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Bel-Robinson Tensor in empty spacetime

  1. Mar 18, 2016 #1

    TerryW

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    1. The problem statement, all variables and given/known data

    This is Exercise 15.2 in MTW - See attachment

    2. Relevant equations

    See attachment

    3. The attempt at a solution

    My attempt at a solution is also in the attachment.

    Are my initial assumptions OK? If not can someone nudge me in the right direction.

    If my initial assumptions are OK, then I've gone wrong somewhere in my workings. Can anyone give me a pointer as to where I have gone wrong.


    TerryW
     

    Attached Files:

  2. jcsd
  3. Mar 18, 2016 #2

    Drakkith

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    Hi Terry!

    Normally we like members to type in their equations and the work they've done into the post itself, but due to the sheer amount of work involved here I think we can let that slide this time. However, could you type up what the last paragraph in your attachment says? The other mentors and I are having a difficult time reading it (which is one of the reasons we normally like people to type up everything into a post instead of attaching a handwritten attachment).

    Thanks!
     
  4. Mar 19, 2016 #3

    TerryW

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    Hi Drakkith,

    My last paragraph sums up the sticky position I have arrived at. I've rewritten it to include some of the expressions referred to in the manuscript.


    So - the process has delivered one part of the answer required, i.e. RανδσRβνγσ, but the second part (-½gαβgκδRωλγσRωλκσ) doesn't look very promising.

    I have reworked this expression into:
    -½gαβgκδRωλγσRλκσ = -½gαβgγδRωλγσRωλκσ + ½gαβgγδRωλτσRωλτσ(τ≠γ)
    -½gαβgκδRωλτσRωλκσ(τ≠κ).​

    It would be nice if I could use the empty spacetime condition to make ½gαβgγδRωλτσRωλκσ(τ≠γ) and ½gαβgκδRωλτσRωλκσ(τ≠κ) disappear but I haven't so far been able to do this. Even if I do, I am still left with -½gαβgγδRωλκσRωλκσ instead of -(½*¼)gαβgγδRωλκσRωλκσ.

    I hope this helps and look forward to your next post.


    Regards

    TerryW
     
  5. Mar 20, 2016 #4

    TSny

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    Terry,

    I looked at your work and it looks good to me all the way through to your expressions (A) and (B). In the attachment below, I noted a couple of minor typos.

    I believe your results for (A) and (B) are correct. They appear to be corroborated here: http://arxiv.org/pdf/1006.3168.pdf (see equation 1).

    I don't follow how you got your expressions for (C) and (D).

    So, it seems that you are left with showing ##g_{\kappa \delta} R_{\omega \lambda \gamma \sigma} R^{\omega \lambda \kappa \sigma} = \frac{1}{4} g_{\gamma \delta} R_{\omega \lambda \tau \sigma} R^{\omega \lambda \tau \sigma}## for the vacuum.

    This identity is mentioned below equation (1) in the link above. I have spent quite a bit of time trying to prove it but have not been successful. The only vacuum condition that I can think of that would be useful is the condition that the Ricci tensor be zero. But, I haven't been able to use that to get anywhere.

    If you crack this nut, please post it.

    [EDIT: After looking further into the paper in the link given above, it appears that the paper does provide a proof of the identity ##g_{\kappa \delta} R_{\omega \lambda \gamma \sigma} R^{\omega \lambda \kappa \sigma} = \frac{1}{4} g_{\gamma \delta} R_{\omega \lambda \tau \sigma} R^{\omega \lambda \tau \sigma}## for the vacuum. The proof is somewhat sprawled out in the paper, but a key part of the proof is in the discussion surrounding equations (16) - (18).]
     

    Attached Files:

    Last edited: Mar 21, 2016
  6. Mar 21, 2016 #5

    TerryW

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    Hi TSny,

    Many thanks for spotting the typos, but most of all thanks for confirming that my process up to (A) and (B) is correct.

    As for (C) and (D), I was trying to find a way forward and arrived at that equation by the brute force method of writing out the values for the indices for κγκ in the expression (B) and comparing them with the indices γκκ in the expression I need to arrive at. What I observed was that the two sets of indices had a common set of four components (000, 111,222,333) and that the remaining 12 components of expression (B) could be written as gαβgδRωλτσRωλτσ(τ≠γ).
    So I add these 12 values to the (negative) expression (B), but then I only have four of the required components of my target expression so I have to subtract a further twelve components provided by gαβgκδRωλτσRωλκσ(τ≠κ).
    Not very elegant I admit, and it didn't lead anywhere, so let's dump it!

