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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.
     
  22. Mar 8, 2013 #21

    Fredrik

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    There's a typo in the first term of the first equality. One of the μ should be ρ. The last term in the second equality is obtained the same way as all the others, by replacing σ with ρ in the corresponding term of the first equality. The only term that requires some thought is the third one. What's going on there is just that when an index appears twice, you can always raise one and lower the other like this: ##T^\mu{}_\mu = T_\mu{}^\mu##. The reason is this:
    $$T^\mu{}_\mu = T^{\mu\nu}g_{\nu\mu} = T_\nu{}^\nu.$$
     
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