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Calculation of double dual of Riemann tensor

  1. Jul 6, 2012 #1
    Hi all, I encounter a technical problem about tensor calculation when studying general relativity. I think it should be proper to post it here.

    Riemann curvature tensor has Bianchi identity: [itex]R^\alpha[/itex][itex]_{[\beta\gamma\delta;\epsilon]}=0[/itex]

    Now given double (Hodge)dual of Riemann tensor: G = *R*, in component form:
    [itex]G^{\alpha\beta}[/itex][itex]_{\gamma\delta}=1/2\epsilon^{\alpha\beta\mu\nu}R_{\mu\nu}[/itex][itex]^{\rho\sigma}1/2\epsilon_{\rho\sigma\gamma\delta}[/itex]

    Show that the Bianchi identity can be simply written in terms of divergence of G as
    [itex]\nabla\cdot G=0[/itex].

    In component form:
    [itex]G_{\alpha\beta\gamma}[/itex][itex]^{\delta}[/itex][itex]_{;\delta}=0[/itex]

    PS: [tex]\nabla[/tex] and ";" represent covariant derivative in abstract and component form respectively.

    I've never done such calculation and is overwhelmed by so much super- and subscripts. Can anyone show me step by step how to get the final answer from the beginning? Thanks very much.
     
    Last edited: Jul 6, 2012
  2. jcsd
  3. Jul 6, 2012 #2
    Heh, this would be really quite easy with some geometric algebra (or at least an exterior derivative). Still, have you tried simply going with the brute force approach and applying the product rule to the expression for G?
     
  4. Jul 6, 2012 #3
    Hi Muphrid, I tried to use brute force, but I don't know how to deal with the two Levi-Civita symbol, because as you might see, the divergence actually sum with one of the Levi-Civita's subscript on the RHS.

    Would mind show me how to do it if it's not too lengthy? Thanks very much.
     
  5. Jul 6, 2012 #4

    PeterDonis

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    The Einstein tensor G is the *contracted* double dual of the Riemann tensor. So you need to contract two of the indices after taking the double dual, so G has only two indices, not four. Then you take the covariant divergence on the second index.
     
  6. Jul 6, 2012 #5
    Hi Peter, this is not Einstein tensor. You obtain Einstein tensor by contracting two indices of "this" G. This is exercise 13.11 of book "Gravitation".
     
  7. Jul 6, 2012 #6

    PeterDonis

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    Staff: Mentor

    Ah, I see. I'll check my copy of the book when I get a chance. MTW does have a lot of exercises involving "index gymnastics".
     
  8. Jul 6, 2012 #7

    robphy

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    You'll probably need identities like

    $$\epsilon_{abmn}\epsilon^{pqcd}
    =(-1)^\sigma\delta_{abmn}^{pqcd}
    =(-1)^\sigma (4!)\delta_{[a\vphantom{b}}^{p}
    \delta_{b\vphantom{]}}^q
    \delta_{m\vphantom{]}}^{c\vphantom{p}}
    \delta_{n]}^{d\vphantom{p}},$$
    where [itex]\sigma[/itex] is the number of negative signs in the metric tensor,
    and
    $$\delta_{abmn}^{pqcd}
    =\delta_{ab}^{pq}\delta_{mn\vphantom{b}}^{cd}
    +\delta_{mn\vphantom{b}}^{pq}\delta_{ab}^{cd}
    +(2!)^2\delta_{[m[a}^{pq}\delta_{b]n]}^{cd\vphantom{p}},
    $$
    where [itex]\delta_{ab}^{pq}=(2!)\delta_{[a}^p\delta_{b]}^q
    =(2!)\delta_{a\vphantom{b}}^{[p}\delta_{b}^{q]}[/itex].
     
  9. Jul 6, 2012 #8

    Mentz114

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    Gold Member

    Isn't the action of the Levi-Civita's here

    Gαβγδ = 1/2ϵαβμνRμνρσ1/2ϵρσγδ

    anti-symmetrizing two pairs of indices so that

    Gαβγδ = (1/4)R[αβ][γδ]

    which seems to go part of the way.
     
    Last edited: Jul 6, 2012
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