Electromagnetism equivalence theorem

[EmilyRuck]

Hello!
In http://my.ece.ucsb.edu/York/Bobsclass/201C/Handouts/Chap1.pdf, pages 19-20, the Love's Theorem in Electromagnetism is declared. In presence of some electric sources $\mathbf{J}$ and magnetic sources $\mathbf{M}$ enclosed by an arbitrary geometrical surface $S$, which produce outside $S$ a field $\mathbf{E}, \mathbf{H}$ and on $S$ a field $\mathbf{E}_S, \mathbf{H}_S$, it is possible to find another solution to Maxwell's equations with:

• zero sources and zero fields inside $S$;
• the same field $\mathbf{E}, \mathbf{H}$ outside $S$;
• impressed $\mathbf{J}_S = \mathbf{\hat{n}} \times \mathbf{H}_S$ and $\mathbf{M}_S = \mathbf{E}_S \times \mathbf{\hat{n}}$ surface currents on $S$.

But there is an alternative solution, in which the volume $V$ enclosed by $S$ is filled with a Perfect Electric Conductor (PEC). In this case we should have:

• again zero sources and zero fields inside $S$ due to the PEC;
• a field $\mathbf{E}, \mathbf{H}$ outside $S$;
• only impressed $\mathbf{M}_S = \mathbf{E}_S \times \mathbf{\hat{n}}$ surface currents on $S$.

According to the Uniqueness Theorem, a solution to Maxwell's equation is unique if the sources and the boudary conditions (relative to $\mathbf{E}$ OR to $\mathbf{H}$) are provided and this fact is exploited in order to prove the Equivalence Theorem just written in its two forms.

My questions are:

1. If the boundary conditions of $\mathbf{E}$ OR to $\mathbf{H}$ are required, why in the first case we need to specify both the surface currents?
2. How can the only $\mathbf{M}_S$ currents generate both the fields in the second case?

Thank you for having read.

Emily

 

[vanhees71]

I'm not sure, how to prove these equivalence theorems. Perhaps you can achieve it, using the retarded potentials.

 

[jasonRF]

Hello!
In http://my.ece.ucsb.edu/York/Bobsclass/201C/Handouts/Chap1.pdf, pages 19-20, the Love's Theorem in Electromagnetism is declared. In presence of some electric sources $\mathbf{J}$ and magnetic sources $\mathbf{M}$ enclosed by an arbitrary geometrical surface $S$, which produce outside $S$ a field $\mathbf{E}, \mathbf{H}$ and on $S$ a field $\mathbf{E}_S, \mathbf{H}_S$, it is possible to find another solution to Maxwell's equations with:

• zero sources and zero fields inside $S$;
• the same field $\mathbf{E}, \mathbf{H}$ outside $S$;
• impressed $\mathbf{J}_S = \mathbf{\hat{n}} \times \mathbf{H}_S$ and $\mathbf{M}_S = \mathbf{E}_S \times \mathbf{\hat{n}}$ surface currents on $S$.

But there is an alternative solution, in which the volume $V$ enclosed by $S$ is filled with a Perfect Electric Conductor (PEC). In this case we should have:

• again zero sources and zero fields inside $S$ due to the PEC;
• a field $\mathbf{E}, \mathbf{H}$ outside $S$;
• only impressed $\mathbf{M}_S = \mathbf{E}_S \times \mathbf{\hat{n}}$ surface currents on $S$.

According to the Uniqueness Theorem, a solution to Maxwell's equation is unique if the sources and the boudary conditions (relative to $\mathbf{E}$ OR to $\mathbf{H}$) are provided and this fact is exploited in order to prove the Equivalence Theorem just written in its two forms.

My questions are:

1. If the boundary conditions of $\mathbf{E}$ OR to $\mathbf{H}$ are required, why in the first case we need to specify both the surface currents?
2. How can the only $\mathbf{M}_S$ currents generate both the fields in the second case?

Thank you for having read.

Emily

Here are my best answers for the moment ...

1. Because you were specifying the field both inside and outside of $S$, so in some sense you need conditions for both the internal and external problems. If you don't care about the fields inside of S you should be able to construct a solution using either a magnetic or an electric surface current that gnerates the correct fields outside of S, but nonzero fields inside of S. For example, what if you assume there is no discontinuity of the electric field across the S?

2. In the second case the magnetic current is radiating in the presence of the PEC; this is completely different than radiating in a homogeneous space.

jason