# What's the Lagrangian of Electromagnetic field with magnetic and electronic charge?

Lagrangian for electromagnetic field without magnetic charge is $$\mathcal{L} \, = \, - \frac{1}{4 \mu_0} F^{\alpha \beta} F_{\alpha \beta} - A_{\alpha} J^{\alpha}$$

And we know that Maxwell's equation with magnetic charge is $$\nabla \cdot \mathbf{E} = \frac{\rho_{\mathrm e}}{\epsilon_0}$$ $$\nabla \cdot \mathbf{B} = \mu_0\rho_{\mathrm m}$$ $$-\nabla \times \mathbf{E} = \frac{\partial \mathbf{B}} {\partial t} + \mu_0\mathbf{j}_{\mathrm m}$$ $$\nabla \times \mathbf{B} = \mu_0 \epsilon_0 \frac{\partial \mathbf{E}} {\partial t} + \mu_0 \mathbf{j}_{\mathrm e}$$

\begin{aligned}\mathbf {F} ={}&q_{\mathrm {e} }\left(\mathbf {E} +\mathbf {v} \times \mathbf {B} \right)+\\&q_{\mathrm {m} }\left(\mathbf {B} -\mathbf {v} \times {\frac {\mathbf {E} }{c^{2}}}\right)\end{aligned}

How to construct the 4-vector potential to describe the $E, B$? Because if $F=dA$ then $F$ is closed and cannot admit magnetic charge. How to construct the lagrangian for this system.

• – ACuriousMind Nov 17 '16 at 4:02
• @ACuriousMind How to express $E,B$ explicitly in your $F , G$? – 346699 Nov 17 '16 at 4:04
• I have corrected my old answer with the correct Lagrangian. $E$ and $B$ are your usual temporal and spatial components of $F_{\mu\nu}$ now. – ACuriousMind Nov 17 '16 at 5:26