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Notation of General Relativity

  1. Dec 22, 2003 #1
    I'm reading, General Relativity, by Robert M. Wald. On page 39 he has the following notation:

    [tex]
    \displaystyle{
    {\rm R}_{{\rm [abc]}}^{\,\,\,\,\,\,\,\,\,\,\,\,{\rm d}} = 0
    }

    [/tex]

    and

    [tex]
    \displaystyle{
    \nabla _{{\rm [a}} {\rm R}_{{\rm bc]d}}^{\,\,\,\,\,\,\,\,\,\,{\rm e}} = 0
    }
    [/tex]

    What do the square brakets [ ] mean? How can the subscript of the [tex]\nabla[/tex] be included in these brakets? I've not seen this notation in the other books I have, so I could use a little help.

    Thanks.
     
    Last edited: Dec 22, 2003
  2. jcsd
  3. Dec 22, 2003 #2

    chroot

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    Gold Member

    The brackets denote that tensor is symmetric in the indices inside the brackets.

    The reason why the index of the [itex]\nabla[/itex] can be included in the brackets is simple: the entire construction [itex]\nabla_a R_{bcd}{}^{e}[/itex] is itself another tensor. It has five indices, under four of which it is symmetric.

    You could think of it as a new tensor T:

    [tex]T_{abcd}{}^{e} = \nabla_a R_{bcd}{}^{e}[/tex]

    where all of the lower indices of T are symmetric:

    [tex]T_{[abcd]}{}^{e} = 0[/tex]

    - Warren
     
    Last edited: Dec 22, 2003
  4. Dec 22, 2003 #3
    So this bracket notation is the same as the sum of all permutations with a coefficient of +1 for even permutations and -1 for odd permutations?

    The subscript on the [itex]\nabla[/itex] confuses me because on page 31, Wald writes, "It is often notationally convenient to attach an index directly to the derivative operator and write it as [itex]\nabla[/itex]a, although this is to some extent an abuse of the index notation since [itex]\nabla[/itex]a is not a dual vector."

    So what is he doing using it like this?
     
  5. Dec 25, 2003 #4

    jeff

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    This notation is defined on p26.

    It wasn't stated explicitly by either of you so, just in case, I'll point out that it's the vanishing of the antisymmetric part that means it's completely symmetric in those indices. In other words

    [itex]R_{[abc]}{}^d=0 \Leftrightarrow
    R_{abc}{}^d=R_{(abc)}{}^d[/itex].

    It doesn't matter that it's a tensor.

    To make operations on the components of [itex]\nabla[/itex] explicit.
     
  6. Dec 27, 2003 #5
    Does the subscript on [itex]\nabla[/itex] stand for differentiation with respect to the a'th coordinate curve of some as yet unspecified system?

    Is that system always orthogonal?

    Thanks.
     
  7. Dec 28, 2003 #6

    jeff

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    If you mean covariant differentiation with respect to the ath coordinate of some as yet unspecified coordinate system, then yes.

    Again, by "system" I'll assume you mean coordinate system. A coordinate system S is orthogonal with respect to some inner product ( , ) if S's basis vectors are mutually orthogonal with respect to ( , ). In GR, ( , ) is the spacetime metric, and S needn't be chosen orthogonal with respect to it. On the other hand, given any point p on a manifold, one may always choose a coordinate system which is orthogonal at p.
     
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