In the follow Planck units have been used.
The Electromagnetic Energy-Momentum Tensor is $T^{\mu\nu} = \frac{1}{4 \pi} \left[ F^{\mu \alpha}F^\nu{}_{\alpha} - \frac{1}{4} g^{\mu\nu}F_{\alpha\beta} F^{\alpha\beta}\right]$ where $F^{\mu \nu}$ denotes the Faraday Tensor.
From the second Bianchi identity we know that the 4-divergence of the Einstein Tensor $G^{\mu \nu}$ is null, i.e. $\nabla_{\mu} G^{\mu \nu}=0$.
Putting this in the Einstein Equations $G^{\mu \nu}=8\pi T^{\mu \nu}$ we also have the conservation of the energy-momentum tensor: $\nabla_{\mu} T^{\mu \nu}=0$.
Now, the problem is that for the Electromagnetic Energy-Momentum Tensor the 4-divergence is not null, but we have instead $\nabla_{\mu} T^{\mu \nu}=F^{\mu \nu}j_{\nu}$, with $j^{\nu}$ the 4-current.
So is it not compatible with the Einstein Equations? Am I missing something?
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$\begingroup$ Have you read Dirac's 75 page booklet on GR and electrodynamics? $\endgroup$– DanielCCommented Aug 22, 2020 at 18:02
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$\begingroup$ Related: physics.stackexchange.com/q/439753/2451 $\endgroup$– Qmechanic ♦Commented Aug 22, 2020 at 18:12
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$\begingroup$ @DanielC nope, what's the specific name of that book? I can't find it $\endgroup$– Aleph12345Commented Aug 22, 2020 at 20:54
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1$\begingroup$ The great answer by professore Moretti can be supplemented by reading the relevant paragraphs of Dirac's book. I won't give you the amazon link (Jeff Bezos has enough money already), but he LoC one. lccn.loc.gov/75008690 $\endgroup$– DanielCCommented Aug 23, 2020 at 0:34
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1 Answer
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As soon as you have a non-zero current $J$, the EM field is not isolated and it is not the complete source of the gravitational field. On the right-hand side of the Einstein equations you should insert the whole stress-energy tensor and not only the EM one. With the added part you find a consistent system of equations.
answered Aug 22, 2020 at 19:47
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$\begingroup$ Oh, then what is the complete stress-energy tensor? $\endgroup$ Commented Aug 22, 2020 at 21:22
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3$\begingroup$ It depends on what is the type of matter interacting with the EM field. The simplest model is a gas of non interacting charged particles...In this case, the complete stress-energy tensor is the sum of the EM one and the gas one (no added interacting term arises in this case). $\endgroup$ Commented Aug 22, 2020 at 21:54
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3$\begingroup$ $T^{ab}=T_{EM}^{ab}+ \mu V^aV^b$ and where $J^a=\rho V^a$ with $\rho = g \mu$ for some constant $g$. $\endgroup$ Commented Aug 22, 2020 at 21:59