# Electromagnetism duality theorem

Concerning Electromagnetism, textbooks often refer to the Duality Theorem. Sometimes it is presented like this:

«Consider the Maxwell's Equations (with phasors) and a known field $\mathbf{E}_1$, $\mathbf{H}_1$:

$\nabla \times \mathbf{E}_1 = - j \omega \mu \mathbf{H}_1$

$\nabla \times \mathbf{H}_1 = j \omega \epsilon \mathbf{E}_1$

If $\mathbf{E}_1$ is replaced with $\mathbf{H}_2$ (the magnetic field of another electromagnetic field: $\mathbf{E}_2$, $\mathbf{H}_2$) and $\mathbf{H}_1$ is replaced with $-\mathbf{E}_2$, $\mu$ is replaced with $\epsilon$ and $\epsilon$ with $\mu$, then the above equations become respectively

$\nabla \times \mathbf{H}_2 = j \omega \epsilon \mathbf{E}_2$

$\nabla \times \mathbf{E}_2 = - j \omega \mu \mathbf{H}_2$»

They are valid Maxwell's Equations too. But now what does follow from this substitution?

1) Should be $\mathbf{E}_1 = \mathbf{H}_2$ and $\mathbf{H}_1 = -\mathbf{E}_2$? It is dimensionally incorrect.

2) I read also $\mathbf{E}_1 = \eta \mathbf{H}_2$ and $\eta \mathbf{H}_1 = -\mathbf{E}_2$.

My question is double: what is the advantage of Duality Theorem and which is the correct form between the two just written?

Thank you anyway!

• regarding the dimensions, there are many different unit conventions you can use, and although E and H have different dimensions in SI units, they are the same in cgs. – George G Mar 25 '14 at 20:11