Is there a way to derive the Gauss-Faraday Law in the Covariant formulation of classical electromagnetism from the Lagrangian density?

Question

everywhere I look I can only find derivations of the Gauss–Ampère law $\partial_\alpha F^{\alpha\beta}=\mu_0 J^{\beta}$, and this follows quite simply from the variational method with the Lagrangian Density, but I cannot seem to find any way to derive the Gauss-Faraday law $\partial_\alpha\left(\frac{1}{2}\epsilon^{\alpha\beta\gamma\delta}F_{\gamma\delta}\right)=0$ (equivalently as $\partial_\gamma F_{\alpha\beta} + \partial_\alpha F_{\beta\gamma} + \partial_\beta F_{\gamma\alpha}=0$ "Bianchi identity") also from the Lagrangian Density and Euler-Lagrange Equations.

As far as I know the Gauss–Ampère law represents only half of the Maxwell's equations, the other half being represented by the Gauss-Faraday law, does this mean that Faraday's law of induction and Gauss's law for magnetism are not equations of motion of the fields in question?

Answer 1

For $F^{\mu\nu} = \partial ^\mu A^\nu - \partial^\nu A ^\mu$ the Bianchi identity follows trivially.

Answer 2

The Lagrangian density is built from invariants of the theory (under certain restrictions of the number of derivatives) and can only lead to one set of field equations, namely $\partial_\mu F^{\mu\nu} = j^{\nu}$. In simplified geometric form, the Euler - Lagrange EOM for the em field are:

$$\delta F = j,$$ with $\delta$ the space-time exterior codifferential. Since $F$ is a two-form obeying $F=dA$, then it is trivial that $dF =0$, as the space-time exterior differential is nilpotent in order 2. So you see, it would be a little circular to derive $dF=0$ as the (other) EOM for the Lagrangian density since this is a constraint for the $F$ entering it in the first place.