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?


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.

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    – Qmechanic
    Commented Dec 28, 2023 at 11:51

1 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.

  • $\begingroup$ I still don't understand how to proceed, I've tried to expand like Prahar suggested but then I'm left with 2 levi civita tensors, one with 4 indexes and the other with 3 indexes, and a lot of $F_{\mu\nu}$ terms. $\endgroup$
    – Salmon
    Commented Dec 27, 2023 at 19:39
  • $\begingroup$ I still don't understand... I've also edited my question inserting the expansion but I don't know how to proceed. $\endgroup$
    – Salmon
    Commented Dec 27, 2023 at 19:55
  • $\begingroup$ All the terms with sigma are not needed and were taken care off by the very first expansion. All the terms with rho also disapear after the second expansion. The sigma sum became a sum over $i$ and the zero term written out on its own $\endgroup$
    – JohnA.
    Commented Dec 27, 2023 at 19:59
  • $\begingroup$ I'm sorry but I'm not able to compute the demonstration, I the only thing I have in mind is to write explicitly all the terms but I would like to write, if possible, a more general formula. $\endgroup$
    – Salmon
    Commented Dec 27, 2023 at 20:48

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