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Contracting a Tensor

  1. Feb 16, 2015 #1
    contract1.png


    What do they mean by contracting ##\mu## with ##\alpha## ?
     
  2. jcsd
  3. Feb 16, 2015 #2

    PeterDonis

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    2016 Award

    Staff: Mentor

    Contraction is an operation that can be applied to any tensor or product of tensors with an upper and a lower index free. (In this case the upper index is ##\mu## and the lower index is ##\alpha##.) The contraction is just a sum over all tensor components for which ##\mu## and ##\alpha## take the same value. So, for example, a tensor ##T^{\mu}{}_{\alpha}## with one upper and one lower index can be contracted to a scalar ##T = T^0{}_0 + T^1{}_1 + T^2{}_2 + T^3{}_3##.

    The contraction of the Bianchi identity has more terms because the identity itself has three terms, and each one becomes a sum of four terms when contracted. Note that the second step, multiplying by ##g^{\nu \gamma}##, is also a contraction, because the indexes ##\nu## and ##\gamma## appear as lower indexes in the Bianchi identity.
     
  4. Feb 16, 2015 #3
    Do you mind showing the steps leading to the final result ## 2\nabla_v R_\beta^v -\nabla_\beta R = 0##?
     
  5. Feb 16, 2015 #4

    PeterDonis

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    We start with

    $$
    \nabla_{\gamma} R^{\mu}{}_{\nu \alpha \beta} + \nabla_{\beta} R^{\mu}{}_{\nu \gamma \alpha} + \nabla_{\alpha} R^{\mu}{}_{\nu \beta \gamma} = 0
    $$

    Contracting ##\mu## and ##\alpha## gives (note that an index that is repeated, once upper and once lower, is summed over as I described; this is called the "Einstein summation convention" and is extremely useful):

    $$
    \nabla_{\gamma} R^{\alpha}{}_{\nu \alpha \beta} + \nabla_{\beta} R^{\alpha}{}_{\nu \gamma \alpha} + \nabla_{\alpha} R^{\alpha}{}_{\nu \beta \gamma} = 0
    $$

    Completing the contractions (meaning, collapsing the contracted indexes and using the fact that contracting the Riemann tensor on the upper and the middle lower index gives the Ricci tensor) gives:

    $$
    \nabla_{\gamma} R_{\nu \beta} - \nabla_{\beta} R_{\nu \gamma} + \nabla_{\alpha} R^{\alpha}{}_{\nu \beta \gamma} = 0
    $$

    where the minus sign in the second term comes in because the ##\alpha## index was the last lower index, not the middle one; swapping the indexes flips the sign (because the Riemann tensor is antisymmetric in the last two lower indexes). Now we contract with ##g^{\nu \gamma}## to give:

    $$
    g^{\nu \gamma} \nabla_{\gamma} R_{\nu \beta} - g^{\nu \gamma} \nabla_{\beta} R_{\nu \gamma} + g^{\nu \gamma} \nabla_{\alpha} R^{\alpha}{}_{\nu \beta \gamma} = 0
    $$

    Completing the contractions gives (note that ##R = g^{\nu \gamma} R_{\nu \gamma}## is the Ricci scalar):

    $$
    \nabla^{\nu} R_{\nu \beta} - \nabla_{\beta} R + \nabla_{\alpha} R^{\alpha}{}_{\beta} = 0
    $$

    The first and third terms are really the same thing, because the contracted index is a "dummy" index and we can relabel it freely, and we can also freely "flip" the indexes in the contraction (to put the upper index on ##R## and the lower index on ##\nabla## in the first term). This gives what we were looking for:

    $$
    2 \nabla_{\nu} R^{\nu}{}_{\beta} - \nabla_{\beta} R = 0
    $$
     
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