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Draft re Ricci vs Riemann tensors

  1. Aug 23, 2014 #1

    andrewkirk

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    Draft re Ricci vs Riemann tensors

    This one is really just the beginning of a musing. I can't even remember if I came to any conclusion or just forgot about it. I started a thread in Jan 2014, a couple of months after this blog post, on the related issue of what the physical significance of the Ricci tensor is, and got some answers. I can't remember now whether they would have also covered off the questions raised in what is below.

    So, despite its incompleteness and uncertainty, I'll save it here so I have the option of reactivating it.

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    I have been thinking about the fact that Einstein's field equation uses the Ricci rather than the Riemann tensor. The Ricci tensor has ten degrees of freedom and the Riemann tensor has 24, so some information has been lost in the process of contracting the Riemann tensor to get the Ricci. It seems to follow that that information has no impact on gravitation. I've been wondering what sort of information that is, and whether it can be described in some physical, intuitive way.

    One way to think of it is as follows: Let ##TR=_{def}T_P^*M\otimes(T_PM)^3## be the space of (1 3) tensors at a point P in spacetime M, and let ##RiccContract(R)## denote the contraction of a (1 3) tensor on the 1st and 3rd indices.

    Then, for a given Ricci tensor ##R^{Ricci}## at P, is there some pattern of the following set of vectors, which share a common Ricci contraction?

    $$S(R^{Ricci},i,j,k) = \{R(\vec{e}_i, \vec{e}_j, \vec{e}_k) : R\in TR, RiccContract(R)=R^{Ricci}\}$$

    The set has two degrees of freedom and so might trace out a two-dimensional surface in the 4-dimensional space-time. If that surface were a [yes, it really did stop in the middle of a sentence!]
     
  2. jcsd
  3. Aug 28, 2014 #2
    I would not really put it that way. I think what happens is that the Einstein equations connect the local geometry of space-time to local sources of energy-momentum; as such, the energy-momentum tensor only specifies a very specific combination of components of the Riemann tensor, known as the Einstein tensor. This tensor is formed from the trace of Riemann ( being the Ricci tensor ) as well as its contraction, the Ricci scalar, and the metric. However, the information that is not specified here is contributions by distant sources of energy-momentum, i.e. gravitational radiation of distant sources, which is encapsulated in the Weyl tensor, being the traceless part of Riemann. So, in order to completely "fix" the Riemann tensor, one needs to specify other constrains over and above the Einstein equations.

    I am just an amateur, so the real experts on this will probably correct me - which I am looking forward to, as it will allow me to learn also :)
     
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