# [Physics] Why does the divergence of the Poynting vector have energy flux density

electromagnetismelectrostaticspoynting-vectorVector Fieldsvectors

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?

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.