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Ricci Tensor is the contraction of the Riemann Tensor. Even if all the components of the Ricci Tensor is zero, that doesn't mean that the spacetime is flat. If all the components of the Riemann Tensor is zero, then only the curvature is zero. In order for the field equations to be linear, the contraction is necessary. But, the contraction also neglects some components of the Riemann Tensor. Does it mean that the full Riemann curvature is not represented in the field equations?

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    $\begingroup$ yes. that is what it means. The rest of the Riemann tensor is captured by the Weyl tensor and this is not fixed by Einstein's equations at all. For example, gravitational waves are described by the Weyl tensor, not by the Ricci tensor. $\endgroup$
    – Prahar
    Commented Mar 20, 2023 at 14:23
  • $\begingroup$ Differential equations are meaningless without boundary conditions. And with the boundary conditions the gravitational equations do define the full Riemann tensor. $\endgroup$
    – safesphere
    Commented Mar 26, 2023 at 15:23

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You are basically correct: the field equations do not themselves fix the curvature (and hence the spacetime) in full. This implies that a given matter content does not uniquely fix what spacetime is doing. A good example of this is the case of zero matter content (where I mean the complete stress-energy tensor $T^{ab}$ is zero). In this case the solutions include flat spacetime, and they also include spacetime with all sorts of gravitational waves.

More generally, for any solution to the field equation with a given $T^{ab}$ and no gravitational waves, there are in principle further solutions with gravitational waves.

Similar statements apply to solutions of Maxwell's equations in classical electromagnetism.

To completely fix the solution, under a given $T^{ab}$, one can add some further constraints, such as symmetry or boundary conditions. This is what is done when finding standard solutions such as Schwarzschild-Droste, Kerr, FLRW etc.

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