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Riemann Tensor

  1. Oct 15, 2009 #1
    i need to show that [itex]R_{abc}{}^{e} g_{ed} + R_{abd}{}^{e} g_{ce}=(\nabla_a \nabla_b - \nabla_b \nabla_a) g_{cd} = 0[/itex]

    ok well i know that [itex]R_{abc}{}^{d} \omega_d=(\nabla_a \nabla_b - \nabla_b \nabla_a) \omega_c[/itex]

    so i reckon that [itex]R_{abc}{}^{e} g_{ed} = (\nabla_a \nabla_b - \nabla_b \nabla_a) g_{cd}[/itex]
    and
    [itex]R_{abd}{}^e g_{ce}=(\nabla_a \nabla_b - \nabla_b \nabla_a) g_{dc}[/itex]

    but [itex]g_{cd}=g_{dc}[/itex] so when i add those terms, surely is should get
    [itex]R_{abc}{}^{e} g_{ed} + R_{abd}{}^{e} g_{ce}= 2(\nabla_a \nabla_b - \nabla_b \nabla_a) g_{cd}[/itex]
    so why is there no factor of 2 in the book's answer?

    and then, how do i get the whole thing to go to 0?
     
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  3. Oct 15, 2009 #2

    gabbagabbahey

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    Are you sure about this identity? When I calculate it out I get

    [tex]2\nabla_{[a}\nabla_{b]}\omega_c=(\nabla_a \nabla_b - \nabla_b \nabla_a) \omega_c=-R^{d}{}_{cab}\omega_d[/tex]

    I don't think it works like that; [itex]g_{cd}[/itex] is a symmetric tensor, not a vector.
     
  4. Oct 16, 2009 #3
    it says in the book "General Relativity" by Wald that

    [itex]\nabla_a \nabla_b \omega_c - \nabla_b \nabla_a \omega_c=R_{abc}{}^{d} \omega_d[/itex]?
     
  5. Oct 16, 2009 #4

    gabbagabbahey

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    Hmmm... yes, it seems Wald uses that as the definition of the curvature tensor. Most texts I've seen define the Riemann tensor by the equation:

    [tex]R^a{}_{bcd}=\partial_c\Gamma^a_{bd}-\partial_d\Gamma^a_{bc}+\Gamma^e_{bd}\Gamma^a_{ec}-\Gamma^e_{bc}\Gamma^a_{ed}[/tex]

    From this definition, and a straight forward computation of the covariant derivatives I get:

    [tex](\nabla_a \nabla_b - \nabla_b \nabla_a) \omega_c=-R^{d}{}_{cab}\omega_d
    [/tex]

    So it appears, that for the two definitions to be equivalent (assuming I haven't messed up the calculation), we require [itex]R_{abc}{}^{d}=-R^{d}{}_{cab}[/itex]...I guess it might be easily provable using the metric to raise and lower the appropriate indices, together with equation 3.2.15 from Wald....

    Anyways, it seems like a straight forward application of equation 3.2.12 is all that is required for your problem ([itex]g_{cd}[/itex] is a second rank covariant tensor, not a covariant vector)....
     
  6. Oct 16, 2009 #5
    the sums on the RHS of 3.2.12 are confusing me, surely i just want that RHS to be [itex]R_{abc}{}^e g_{ed} + R_{abd}{}^e g_{ce}[/itex].

    could i use 3.2.13 to re write [itex]R_{abc}{}^e g_{ed}[/itex] as [itex]-R_{bac}{}^e g_{ed}[/itex]? then it would be in the form of the RHS of 3.2.13 and both the sums would just run form 1 to 1.
     
  7. Oct 16, 2009 #6

    gabbagabbahey

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    Well, let's see if I can clarify equation 3.2.12 for you a little;

    [tex]\begin{aligned}(\nabla_a \nabla_b - \nabla_b \nabla_a) T^{c_1\ldots c_k}{}_{d_1\ldots d_l}=&-\sum_{i=1}^{k}R_{abe}{}^{c_i}T^{c_1\ldots e\ldots c_k}{}_{d_1\ldots d_l}\\&+\sum_{j=1}^{l}R_{abd_j}{}^{e}T^{c_1\ldots c_k}{}_{d_1\ldots e\ldots d_l}\end{aligned}[/tex]

    The first sum looks like

    [tex]\sum_{i=1}^{k}R_{abe}{}^{c_i}T^{c_1\ldots e\ldots c_k}{}_{d_1\ldots d_l}=R_{abe}{}^{c_1}T^{ec_2c_3\ldots c_k}{}_{d_1\ldots d_l}+R_{abe}{}^{c_2}T^{c_1ec_3c_4\ldots c_k}{}_{d_1\ldots d_l}+R_{abe}{}^{c_3}T^{c_1c_2ec_4c_5\ldots c_k}{}_{d_1\ldots d_l}+\ldots +R_{abe}{}^{c_k}T^{c_1c_2c_3\ldots e}{}_{d_1\ldots d_l}[/tex]

