In Heaviside-Lorentz units the Maxwell's equations are:
∇⋅→E=ρ
From EM Lagrangian density: L=−14FμνFμν−JμAμ
I can derive the first two equations from the variation of the action integral: S[A]=∫Ld4x. 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 Aμ=(ϕ,→A) alone implies that the source-free Maxwell equations →∇⋅→B = 0‘‘no magnetic monopole"
→∇×→E+∂→B∂t = →0‘‘Faraday's law"
are already identically satisfied. To prove them, just use the definition of the electric field
→E := −→∇ϕ−∂→A∂t,
and the magnetic field
→B := →∇×→A
in terms of the gauge 4-potential Aμ=(ϕ,→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 Aμ 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.
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