Suppose that there exists manifold $M$ s.t $T_{\mu\nu} \in M$, where $T_{\mu\nu} $ is the stress-energy tensor. The einstein field equations are given by $$R_{\mu\nu}-\frac12Rg_{\mu\nu}+\Lambda g_{\mu\nu}=\frac{8\pi G}{c^2}T_{\mu\nu}.$$ One can define the stress-energy tensor for a perfect fluid $T_{\mu\nu}$ as $$T_{\mu\nu}=pg_{\mu\nu}+(p+\rho)U_{\mu}U_{\nu} ,$$ where $U$ is the four-velocity. The stress-energy tensor has also been derived as the E-H action using functional derivatives (confusing). I have seen similar questions on the site, however they do not answer this question. What definition would an individual use to set up the stress-energy tensor specifically for the Einstein Field Equations? Is there a definition true for any type of fluid and or any type of situation one could encounter in GR? Is the Einstein-Hilbert action the correct definiton?
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Briefly speaking, the stress-energy-momentum (SEM) tensor that constitutes the matter side of the EFE is the symmetric/Hilbert SEM tensor for the matter (=non-gravitational) fields of the physical system.
The functional derivative of the Einstein-Hilbert (EH) action yields the Einstein tensor, i.e. the gravitational/geometric side of the EFE.
Interestingly, there's no well-defined gravitational SEM tensor, cf. e.g. this, this, this, this, this Phys.SE posts and links therein.
answered Dec 3, 2021 at 23:32
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$\begingroup$ In the definition, using the Einstein-Hilbert action, is defined using the functional derivative. The definiton follows as $$T_{\mu\nu}=\frac{-2}{\sqrt{-g}}\frac{\delta S_{matter}}{\delta g^{\mu\nu}}=T_{\mu\nu}=\frac{-2}{\sqrt{-g}}\frac{\partial(\sqrt{-g}L_{matter})}{\partial g^{\mu\nu}}=-2\frac{\partial (L_{matter})}{\partial g^{\mu\nu}}+g_{\mu\nu}L_{matter}$$, is $$2\frac{\partial (L_{matter})}{\partial g^{\mu\nu}}+g_{\mu\nu}L_{matter}$$ now the partial derivative or still the functional derivative with a switch in notation? $\endgroup$– aygxCommented Dec 3, 2021 at 23:46
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$\begingroup$ The formulas using partial derivatives are not correct if $L$ depends on derivatives of the metric field. $\endgroup$– Qmechanic ♦Commented Dec 3, 2021 at 23:52
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$\begingroup$ Thats makes sense. Now, for the functional derivative with respect to the metric $$\frac{\delta L_{matter}}{\delta g^{\mu\nu}}$$, which definition of the functional derivative would It follow? Would It follow the same-spacetime definition and just give me the EL equations? $\endgroup$– aygxCommented Dec 4, 2021 at 0:06
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$\begingroup$ 1. How would one determine the matter part of the Lagrangian? 2. Would It be the normal formula i.e. $$L=\frac12g_{\mu\nu}\dot x^{\mu}\dot x^{\nu}?$$ $\endgroup$– aygxCommented Dec 6, 2021 at 1:01