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

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?

• Hint: $u$ is not energy, is energy density. You have derived the continuity equation en.wikipedia.org/wiki/Continuity_equation#Differential_form Jun 13, 2016 at 8:53
• Indeed, I am aware that I have derived the continuity equation. And yes, $u$ was a mistake. Correction has been made, Jun 13, 2016 at 8:56
• Jun 13, 2016 at 8:56
• I have an insight. Clearly, I made the mistake of performing dimensional analysis on the left hand side. I could have just work on the right hand side. The right hand side has units of joules per unit volume. Taking the time derive once, a unit of time in second is introduced into the denominator. This gives joules per unit volume per second. However, it should, correctly be joules per unit area per unit time.. Jun 13, 2016 at 8:59
• You need to concentrate on the work done by the field in any matter; add $\bf E\cdot J$ to complete the continuity equation.
– user36790
Jun 13, 2016 at 9:02

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.