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Raising indices in curved space

  1. May 1, 2009 #1
    In curved space, can I raise an index on a tensor that is being differentiated? Ie, is the following true?

  2. jcsd
  3. May 1, 2009 #2


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    Only if the metric tensor is independent of [itex]x_{\nu}[/itex].
  4. May 1, 2009 #3
    This is troubling. I'm trying to obtain Maxwell's Equations from the Lagrangian, and since we have indices both up and down ie [itex]F^{\mu\nu}F_{\mu\nu}[/itex], varying with respect to [itex]\delta A_\lambda[/itex] inevitably introduces metrics:

    [tex] \delta F^{\mu\nu}=\partial^\mu g^{\nu\lambda}\delta A_\lambda - \partial^\nu g^{\mu\lambda}\delta A_\lambda[/tex]

    Integrating a term:

    [tex] \int d^4x\sqrt{-g}(\delta F^{\mu\nu} \delta F_{\mu\nu}) [/tex]
    [tex] = \int d^4x\sqrt{-g}\left[\partial^\mu(g^{\nu\lambda}\delta A_\lambda)F_{\mu\nu} - \partial^\nu(g^{\mu\lambda}\delta A_\lambda)F_{\mu\nu}\right] [/tex]
    [tex] = \int d^4x\sqrt{-g}(g^{\mu\lambda} \partial^\nu F_{\mu\nu} - g^{\nu\lambda} \partial^\mu F_{\mu\nu})\delta A_\lambda[/tex]

    And now I have these terms which are confusing me.
  5. May 1, 2009 #4


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    What happens if you vary

    [tex]g_{\mu b}g_{\nu a}F^{ab}F^{\mu\nu}[/tex]

    instead ?
    Last edited: May 1, 2009
  6. May 1, 2009 #5
    I will try that, but I suspect it is equivalent.
    If I had used covariant derivatives the whole time, there would be no problem. But here
    it states that the extra terms introduced by using covariant derivatives would cancel out. So maybe my original method is fine.
    Of course, I was stupid to overlook look this fact:
    [tex]\delta F^{\mu\nu} F_{\mu\nu}=\delta F_{\mu\nu} F^{\mu\nu}[/tex]
  7. May 1, 2009 #6


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    This may interest you.

    "Maxwell’s Equations In a Gravitational Field" by Andrew E. Blechman

    Attached Files:

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