    And the second sum looks like

    [tex]\sum_{j=1}^{l}R_{abd_j}{}^{e}T^{c_1\ldots c_k}{}_{d_1\ldots e\ldots d_l}=R_{abd_1}{}^{e}T^{c_1\ldots c_k}{}_{ed_2d_3\ldots d_l}+R_{abd_2}{}^{e}T^{c_1\ldots c_k}{}_{d_1ed_3d_4\ldots d_l}+R_{abd_3}{}^{e}T^{c_1\ldots c_k}{}_{d_1d_2 ed_4d_5\ldots d_l}+\ldots R_{abd_l}{}^{e}T^{c_1\ldots c_k}{}_{d_1d_2d_3\ldots e}[/tex]

    Does that clarify things for you?

    If so, just apply that to the tensor [itex]T^{c_1\ldots e\ldots c_k}{}_{d_1\ldots d_l}=g_{cd}[/itex], which has zero contravariant indices ([itex]k=0[/itex], making the first sum zero) and two covariant indices ( [itex]l=2[/itex] and [itex]d_1=c[/itex] and [itex]d_2=d[/itex])....
     
  8. Oct 16, 2009 #7
    ok. i understand the expansion of the sum. i guess i didn't realise that e was adopting the position of [itex]c_i[/itex] and [itex]d_j[/itex] respectively.

    why does the first sum vanish? i get
    [itex]\sum_{i=1}^{k}R_{abe}{}^{c_i}g_{cd}[/itex]
    is it because there are no contravariant indices and so the e index can't be placed anywhere on [itex]g_{cd}[/itex] and so there's nothing to sum over? if so, how do i explain this in maths?

    if i did expand the sum however i'd get something like
    [itex]R_{abe}{}^{c_1}g_{cd}+R_{abe}{}^{c_2}g_{cd}+...+R_{abe}{}^{c_k}g_{cd}[/itex]
    now i think im ok to say that [itex]c \neq c_i \forall i \in [1,k] \cap \mathbb{Z}[/itex]. this would mean there would be no contraction in each of the terms. if there's no contraction, does the action of the metric just cause things to vanish? i.e. [itex]X^a g_{bc}=0[/itex] because there's no contraction?
    are either of the above correct? if not could you please explain why this sum disappears?

    would the "j sum" come out to be:
    [itex]R_{abd_1}{}^eg_{ed}+R_{abd_2}{}^eg_{ce}[/itex]?
    then contraction with the metric gives
    [itex]=R_{abd_1d}+R_{abd_2c}[/itex]
     
  9. Oct 16, 2009 #8

    gabbagabbahey

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    [itex]k[/itex] is the number of contravariant indices....if [itex]k=0[/itex], then surely any sum of the form [tex]\sum_{i=1}^k \text{some stuff}[/tex] is automatically zero; since there are no terms in the sum.

    Why do you have a [itex]d_1[/itex] and [itex]d_2[/itex] in that expression?
     
  10. Oct 16, 2009 #9
    that's what is in the eqn 3.2.12. should i just be d? im confused as to why it would be though?
     
  11. Oct 16, 2009 #10

    gabbagabbahey

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    You changed the [itex]d_1[/itex] and [itex]d_2[/itex] to [itex]c[/itex] and [itex]d[/itex] respectively for the metric on the RHS (since [itex]c[/itex] is the first covariant index and [itex]d[/itex] is the second for the [itex]g_{cd}[/itex] on the LHS), so why wouldn't you do the same for the indices of the Riemann tensor?
     
  12. Oct 16, 2009 #11
    ok. so the j sum is:

    [itex]\nabla_a g_{cd}=-R_{abc}{}^e g_{ed} - R_{abd}{}^e g_{ce}=-R_{abcd} - R_{abdc}[/itex]

    what would be the next step in the proof from here?
     
  13. Oct 16, 2009 #12

    gabbagabbahey

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    Errm... Don't you mean [itex](\nabla_a \nabla_b - \nabla_b \nabla_a) g_{cd}=R_{abc}{}^e g_{ed} + R_{abd}{}^e g_{ce}=R_{abcd} +R_{abdc}[/itex]?

    If so, just use the fact that [itex]\nabla_a g_{cd}=0[/itex] (equation 3.1.22).
     
