Poynting relationship

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### Problem:

Given

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

\spacegrad \cross \BE

= -\BM_i – \PD{t}{\BB},

\end{equation}

and

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

\spacegrad \cross \BH

= \BJ_i + \BJ_c + \PD{t}{\BD},

\end{equation}

expand the divergence of \( \BE \cross \BH \) to find the form of the Poynting theorem.

### Solution:

First we need the chain rule for of this sort of divergence. Using primes to indicate the scope of the gradient operation

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

\begin{aligned}

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

&=

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

–

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

&=

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

–

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

&=

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

–

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

\end{aligned}

\end{equation}

In the second step, cyclic permutation of the triple product was used.

This checks against the inside front cover of Jackson [1]. Now we can plug in the Maxwell equation cross products.

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

\begin{aligned}

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

&=

\BH \cdot \lr{ -\BM_i – \PD{t}{\BB} }

–

\BE \cdot \lr{ \BJ_i + \BJ_c + \PD{t}{\BD} } \\

&=

-\BH \cdot \BM_i

-\mu \BH \cdot \PD{t}{\BH}

–

\BE \cdot \BJ_i

–

\BE \cdot \BJ_c

–

\epsilon \BE \cdot \PD{t}{\BE},

\end{aligned}

\end{equation}

or

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

\boxed{

0

=

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

+ \frac{\epsilon}{2} \PD{t}{} \Abs{ \BE }^2

+ \frac{\mu}{2} \PD{t}{} \Abs{ \BH }^2

+ \BH \cdot \BM_i

+ \BE \cdot \BJ_i

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

}

\end{equation}

# References

[1] JD Jackson. *Classical Electrodynamics*. John Wiley and Sons, 2nd edition, 1975.