# Is $F_{\mu\nu}F^{\mu\nu}$ the only possible gauge-invariant Lagrangian for the electromagnetic field?

Maxwell's equations can be derived from a Lagrangian formulation using the Lagrangian term (modulo some constants) $$\mathcal L=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}-A_\mu J^\mu.$$ Focusing on the free term for the moment, I've seen mentioned (though I can't find a source right now) that the $$F^2$$ term is the only possible gauge-invariant Lagrangian for electromagnetism.

Is this the case? How can we prove it?

• There certainly are other conditions. For example, $F^4$ is also gauge-invariant. – knzhou Jan 14 at 20:36
• You can have many other higher order terms. But even at this order, you can have terms like $F\wedge F$ (in 3+1 dimensions). But this term is a total derivative, can play a big role sometimes. In 2+1 dimensions you can also have Chern-Simons terms which are lower order $A\wedge dA$, this is a topological field theory (if $F\wedge\star F$ is not added). – Heidar Jan 14 at 20:39
• There is also $\varepsilon^{\alpha\beta\mu\nu} F_{\alpha\beta} F_{\mu\nu}$ (I think that's $\star F \wedge F$), but it is not invariant under P and T symmetry – Slereah Jan 14 at 20:40
• Related: physics.stackexchange.com/q/87817/2451 and links therein. – Qmechanic Jan 14 at 21:00
• I should've been more explicit $F\wedge\star F$ is $F^{\mu\nu}F_{\mu\nu}$ this term works in any dimension (the $\star$ essentially means contraction with the metric tensor). The $F\wedge F$ term is $\epsilon^{\alpha\beta\mu\nu}F_{\alpha\beta}F_{\mu\nu}$, but it only works in $3+1$ dimensions and it is a total derivative $F\wedge F = d(A\wedge dA)$ (atleast locally). – Heidar Jan 14 at 21:10

It is up to a trivial constant the only gauge invariant Lagrangian that reproduces the Maxwell equations. Since the Maxwell equations are linear in $$F^{\mu\nu}$$ only quadratic functions of $$F^{\mu\nu}$$ are possible. There are two possibilities, namely $$F_{\mu\nu}F^{\mu\nu}$$ and $$\epsilon_{\mu\nu\rho\sigma} F^{\mu\nu} F^{\rho\sigma}$$. The Maxwell equations can be recovered if the first but not the second term is present in the Lagrangian.
Not sure if this is what you're looking for, but here is a exhaustive search for all gauge invariant objects in electromagnetism: Say $$f[A_{\mu},\partial_{\mu} A_{\nu},\partial_{\mu}\partial_{\nu} A_{\rho},...]$$ is a gauge invariant object. Here $$f$$ could itself have indices which have been suppressed. Under a gauge transform:
$$A'_{\mu}(x)=A_{\mu}(x)+\partial_{\mu}\theta$$ Gauge invariance means that: $$f[A'_{\mu},\partial_{\mu} A'_{\nu},\partial_{\mu}\partial_{\nu} A'_{\rho},...]=f[A_{\mu},\partial_{\mu} A_{\nu},\partial_{\mu}\partial_{\nu} A_{\rho},...]$$ One can of course expand the left hand side to first order in $$\theta$$ $$f[A'_{\mu},\partial_{\mu} A'_{\nu},\partial_{\mu}\partial_{\nu} A'_{\rho},...]=f[A_{\mu},\partial_{\mu} A_{\nu},\partial_{\mu}\partial_{\nu} A_{\rho},...]+\frac{\delta f}{\delta A_{\mu}}\partial_\mu\theta+\frac{\delta f}{\delta \partial _\mu A_{\nu}}\partial_\mu\partial_\nu \theta+\frac{\delta f}{\delta \partial _\mu \partial _\nu A_{\rho}}\partial_\mu\partial_\nu\partial_\rho\theta+....$$ Since $$\theta$$ is an arbitrary function of $$x$$, each of the offending terms should separately vanish. $$\frac{\delta f}{\delta A_{\mu}}=0$$ For the next term: $$\frac{\delta f}{\delta \partial _\mu A_{\nu}}\partial_\mu\partial_\nu \theta=0$$ This means $$\frac{\delta f}{\delta \partial _\mu A_{\nu}}$$ is antisymmetric in $$\mu, \nu$$. i.e. $$f=f[F_{\mu,\nu},....]$$ For the next term, $$\frac{\delta f}{\delta \partial _\mu \partial _\nu A_{\rho}}\partial_\mu\partial_\nu\partial_\rho\theta=0$$ This means $$\frac{\delta f}{\delta \partial _\mu \partial _\nu A_{\rho}}$$ is antisymmetric in $$\nu, \rho$$(it can't be antisymmetric in any other pair of indices). i.e. $$f=f[F_{\mu,\nu},\partial_{\rho}F_{\mu \nu}....]$$ So to first order in the gauge parameter, $$f$$ must be a function of the field strength and its derivatives. Since we know that these quantities are gauge invariant to all orders in the gauge parameter, this is an exact result.
So the upshot is: any function of $$F_{\mu \nu}$$ and its derivatives is gauge invariant. This is the only way to form a gauge invariant object.
A sensible quantum theory requires we exclude higher derivatives of $$F_{\mu \nu}$$ and exclude higher powers of $$F_{\mu \nu}$$(higher than 2). Poincare invariance requires us to soak up all the indices, therefore requiring at least two powers of $$F_{\mu \nu}$$. This leaves us with only two terms in the Lagrangian: $$\mathcal{L} = c_1F_{\mu \nu}F^{\mu \nu}+c_2 \epsilon_{\mu \nu \rho \sigma}F^{\mu \nu}F^{\rho \sigma}$$
• Very nice approach. But I think it might need some modification to be general, as you can in principle have actions that are invariant up to surface terms (which are then zero for appropriate boundary conditions) or even actions that are not invariant but $e^{iS}$ is invariant (thus quantum mechanically invariant). For example, what would you get in $2+1$ dimensions from this analysis? – Heidar Jan 15 at 8:00
• @Heidar Clearly, this process is looking only for strictly gauge invariant objects. This is the same in all dimensions $d>2$. some modification to the process may enable us to find functions that are invariant upto surface terms. But this seems hard, since, as you rightly pointed out, this depends on the dimension. The $2+1$ dimensional theory has the surface term $A \wedge dA$ that would be nice to get from an exhaustive search. – Anonjohn Jan 15 at 16:40