- #1

Unconscious

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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 ##?