# A Electromagnetic action in differential forms

1. Oct 30, 2016

### 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?

2. Oct 31, 2016

### 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.

3. Oct 31, 2016

### 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?

4. Nov 2, 2016

### 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 ${^\dagger}F^{\mu \nu} F_{\mu \nu} \propto \vec{E} \cdot \vec{B}$.