# Power transferred from the sources to the electromagnetic field

1. Feb 25, 2013

### EmilyRuck

Hello!!
In an electro-magnetic context, the power that an electric source of field delivers to the field itself may be written as

$p_S = - \mathbf{E} \cdot \mathbf{J}$

where $\mathbf{E}$ is the electric field produced by the source and $\mathbf{J}$ is the corrent flowing on the source, forced by the generator. I have some doubts about the procedure to obtain the above expression.
The current $J$ may be seen as a charge density $\rho$ which moves in a region with velocity $\mathbf{v}$. If in the region is present an electromagnetic field $\mathbf{E}, \mathbf{H}$, the charge will experience a force per unit volume

$\mathbf{f} = \rho \mathbf{E} + \rho \mathbf{v} \times \mathbf{H}$

The charge will also experience an infinitesimal displacement $d\mathbf{r}$, in a time $dt$. If we had $d\mathbf{r} / dt = \mathbf{v}$, then we could say that the power per unit volume from the field is

$\mathbf{f} \cdot \mathbf{v} = \rho \mathbf{E}\cdot \mathbf{v} + \rho \mathbf{v} \times \mathbf{H} \cdot \mathbf{v} = \rho \mathbf{E}\cdot \mathbf{v} = \mathbf{E}\cdot \mathbf{J}$

This is the power delivered from the field to the charge density. If we are interested in the power delivered from the charge density to the field, we have just to change the sign and we will obtain $- \mathbf{E}\cdot \mathbf{J}$.

But if the moving charge $\rho$ is the source of the field, how can she "suffer" the effect of the field? If the source was an electric dipole, I imagine the field as something which goes away from the source, without interacting with the source. How can a field exercise a force on its source, minute by minute?
Moreover, the displacement $d\mathbf{r}$ is due to the generator and not the field. Is it correct to calculate the work, and consequently the power, considering a force and a displacement which is not caused from the force?
Thank you for having read it.

Emily

2. Feb 25, 2013

### marcusl

In essence, this is equivalent to P = V * I in a conventional circuit, where voltage V and current I are independent of each other as determined by the circuit and the source drive. So the coupling that you are worried about is not really an issue.

Being careful reveals that $\vec{E}\cdot\vec{J}$ is power density, not power. It comes from applying the divergence theorem to the Poynting vector. The derivation starts from two Maxwell equations $$\nabla\times\vec{E}=-\vec J_m,\\ \nabla\times\vec{H}=\vec J_e$$ where J_e is current density and J_m is effective magnetic current density. Multiply the first equation by H* and the second by E, take their difference and apply a vector identity to get $$\nabla\cdot(\vec E\times\vec H^*) + \vec E\cdot\vec J_e^*+\vec H^*\cdot\vec J_m=0.$$ Integrate over all space and apply the divergence theorem to get
$$\oint \vec E \times \vec H^* \cdot d\vec A + \int(\vec E\cdot\vec J_e^*+\vec H^*\cdot\vec J_m)dV=0.$$ The last term is zero in the absence of magnetic material, so the volume integral of $\vec E \cdot \vec J_e$ equals the surface integral of the Poynting vector, or, in other words, radiated power. That makes EJ itself a power density.

3. Feb 28, 2013

### EmilyRuck

Ok! They are indipendent because in this case only one of them is forced by the source.

Yes, obviously, maybe I wrote in a rush, but I know that it is a power density. My problem was not about the $\mathbf{E} \cdot \mathbf{J}$ product, but about the above procedure to associate it to a power density. In other words, why the generator must win the force

$\mathbf{f} = \rho \mathbf{E} + \rho \mathbf{v} \times \mathbf{H}$ ?

$\mathbf{E}, \mathbf{H}$ is the field just produced by the accelerated charge. How can this field influence the accelerated charge itself? Shouldn't the field just go away from the source?

Emily

4. Feb 28, 2013

### marcusl

Ok, I see where the confusion is--you are speaking of expressions related to the Lorentz force but are still talking about them as power quantities, which they aren't. You'll notice that I wrote power density expressions that had a complex conjugation. (Oops--I see I left the star off of $\vec E \cdot \vec J_e^*$ in the second and the second-to-last sentences of post #2.) Conjugation matters, of course, and here it produces a real power quantity.

Since your expressions relating to the Lorentz force have no conjugation, they do not have units of power density and cannot be related to the Poynting theorem. We are allowed to examine them, which I'll do next, so long as we stop talking about power.

The E field generated by J acts back upon J, producing some effect that is captured, in the absence of magnetic material, by a volume integral over $\vec E \cdot \vec J$ (I'll drop the subscript "e"). Rumsey symbolized this quantity for a system "a" by <a,a> in a paper from the 50's, and called it "reaction." The term is appropriately descriptive of the effect the current-generated E field has on the current itself. If we consider the current density to flow in a device such as an antenna, as an example where it is easy to define the total current I at the antenna port, then the quantity $$Z_{aa}=\frac{<a,a>}{I^2}$$ is called the impedance or, more properly, self-impedance. It is generally a complex quantity. The imaginary part is called "reactance" which, again, supports the notion that the device produces an electromagnetic reaction onto the source. So the forces that you are looking at do, indeed, cause a back reaction onto the current source and we capture it for devices like antennas, coils, capacitors, etc. through their self impedances.

The concept can be extended to more complicated multi-port systems as well. In these cases the reaction leads to reciprocity relations (the Lorentz reciprocity theorem may have been the first) and to mutual impedances.

5. Mar 1, 2013

### EmilyRuck

No, I consider them (density) power quantities only after the multiplication by $\mathbf{v}$.

Yes, it is essential with phasors. Anyway consider my expressions in the time domain, say in a particular instant (with time-varying fields, the only one that can produce a radiation), not with phasors.

Ok! Thank you, so this is true. I just couldn't figure out how it could happen.

Emily