    Thanks particularly for the reference to the paper. As you have pointed out in your edit, it does appear to provide a proof of the Lanczos identity, and given that Lanczos died in 1974, aged 81, it is likely that this identity was around when MTW was being written (1972), though I don't think MT&W would have been expecting the reader to know it!

    All considered, a fairly tough problem, so I'm not too upset about not finishing it!


    Regards



    TerryW
     
  7. Mar 21, 2016 #6

    TSny

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    I think it is uncharacteristic of MTW to expect the reader to bring this much to the table to solve one of their exercises. For their tougher problems, they usually provide hints or guidance (by breaking up the problem into multiple parts). Makes me wonder if there is a better way.
     
  8. Mar 22, 2016 #7

    TerryW

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    I agree. I am working my way through MTW and have completed almost all of the exercises (with the occasional bit of help from some great guys (and gals?) on PF, including your good self). This is the first exercise where the key is a fairly obscure formula, though the paper you referenced does say "there is a well known equation in empty space"!

    I will probably have another look at this when I come to do my write up. If I find anything, I'll post it.


    Regards


    TerryW
     
  9. Mar 22, 2016 #8

    TSny

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    Thanks.
     
  10. Aug 11, 2016 #9

    TerryW

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    Hi TSny,

    I've been doing my write up and have had a look at the outstanding bit of this problem again, i.e. to prove:

    RγσλμRδσλμ = ¼gγδRρσλμRρσλμ

    As this expression is only valid in empty spacetime, I decided to see if Gσμ = Rσμ - ½gσμR = 0 could help.

    So

    Rσμ = Rλσλμ = gγλRγσλμ = ½gσμR

    Therefore Rγσλμ = ½gσμgγλR

    Rσμ = gσαgβμRαβ = gσαgβμRλαλβ = gσαgβμgδλRδαλβ = gδτRδστμ = gδτgλτRδσλμ = δδλRδσλμ = ½gσμR

    Therefore Rδσλμ = ½gσμδλδR

    Now multiply the two expressions together to get:

    RγσλμRδσλμ = ¼gσμgγλgσμδλδRR

    It is then simple to show that RR = RρσλμRρσλμ

    Hope you like it!


    Regards and thanks for all th help


    TerryW
     
  11. Aug 15, 2016 #10

    TSny

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    Hi Terry,
    You'll need to walk me through how you got the above.

    The equation gγλRγσλμ = ½gσμR may be written as

    gαβRασβμ = ½gσμR

    Then, multiplying both sides by gγλ gives

    gγλgαβRασβμ = ½gσμgγλR

    I don't see how to simplify the left side to Rγσλμ.

    EDIT: I don't see how the equation Rγσλμ = ½gσμgγλR can be true in general for the vacuum. Rγσλμ need not be zero in the vacuum while R must be is zero in the vacuum.
     
    Last edited: Aug 16, 2016
  12. Aug 16, 2016 #11

    TerryW

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    Hi TSny,


    Well, I prefer to stick with my own formulation of

    gγλRγσλμ = ½gσμR

    So that I can multiply both sides by gγλ to get

    gγλgγλRγσλμ = ½gγλgσμR

    And as gγλgγλ = 1

    then Rγσλμ = ½gγλgσμR

    If my maths is correct, then maybe this is not true. I've attached an example from MTW in which R is non-zero in empty spacetime.


    Regards

    Terry
     

    Attached Files:

  13. Aug 16, 2016 #12

    TSny

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    As you know, the indices γ and λ are being summed over on the left hand side. γ and λ will take on all values.

    This is where the trouble starts.

    Note the confusion on the left where the red γ and λ are being summed over but the blue γ and λ are not. The blue indices have some fixed values.

    This does not follow since the blue indices have fixed values. They are not summed with the red indices.

    (In the case where you are summing indices, then I believe gγλgγλ = 4.)

    I don't believe this is an example of working with the Einstein equations in a vacuum.

    For a vacuum you always have Gσμ = Rσμ - ½gσμR = 0.

    As a little exercise, start with Rσμ - ½gσμR = 0 and raise the σ index. Then contract σ with μ to show that R = 0.
    Then conclude from Rσμ - ½gσμR = 0 that Rσμ = 0 for all values of σ and μ
     
    Last edited: Aug 16, 2016
  14. Aug 19, 2016 #13

    TerryW

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    Hi TSny,

    Yes, my critical faculties were overridden by the tantalising prospect of a solution to the problem!

