# Electromagnetic action in differential forms

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

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

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

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

since $**=(-1)^{p(n+p)}$ in Euclidean space, where $\star$ is applied on a $p$-form and $n$ is the number of spacetime dimensions, so that, in four dimensions for the $4$-form $dt\wedge dx^{j}$, $**=(-1)^{p(n+p)}=1$.

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?

• Have you tried just expanding the first equation in components? Oct 30, 2016 at 18:47
• see my edit, please Oct 30, 2016 at 18:54
• Okay, why don't you just keep going? You're getting there! Oct 30, 2016 at 18:58
• What is the hodge star acting on the volume form equal to? Oct 30, 2016 at 19:00
• The Maxwell Lagrangian should be proportional $E^2-B^2$. I don't know why you've accepted the wrong answer here. May 7, 2019 at 16:45

Expand the wedge product and notice that the non vanishing terms are only the $EE$ terms and $BB$ terms and more over, $dt\wedge dx^i\wedge \star(dt\wedge dx^j)=\delta^{ij}dV$. There is a mistake in your second line: $\star\star=(-1)^{s+p(n-p)}$, where $s$ is the number of minus sign in the signature of your metric. Also there should be a extra factor of minus two as, for example, the coefficient of $dt\wedge dx^i$ should be $F_{0i}-F_{i0}=-2E_i$. So one should have $$\frac{1}{2}\int \left(\sum_i E_i\,{\rm d}t\wedge{\rm d}x^i - \star\sum_i B_i\,{\rm d}t\wedge{\rm d}x^i\right)\wedge \left(\star \sum_j E_j\,{\rm d}t\wedge{\rm d}x^j + \sum_j B_j\,{\rm d}t\wedge{\rm d}x^j\right)$$ $$=\frac{1}{2}\int \left(\sum_{i,j} E_iE_j\,{\rm d}t\wedge{\rm d}x^i\wedge\star(dt\wedge dx^j)+\sum_j B_iB_j\,{\rm d}t\wedge{\rm d}x^j\wedge\star(dt\wedge dx^i)\right)$$ $$=\frac{1}{2}\int (E^2+B^2) dV,$$ where in the second to last line we used $\star(dt\wedge dx^i)\wedge(dt\wedge dx^j)=-(dt\wedge dx^j)\wedge\star(dt\wedge dx^i)$.