electromagnetism - Significance of the Dual Electromagnetic Tensor $tilde{mathbf{F}}$/its derivation

In the context of Maxwell's equations, I was wondering whether there was any physical significance to the dual EM Field Tensor and/or its various derivations. It has components: $$\tilde{\textbf{F}} = \begin{bmatrix} 0 & B_1 & B_2 & B_3 \\ -B_1 & 0 & -E_3 & E_2 \\ -B_2 & E_3 & 0 & -E_1 \\ -B_3 & -E_2 & E_1 & 0 \end{bmatrix}.$$ I have seen two ways of deriving it, one which uses index lowering via the Minkowski Metric [1]: $$\tilde{F}_{\mu\nu} = \eta_{\mu\alpha}F^{\alpha\beta}\eta_{\beta\nu}$$ and another which uses the Levi-Civitia Tensor [2, P.111]: $$\tilde{F}_{\mu \nu} = \frac{1}{2}\epsilon_{\mu \nu \alpha \beta}F^{\alpha \beta}.$$ [I am also aware it can be reached via the substitution $ E_m \to -B_m$ and $B_m \to E_m$ however it seems as though this is just a consequence of the above definitions.]

Are both these definitions just formalisms so the dual EM tensor has the form it does so it can be used in certain physical equations (such as the simplification of Maxwell's equations: $\partial_\mu\tilde{F}_{\mu \nu}$) or is there any physical intuition to the way(s) in which it is defined?

Answer

To understand the reason for defining the dual electromagnetic tensor, you need to understand electromagnetism in the language of differential forms. A highly recommended book for that is Gauge Fields, Knots and Gravity by Baez and Muniain.

When you read that book you will see that the electromagnetic tensor $F$ is a 2-form. In simple terms, an $n$-form can be described as a tensor with $n$ antisymmetric indices, and so it makes sense that $F_{\mu\nu}$ is a 2-form if it's antisymmetric in $\mu$ and $\nu$ (which it is).

In $d$ dimensions, there is an operation called the Hodge dual and denoted by a star $\star$, which takes an $n$-form $A$ to a $(d-n)$-form $\star A$. So, in the special case of the electromagnetic tensor $F$ in 2 dimensions, we have $n=2$ and $d=4$, and the dual $\star F$ of the electromagnetic tensor is another 2-form defined as follows:

$$(\star F)_{\mu\nu} \equiv \frac{1}{2}\epsilon_{\mu\nu\rho\sigma}F^{\rho\sigma}.$$

Now, in the language of differential forms, Maxwell's equations can be very expressed elegantly as:

$$dF=0,$$

$$d \star F = \star J,$$

where $J$ is the current, which is a 1-form. The first line gives you the first two Maxwell's equations in terms of $E$ and $B$, and the second line gives you the two other Maxwell's equations in terms of $E$ and $B$.

Crucially, it is impossible to get all 4 Maxwell's equations using $F$ alone; its dual $\star F$ must be used in the second equation.

I urge you to read the book I recommended above for more information; all of this, and much more, is explained very clearly there.