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Divergence of product of killing vector and energy momentum tensor vanishes. Why?

  1. Apr 10, 2010 #1
    Hi,

    in my book, it says:
    -----------------------
    Beacause of [tex]T^{\mu\nu}{}{}_{;\nu} = 0[/tex] and the symmetry of [tex]T^{\mu\nu}[/tex], it holds that

    [tex]\left(T^{\mu\nu}\xi_\mu\right)_{;\nu} = 0[/tex]
    -----------------------

    (here, [tex]T^{\mu\nu}[/tex] ist the energy momentum tensor and [tex]\xi_\mu[/tex] a killing vector. The semicolon indicates the covariant derivative, i.e. [tex]()_{;}[/tex] is the generalized divergence)


    I don't understand, why from

    " [tex]T^{\mu\nu}{}{}_{;\nu} = 0[/tex] and the symmetry of [tex]T^{\mu\nu}[/tex] "

    it follows, that

    [tex]\left(T^{\mu\nu}\xi_\mu\right)_{;\nu} = 0[/tex]

    must hold.


    ---
    derivator
     
  2. jcsd
  3. Apr 10, 2010 #2

    George Jones

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    What results when

    [tex]\left(T^{\mu\nu}\xi_\mu\right)_{;\nu} = 0[/tex]

    is expanded?
     
  4. Apr 10, 2010 #3
    [tex]
    \left(T^{\mu\nu}\xi_\mu\right)_{;\nu} = \left(T^{\mu\nu})_{;\nu}\xi_\mu\right + T^{\mu\nu} \left(\xi_\mu\right)_{;\nu} = 0 + T^{\mu\nu} \left(\xi_\mu\right)_{;\nu}
    [/tex]

    So [tex]
    T^{\mu\nu} \left(\xi_\mu\right)_{;\nu}
    [/tex] should be equal to 0. But why?
     
  5. Apr 10, 2010 #4

    nicksauce

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    Use the fact that T^{\mu\nu} is symmetric.
     
  6. Apr 10, 2010 #5
    Lets write [tex]

    T^{\mu\nu} \left(\xi_\mu\right)_{;\nu}

    [/tex] without Einstein summation convention:

    [tex]

    \sum_\mu\sum_\nu T^{\mu\nu} \left(\xi_\mu\right)_{;\nu} = \sum_\mu\sum_\nu T^{\nu\mu} \left(\xi_\mu\right)_{;\nu} = - \sum_\mu\sum_\nu T^{\nu\mu} \left(\xi_\nu\right)_{;\mu}

    [/tex]

    I see no chance to get it =0
    :-(
     
  7. Apr 10, 2010 #6

    George Jones

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    Is

    [tex]\sum_\mu\sum_\nu T^{\mu\nu} \left(\xi_\mu\right)_{;\nu} = \sum_\alpha\sum_\beta T^{\alpha\beta} \left(\xi_\alpha\right)_{;\beta}[/tex]

    correct?
     
  8. Apr 10, 2010 #7
    why shouldn't it be correct? You only changed the names of the indices?!
     
  9. Apr 10, 2010 #8

    George Jones

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    What about

    [tex]\sum_\mu\sum_\nu T^{\nu\mu} \left(\xi_\nu\right)_{;\mu} = \sum_\beta\sum_\alpha T^{\alpha\beta} \left(\xi_\alpha\right)_{;\beta}[/tex]
     
  10. Apr 10, 2010 #9

    samalkhaiat

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    [tex]
    T^{\mu\nu}(\xi_\mu)_{;\nu} = (1/2)\left( T^{\mu\nu}\xi_{\mu;\nu} + T^{\nu\mu}\xi_{\nu;\mu} \right) = (1/2)T^{\mu\nu}\left(\xi_{\mu;\nu} + \xi_{\nu;\mu} \right) = 0
    [/tex]
     
    Last edited: Apr 10, 2010
  11. Apr 10, 2010 #10

    Yes, it's also correct. But I don't see your point.
     
  12. Apr 10, 2010 #11

    George Jones

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    You wrote
    Substitute the relabeled expressions into the left and right sides of the above.
     
  13. Apr 10, 2010 #12
    Oh I see! Thanks!

    ---------

    @samalkhaiat

    how do you justify this step:
    [tex]

    T^{\mu\nu}(\xi_\mu)_{;\nu} = (1/2)\left( T^{\mu\nu}\xi_{\mu;\nu} + T^{\nu\mu}\xi_{\nu;\mu} \right)

    [/tex]

    ?

    is it true, that you only relabled the summation indices in
    [tex]

    T^{\nu\mu}\xi_{\nu;\mu} \right)

    [/tex]
     
    Last edited: Apr 10, 2010
  14. Apr 10, 2010 #13

    nicksauce

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    That is correct.
     
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