One hard theoretical question

1. Sep 3, 2010

Petar Mali

The problem is with energy of electromagnetic field

$$-\frac{d}{dt}\int_V\frac{1}{2}(\vec{E}\cdot\hat{\epsilon}\vec{E}+\vec{B}\hat{\mu^{-1}}\vec{B})dV=\oint(\vec{E}\times\vec{H})\cdot d\vec{S}+\int_V\vec{j}\cdot\vec{E}dV$$

I have this relation

$$\hat{\epsilon}, \hat{\mu}^{-1}$$ are symmetric tensors. Now we look total field. This is or finite area in which bounaries electric and magnetic field are equal to zero, or whole space with a condition that electric and magnetic field goes to zero at least as $$\frac{1}{r^2}$$ in infinity.

So electric and magnetic field must be functions of $$\frac{1}{r^{2+\epsilon}}$$ where $$\epsilon \geq 0$$

Why this condition must be satisfied?

In first case of finite area $$\oint(\vec{E}\times\vec{H})\cdot d\vec{S}\equiv 0$$. Why is that? I don't understand?

And in second case with take sphere infinitely long away and have

$$lim_{S \rightarrow \infty}\oint(\vec{E}\times\vec{H})\cdot d\vec{S}=lim_{S\rightarrow \infty} [\overline{(\vec{E}\times\vec{H})}_n 4\pi r^2]=0$$

because $$\overline{(\vec{E}\times\vec{H})}_n$$ goes to zero at least as $$\frac{1}{r^4}$$.

2. Sep 3, 2010

gabbagabbahey

If it weren't satisfied, then no matter how far away you got from the source of the fields, you would still feel their effects. This is unphysical, since all charge and current distributions are localized.

You do understand that when you compute the surface integral of some vector field ( $\textbf{E}\times\textbf{H}$ is this case ), the field is going to have some definite value at every point on the surface, right? In this case, $\textbf{E}$ and $\textbf{H}$ are both zero everywhere on the surface you are integrating over (the boundary of whatever finite region you are interested in), and so you are integrating zero times $d\textbf{S}$ over the surface, which clearly results in zero.

3. Sep 4, 2010

Petar Mali

Can you be more specific with bold part?

Right.Why?

4. Sep 19, 2010

Petar Mali

Can anyone explain to me the easy way what is completely field? What is that area? Obviously I have a problem with that!

5. Sep 19, 2010

Petr Mugver

In a finite region, there is no need for the fields to vanish on the boundary. That's exactly the Poynting vector term in your equation: energy variations are due to the flux of the Poynting vector on the boundary (energy that leaves the region) plus energy dissipation inside the region. If the region considered is all space, then yes, the fields must vanish at infinity. But that's always the case for source charge-currents placed in a finite region (and it's not valid, for example, for a uniformly charged wire or plane): you can expand the charge-currents in series of multipoles. The first term, the monopole, causes fields that behave like 1/r^2, the second term, the dipole, has fields like 1/r^3, and so on, so the flux of the fields at large distances in always zero.

6. Sep 19, 2010

Born2bwire

That term is the power flow over the surface of your integrating volume. This has to go to zero at infinity otherwise it implies that the sources are radiating infinite power. The energy of the wavefront must remain constant over the surface that encloses all of the sources. This is due to the fact that the volume is not an active medium. So if we were to spread out this same energy density across an infinite surface then it must become zero for a finite density. A more mathematical relationship can be found via the Sommerfeld radiation condition. It turns out that for the vector wave equation, the fields must go to zero at infinity to ensure a unique solution (an equivalent condition is to introduce an infinitesimal loss to the volume)