Why does the divergence of the Poynting vector have energy flux density?

Question

The poynting vector is defined as

$\vec{S}=\mu_{0}^{-1}\vec{E}\times \vec{B}$

Taking the divergence of the poynting vector, one arrives at

$\vec{\nabla} \cdot \vec{S}=-\frac{\partial u}{\partial t}=0$

after some algebraic manipulation.

Note $u$ is the electromagnetic energy density.

The claim is that the $\vec{\nabla} \cdot \vec{S}$ is an energy flux density.

How do I see this is true?

Thanks in advance.

Answer 1

The converse is not a "theorem" in that you can't prove that it is true, even assuming Maxwell's equations as axioms. It is simply a "hunch" that $|\vec{S}|$ represents the power intensity and $U=\frac{1}{2}\,\epsilon\,|\vec{E}|^2+\frac{1}{2}\,\mu\,|\vec{H}|^2$ the energy density. What you can prove (and what you already understand) from Maxwell's equations is that:

1. $\vec{S}$ is the flux of $U$ since the pair together fulfill a continuity equation in source free zones: where there are sources, the difference between $\vec{S}$ and the time rate of change of $U$ can be shown to be the rate of working of electric fields on currents (the $\vec{E}\cdot\vec{J}$ term);
2. If $U$ integrated over all space can be defined (i.e. a convergent integral) then $U$ is constant if $\epsilon$ and $\mu$ are real (i.e. no absorbing materials) and there are no sources, or that its rate of change equals the rate of working of the $\vec{E}$ field (the integrated $\vec{E}\cdot\vec{J}$ term) if there are sources.

So the pair together behave exactly as an energy density and its flux vector ought to. We therefore postulate that they are the energy density and its flux vector and, so far, no experimental result has failed to agree with calculations made assuming this postulate.