I Equivalence principle and the Uniqueness theorem

[Unconscious]

We work with Maxwell's equations in the frequency domain.
Let's consider a bounded open domain $V$ with boundary $\partial V$.

1. The equivalence theorem tells me that if the field sources in $V$ are assigned and if the fields in the points of $\partial V$ are assigned, then I can compute the field in each point of $V$ as ( Stratton-Chu solution):

$$\mathbf{E}(\mathbf{r})=\frac{1}{4\pi}\int_V\left( \frac{\rho}{\epsilon}\nabla'\psi-j\omega\mu\psi\mathbf{J} \right)dV+\frac{1}{4\pi}\int_{\partial V}(\mathbf{n}_0\cdot\mathbf{E})\nabla'\psi-j\omega\psi(\mathbf{n}_0\times\mathbf{B})+(\mathbf{n}_0\times\mathbf{E})\times\nabla'\psi)dS$$
$$\mathbf{B}(\mathbf{r})=\frac{1}{4\pi}\int_V\left( \mu\mathbf{J}\times\nabla'\psi \right)dV+\frac{1}{4\pi}\int_{\partial V}(\mathbf{n}_0\cdot\mathbf{B})\nabla'\psi-j\frac{\omega\psi}{c^2}(\mathbf{n}_0\times\mathbf{E})+(\mathbf{n}_0\times\mathbf{B})\times\nabla'\psi dS$$

where $\psi=\frac{e^{-jkR}}{R}$ is the Green function.

2. The uniqueness theorem tells me that if only the tangential component of only the electric field (or only the magnetic field) on $\partial V$ is assigned, then the field in the points inside $V$ is uniquely determined.

I wonder then: why it seems that for the equivalence theorem it is necessary to know the entire field (both electric and magnetic and both normal and tangential components) on $\partial V$, while the uniqueness theorem needs much less information ? Is it just a question of calculation? In the sense that perhaps it is true that less information is needed to uniquely determine the field, but then to actually calculate it we do not know how to do it if we do not have all the information that the equivalence theorem requires on $\partial V$?

 

[vanhees71]

I think in this case you need the boundary conditions as well as initial conditions for the em. field. You can also work in the boundary conditions into the Green's function. The Green's function you quote should be the usual retarded Green's function (i.e., the out-going wave solutions of the Helmholtz equation, i.e., the temporal part in your ansatz should be $\exp(+\mathrm{j} \omega t)$, i.e., the engineering convention, which is opposite from the physics convention! This Green's function holds if you take the entire $\mathbb{R}^3$, where the complete sources $\rho$ and $\vec{j}$ are defined ("microscopic electrodynamics"), i.e., in your boundary-value problem you need to add an arbitrary solution of the source-free Helmholtz equation, which then has to be found by employing the initial and boundary conditions.

 

[Unconscious]

Why do you talk about 'initial conditions'? We are working in frequency domain, so there is no 'initial time', there are only boundary conditions. No?

The eventually presence of an additional source-free solution will modify the field value on the boundary points, so this fact is automatically taken into account:
1. in the Stratton-Chu solution, by the terms in the surface integral (boundary conditions),
2. in the uniqueness theorem, by the boundary conditions again.

I don't know if you wanted to tell me anything else that I didn't understand.

 

[vanhees71]

I'm not so familiar with the engineering approach. So if you just look at the harmonic time dependence, i.e., the Helmholtz equations boundary values are sufficient.

 

[Unconscious]

When you talk about Helmoltz equation, what I think is this:

$$\nabla^2 \mathbf{A}(\mathbf{r})+k^2\mathbf{A}(\mathbf{r})=-\mathbf{J}(\mathbf{r})$$

where $\mathbf{A}(\mathbf{r})$ is the magnetic vector potential and $\mathbf{J}(\mathbf{r})$ is the current density (the source).
Se, when you say that ' Helmholtz equations boundary values are sufficient ' I agree, because assigning the values of the magnetic vector potential on $\partial V$ is equivalent (in the sense that then the field is uniquely determined) to assign the values of the field on $\partial V$.
But this does not solve the problem, it still seems that the extra-information needed by Stratton-Chu integration are only for effective calculus reasons.