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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?

Qmechanic
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Sito
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2 Answers2

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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).

Prof. Legolasov
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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.

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

Phoenix87
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