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In general relativity, when solving for the schwartzchild solution, we set $T=0$.

1) Is it possible for the stress energy tensor to have nonzero value in a vacuum region?

2) Is the stress $T=0$ in a vacuum region in $f(R)$ modified gravity?

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In the original $f(R)$ gravity, the action is $$ S = \int \sqrt{-g} \left[ \frac{f(R)}{16 \pi G} + \mathcal{L}_\text{mat}(g^{ab}, \psi) \right] \, d^4x. $$ Here, $\psi$ represents the collection of matter fields present. The Euler-Lagrange equation for the metric resulting from this action is $$ f'(R) R_{ab} - \frac{1}{2} f(R) g_{ab} + \left[ g_{ab} \nabla_c \nabla^c - \nabla_{(a} \nabla_{b)} \right] f'(R) = 8 \pi G T_{ab}, $$ where $$ T_{ab} = - \frac{\delta \mathcal{L}_\text{mat}}{\delta g^{ab}}. $$ Now, when we're constructing these models, we usually think of them as "modifying the gravitational dynamics" rather than "modifying the matter dynamics"; so usually we just take the standard matter Lagrangian for whatever matter want to study (electromagnetic fields, perfect fluids, etc.) and drop it into the modified Lagrangian. So in that sense, $T_{ab} = 0$ in the absence of matter ($\psi = 0$) in $f(R)$ gravity if and only if the same statement holds in conventional gravity.

However, there's one small ambiguity here. Consider adding a cosmological constant to the above action: $$ S' = \int \sqrt{-g} \left[ \frac{f(R)}{16 \pi G} + \mathcal{L}_\text{mat}(g^{ab}, \psi) + \Lambda \right] \, d^4x. $$ Do we view this $\Lambda$ as part of $f(R)$? Or as part of $\mathcal{L}_\text{mat}$? It could equally well be viewed as a trivial function of $R$, or as a trivial function of $g^{ab}$ and $\psi$. But depending on the meaning we assign to $\Lambda$, it will end up contributing to the $f(R) g_{ab}$ term in the first case, or contributing to $T_{ab}$ in the second case.

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No.

But be careful about how you define the word vacuum. For example the Reissner-Nordström black hole has a non-zero stress-energy tensor because the SET includes a contribution from the electrostatic field. So although the vacuum contains no matter it isn't really a vacuum.

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  • $\begingroup$ Is the answer still No if we are working in f(R) gravity. $\endgroup$
    – MrDi
    Jul 7, 2015 at 7:21
  • $\begingroup$ @MrDice: don't know I'm afraid. $\endgroup$ Jul 7, 2015 at 7:22
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    $\begingroup$ Isn't that usually called electro-vacuum. $\endgroup$
    – MBN
    Jul 7, 2015 at 12:58
  • $\begingroup$ Technically yes, but most people usually call it only vacuum. $\endgroup$ Jul 7, 2015 at 15:23

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