On page 8: http://local.eleceng.uct.ac.za/courses/EEE3055F/lecture_notes/2011_old/eee3055f_Ch4_2up.pdfele

I don't understand why $E_{t1} = E_{t2}$ is equivalent to $\hat n \times (\vec E_1 - \vec E_2) =0 $.

How can I see that the cross product produces a tangential term when it clearly produces a term that goes out of the page?


If the cross product is zero, the two vectors are parallel (here $\hat n$ and $\vec E_1 - \vec E_2$). If the difference of the two fields has only a normal component the tangential component (of the difference) is zero or: $\vec E_{1,t}-\vec E_{2,t} = 0$ or $\vec E_{1,t} = \vec E_{2,t}$.

A more detailed way to look at it, is by splitting the vectors into normal and tangential parts: $\vec E_1 = E_{1,n} \hat n + E_{1,t} \hat t$ and $\vec E_2 = E_{2,n} \hat n + E_{2,t} \hat t$ Note that $\hat n \times \hat n = 0$ and $\hat n \times \hat t\neq 0$).

\begin{align} \hat n \times (\vec E_1 - \vec E_2) &= \hat n \times \vec E_1 -\hat n \times \vec E_2\\ &=\hat n \times (E_{1,n} \hat n + E_{1,t} \hat t) -\hat n \times (E_{2,n} \hat n + E_{2,t} \hat t)\\ &=\hat n \times E_{1,t} \hat t -\hat n \times E_{2,t} \hat t\\ &=( E_{1,t} -\ E_{2,t} ) \hat n \times \hat t \end{align} And if $E_{1,t}$ and $\ E_{2,t}$ are equal, this is zero ($E_{1,t} -\ E_{2,t} = 0 \Leftrightarrow E_{1,t} =\ E_{2,t}$).


If the cross product of two vectors is zero, that means that the two vectors are parallel. That is, $\hat{n}$ is parallel to $\vec{E_1} - \vec{E_2}.$ Since $\hat{n}$ is normal to the surface, that means that $\vec{E_1} - \vec{E_2}$ is normal to the surface. That is, the only difference between the two fields lies normal to the surface.

Therefore there is no difference between the two fields in the direction tangential to the surface (since this direction is orthogonal to $\hat{n}$).

  • $\begingroup$ okay, but the "quantity" of $\hat n \times (\vec E_1 - \vec E_2)$ is still out of the page, no? $\endgroup$ – Carlos - the Mongoose - Danger Nov 18 '14 at 20:39
  • $\begingroup$ @IllegalImmigrant, let $\hat n = \hat z$ then $\hat n \times \vec E$ is a vector in the $xy$ plane. As the footnote on that page cautions: "These vector forms require careful 3-D visualization." $\endgroup$ – Alfred Centauri Nov 18 '14 at 22:36

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