  14. Oct 17, 2009 #13
    so the final proof that it all equals 0 just follows from
    [itex]\nabla_a \nabla_b g_{cd} - \nabla_b \nabla_a g_{cd}=\nabla_a (0) - \nabla_b (0) = 0[/itex]
     
  15. Oct 17, 2009 #14

    gabbagabbahey

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  16. Oct 17, 2009 #15
    kl. thanks.

    i thought i'd post another question here as its on the same stuff but that way i wouldnt have to start a new thread.

    if you look at p45 of Wald, i've got to eqn 3.3.13 and then it says we use the formula for the Christoffel symbol to show that 3.3.13 is just the geodesic eqn.
    clearly Wald things this is just so blatantly obvious to the reader that there was no point in doing the calculation...i hate it when textbooks do this.
    i can't get anywhere with this.

    any advice?
     
  17. Oct 17, 2009 #16

    gabbagabbahey

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    Hmmm... You might want to multiply 3.3.13 by [itex]g^{\sigma\beta}[/itex] and sum over [itex]\beta[/itex]....

    Textbooks aren't the only place you find this type of thing. All throughout the scientific literature, it is a common occurrence for authors to only provide important details of how they did a calculation (just enough for the reader to reproduce the calculation themselves if they are so inclined) and not include all the nitty gritty details. It's really only in introductory level texts that detailed calculations are frequently given, in higher level material it is assumed that the reader/student will reproduce any important calculations themselves; the purpose of which is two-fold:

    (1) It saves space and writing for the author

    (2)It keeps the reader/student actively involved in the studying process and allows them to slowly master the material as they go along simply by reproducing calculations that are omitted.

    So, I think its unfair to say that Wald simply thins the result is blatantly obvious. He merely assumes that at this point in the readers studies, they have enough sense to take the time and work out the details for themselves.

    I'm disappointed with the number of professors who never preach this practice of working out the details of a calculation to their students.
     
  18. Oct 18, 2009 #17
    ok. that gives

    [itex]0=- \sum_{\alpha,\beta} g_{\alpha \beta} g^{\alpha \beta} \frac{d^2 x^{\alpha}}{d t^2} - \sum_{\alpha, \beta, \lambda} g^{\alpha \beta} \frac{ \partial g_{\alpha \beta}}{\partial x^{\lambda}} \frac{d x^{\lambda}}{dt} \frac{d x^{\alpha}}{dt} + \frac{1}{2} \sum_{\alpha, \beta, \lambda} \frac{\partial g_{\alpha, \lambda}}{\partial x^{\beta}} \frac{d x^{\alpha}}{dt} \frac{d x^{\lambda}}{dt} g^{\alpha \beta}[/itex]
    now if im summing over alpha and beta [itex]g_{\alpha \beta} g^{\alpha \beta}=n[/itex] and so the first term will be
    [itex]-\sum_{\alpha} n \frac{d^2 x^{\alpha}}{d t^2}[/itex]
    im not sure how to simplify the other two terms though. how does the metric act on derivatives?
     
  19. Oct 18, 2009 #18

    gabbagabbahey

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    No, you don't want to multiply by [tex]g^{\alpha\beta}[/tex]. Since [itex]\alpha[/itex] is already being summed over, it is essentially a dummy variable. You can't take [tex]g^{\alpha\beta}[/tex] inside the sums like that.

    Instead, multiply by [tex]g^{\sigma\beta}[/tex] or [tex]g^{\rho\beta}[/tex] or something like that...and then sum over [itex]\beta[/itex]...
     
  20. Oct 18, 2009 #19
    sorry. i should have spooted i had three [itex]\alpha[/itex] indices present. looking at this i'd get

    [itex]- \sum_{\beta} g^{\sigma \beta} \sum_{\alpha} g_{\alpha \beta} \frac{d^2 x^{\alpha}}{dt^2}-\sum_{\beta} g^{\sigma \beta} \sum_{\alpha, \lambda} \frac{\partial g_{\alpha \beta}}{\partial x^{\lambda}} \frac{d x^{\lambda}}{dt} \frac{d x^{\alpha}}{dt} + \frac{1}{2} \sum_{\beta} g^{\sigma \beta} \sum_{\alpha, \lambda} \frac{\partial g_{\alpha \lambda}}{\partial x^{\beta}} \frac{d x^{\alpha}}{dt} \frac{d x^{\lambda}}{dt} [/itex]

    how can i sum over [itex]\beta[/itex] unless im able to move the [itex]g^{\sigma \beta}[/itex] metric inside the other summations - that way i can put the metrics together and do something constructive...
     
    Last edited: Oct 18, 2009
  21. Oct 18, 2009 #20

    gabbagabbahey

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    Why wouldn't you be able to move [tex]g^{\sigma\beta}[/itex] inside the other sums? None of them are summing over either [itex]\sigma[/itex] or [itex]\beta[/itex].
     
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