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The Ricci tensor, for the spacetime surrounding the Earth, is zero, so the spacetime around the Earth is Ricci-flat.

The Riemann tensor though is not zero since spacetime certainly is curved. This means that the Weyl tensor, the off-diagonal part of the Riemann tensor (the Ricci tensor being the trace of the Riemann tensor), is non-zero.

In the Einstein field equations, the Weyl tensor is absent, since the Ricci tensor (and scalar) contains enough information for a point-wise solution for the field when proper boundary conditions are given.

Does this mean that the Weyl tensor harbors non-local tidal effects (the only real force effects of gravity)? If so, how? If not, what does it stand for, say, for example, in the spacetime surrounding Earth?

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  • $\begingroup$ Yes, the Weyl tensor conveys information about the local tidal effects (not global). The Wikipedia page has a nice description of it: en.wikipedia.org/wiki/Weyl_tensor $\endgroup$
    – Cham
    Commented Apr 5 at 0:35

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A nice way of understanding why the Weyl tensor is of importance in this context is to see that you can define a ``gravitational entropy" (NOT to be confused with the area term in generalized entropy like $\frac{A}{4G}+S_{\text{bulk}}$). In a black hole spacetime, we want this to be precisely $\frac{A}{4G}$ (although without the more holographic aspect), which naturally comes from the Weyl curvature calculation -- see Rudjord and Gron's paper on this. In more general settings, if the spacetime is Petrov type D or N, you can use the Newman-Penrose formalism to define the estimator that Clifton, Ellis and Tavakol described. The reason the Weyl tensor is used here instead of other curvature invariants is because it describes the tidal forces locally, so that given the spacetime isn't conformally flat (i.e. conformally Minkowski or type O), the Weyl curvature has a kind of monotonicity and non-vanishing property that allows us to describe gravitational thermodynamics, so to speak. This is also a reason why the Weyl curvature conjecture is interesting, because this uses the Weyl curvature (which in the cosmological setting tells us how the anisotropies or structure formations form and evolve over time) as an entropic quantity. I'm not sure what you mean by non-local effects (presuming you mean some kind of global information), but gravity is not non-local -- what you said (with the spacetime around the Earth) is just a local measurement of the tidal forces.

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  • $\begingroup$ But isn't the tidal force a non-local force? I mean it only exists if you consider two separate points. The tidal force only makes sense if you only look at one point, it seems to me. Only for the three basic forces of nature, it makes sense to look at one point only. $\endgroup$
    – Il Guercio
    Commented Apr 5 at 6:48
  • $\begingroup$ @IlGuercio Gravity is not non-local, since two spacelike separated points cannot feel the effect of each other's gravitation. What does it mean to only look at one point? You take two points and try to see if there are tidal forces affecting the motion of each particle; one point or a particle on its own is not enough to gauge gravitation. $\endgroup$
    – VaibhavK
    Commented Apr 5 at 9:55

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