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  • $\begingroup$ Thanks for the answer. It seems that $\mathcal{L}_e-\mathcal{L}_m$ does not give an EOM with the correct sign for the charge, also if I take $m^\mu=0$ the final Lagrangian does not give $-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-j^\mu A_\mu$. I think simply adding up $-j^\mu A_\mu$ and $-m^\mu C_\mu$ works if both $A$ and $C$ can be found for a given EM field, but this seems to be not generally possible. For example, if the background EM field is generated by some static electric charges, then $\nabla \cdot\mathbf{E}=\rho_e$, but then $\rho_e=\nabla\cdot(-\nabla \times \mathbf{C})=0$. $\endgroup$
    – hao123
    Commented Jun 14 at 10:58