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Tensors create form metric

  1. Jul 4, 2010 #1
    Is it true that the only combination of second order derivative of metric which
    transforms tensorially is Riemann tensor (and its traces)?
     
  2. jcsd
  3. Jul 4, 2010 #2
    Yes.

    AB
     
  4. Jul 4, 2010 #3
    Thanks for answer.
    Do you know any proof of it.
     
  5. Jul 4, 2010 #4
    Actually this is based on the fact that until today there we have only Riemann tensor created from the combination of the first and second order derivatives of metric tensor and the metric tensor itself. But for an informal proof, see

    GRAVITATION AND COSMOLOGY: PRINCIPLES AND APPLICATIONS OF THE GNERAL RELATIVITY by S. Weinberg. John Wiley & Sons, Inc., 1972, pp 133-34.

    AB
     
  6. Jul 4, 2010 #5

    atyy

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    How about the Einstein tensor?
     
  7. Jul 4, 2010 #6
    It's lieanr combination of traces of Riemann tensor.
     
  8. Jul 4, 2010 #7
    Of course!

    AB
     
  9. Jul 4, 2010 #8

    atyy

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    Thanks guys!
     
  10. Jul 4, 2010 #9

    atyy

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    I followed Altabeh's suggestion to look at Weinberg, and there he states uniqueness with all the conditions previously mentioned in this thread, plus the requirement that it be linear in second derivatives.

    There is an interesting comment in Berger's http://books.google.com/books?id=d_...&resnum=3&ved=0CCAQ6AEwAg#v=onepage&q&f=false "A important remark is in order: many people think that the curvature and its derivatives are the only Riemannian invariants. This is true and classical when looking for algebraic invariants which stem from the connection, see page 165 of Schouten 1954 [1109] and the references there. But things are dramatically different if one asks only for tensors which are invariant under isometries (called natural ). Then there is no hope to get any kind of classification, as explained in Epstein 1975 [491]. For more see Munoz & Valdes 1996 [952]."
     
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