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Linearised gravity contraction

  1. Mar 7, 2013 #1
    Hi is this a correct contraction:

    [itex] R_{\mu \nu}=\eta^{\rho \alpha} \eta_{\alpha \sigma} R^{\sigma}_{\mu \nu \rho}[/itex]
     
  2. jcsd
  3. Mar 7, 2013 #2

    Mentz114

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    The indexes look OK but shouldn't it be
    [tex]
    R_{\mu \nu}=g^{\rho \alpha} g_{\alpha \sigma} R^{\sigma}_{\mu \nu \rho}
    [/tex]
    because [itex]\eta[/itex] usually means the Minkowski metric.
     
  4. Mar 7, 2013 #3
    I thought you could also contract with [itex] \eta [/itex]
     
  5. Mar 7, 2013 #4
    also if one has [itex] g^{\alpha \beta} \partial_{\beta} T_{\beta \rho} [/itex] why would it become [itex] \partial^{\alpha} T_{\beta \rho}[/itex] and not [itex] \partial^{\alpha} T_{\rho}^{\alpha}[/itex]
     
  6. Mar 7, 2013 #5

    Mentz114

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    Sure, you can. You can contract in a lot of ways.
     
  7. Mar 7, 2013 #6

    Fredrik

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    The right-hand side is equal to ##\delta^\rho_\sigma R^{\sigma}_{\mu \nu \rho}##, and this is equal to ##R^\rho{}_{\mu\nu\rho}##. But Wikipedia defines the Ricci tensor by ##R_{\mu\nu} = {R^\rho}_{\mu\rho\nu}##.
     
  8. Mar 7, 2013 #7
    This contraction is not correct. You have ##\beta## three times, which is very ambiguous at least, incorrect at most. If you want to contract more than two indices, use more metric tensors.
     
  9. Mar 7, 2013 #8
    The Riemann tensor is antisymmetric in the last two indices, so that's simply minus the Ricci tensor.
     
  10. Mar 7, 2013 #9

    WannabeNewton

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    Your first one is almost there mate: [itex]\eta ^{\rho \alpha }\eta _{\alpha \sigma }R^{\sigma }_{\mu \nu \rho } = R^{\alpha} _{\mu \nu \alpha } = -R^{\alpha} _{\mu \alpha\nu } = - R_{\mu \nu }[/itex].
     
  11. Mar 7, 2013 #10

    WannabeNewton

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    It is linearized gravity on a background minkowski space -time so one uses the minkowski metric in place of the general metric tensor.
     
  12. Mar 7, 2013 #11
    yeah but is it correct with eta in place?
     
  13. Mar 7, 2013 #12

    WannabeNewton

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    Are you asking if you can use eta to contract? If so then yes assuming you are working with a metric perturbation tensor field propagating on a background flat space - time as per the linearized approximation as your title suggests.
     
  14. Mar 7, 2013 #13
    I know that you can use eta to contract but is that contraction correct?

    also when going from [itex] R^{\sigma}_{\mu \nu \rho}= \frac{1}{2}(\partial_{\nu}\partial_{\mu}h^{\sigma}_{\mu}+\partial_{\rho}\partial^{\sigma}h_{\mu \nu}-\partial_{\nu}\partial^{\sigma} h_{\mu \rho} - \partial_{\rho}\partial_{\mu}h^{\sigma}_{\nu} ) [/itex]
    to

    [itex]R_{\mu \nu}=\frac{1}{2}(\partial_{\nu}\partial_{\mu}h +\partial^{\sigma}\partial_{\sigma}h_{\mu \nu} - \partial_{\nu}\partial_{\rho}h^{\rho}_{\mu}-\partial_{\rho}\partial_{\mu}h^{\rho}_{\nu}) [/itex]

    I am confused as to how one gets the last two terms, the first two are straightforward, but am stuck on the last two.
     
  15. Mar 7, 2013 #14

    robphy

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    Since the Riemann Curvature tensor [tex]R^\sigma{}_{\mu\nu\rho}[/tex] you use already has an upper index and a lower index, which you are contracting together, you actually don't require a metric to do this. You're essentially using a Kronecker delta (an index substitution operator) then contracting [tracing]. That is to say, the Ricci tensor is automatically a contracted form of Riemann (up to sign conventions).

