For the electromagnetic field strength, $F^{(2)}$, which is an exact $2$-form, i.e. $F^{(2)}=dA^{(1)}$ for a $1$-form $A^{(1)}$, we can define its Hodge dual, $*F^{(2)}$ and then define the action $$\int *F^{(2)} \wedge F^{(2)}\propto \int d^4x\ F^{\mu\nu}F_{\mu\nu}$$

My question is what is the intuition behind the geometry of the volume form we are integrating over? Also, if one started from differential forms, how could one predict that such a volume form would be relevant to physics?

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    $\begingroup$ I don't understand what sort of "intuition" you're asking for here. We notice that the action of classical electromagnetism can be written this way, and so we use this action to consider "generalized electromagnetism". Why should there be any intuition in this? $\endgroup$
    – ACuriousMind
    Mar 9, 2019 at 12:56
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    $\begingroup$ @ACuriousMind Even if there is no "intuition" behind this, I don't think that this i not a sensible question to ask. It is simply revering the logic. I.e., why would one expect that such a volume form would be relevant to physics and what is the geometric meaning of such a volume form. Cheers.. $\endgroup$ Mar 10, 2019 at 2:49

1 Answer 1


Here's a very heuristic argument for how one might come up with such an action. First of all, by looking at the typical Maxwell's equations in terms of $E$ and $B$, we see that the PDEs are linear in the electric and magnetic fields. As such, if we want to encapsulate the information of $E$ and $B$ into the field strength 2-form $F$, then the equations involving $F$ also better be linear.

As such, if we want to start from an action, we need $F$ to appear in there "quadratically", because once we perform the variation, we would then arrive at equations linear in $F$. Now, since the spacetime manifold $M$ is 4-dimensional, we need to be integrating a 4-form... this is just basic integration on manifolds.

So, how can we construct a $4$-form using $F$ which is also "quadratic"? Well, the most natural choices are:

  • $F\wedge F$
  • $F\wedge \star F$
  • $(\star F)\wedge (\star F)= F\wedge (\star \star F)=F\wedge F$ (because in general $\alpha \wedge \star \beta = \beta \wedge \star \alpha$)

Now, $F\wedge F= dA\wedge dA = d(A\wedge dA)$, which means by Stokes' theorem, this term shouldn't affect the equations of motion. Therefore, the first and the third options are out, thereby leaving us only with the second one. Of course, from here, we can put in factors of $\pm\frac{1}{2}$, and then add on coupling terms like $A\wedge J$, where $J$ is the current 3-form etc to get the action \begin{align} S[A,dA]&=\int_M \frac{1}{2}(dA)\wedge (\star dA) \pm A\wedge J. \end{align} Performing the variation then yields the field equation $d(\star dA)= \mp J$ (and the trivially true equation $d(dA)=0$). Written in terms of $F$, this reads: \begin{align} \begin{cases} d(\star F)&= \mp J\\ dF&= 0. \end{cases} \end{align} (I'm not sure about the sign because $\star\star$ may be $\pm 1$ depending on the degree of the form and right now I can't remember which way it should be).

This is as best as I can motivate the situation (though I'd love to hear other opinions); because all we've done here is start out with some general principles (in this case linearity of the field equations, and of course the obvious thing that we should be integrating a 4-form on a 4-manifold), then proposed some possible actions based on these principles, and then narrowed down to a particular choice then verify that our choice does indeed give the right field equations. After all, there is no "cookie-cutter" procedure for obtaining actions.

Side comment: you use the terminology "volume form", yet there is no volume form here. A volume-form is a top-degree differential form which is nowhere vanishing. In our situation $F$ may very well vanish at some points. What you probably meant is simply a top-degree differential form.

  • $\begingroup$ You might add that you also want an action that is quadratic in the derivatives (from Maxwell's equations for $A$), so, e.g., $A\wedge F$ is not desirable. $\endgroup$
    – Toffomat
    Feb 22, 2021 at 11:53
  • $\begingroup$ @Toffomat Well, $A\wedge F$ certainly would not make sense since it's a 3-form on a 4-dimensional manifold, so we shouldn't expect to put it in the integral. But you're right; something like $A\wedge (\star A)$ which is indeed a 4-form would be undesirable since it's not quadratic in the derivatives $dA=F$. Next of course we have trivial things like $A\wedge F\wedge A$ or whatever which are 4-forms but these are $0$ by basic properties of wedge product. So really, we're stuck with $F\wedge \star F$. $\endgroup$
    – peek-a-boo
    Feb 22, 2021 at 12:35
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    $\begingroup$ Of course. my point was just that by requiring the action to be be quadratic in the field and of second order in derivatives, we have two $A$'s, two $\text{d}$'s and maybe some $\star$s, and then there's essentially just one choice $\endgroup$
    – Toffomat
    Feb 22, 2021 at 14:07

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