Local construction of Maxwell's action

[gentsagree]

What do we mean by "there is no local construction for an action in terms of $F^{\mu\nu}$, or E and B"?

So, I understand the construction "on-shell", once we solve Maxwell's equations to find $F^{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$, and how we can then write an action which is both gauge and Lorentz invariant.

But before we know anything about gauge symmetry, which comes about once we solve for F and express it in terms of A, what tells us that our action in not good enough (i.e. there is no local action to be constructed)?

Thanks

 

[gtoguy97]

for your help!What this means is that, before we know anything about gauge symmetry or how to express F^{\mu\nu} in terms of A, there is no way to construct a local action (an action that is specific to a certain point in spacetime) with just F^{\mu\nu}, E, and B. This is because these quantities are not Lorentz invariant, and any action constructed with them will not be invariant either. To construct a local action that is Lorentz invariant, we must solve for F^{\mu\nu} and express it in terms of A, which is the only way to make an action both gauge and Lorentz invariant.