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Index Raising in Linearized General Relavitiy

  1. Apr 17, 2012 #1
    I'm reading a few textbooks (Straumann, Schutz, Hartle) on GR and am a little confused working through a small part of each on linearized GR.

    1. Relevant equations

    Using Straumann, the Ricci tensor is given by

    [tex]
    R_{\mu\nu} =
    \partial_{\lambda} \Gamma^{\lambda}_{\phantom{k}\nu\mu} -
    \partial_{\nu} \Gamma^{\lambda}_{\phantom{k}\lambda\mu}
    [/tex]

    with the Christoffel symbols given by

    [tex]
    \Gamma^{\alpha}_{\phantom{k}\mu\nu}
    =
    \frac{1}{2}\eta^{\alpha\beta}
    (
    h_{\mu\beta,\nu} +
    h_{\beta\nu,\mu} -
    h_{\mu\nu,\beta}
    )
    [/tex]

    2. The problem

    My problem is that the book is confusing me on the next equality. This what I expected when applying the flat metric:

    [tex]
    \Gamma^{\alpha}_{\phantom{k}\mu\nu}
    =
    \frac{1}{2}
    (
    \eta^{\alpha\beta}h_{\mu\beta,\nu} +
    \eta^{\alpha\beta}h_{\beta\nu,\mu} -
    \eta^{\alpha\beta}h_{\mu\nu,\beta}
    )
    \\
    \Gamma^{\alpha}_{\phantom{k}\mu\nu}
    =
    \frac{1}{2}
    (
    h_{\mu\phantom{\alpha},\nu}^{\phantom{k}\alpha} +
    h^{\alpha}_{\phantom{\alpha}\nu,\mu} -
    h_{\mu\nu}^{\phantom{\mu\nu},\alpha}
    )
    [/tex]

    i.e. the flat metric raises all [itex]\beta[/itex]'s to [itex]\alpha[/itex]'s.

    However, the book gets this

    [tex]
    \Gamma^{\alpha}_{\phantom{k}\mu\nu}
    =
    \frac{1}{2}
    (
    h^{\alpha}_{\phantom{\alpha}\mu,\nu} +
    h^{\alpha}_{\phantom{\alpha}\nu,\mu} -
    h_{\mu\nu}^{\phantom{\mu\nu},\alpha}
    )
    [/tex]

    So, the problem is in the first term: how come the book is able to swap the [itex]\alpha[/itex] and [itex]\mu[/itex] like that?
     
  2. jcsd
  3. Apr 17, 2012 #2

    George Jones

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    Because [itex]h_{\mu \beta , \nu}[/itex] is symmetric in [itex]\mu[/itex] and [itex]\beta[/itex], [itex]h_{\mu\phantom{\alpha},\nu}^{\phantom{k}\alpha} = h^{\alpha}_{\phantom{\alpha}\mu,\nu}[/itex].


    [tex]\eta^{\alpha\beta} h_{\mu\beta,\nu} = \eta^{\alpha \beta} h_{\beta \mu , \nu}[/tex]
     
  4. Apr 17, 2012 #3
    How do we know that [itex]h_{\alpha\beta}[/itex] is symmetric? I can't see it mentioned anywhere. The only condition I see is [itex]\lvert h_{\alpha\beta}\rvert \ll 1[/itex].
     
  5. Apr 17, 2012 #4

    George Jones

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    [itex]h_{\alpha\beta} = g_{\alpha\beta} - \eta_{\alpha\beta}[/itex], and [itex]g[/itex] and [itex]\eta[/itex] are both symmetric.
     
  6. Apr 17, 2012 #5
    Why do we assume [itex]g_{\alpha\beta}[/itex] is symmetric then? Is that a property we assume of all metrics? I didn't think we did. Do we assume symmetry of [itex]g_{\alpha\beta}[/itex] as it deviates only slightly from the Minkowski metric?
     
    Last edited: Apr 17, 2012
  7. Apr 17, 2012 #6

    George Jones

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    In standard general relativity, yes.

    No, a symmetric [itex]g[/itex] can differ substantially from the Minkowski metric.

    If the metric weren't symmetric, then it would not always have a tangent space isomorphic to Minkowski spacetime. If a metric tensor field is not symmetric, then there exists at least one point (event) at which the metric tensor for the tangent space is not symmetric.
     
  8. Apr 17, 2012 #7
    Got it now, thank you very much!
     
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