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Einstein tensor as coarse-grained Weyl tensor

  1. Aug 10, 2015 #1

    stevendaryl

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    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?)

    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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  3. Aug 10, 2015 #2

    Ben Niehoff

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    The Riemann tensor can always be recovered from the Ricci tensor and the Weyl tensor. Essentially, the Weyl tensor is just the Riemann tensor with all its traces removed, and the Ricci tensor contains all of the traces.

    Whether the Einstein tensor can replace the Ricci tensor in the above sentence depends on whether it contains information about all of the traces. The key point is whether the Ricci scalar can be recovered from the Einstein tensor; if so, then the Ricci tensor can be recovered. Since the Einstein tensor is given by

    $$G_{\mu\nu} = R_{\mu\nu}- \frac12 R g_{\mu\nu},$$
    it turns out that one can recover the Ricci scalar by taking the trace of the above, except in dimension 2, where the trace is always zero. It turns out Einstein's equations are particularly degenerate in dimension 2.

    This sounds like an interesting question in mathematical relativity. I'm not aware of any papers on this, but I know people who might know.
     
  4. Aug 10, 2015 #3
    The Riemann tensor can be decomposed into the Weyl tensor, plus a combination of the Ricci tensor and its contractions. The Ricci tensor itself in turn is completely determined by knowledge of the Einstein tensor, so it seems the answer to the above is yes.

    This is a very good question, which I better leave to more knowledgeable members to answer. Personally my feeling is that in the context of GR, a collection of massive gravitationally interacting particles is not physically equivalent to a continuous mass distribution; on the other hand though, this paper would suggest otherwise, at least for the case of FLRW :

    http://arxiv.org/abs/1208.1411

    I'm looking forward to reading what the experts have to say on this one !
     
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