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marcusl
#2
Feb25-13, 02:19 PM
Sci Advisor
PF Gold
P: 2,081
In essence, this is equivalent to P = V * I in a conventional circuit, where voltage V and current I are independent of each other as determined by the circuit and the source drive. So the coupling that you are worried about is not really an issue.

Being careful reveals that [itex]\vec{E}\cdot\vec{J}[/itex] is power density, not power. It comes from applying the divergence theorem to the Poynting vector. The derivation starts from two Maxwell equations [tex]\nabla\times\vec{E}=-\vec J_m,\\
\nabla\times\vec{H}=\vec J_e[/tex] where J_e is current density and J_m is effective magnetic current density. Multiply the first equation by H* and the second by E, take their difference and apply a vector identity to get [tex]\nabla\cdot(\vec E\times\vec H^*) + \vec E\cdot\vec J_e^*+\vec H^*\cdot\vec J_m=0.[/tex] Integrate over all space and apply the divergence theorem to get
[tex]\oint \vec E \times \vec H^* \cdot d\vec A + \int(\vec E\cdot\vec J_e^*+\vec H^*\cdot\vec J_m)dV=0.[/tex] The last term is zero in the absence of magnetic material, so the volume integral of [itex]\vec E \cdot \vec J_e[/itex] equals the surface integral of the Poynting vector, or, in other words, radiated power. That makes EJ itself a power density.