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The time domain Poynting relationship was found to be

\begin{equation}\label{eqn:poyntingTimeHarmonic:20}

0

=

\spacegrad \cdot \lr{ \BE \cross \BH }

+ \frac{\epsilon}{2} \BE \cdot \PD{t}{\BE}

+ \frac{\mu}{2} \BH \cdot \PD{t}{\BH}

+ \BH \cdot \BM_i

+ \BE \cdot \BJ_i

+ \sigma \BE \cdot \BE.

\end{equation}

Let’s derive the equivalent relationship for the time averaged portion of the time-harmonic Poynting vector. The time domain representation of the Poynting vector in terms of the time-harmonic (phasor) vectors is

\begin{equation}\label{eqn:poyntingTimeHarmonic:40}

\begin{aligned}

\boldsymbol{\mathcal{E}} \cross \boldsymbol{\mathcal{H}}

&= \inv{4}

\lr{

\BE e^{j\omega t}

+ \BE^\conj e^{-j\omega t}

}

\cross

\lr{

\BH e^{j\omega t}

+ \BH^\conj e^{-j\omega t}

} \\

&=

\inv{2} \textrm{Re} \lr{ \BE \cross \BH^\conj + \BE \cross \BH e^{2 j \omega t} },

\end{aligned}

\end{equation}

so if we are looking for the relationships that effect only the time averaged Poynting vector, over integral multiples of the period, we are interested in evaluating the divergence of

\begin{equation}\label{eqn:poyntingTimeHarmonic:60}

\inv{2} \BE \cross \BH^\conj.

\end{equation}

The time-harmonic Maxwell’s equations are

\begin{equation}\label{eqn:poyntingTimeHarmonic:80}

\begin{aligned}

\spacegrad \cross \BE &= – j \omega \mu \BH – \BM_i \\

\spacegrad \cross \BH &= j \omega \epsilon \BE + \BJ_i + \sigma \BE \\

\end{aligned}

\end{equation}

The latter after conjugation is

\begin{equation}\label{eqn:poyntingTimeHarmonic:100}

\spacegrad \cross \BH^\conj = -j \omega \epsilon^\conj \BE^\conj + \BJ_i^\conj + \sigma^\conj \BE^\conj.

\end{equation}

For the divergence we have

\begin{equation}\label{eqn:poyntingTimeHarmonic:120}

\begin{aligned}

\spacegrad \cdot \lr{ \BE \cross \BH^\conj }

&=

\BH^\conj \cdot \lr{ \spacegrad \cdot \BE }

-\BE \cdot \lr{ \spacegrad \cdot \BH^\conj } \\

&=

\BH^\conj \cdot \lr{ – j \omega \mu \BH – \BM_i }

– \BE \cdot \lr{ -j \omega \epsilon^\conj \BE^\conj + \BJ_i^\conj + \sigma^\conj \BE^\conj },

\end{aligned}

\end{equation}

or

\begin{equation}\label{eqn:poyntingTimeHarmonic:140}

0

=

\spacegrad \cdot \lr{ \BE \cross \BH^\conj }

+

\BH^\conj \cdot \lr{ j \omega \mu \BH + \BM_i }

+ \BE \cdot \lr{ -j \omega \epsilon^\conj \BE^\conj + \BJ_i^\conj + \sigma^\conj \BE^\conj },

\end{equation}

so

\begin{equation}\label{eqn:poyntingTimeHarmonic:160}

\boxed{

0

=

\spacegrad \cdot \inv{2} \lr{ \BE \cross \BH^\conj }

+ \inv{2} \lr{ \BH^\conj \cdot \BM_i

+ \BE \cdot \BJ_i^\conj }

+ \inv{2} j \omega \lr{ \mu \Abs{\BH}^2 – \epsilon^\conj \Abs{\BE}^2 }

+ \inv{2} \sigma^\conj \Abs{\BE}^2.

}

\end{equation}