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If we defined spacetime as a purely geometrical (not physical) structure of the kind that is in general relativity (a 4-dimensional Lorentzian manifold), would it automatically have properties that would behave like energy and momentum in Einstein field equations?

I am wondering whether the purely geometrical properties of a 4D Lorentzian manifold impose existence of matter (that is, properties that behave like energy and momentum).

From what I have read, it seems that the answer is no, and so energy and momentum seem to be encoded in the points of the manifold rather than in its geometry.

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Given a Lorentzian manifold, one can calculate $$ R_{\mu\nu} - \frac{1}{2}Rg_{\mu\nu}$$ If you want, you can declare that this quantity represents energy and momentum, and then the Einstein field equations are satisfied. Is this what you are asking?

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  • $\begingroup$ No, I am asking whether this manifold necessarily has the properties of energy and momentum. It seems that energy and momentum are just added to the manifold by choice. $\endgroup$ Jul 12, 2020 at 22:37
  • $\begingroup$ Which properties? This has the property "satisfies Einstein's field equations." What other equations do you want satisfied? $\endgroup$
    – Daniel
    Jul 12, 2020 at 22:39
  • $\begingroup$ Does this manifold necessarily satisfy Einstein's field equations? Do Einstein's field equations logically follow from the geometrical properties of this manifold? $\endgroup$ Jul 12, 2020 at 23:05
  • $\begingroup$ All the manifold gives you is topological structure and $g$. The point is that for any $g$, there necessarily exists a $T$ that satisfies the field equations - the manifold doesn't come with this $T$ built-in, but you are free to define it. When people say "geometrical properties", they usually mean equations only involving $g$. So no, in that sense Einstein's equations don't follow from geometrical properties. $\endgroup$
    – Daniel
    Jul 13, 2020 at 0:30
  • $\begingroup$ I was told that the manifold has energy and momentum also when T=0 because there are gravitational waves defined purely by geometrical properties of the manifold. So it seems that for this special case, the manifold indeed has energy and momentum necessarily? $\endgroup$ Jul 13, 2020 at 9:09

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