I refer to the time-domain version of the Poyinting theorem in electro-magnetism:

$- \displaystyle \oint_S (\mathbf{E} \times \mathbf{H}) \cdot d\mathbf{S} - \int_V \mathbf{E} \cdot \mathbf{J}_i \ dV = \int_V \frac{\partial}{\partial t} \frac{\mu |\mathbf{H}|^2}{2} \ dV + \int_V \frac{\partial}{\partial t} \frac{\epsilon |\mathbf{E}|^2}{2} \ dV + \int_V \sigma |\mathbf{E}|^2 dV$

where $S$ is the boundary surface of the integration volume $V$. With $\mathbf{J}_i$ I mean impressed currents (imposed by some field source, like antennas) and with $\mathbf{J}$ I mean induced currents (induced on some conductor, if present, into $V$).

The product $\mathbf{E} \cdot \mathbf{J}_i$ must be negative if $\mathbf{J}_i$ is the current that generates $\mathbf{E}$. But if some current $\mathbf{J}_i$ generates power into $V$, some power should enter $V$ to feed $\mathbf{J}_i$.

If there is a Poynting vector entering $V$ and providing power to $\mathbf{J}_i$, can we consider both (the Poynting vector and the current $\mathbf{J}_i$) as increasing the power into $V$, and so both negative? Should not the Poynting vector entering $V$ be considered the only source of power in $V$?

As an example we can consider a volume $V$ enclosing an antenna. There will be a term $\mathbf{E} \cdot \mathbf{J}_i$ due to the radiation; there will be a Poynting vector exiting from the volume (radiated power); but what about the Poyting vector related to the feed line of the antenna? It is entering the volume and it has a different form than the previous one, because it is the Poyting vector of a transmission line and not of a radiated field. How is it represented in the above equation?

Thank you anyway!


1 Answer 1


$\def\vH{{\mathbf{H}}}$ $\def\vB{{\mathbf{B}}}$ $\def\vE{{\mathbf{E}}}$ $\def\rot{\operatorname{rot}}$ $\def\div{\operatorname{div}}$ $\def\rmi{{\mathrm i}}$ Your problem is that you count the ohmic losses per volume $\mathbf{E}\cdot\mathbf{J}_i$ twice. On the left hand side as $\mathbf{E}\cdot\mathbf{J}_i$ and on the right hand side as $\sigma|\mathbf{E}|^2$. It is just wrong that $\mathbf{E}\cdot\mathbf{J}_i$ would provide power. The (input) power flux is only provided by the Poynting vector $\mathbf{E}\times\mathbf{H}$.

This is even the case if you have a generator in your volume that is provided with power from a DC-power supply and the DC-wires enter your volume.

Also the DC-power is transported through the (constant) power flux $\mathbf{E}\times\mathbf{H}$ in the dielectric of the DC-wire. The part $\mathbf{E}\cdot\mathbf{J}_i$ are the ohmic losses (mostly in the metal) that are transformed into heat.

Induction corresponds to a decrease of magnetic power $\vH\cdot\dot\vB=-\vH\cdot\rot\vE$. This becomes a part of Poynting's power flow.

What I could understand is the transformation of chemical or thermic energy in to electrical energy. This would impose an electrical field $\vE_\rmi$ through the difference of the electrode potentials or diffusion, respectively.

  • $\begingroup$ Maybe my notation was confusing. I considered $\mathbf{J}$ as induced currents (in some conductor present in the volume $V$), while $\mathbf{J}_i$ as impressed currents. How can $\mathbf{E} \cdot \mathbf{J}_i$ be ohmic, if $\mathbf{J}_i$ is an impressed current? $\endgroup$
    – BowPark
    Mar 12, 2014 at 19:10
  • $\begingroup$ @BowPark I have added a comment on induced and imposed fields. For a discussion we should switch to chat.stackexchange.com. Best regards, Tobias. $\endgroup$
    – Tobias
    Mar 13, 2014 at 0:00

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