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What is the trace of the second-Rank tensor?

  1. Feb 23, 2009 #1
    In Lorentz group in QFT, why the trace of the symmetric second-Rank tensor [tex]S^{\mu\nu}[/tex] is defined as follows?
    [tex] S=g_{\mu\nu}S^{\mu\nu}[/tex].
    Is it just a definition or the genuine trace of the second-Rank tensor, and why?
     
  2. jcsd
  3. Feb 23, 2009 #2

    cristo

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    What's the trace of a matrix?
     
  4. Feb 23, 2009 #3
    my opinion,The trace of [tex]S^{
    \mu\nu}[/tex] should be [tex]S^{
    00}+S^{11}+S^{22}+S^{33}[/tex]
    , i.e. the sum over the diagonal elements.
    Now the metric tensor [tex]g_{
    \mu\nu}[/tex] is taken into account, the trace becomes
    [tex]S^{
    00}-S^{11}-S^{22}-S^{33}[/tex].
    The problem is that I don't know why
     
  5. Feb 23, 2009 #4
    This seems to be a common point of misunderstanding:

    Matrices properly correspond to tensors with one index up and one index down.

    The fact that books often just write the metric with all up or all down as a matrix is just confusing.
     
  6. Feb 23, 2009 #5

    clem

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    The basic definition of the trace of a matrix is simply the sum of its diagonal elements.
    However, in order to make the trace invariant under a generalized rotation, the metric is included.
     
  7. Feb 25, 2009 #6
    Thank Cristo, genneth, and clem.
    You are correct, it is described in general relativity.
     
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