2
$\begingroup$

In the electromagnetic Uniqueness theorem, we consider a volume $V$ enclosed by a surface $S$. It is initially assumed that two different fields are valid solutions for the Maxwell's equations with the same sources, $\mathbf{E}^a, \mathbf{H}^a$ and $\mathbf{E}^b, \mathbf{H}^b$.

Then we consider the difference fields $\delta \mathbf{E} = \mathbf{E}^a - \mathbf{E}^b$ and $\delta \mathbf{H} = \mathbf{H}^a - \mathbf{H}^b$ and the solution is unique if $\delta \mathbf{E} = 0$, $\delta \mathbf{H} = 0$.

Any possible proof needs that (with phasors)

$$\displaystyle \oint_S \delta \mathbf {E} \times \delta \mathbf{H}^* \cdot d\mathbf{S} = 0$$

(then with other considerations it will lead to $\delta \mathbf{E} = 0$, $\delta \mathbf{H} = 0$. An example of the complete procedure may be found here, NB: pdf).

$\mathbf{\hat{n}}$ is the unit vector normal to $S$ and in the integral $d \mathbf{S} = dS \mathbf{\hat{n}}$.

The above integral vanishes if $\mathbf{\hat{n}} \times \mathbf{E}$ or $\mathbf{\hat{n}} \times \mathbf{H}$ (or both) is specified and this is possible in a finite volume $V$ if are known the boundary conditions.

But what if the volume is infinite? We only have the Sommerfeld conditions (assuming that $-\mathbf{\hat{n}}$ is the direction of propagation of the fields):

$$\displaystyle \lim_{r \to \infty} r|\mathbf{E}| < q_1$$

$$\displaystyle \lim_{r \to \infty} r|\mathbf{H}| < q_2$$

$$\displaystyle \lim_{r \to \infty} r(\mathbf{E} + \eta \mathbf{H} \times \mathbf{\hat{n}}) = 0$$

$$\displaystyle \lim_{r \to \infty} r \left( \mathbf{H} + \frac{\mathbf{\hat{n}} \times \mathbf{E}}{\eta} \right) = 0$$

with $q_1, q_2$ arbitrary real constants. They don't give a direct value for $\mathbf{\hat{n}} \times \mathbf{E}$; they just relate $\mathbf{\hat{n}} \times \mathbf{E}$ with $\mathbf{\hat{n}} \times \mathbf{H}$ (which are the tangential components on the infinite $S$) and they state that the power of the fields is exiting from $V$. So the fields are not constrained to have a fixed value and direction: it is sufficient that $\mathbf{E} \times \mathbf{H}^*$ is exiting from $S$ and there may be many $\mathbf{E}^b, \mathbf{H}^b$ which satisfy the Maxwell's equations in addition to $\mathbf{E}^a, \mathbf{H}^a$.

If the sources are the same, we also know that both $\mathbf{E}^a, \mathbf{H}^a$ and $\mathbf{E}^b, \mathbf{H}^b$ carry the same power to infinity; but the power of the difference field $\delta \mathbf{E}, \delta \mathbf{H}$ does not seem to be the difference of the single powers! So, how to prove even in this case that the above integral vanishes?

The Sommerfeld conditions were exactly established in order to prove the uniqueness of the solution in this case, with an infinite volume. So, what is the trick? How can we use the Sommerfeld condition to vanish the above integral?

$\endgroup$
  • $\begingroup$ according to this image, the normal vector is outgoing from the surface, but in my question I considered it in the opposite direction (this is the reason why the fields are propagating along $-\mathbf{\hat{n}}$). But it does really matter? In the link in the question it is outlined that for Uniqueness does matter that the field is outgoing from $S$, but I don't know how a minus sign could be important in this proof. $\endgroup$ – BowPark Apr 10 '14 at 12:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.