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The Lagrangian for ED without Gauge fixing term is given by $$\mathcal{L}=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu},\quad \text{where}\quad F_{\mu\nu}:=\partial_\mu A_\nu-\partial_\nu A_\mu.$$

I was wondering if this step, defining $F_{\mu\nu}$ over the $4$-potential $A_\mu$, is necessary. Can't we just formulate electrodynamics in terms of the tensor $F_{\mu\nu}$? That is, set $$F_{\mu\nu}:=\begin{bmatrix}0&E_{x}&E_{y}&E_{z}\\-E_{x}&0&-B_{z}&B_{y}\\-E_{y}&B_{z}&0&-B_{x}\\-E_{z}&-B_{y}&B_{x}&0\end{bmatrix}$$ and then derive the Maxwell equations directly from here, without going over the $4$-potential? If this doesn't work, what exactly is the problem?

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The second and third Maxwell's equations can be written covariantly as $$ \varepsilon^{\mu \nu \sigma \tau} \partial_{\mu} F_{\nu \sigma} = 0. $$

In gauge geometry, this equation is known as the (Abelian) Bianchi identity (not to be confused with the Bianchi identity from Riemannian geometry, which is related, but different).

In Minkowski space-time, any electromagnetic field strength tensor field that satisfies the electromagnetic Bianchi identity can always be written as $$ F_{\mu \nu} = \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu} $$ for some $A_{\mu}$.

In the language of differential forms, this is a statement about the 2nd de Rham cohomology of the Minkowski space being trivial: for any $2$-form $F$ such that $dF = 0$, there exists a $1$-form $A$ such that $F = dA$.

This proves that treating $F_{\mu \nu}$ as fundamental is completely equivalent to treating $A_{\mu}$ as fundamental (up to gauge transformations).

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  • $\begingroup$ Can one show that $\varepsilon^{\mu \nu \sigma \tau} \partial_{\mu} F_{\nu \sigma} = 0$ holds without defining the $4$-potential? $\endgroup$ – Sito Jul 16 '20 at 7:50
  • $\begingroup$ @Sito it is the covariant form of the 2-nd ($\nabla \vec{B} = 0$) and 3-rd ($\nabla \times \vec{E} = - \partial \vec{B} / \partial t$) Maxwell equations. $\endgroup$ – Prof. Legolasov Jul 16 '20 at 9:14
  • $\begingroup$ I get that. But when you start with the Lagrangian you don't really know the eom, so you can't really know that $\varepsilon^{\mu \nu \sigma \tau} \partial_{\mu} F_{\nu \sigma} = 0$ holds. Therefore the question, can you show that said equation holds without defining the $4$-potential or knowing the Maxwell equations? $\endgroup$ – Sito Jul 17 '20 at 13:34
  • $\begingroup$ @Sito there was nothing in your question that indicated that you were looking for a Lagrangian formulation. I'm not aware of a Lagrangian approach to ED that doesn't treat $A$ as the field variable. $\endgroup$ – Prof. Legolasov Jul 17 '20 at 14:39
  • $\begingroup$ I'm sorry, in that case I formulated my question poorly, because this is exactly what I was looking for. Could you maybe suggest an edit to the question to make it more clear? Also, do you know why there is no approach where $F_{\mu\nu} $ is the field variable? $\endgroup$ – Sito Jul 17 '20 at 15:20
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Why don't you see for yourself? Assume that the components of $F$ are the generalised coordinates of the Lagrangian, together with $\partial F$, and see what equations you get out.

This, in a sense, would be equivalent to considering the Lagrangian

$$\mathcal L(q,\dot q) = \frac12 q^2$$

from which you want to derive the equation of motion that yield $q(t)$ as a solution. You'd get more interesting equations if you had a more "classical" Lagrangian, e.g.

$$\mathcal L(q,\dot q) = \frac12 \dot q^2$$

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