    I've had another look and this time got as far as

    gγδRγσλμRδσλμ = ¼RR

    but no further because of the original problem with summation indices.



    I agree with your argument that leads to R = 0 for empty spacetime, so why have we been bothering with this latter part of the exercise which is essentially creating a complex expression for something which has the value of 0!?


    Regards



    TerryW
     
  15. Aug 19, 2016 #14

    TSny

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    I don't see how you get the above result.

    We need to show RγσλμRδσλμ = ¼gγδRρσλμRρσλμ. Here, the two sides need not be zero in a vacuum.

    In post #9 it looked like you were going to construct a proof of this based on Gσμ = Rσμ - ½gσμR = 0, which you rearranged as Rσμ = ½gσμR. Here, both sides are each equal to zero in a vacuum. I wasn't sure where you were going with it. This equation doesn't seem to contain any more information than the equation Rσμ = 0.

    As I mentioned back in post #4, I tried using Rσμ = 0 to derive RγσλμRδσλμ = ¼gγδRρσλμRρσλμ but had no luck. I don't think I have my notes any more.
     
  16. Aug 20, 2016 #15

    TerryW

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    Hi TSny,

    My attempt to get a resolution to this problem by using Rαβ = ½gαβ was based as much as anything on the fact that I could (or at least I thought I could) generate expressions for Rγσλμ and Rδσλμ, which I could multiply together and (inter alia) produce the factor of ¼ which was needed.

    It is now pretty clear that this just doesn't work.

    I've just revisited my earlier workings and realised that A is only one third of the answer! I need to wrangle B into a sum of two more expressions -

    + RαργσRβρδσ and - 1/8gαβgγδRρσλμRρσλμ

    So it's back to the drawing board. I'll post again if I make any progress

    BTW, I've realised that my earlier supposition that RρσλμRρσλμ = RR isn't true!

    So thanks for all your help so far


    TerryW
     
  17. Aug 20, 2016 #16

    TSny

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    OK. Good luck with it.
     
  18. Aug 22, 2016 #17

    TerryW

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    Hi TSny,

    Is this a proof?

    Suppose
    RγσμλRδ σλμ = ¼gδγRρσλμRρσλμ

    Then

    gγδRγσλμRδσλμ = ¼gγδgγδRρσλμRρσλμ

    So RγσλμRγσμ = RρσλμRρσλμ


    TerryW
     
  19. Aug 22, 2016 #18

    TSny

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    You've shown that if RγσμλRδσλμ = ¼gδγRρσλμRρσλμ is assumed to be true, then it follows that RγσλμRγσλμ = RρσλμRρσλμ is true. (You inadvertently left out the superscript λ on the left side.)

    The conclusion RγσλμRγσλμ = RρσλμRρσλμ is clearly a true statement. But this argument doesn't prove that your initial assumption RγσμλRδσλμ = ¼gδγRρσλμRρσλμ is true.

    If you could show the converse, then you would have a proof of what you want. Starting with the true statement RγσλμRγσλμ = RρσλμRρσλμ you can easily work back to gγδRγσλμRδσλμ = ¼gγδgγδRρσλμRρσλμ. But I don't see how to go from this to the conclusion
    RγσμλRδσλμ = ¼gδγRρσλμRρσλμ .
     
  20. Aug 23, 2016 #19

    TerryW

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    It would be nice to be able to divide both sides by gγδ but we can't do that. Also, this line of attack hasn't involved the empty spacetime consideration so something is missing along the way.

    It's all very frustrating!


