1

The electromagnetic action is given in the language of differential forms by $$S[A]=-\frac{1}{4}\int F\wedge \star F$$ The variation of the electromagnetic action $S$ gives us Maxwell's equations $$d\star F=0.$$

How do you take the variation $\delta S = S[A+\delta A]-S[A]$ of the above action $S$ to obtain Maxwell's equations?

Qmechanic
  • 201,751
nightmarish
  • 3,183
  • 1
    you need to compute $\frac{\mathrm d}{\mathrm dt}\big|_{t=0}S(A+t\eta)$ where $F=\mathrm dA$ and $\eta$ arbitrary – Christoph Oct 30 '16 at 19:51
  • How would you do this by taking the variation of the action? – nightmarish Oct 30 '16 at 19:53
  • Possible duplicates: http://physics.stackexchange.com/q/3005/2451 , https://physics.stackexchange.com/q/34241/2451 , http://physics.stackexchange.com/q/51169/2451 , http://physics.stackexchange.com/q/64272/2451 and links therein. – Qmechanic Oct 30 '16 at 21:30

1 Answers1

3

Here we only consider the abelian case. Let $A\to A+\delta A$, and $F\to d(A+\delta A)=dA+d\delta A$.The action becomes

$$S[A+\delta A]=-\frac{1}{4}\int (dA+d\delta A)\wedge\star(dA+d\delta A).$$

So up to terms linear in $\delta A$, $$S[A+\delta A]-S[A]=\frac{1}{2}\int \delta A\wedge d(\star F)+\mathcal{O}(\delta A^2),$$ where we integrated by parts and used the symmetric property of the inner product on $p$ form $(\alpha,\beta)=\int\alpha\wedge\star\beta=\int \beta\wedge \star\alpha=(\beta,\alpha).$

Now $\frac{\delta S}{\delta A}=0$ gives you $d\star F=0$ as $\delta A$ is arbitrary.

user110373
  • 1,349