An electromagnetic wave in the free space satisfy the following relationships:

$\vec{K}\times \vec{E}=Z\vec{H}$

$\vec{E}=Z\vec{H}\times \vec{K}$

$Z=\sqrt{\frac{\mu}{\epsilon} } $

Find local relationship between electromagnetic energy density and magnetic flux density (poynting vector)?


$\vec{S}=\vec{E}\times \vec{H}=\frac{1}{Z}\vec{E}\times (\vec{K}\times \vec{E})=\frac{1}{Z}E^{2}\vec{K}=Z(\vec{H}\times \vec{K})\times \vec{H}=ZH^{2}\vec{K}$

$w=\frac{1}{2}(\epsilon E^{2}+\mu H^{2})=\epsilon E^{2}=\mu H^{2}$

$\vec{S}=\omega c \vec{K} $

I am really confused here, especially with first step with vectors. Can someone explain how they from $\vec{S}$ came to $ZH^{2}\vec{K}$?


It is easy to see using the "BAC-CAB"-rule: $a \times (b \times c) = b (a \cdot c) - c (a \cdot b)$.

Additionally, you need $\vec{E} \perp \vec{K} \perp \vec{H}$ which follows from the definition of $\vec{E}$ and $\vec{K}$.

For example, one finds: $(\vec{H} \times \vec{K})\times \vec{H} = \vec{K} (\vec{H} \cdot \vec{H}) - \vec{H} (\vec{K} \cdot \vec{H}) = \vec{K} H^2$

  • $\begingroup$ Okay. One more probably stupid question, why the result of your example is not 0? I am probably missing some vector rule.. $\endgroup$ – Ana Matijanovic Nov 17 '16 at 16:34
  • $\begingroup$ Why do you think it should be 0? $\endgroup$ – Virft Nov 17 '16 at 16:40
  • $\begingroup$ because of the KH^2-KH^2 ? . $\endgroup$ – Ana Matijanovic Nov 17 '16 at 16:47
  • $\begingroup$ $\vec{K} \cdot \vec{H} = 0$ because $\vec{K} \perp \vec{H}$ while $\vec{H} \cdot \vec{H} = ||\vec{H}||^2 = H^2$ and thus the result $\vec{K} H^2$ $\endgroup$ – Virft Nov 17 '16 at 16:50
  • $\begingroup$ Can you tell me how they in first row got 5 step from 4 step? I am not sure how I should represent E² here. $\endgroup$ – Ana Matijanovic Nov 17 '16 at 23:48

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