    (If you insist on using something, then if its anything other the spacetime metric [or its equivalent],
    the contracted object isn't the Ricci tensor... but possibly something related to it.)


    To form the Ricci Scalar Curvature, however, you need to use the metric [to raise one index then contract] since Ricci's indices are both down.
     
    Last edited: Mar 7, 2013
  16. Mar 7, 2013 #15

    WannabeNewton

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    Are you working on varying the linearized gravity action? Be really careful with the partials and indices on that one. To go from [itex]R^{\sigma }_{\mu \nu \rho }[/itex] to [itex]R_{\mu \nu }[/itex], simply contract [itex]\sigma [/itex] with [itex]\rho[/itex] (AFTER swapping the last two indices in the riemann tensor, which will give you an overall negative sign). In particular, [itex]\partial _{\rho }\partial _{\mu}h_{\nu }^{\sigma }\rightarrow \partial _{\rho }\partial _{\mu}h_{\nu }^{\rho }[/itex] and [itex]\partial _{\nu }\partial ^{\sigma}h_{\mu\rho }\rightarrow \partial _{\nu }\partial ^{\rho}h_{\mu\rho } = \partial _{\nu }\partial _{\rho}h_{\mu}^{\rho } [/itex] (I'm guessing these were the last two terms you were talking about).
     
  17. Mar 7, 2013 #16

    Mentz114

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    Yes, thanks. But I think the contraction given is not the Ricci tensor as the notation implies.
     
  18. Mar 7, 2013 #17
    I am trying to derive the bottom equation, and don't fully understand why the last two terms are so, for example why does [itex] \eta^{\rho \alpha}\eta_{\alpha \sigma} \partial_{\nu} \partial^{\sigma}h_{\mu \rho} = \partial_{\nu}\partial_{\alpha}h^{\alpha}_{\mu}[/itex]

    WHY doesn't it equal [itex]\partial_{\nu}\partial_{\alpha}h^{\alpha}_{\mu} + \partial_{\nu}\partial^{\sigma}h^{\alpha \rho }_{\mu \sigma} [/itex]
     
  19. Mar 7, 2013 #18

    WannabeNewton

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    Oh sure I don't disagree with you; as noted it was off by a sign. And you were fully right for questioning the validity of using the minkowski metric to contract in general; I just assumed from the title he was working in the linearized approximation.
     
  20. Mar 7, 2013 #19

    Mentz114

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    [itex] \eta^{\rho \alpha}\eta_{\alpha \sigma} \partial_{\nu} \partial^{\sigma}h_{\mu \rho} = \partial_{\nu}\partial_{\alpha}h^{\alpha}_{\mu}[/itex]

    The first η raises the ρ index of h to upper α. The second η lowers the σ of the second diff operator to lower α.
     
  21. Mar 8, 2013 #20
    What is the logic behind
    First question is how one plays with indexes ?
    By the metric [itex] g [/itex]: one of the ways to look at it is like mapping from the vector ( [itex] А^{\mu} [/itex] ) to dual vectors ( [itex] A_{\mu} [/itex] ).
    Which is the metric we use in linear approximation of GR?(a assume you are struggling in that case )
    The metric is [itex] g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu} [/itex] but for some reason in this space the juggling of indexes is done by [itex] \eta_{\mu \nu} [/itex]. This is by definition.
    How a partial derivative looks like when we act with metric tensor on it?
    Lets look at the simple case of [itex] \eta^{\mu \nu} \partial_{\nu} h_{\alpha} = \eta^{\mu \nu} \frac{\partial h_{\alpha} }{\partial x^{\nu} } = \frac{1}{\eta_{\mu \nu}}\frac{ \partial h_{\alpha} }{ \partial x^{\nu} } = \frac{ \partial h_{\alpha} }{ \partial \eta_{\mu \nu} x^{\nu} } = \frac{\partial h_{\alpha}}{\partial x_{\mu}} = \partial^{\mu} h_{\alpha} [/itex]
    This jumping above the derivative is allowed because here the metric eta is a constant; in general case it is true too, but for now it is not important.
     
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