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Quantitative Meaning of Ricci Tensor

  1. May 4, 2014 #1
    Hello,

    I am studying general relativity right now and I am very curious about the Ricci tensor and its meaning. I keep running into definitions that explain how the Ricci tensor describes the deviation in volume as a space is affected by gravity. However, I have yet to find any quantitative explanation of this definition. How can volume differences be calculated with the Ricci tensor? What role does the Ricci tensor play in the volume changes?

    Thanks!
     
  2. jcsd
  3. May 4, 2014 #2

    Bill_K

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    In general there's no such relationship. Maybe you're talking about a cosmological situation, in which the expansion rate of the universe can be related to the matter density? Please give us a quote/reference where you found this.
     
  4. May 4, 2014 #3
    Hi Bill,

    Thanks for the response. In this paper:
    http://arxiv.org/pdf/gr-qc/0401099v1.pdf
    the author writes, "So in roughly the same sense that the Riemann tensor governs the evolution of a vector or a displacement parallel propagated along a geodesic, the Ricci tensor governs the evolution of a small volume parallel propagated along a geodesic."
     
  5. May 4, 2014 #4

    WannabeNewton

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    That is not true in general. It is only true if the geodesic belongs to an irrotational, shear-free time-like geodesic congruence in which case the claim follows from the Raychaudhuri equation. Otherwise the vorticity and shear of the congruence will both contribute to the evolution of the volume of the geodesic ball, in which case the Ricci tensor won't be the only thing governing the evolution.
     
  6. May 4, 2014 #5
    Ok, excellent. So what, if anything, is the geometric or physical meaning of the Ricci tensor if it has no general relationship with volume?
     
  7. May 4, 2014 #6

    WannabeNewton

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  8. May 7, 2014 #7
    You can interpret it that way, but you have to be careful.

    http://math.ucr.edu/home/baez/einstein/node3.html

    The rate at which a ball BEGINS to shrink, not just shrink, for one thing. Secondly, notice Baez's "fine print". This interpretation only works in a local reference frame in which the ball is initially at rest.



    The mathematical justification can be found here:

    http://math.ucr.edu/home/baez/einstein/node10.html

    Just a little index-gymnastics directly from Einstein's equation, so I don't think it's in question.
     
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