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Bianchi Identity-where did I go wrong?

  1. Sep 2, 2012 #1
    hello ,
    In the Bianchi Identities of the second kind, we have ∇a Rbcde + ∇b Rcade + ∇c Rabde ≡ 0
    but since ∇c Rabde = - ∇c Rbade
    we get :
    ∇a Rbcde + ∇b Rcade - ∇c Rbade = 0
    in the last term, we exchange the c and the b indices and we would arrive at:
    ∇a Rbcde + ∇b Rcade - ∇b Rcade = 0
    which leads to :
    ∇a Rbcde = 0 .
    but this incorrect ? did I do something wrong ?
    I would like actually to arrive at the formula :
    2 ∇b Racde − ∇a Rbcde ≡ 0.

    see formula 9 here from original Bel's article from 1938:


    (see formula 5 here : https://docs.google.com/viewer?a=v&...9JJLhB&sig=AHIEtbRUONCiZUSV_8erdxK9YSMiouUrjA)

    Last edited: Sep 2, 2012
  2. jcsd
  3. Sep 2, 2012 #2


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    Only "dummy" indices can be freely changed like this, i.e. indices that are being summed over.
  4. Sep 2, 2012 #3
    Thanks I see , I must have been on a rush to get to the formula.
    My problem is then : how do i get rid of the last variable on which the divergence is acting I mean c ? is there a formula I must have have overlooked perhaps ?
  5. Sep 2, 2012 #4


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    I don't think what you're trying to prove is true. Bel's result follows only after he's multiplied by another factor of the Riemann tensor.
  6. Sep 2, 2012 #5
    I see, now I think I found it . I have tried it out indeed with the additional tensor factor. it goes like this :
    we have :
    a Rbcde + ∇b Rcade + ∇c Rabde ≡ 0
    multiply out by Rbcdf , we get :
    Rbcdfa Rbcde + Rbcdfb Rcade + Rbcdfc Rabde ≡ 0

    =>in the last term , we exchange the b and the c since now they are dummy indices I presume and in the second term, we use the antisymmetry of a and c:

    Rbcdfa Rbcde - Rbcdfb Racde + Rcbdfb Racde = Rbcdfa Rbcde - 2 Rbcdfb Racde

    Thank you.

    PS : Sorry the Bel's article above is from 1958. I must have confused it with the article of Lanczos .
    Last edited: Sep 2, 2012
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