The Einstein Maxwell Action with sources

[center o bass]

The free field Einstein-Maxwell action is often states as
$$S[A, g] = - \frac{1}{4}\int_M F^{\mu \nu} F_{\mu \nu}d^4 x + \int_M R(M) d^4x$$
where $M$ is the spacetime manifold $F$ is the field strength and $R(M)$ is the curvature of the spacetime manifold as dictated by the metric tensor $g$.

The question is, how are this modified in the presence of sources?

Individually, the maxwell action with sources are obtained by adding a term
$$\int_M J_\mu A^\mu d^4x$$
such that
$$S[A] = - \frac{1}{4}\int_M F^{\mu \nu} F_{\mu \nu}d^4 x - \int_M J_\mu A^\mu d^4x$$,
while for the Einstein Hilbert action we have to add
$$\int_M \mathcal{L}_M d^4 x$$
for some matter Lagrangian $\mathcal{L}_M$.

The question is whether, for the combined action, the maxwell source term should be included in $\mathcal{L}_M$? This seems logical because charged currents also contribute to the curvature of spacetime. On the other hand, the free field action above is varied with respect to $A$ and $g$ individually, so it seems that if the action with sources are going to reproduce the Maxwell equation
$$\partial_\nu F^{\mu \nu} = J^\mu$$
then we need a $J_\mu A^\mu$ term for the source -- and I've never seen such a term in the matter Lagrangian $\mathcal{L}_M$.

 

[Matterwave]

In the presence of matter, the Hilbert action is modified to:

$$S=\int_M R\tilde{\omega}+\int_M \mathcal{L}_M \tilde{\omega}$$

Where $\tilde{\omega}$ is the volume form on your manifold, equal to $\sqrt{-g}d^4x$ in a given coordinate system.

$\mathcal{L}_M$ includes all the matter terms, and so if you are in the presence of charged particles, it would be all 3 terms, the particle term, the field term, and the interaction term. Therefore, your total action would be:

$$S=\int_M R\tilde{\omega}-\sum_{\text{i=particles}}\int m_ic ds_i-\int_M A^\mu J_\mu \tilde{\omega}-\frac{1}{4}\int F_{\mu\nu}F^{\mu\nu} \tilde{\omega}$$

Every term that is not the gravitational field term should be included in $\mathcal{L}_M$