{1}{\mu_1}\mathbf{B}_1^{\parallel} = \frac{1}{\mu_2}\mathbf{B_2}^{\parallel} \end{align}

"Parallel" and "perpendicular" components can only be measured according to something else, for example a coordinate system. So if these equations do not specify the coordinate system and are given very generally, then how do I interpret the "parallel" and "perpendicular" parts?

If a wave is traveling in the z-direction towards a boundary and is polarized in the x-direction, how does the E-field have a parallel and perpendicular component?

• you do not need a coordinate system to know if something is parallel with or perpendicular to the screen you are staring at. – hyportnex Feb 12 '17 at 23:45
• Ohhh so this is judged by the boundary then? Parallel to the boundary and perpendicular to the boundary? – loltospoon Feb 12 '17 at 23:49
• yes, relative to the boundary where either $\mu$ or $\epsilon$ jumps – hyportnex Feb 12 '17 at 23:54
• So am I correct in thinking of the polarization of a wave as a vector pointing in the x-direction but sliding in the z-direction? I know this is a bit of an over-simplification. But in this case I can see how the vector is parallel to the boundary. I can also say that the vector has no perpendicular component (relative to the boundary). – loltospoon Feb 13 '17 at 0:33
• Even if your wave is propagating along $z$ and polarized along $x$, if the boundary is not parallel to the $xy$-plane, the field will have both parallel and perpendicular components. – Raziman T V Feb 13 '17 at 10:08