1
$\begingroup$

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?

$\endgroup$
5
  • $\begingroup$ Have you tried just expanding the first equation in components? $\endgroup$
    – knzhou
    Oct 30, 2016 at 18:47
  • $\begingroup$ see my edit, please $\endgroup$ Oct 30, 2016 at 18:54
  • 2
    $\begingroup$ Okay, why don't you just keep going? You're getting there! $\endgroup$
    – knzhou
    Oct 30, 2016 at 18:58
  • $\begingroup$ What is the hodge star acting on the volume form equal to? $\endgroup$ Oct 30, 2016 at 19:00
  • $\begingroup$ The Maxwell Lagrangian should be proportional $E^2-B^2$. I don't know why you've accepted the wrong answer here. $\endgroup$
    – level1807
    May 7, 2019 at 16:45

1 Answer 1

1
$\begingroup$

Just continue your second line:

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 $\star(dt\wedge dx^i)\wedge(dt\wedge dx^j)=-(dt\wedge dx^j)\wedge\star(dt\wedge dx^i)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.