Here's a question that has bugged me for a while. The full Riemann curvature tensor [itex]R^\mu_{\nu \lambda \sigma}[/itex] can be split into the Einstein tensor, [itex]G_{\mu \nu}[/itex], which vanishes in vacuum, and the Weyl tensor [itex]C^\mu_{\nu \lambda \sigma}[/itex], which does not. (I'm a little unclear on whether [itex]R^\mu_{\nu \lambda \sigma}[/itex] can be uniquely recovered from [itex]G_{\mu \nu}[/itex] and [itex]C^\mu_{\nu \lambda \sigma}[/itex]. Does someone have a quick answer to that?)(adsbygoogle = window.adsbygoogle || []).push({});

Here's the issue: If the only mass/energy is in the form of point-masses (dust, or maybe uncharged elementary particles, or maybe lots of little black holes), then spacetime would be vacuum almost everywhere. So the solution to Einstein's field equations would be [itex]G_{\mu \nu} = 0[/itex] almost everywhere. In that case, all of the information about spacetime curvature would be carried by the Weyl tensor [itex]C^\mu_{\nu \lambda \sigma}[/itex]. However, it seems to me that it should be possible to approximate a spacetime filled with massive point-particles by a spacetime filled with a continuous mass density. You just pick a coarse-graining size, partition space into little cells of that size, and average the energy/momentum density within each cell. Under this approximation, it would no longer be true that [itex]G_{\mu \nu}[/itex] would be zero.

This makes me think that there is a way to derive an approximate [itex]G_{\mu \nu}[/itex] from [itex]C^\mu_{\nu \lambda \sigma}[/itex] through coarse-graining. Is that true?

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# Einstein tensor as coarse-grained Weyl tensor

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