Electromagnetic action in differential forms

[spaghetti3451]

The electromagnetic action can be written in the language of differential forms as

##\displaystyle{S=-\frac{1}{4}\int F\wedge \star F.}##

The electromagnetic action can also be written in the language of vector calculus as

$$S = \int \frac{1}{2}(E^{2}+B^{2})$$

How can you show the equivalence between the two formulations of the electromagnetic action?Here is my attempt:

##\displaystyle{S=-\frac{1}{4}\int F\wedge \star F}##

##\displaystyle{=-\frac{1}{4}\int \left(\sum_i E_i\,{\rm d}t\wedge{\rm d}x^i - \star\sum_i B_i\,{\rm d}t\wedge{\rm d}x^i\right)\wedge \star \left(\sum_j E_j\,{\rm d}t\wedge{\rm d}x^j - \star\sum_j B_j\,{\rm d}t\wedge{\rm d}x^j\right)}##

##\displaystyle{=-\frac{1}{4}\int \left(\sum_i E_i\,{\rm d}t\wedge{\rm d}x^i - \star\sum_i B_i\,{\rm d}t\wedge{\rm d}x^i\right)\wedge \left(\star \sum_j E_j\,{\rm d}t\wedge{\rm d}x^j - \sum_j B_j\,{\rm d}t\wedge{\rm d}x^j\right),}##

since ##\displaystyle{**=(-1)^{p(n+p)+1}}## in Lorentzian space, where ##\star## is applied on a ##p##-form and ##n## is the number of spacetime dimensions, so that, in four dimensions for the ##2##-form ##\displaystyle{dt\wedge dx^{j}}##, ##\displaystyle{**=(-1)^{p(n+p)+1}=-1}##.

What do you do next?

 

[vanhees71]

First of all your ##S## in (1+3)-form is wrong. The integrand should be ##\propto (\vec{E}^2-\vec{B}^2)##. Then first write down more carefully ##F## and then ##*F## and then multiply out the forms.

 

[spaghetti3451]

So, you mean that

##\displaystyle{F=\left(\sum_i E_i\,{\rm d}t\wedge{\rm d}x^i - \star\sum_i B_i\,{\rm d}t\wedge{\rm d}x^i\right)}##

is wrong?

 

[vanhees71]

It's much easier in components (as usual). The Lagrangian is
$$\mathcal{L}=-\frac{1}{4} F_{\mu \nu} F^{\mu \nu}.$$
Now
$$F_{0j}=E_j, \quad F_{jk}=-\epsilon_{ijk} B_i,$$
and you can easily decompose the Lagrangian in temporal and spatial components to write it in terms of the ##(1+3)##-formalism. You must get something ##\propto (\vec{E}^2-\vec{B}^2)##. The other invariant of the Faraday tensor is ##{^\dagger}F^{\mu \nu} F_{\mu \nu} \propto \vec{E} \cdot \vec{B}##.