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?

  • 1
    $\begingroup$ you do not need a coordinate system to know if something is parallel with or perpendicular to the screen you are staring at. $\endgroup$ – hyportnex Feb 12 '17 at 23:45
  • $\begingroup$ Ohhh so this is judged by the boundary then? Parallel to the boundary and perpendicular to the boundary? $\endgroup$ – loltospoon Feb 12 '17 at 23:49
  • $\begingroup$ yes, relative to the boundary where either $\mu$ or $\epsilon$ jumps $\endgroup$ – hyportnex Feb 12 '17 at 23:54
  • $\begingroup$ 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). $\endgroup$ – loltospoon Feb 13 '17 at 0:33
  • $\begingroup$ 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. $\endgroup$ – Raziman T V Feb 13 '17 at 10:08

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