Basic question in electromagnetic duality

[PhyAmateur]

We originally have $$\overrightarrow{\nabla}\cdot\overrightarrow{B} = 0$$

$$\overrightarrow{\nabla}\times\overrightarrow{E} = -\frac{\partial \overrightarrow B}{\partial t}$$

When electromagnetic duality is concerned this rank 2 tensor kicks in:$$G^{\mu\nu}$$

And most of books and sites define $$D_i = G_{i0}$$ and $$H_i = 1/2 \epsilon_{ijk}G_{jk}$$
My question is why is this $$G^{\mu\nu}$$ related to our previously known electric induction D and magnetic intensity H?
In other words,how comes that $$G^{\mu\nu}$$ this new notion that rised after duality came in related to our old notion of electric induction D and magnetic intensity H?

 

[gtoguy97]

I'm looking for a strong mathematical proof that shows the relation between them.Any help would be appreciated!Thanks in advance!A:In general, if you have a tensor $T^{\mu\nu}$, and you don't already know what it is, then the only way to find out is by looking at the components. If you look at the components of the electromagnetic duality tensor you find that it has the form $$G^{\mu\nu} = \left( \begin{matrix} 0 & D_x & D_y & D_z \\ -D_x & 0 & H_z & -H_y \\ -D_y & -H_z & 0 & H_x \\ -D_z & H_y & -H_x & 0 \end{matrix} \right).$$This should remind you of the components of the electric and magnetic fields. To make this comparison you can let $E_x = -D_x, E_y = -D_y, E_z = -D_z, B_x = H_z, B_y = -H_x, B_z = H_y$. It is easy to check that this definition satisfies the equation $E_i = \epsilon_{ijk}B_j$ and so you can conclude that $G^{\mu\nu}$ is related to the electric and magnetic fields in the way you expected it to be.