Dot product of the electric and magnetic field as the contraction of the electromagnetic tensor and its dual

Question

I've see in some examples, e.g. here, that $$-4i\vec{E}\cdot \vec{B}=\tilde{F}_{\mu\nu}F^{\mu\nu}$$

How would you show such a relation? By inserting terms by terms inside this equation I've seen it is correct, but I would like to prove it in a more general way. I was only able to show this:

$\tilde{F}_{\mu\nu}F_{\mu\nu}=\frac{1} {2}\epsilon_{\mu\nu\rho\sigma}F^{\mu\nu}F^{\rho\sigma}$

which is the definition of the dual but then I don't know how to go on. I know that $E_i=iF_{i4}$ and $F_{ij}=\epsilon_{ijk}B_k$ and that $\vec{E} \cdot \vec{B}=E_iB_i$ but then I have a $4$ dimensions Levi Civita on one side and a $3$ dimensions Levi Civita on the other side, any hint to go on?

[EDIT]

Based on the comments I came to the conclusion, it might be useful for someone:

$\tilde{F}_{\mu\nu}F_{\mu\nu}=\frac{1}{2}\epsilon_{\mu\nu\rho\sigma}F_{\mu\nu}F_{\rho\sigma}=\frac{1}{2}\epsilon_{\mu\nu\rho4}F_{\mu\nu}F_{\rho4}+\frac{1}{2}\epsilon_{\mu\nu\rho i}F_{\mu\nu}F_{\rho i}=\frac{1}{2}\epsilon_{\mu\nu i4}F_{\mu\nu}F_{i 4}+\frac{1}{2}\epsilon_{\mu\nu 4 i}F_{\mu\nu}F_{4i}+\frac{1}{2}\epsilon_{\mu\nu j i}F_{\mu\nu}F_{ji}=\frac{1}{2}\epsilon_{\mu j i 4}F_{\mu j}F_{i4}+\frac{1}{2}\epsilon_{\mu j4i}F_{\mu j}F_{4i}+\frac{1}{2}\epsilon_{\mu 4ji}F_{\mu 4}F_{ji}+\frac{1}{2}\epsilon_{\mu kji}F_{\mu k}F_{ji}=\frac{1}{2}\epsilon_{k j i 4}F_{k j}F_{i4}+\frac{1}{2}\epsilon_{k j4i}F_{k j}F_{4i}+\frac{1}{2}\epsilon_{k 4ji}F_{k 4}F_{ji}+\frac{1}{2}\epsilon_{4 kji}F_{4 k}F_{ji}=\epsilon_{kji4}F_{kj}F_{i4}+\epsilon_{4kji}F_{4k}F_{ji}=\epsilon_{kji}F_{kj}F_{i4}-\epsilon_{kji}F_{4k}F_{ji}=\epsilon_{ikj}F_{kj}F_{i4}-\epsilon_{ikj}F_{kj}F_{4i}=-2iE_iB_i-2iE_iB_i=-4iE_iB_i=-4i\vec{E}\cdot \vec{B}$

P.S.: there may be confusion about signs or imaginary unit since they depend upon the metric used.

Answer 1

To continue down the path you specified please use the following definition of the electromagnetic tensor $$F\,^{\mu\nu} = \partial\,^{\mu}A^{\nu} - \partial\,^{\nu}A^{\mu}$$. The standard definition of it (and how to recover the fields from the potential 4-vector) can be found here.

If you use this instead of the 3-index version you specify at the end of the question, you won't have the problem with the levi-cevitas having fewer indeces than needed.

Alternatively, expand the sum as suggested by Prahar in the comments in the following way: $$F^{\mu\nu}𝐹_{\mu\nu}=\frac{1}{2}\epsilon_{\mu\nu\sigma\rho}F_{\mu\nu}F_{\rho\sigma}=\frac{1}{2}\epsilon_{\mu\nu\rho 0}F^{\mu\nu}F^{\rho 0}+\frac{1}{2}\epsilon_{\mu\nu\rho i}F^{\mu\nu}F^{\rho i}.$$ where you can now used the relationships with the 3-index levi-cevita. Note that you will have to do the same expansion for the sum over all the remaining indices to get an expresion purely consisting of $F_{0i}$'s, $F_{i0}$'s and $F_{ij}$s.

Further expansion: Let’s do the same thing with the $\rho$ index: $$\frac{1}{2}\epsilon_{\mu\nu\rho 0}F^{\mu\nu}F^{\rho 0}+\frac{1}{2}\epsilon_{\mu\nu\rho i}F^{\mu\nu}F^{\rho i}= \frac{1}{2}\epsilon_{\mu\nu 00}F^{\mu\nu}F^{\rho 0}+ \frac{1}{2}\epsilon_{\mu\nu j0}F^{\mu\nu}F^{j0} + \frac{1}{2}\epsilon_{\mu\nu 0 i}F^{\mu\nu}F^{0i} + \frac{1}{2}\epsilon_{\mu\nu j i}F^{\mu\nu}F^{ji} $$ where the term with the two indices set to 0 is just equal to $0$ via how levi-cevitas work. The $i,j$ indices are to be understood as from 1 to 3, rather than 0 to 3.

Another useful formula from the comments is $\epsilon_{0ijk} = \epsilon_{ijk}$. This will be needed towards the end of the algebra.