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Invariant Tensors in GR and SR

  1. Oct 25, 2009 #1
    Hello all, this is my first post on this forum, though I have been perusing it for a while.

    I am currently re-reading through Carroll's text on SR and there is a curious comment on p24 that intrigues me. Carroll says that the *only* tensors in SR which are invariant are the Kronecker delta, the Levi-Civita tensor, and the metric tensor. In GR, the latter is no longer invariant so there are only two invariant tensors.

    Carroll says "we won't prove it" but I'm dying to see the proof, which I've spent a few hours trying to derive. The internet has also been little help, aside from this topic which doesn't really give an answer:
    https://www.physicsforums.com/showthread.php?t=344626

    So how do we show that there are no other invariant tensors?

    Thanks!
     
  2. jcsd
  3. Oct 26, 2009 #2
    Strictly speaking, I think, even the Levi-Civita tensor is not invariant in GR... More generally, the only tensors that can be invariant in GR are the ones where the number of upper indices is equal to the number of lower indices, as can be seen by applying a simple transformation of the form x^i -> c*x^i.
     
  4. Oct 26, 2009 #3
    Well, the Levi-Civita symbol is tensor density, which is artificially made into a tensor by adding a square root of det(g) term to it.

    Maybe ignoring such artificial examples would help in answering the question?
     
  5. Oct 28, 2009 #4
    Progress at last: it turns out that tensors whose components are the same in all coordinate systems are called "isotropic tensors".

    This is helpful since MathWorld has an entry which lists them, and I found some lecture notes which (partially) show that the list is exhaustive:

    http://www.ig.utexas.edu/people/students/classes/spring02/geo391/Lecture1.pdf [Broken]

    The proof needs to be adapted to GR but I'm working on it...
     
    Last edited by a moderator: May 4, 2017
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