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I am writing a paper entitled "On static solutions of Einstein's field equations for fluid spheres". I assume there a diagonal stress-energy tensor, $T_{\mu}^{\nu}=diag~\{\varepsilon,-p,-p,-p\}$, but I would prefer to not use the word "perfect fluid" because of restrictions traditionally associated with this notion (energy conditions, equation of state, etc). In my view the left side of Einstein field equations stands for local geometrical properties and the right side for physical properties of spacetime. Under physical properties I understand energy density and mean hydrostatic stress of spacetime itself, with no a priory relation between them as it is usually anticipated by using notion of perfect fluid matter. In other words, $T_{\mu}^{\nu}$ represents physical properties (or state) of spacetime and not matter. I have found only one publication that seems to support such a view. I wonder if someone knows others.

Luciano Combi, "Spacetime is material", arXiv:2108.01712v1

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JanG
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    Seems like you have made up your mind what answer you want. But I feel quite strongly that the stress-energy tensor measures properties of matter such as energy density, pressure, energy flux, and shear. – Andrew Nov 19 '21 at 17:16
  • @Andrew, I felt always as you but started to question it after deriving the metric: ${\rm d}s^2=(r/R)^{4}~c^2{\rm d}t^2-{\rm d}r^2-r^2{\rm d}\Omega^2$, which describes spacetime without matter but with pressure. Einstein equations result in: $\rho=0,~~~p=2G/c^{4}\cdot1/(4\pi)\cdot r^{-2}$ – JanG Nov 19 '21 at 18:46
  • Hm, can you say a few words as to why this metric convinced you otherwise? Some of my arguments would be that (a) $T_{\mu\nu}$ vanishes if no matter fields are present, (b) for a given matter field, after replacing covariant derivatives with partial derivatives, (and modulo an ambiguity on how to define $T_{\mu\nu}$ on flat space), the $T_{\mu\nu}$ that appears in Einstein's equations is the same $T_{\mu\nu}$ that you derive in flat space without gravity by appealing to Noether's theorem and which gives you the energy density, etc, (c) in the Newtonian limit, $T_{\mu\nu}$ becomes the mass. – Andrew Nov 19 '21 at 18:53
  • Well, lets take another example: a 'fluid' ball with constant energy density. Outside the ball there is no matter, i.e. $\rho=0$. However, if you not impose the condition of asymptotic flatness, you can say that pressure on the boundary is not zero, but say p(R). Such exterior solution have a non vanishing pressure (without matter). – JanG Nov 19 '21 at 19:04
  • Another argument is that for a given metric you can always derive $T_{\mu\nu}$from Einstein equations. See Introduction in [lhttps://arxiv.org/abs/2012.11569], The Return of the Singularities: Applications of the Smeared Null Energy Condition, Ben Freivogel, Eleni-Alexandra Kontou, Dimitrios Krommydas. – JanG Nov 19 '21 at 19:20
  • "Outside the ball there is no matter, i.e. =0." -- How do you define $\rho$? If you are defining it in the usual way as $T^0_{\ 0}$, then it sounds like you actually do believe that $T_{\mu\nu}$ tells you something about the matter distribution. " However, if you not impose the condition of asymptotic flatness, you can say that pressure on the boundary is not zero, but say p(R)." -- I'd have to see the details but it seems possible that the boundary conditions could be interpreted as some kind of matter. – Andrew Nov 19 '21 at 20:09
  • "Another argument is that for a given metric you can always derive from Einstein equations." -- All this proves is that given a metric, you can write down a $T_{\mu\nu}$ so that Einstein's equations are satisfied. That doesn't prove that $T_{\mu\nu}$ is not associated with matter. In fact I'd say it tells you what kind of matter you need to have in order for your metric to solve Einstein's equations. – Andrew Nov 19 '21 at 20:10
  • @Andrew, would you mind to look into https://physics.stackexchange.com/a/679431/281096? There you will find see the details you have asked. – JanG Nov 28 '21 at 18:45

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The stress-energy tensor in Einstein field equations represents matter, not spacetime. More precisely, it models the effect of matter on the spacetime geometry [1]. However, if I understand it properly, the quantity $p$ in $T_{\mu}^{\nu}=diag~\{\varepsilon,-p,-p,-p\}$ is not a scalar quantity, as for example pressure in gas. As a component of normal stress, it can have 3 different values at the same point (fluid with anisotropic pressure for example). It is actually then a vector. Generally, I quote [2], "Its components are related to the matter in the spacetime by

\begin{equation} T_{\mu}^{\nu} = \left( \begin{array}{c | c} \rho & S^{\nu} \\ \hline S_{\mu} & \pi_{i}^{j} \end{array} \right), \label{eq:Tab-compts} \end{equation} where $\rho$ is the energy density, $S_\mu$ is the energy-flux, and $ \pi_{i}^{j}$ is the stress ($i,j=$1,2,3). Typically, $S_{\mu}$ is considered a generalization of the Poynting vector and $ \pi_{i}^{j}$ is considered a generalization of the notion of pressure."

[1] https://arxiv.org/abs/1803.09872v1, Dennis Lehmkuhl, "How Einstein saw the role of the energy-momentum tensor in GR", page 5.

[2] https://arxiv.org/abs/2110.01121, Thomas Berry, Thesis, "Mimicking Black Holes in General Relativity", page 46.

JanG
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