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I What exactly does this expression mean?

  1. Mar 9, 2016 #1
    Given: Ra[bcd] = 0
    What permutations of bcd make this expression 0? TIA for your response.
     
  2. jcsd
  3. Mar 9, 2016 #2

    Samy_A

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    Rabcd+Racdb+Radbc=0
     
  4. Mar 9, 2016 #3

    Ibix

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    I think this is an anti-symmetrised tensor, not a symmetrised one, isn't it?
     
  5. Mar 9, 2016 #4
    Yes it is an antisymmetric tensor.
     
  6. Mar 9, 2016 #5
    Samy, I'm afraid I'm too thick to understand your answer. Can you elaborate further please? IIRC, this expression only eliminates one expression from the 21 possible components.
     
  7. Mar 9, 2016 #6

    Ibix

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    I'm pretty new to index gymnastics too, so check what I am saying with other sources.

    I think Samy has written ##R_{a (bcd)}##, not ##R_{a [bcd]}##. The latter is ##R_{abcd}- R_{acbd} +R_{acdb} -R_{adcb} +R_{adbc} -R_{abdc}=0##. Note that every ordering of the last three indices appears and that any pair of orderings that you can make by swapping two indices have opposite signs.

    Both Samy's and my expressions are interpreted to mean that for any a, b, c, d, that statement must be valid. Because you are permuting b, c and d, though, there is a lot of replication in the expressions. a,b,c,d=0,1,2,3 and 0,1,3,2 yield identical expressions, for example.
     
  8. Mar 9, 2016 #7
    What I'm looking for is a rule to help me understand. For instance, for R[ab][cd], R = 0 when a=b and c=d. Is there similar rule for the antisymmetric part of [bcd]?
     
  9. Mar 9, 2016 #8

    Ibix

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    It's possible, for example, that Samy has used the symmetries of the Riemann tensor to simplify his expression. I'd need to think about it with a pen and paper.

    Perhaps best to wait for Samy or someone else to clarify...
     
  10. Mar 9, 2016 #9

    robphy

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    Since ##R_{pqdc}=-R_{pqcd}##, you can write the left-hand side as
    ##R_{abcd}- (-R_{acdb}) +R_{acdb} -(-R_{adbc}) +R_{adbc} -(-R_{abcd})=2(R_{abcd}+R_{acdb}+R_{adbc})##
     
  11. Mar 9, 2016 #10

    Ibix

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    Ok. The Riemann tensor is anti-symmetric in its last two indices. That is, ##R_{abcd}=-R_{abdc}##. So my expression (correct for a general four index tensor) simplifies to Samy's in the case of the Riemann tensor.

    Edit: ...as robphy just pointed out.

    Samy's expression is not ##R_{a (bcd)}##. Forget I said that.

    I'm going to shut up now before I confuse anything even worse.
     
  12. Mar 9, 2016 #11

    Samy_A

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    My apologies, I thought R was the Riemann tensor.
     
  13. Mar 9, 2016 #12
    No apologies Samy, it is the Riemann tensor.
     
  14. Mar 9, 2016 #13
    Am I not being clear in my question? Nobody has yet answered it. What combinations of [bcd] make Ra[bcd] zero?
     
  15. Mar 9, 2016 #14

    Ibix

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    Indeed. It was me who was confused, not you, @Samy_A
     
  16. Mar 9, 2016 #15

    Samy_A

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    Any combination (if we interpret the expression as I did in post ##2).
     
  17. Mar 10, 2016 #16

    Samy_A

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    To elaborate on this, the first Bianchi identity Rabcd+Racdb+Radbc=0 is true for any combination of indices, but not all combination yield interesting information.

    Let's take the case where a=b.
    We then have Raacd+Racda+Radac=0.
    But Raacd=0.
    Also Racda=Rdaac=-Radac, so the identity ends up telling us that 0=0 in case a=b. And similarly for a=c or a=d.

    Let's take the case where b=c.
    Rabbd+Rabdb+Radbb=0.
    Here Radbb=0, and Rabbd=-Rabdb, so again the identity ends up telling us that 0=0 in case b=c.

    The Bianchi identity is only interesting when the indices are all different, and by the symmetries of the Riemann tensor, it allows you to reduce the number of independent components by one.
     
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