Poynting relationship

### Problem:

Given
\label{eqn:poynting:20}
= -\BM_i – \PD{t}{\BB},

and
\label{eqn:poynting:40}
= \BJ_i + \BJ_c + \PD{t}{\BD},

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

\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}

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.

\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}

or

\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.
}

# References

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