electromagnetism - How to derive Maxwell's equations from the electromagnetic Lagrangian?

In Heaviside-Lorentz units the Maxwell's equations are:

$$\nabla \cdot \vec{E} = \rho$$ $$\nabla \times \vec{B} - \frac{\partial \vec{E}}{\partial t} = \vec{J}$$ $$\nabla \times \vec{E} + \frac{\partial \vec{B}}{\partial t} = 0$$ $$\nabla \cdot \vec{B} = 0$$

From EM Lagrangian density: $$\mathcal{L} = \frac{-1}{4} F^{\mu \nu}F_{\mu\nu} - J^\mu A_\mu$$

I can derive the first two equations from the variation of the action integral: $S[A] = \int \mathcal{L} \, d^4x$. Is it possible to derive the last two equations from it?

Answer

Assume for simplicity that the speed of light $c=1$. The existence of the gauge $4$-potential $A^{\mu}=(\phi, \vec{A})$ alone implies that the source-free Maxwell equations $$\vec{\nabla} \cdot \vec{B} ~=~ 0 \qquad ``\text{no magnetic monopole"}$$

$$\vec{\nabla} \times \vec{E} + \frac{\partial \vec{B}}{\partial t} ~=~ \vec{0}\qquad ``\text{Faraday's law"}$$

are already identically satisfied. To prove them, just use the definition of the electric field

$$\vec{E}~:=~-\vec{\nabla}\phi-\frac{\partial \vec{A}}{\partial t},$$

and the magnetic field

$$\vec{B}~:=~\vec{\nabla}\times\vec{A}$$

in terms of the gauge $4$-potential $A^{\mu}=(\phi, \vec{A})$.

The above is more naturally discussed in a manifestly Lorentz-covariant notation. OP might also find this Phys.SE post interesting.

Thus, to repeat, even before starting varying the Maxwell action $S[A]$, the fact that the action $S[A]$ is formulated in terms the gauge $4$-potential $A^{\mu}$ means that the source-free Maxwell equations are identically satisfied. Phrased differently, since the source-free Maxwell equations are manifestly implemented from the very beginning in this approach, varying the Maxwell action $S[A]$ will not affect the status of the source-free Maxwell equations whatsoever.