The free field Einstein-Maxwell action is often states as(adsbygoogle = window.adsbygoogle || []).push({});

$$S[A, g] = - \frac{1}{4}\int_M F^{\mu \nu} F_{\mu \nu}d^4 x + \int_M R(M) d^4x$$

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##.

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# The Einstein Maxwell Action with sources

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