    TerryW
     
  21. Aug 24, 2016 #20
    In the literature the dual in the Bel Robinson tensor is defined as:
    ##*R_{\alpha \rho \gamma \sigma} := \frac {1} {2} \epsilon_{\alpha \rho} {} ^{\lambda \mu} R_{\lambda \mu \gamma \sigma}##
    ##*R_\beta {} ^ \rho {} _\delta {} ^\sigma := \frac {1} {2} \epsilon_\beta {} ^{\rho \tau} {} _\phi R_\tau {} ^ \phi {} _\delta {} ^ \sigma##
    Thus,
    ##*R_{\alpha \rho \gamma \sigma}*R_\beta {} ^ \rho {} _\delta {} ^\sigma = \frac {1} {4} \epsilon_{\alpha \rho} {} ^{\lambda \mu} \epsilon_\beta {} ^{\rho \tau} {} _\phi R_{\lambda \mu \gamma \sigma} R_\tau {} ^ \phi {} _\delta {} ^ \sigma##
    ## = - \frac {1} {4} \epsilon_{\alpha } {} ^{\lambda \mu}\epsilon_\beta {} ^{ \tau} {} _\phi R_{\lambda \mu \gamma \sigma} R_\tau {} ^ \phi {} _\delta {} ^ \sigma## (contracting on the ## \rho## index)
    ## = - \frac {1} {4} \begin{vmatrix}
    {\delta_{\alpha \beta}} & {\delta_\alpha ^ \tau} & {\delta_{\alpha \phi}} \\
    {\delta_\beta ^ \lambda} & {\delta ^ {\lambda \tau}} & {\delta ^ {\lambda} _\phi} \\
    {\delta_\beta ^ \mu} & {\delta ^{\mu \tau}} & {\delta_ \phi ^ \mu}
    \end{vmatrix}R_{\lambda \mu \gamma \sigma} R_\tau {} ^ \phi {} _\delta {} ^ \sigma##
    ##= - \frac {1} {4} \delta_{\alpha \beta}(\delta^{\lambda \tau}\delta_\phi ^ \mu - \delta^{\mu \tau}\delta_\phi ^\lambda)R_{\lambda \mu \gamma \sigma} R_\tau {} ^ \phi {} _\delta {} ^ \sigma##
    ## + \frac {1} {4} \delta_{\alpha}^\tau(\delta_\beta^\lambda \delta_\phi ^\mu - \delta_\phi ^\lambda \delta_\beta ^\mu)R_{\lambda \mu \gamma \sigma} R_\tau {} ^ \phi {} _\delta {} ^ \sigma##
    ## + \frac {1} {4} \delta_{\alpha \phi}(\delta_\beta ^ \mu \delta^{\lambda \tau}-\delta_\beta^\lambda \delta^{\mu \tau})R_{\lambda \mu \gamma \sigma} R_\tau {} ^ \phi {} _\delta {} ^ \sigma##
    ##= \frac {1} {2}(-g_{\alpha \beta}g_{\gamma \delta}R_{ \gamma \sigma \lambda \mu }R^{ \gamma \sigma \lambda \mu }+ R_{\beta \phi \gamma \sigma} R_\alpha {} ^ \phi {} _\delta {} ^ \sigma + R_{\lambda \beta \gamma \sigma}R^{\lambda \alpha}{} _\delta {} ^ \sigma)##
    In the following I have used the anti-symmetry property of the Riemann tensor for adjacent indices, the symmetry property for index pairs and ## \delta_{\alpha \beta}=g_{\alpha \beta}## (free space).
    Examining the first term in the sum I note,
    ##\delta_\rho ^\gamma \delta _\gamma ^ \rho R_{ \gamma \sigma \lambda \mu }R^{ \gamma \sigma \lambda \mu } = 4R_{ \gamma \sigma \lambda \mu }R^{ \gamma \sigma \lambda \mu } = R_{ \rho \sigma \lambda \mu }R^{ \rho \sigma \lambda \mu } ##
    and the first term in the sum becomes,
    ##- \frac {1}{8}g_{\alpha \beta}g_{\gamma \delta}R_{ \rho \sigma \lambda \mu }R^{ \rho \sigma \lambda \mu }##
    Examining the second term,
    ##\frac {1}{2}(\delta^{\phi \rho} \delta_\sigma ^ \sigma \delta_{\phi \rho} \delta_\sigma ^ \sigma)R_{\beta \phi \gamma \sigma} R_\alpha {} ^ \phi {} _\delta {} ^ \sigma = \frac {1}{2}(1)R_{\beta \phi \gamma \sigma} R_\alpha {} ^ \phi {} _\delta {} ^ \sigma = \frac {1}{2}R_{\alpha \rho \delta \sigma}R_\beta{}^\rho{} _\gamma {} ^\sigma##
    Examining the third term,
    ##\frac {1}{2}(\delta_\alpha ^ \alpha \delta^{\lambda \rho} \delta_\rho ^ \rho \delta_{\lambda \rho} \delta_\rho ^ \rho) R_{\lambda \beta \gamma \sigma}R^{\lambda \alpha}{} _\delta {} ^ \sigma = \frac {1}{2}(1) R_{\lambda \beta \gamma \sigma}R^{\lambda \alpha}{} _\delta {} ^ \sigma = \frac {1}{2}R_{\alpha \rho \delta \sigma}R_\beta{}^\rho{} _\gamma {} ^\sigma##
    and adding the terms we find,
    ##*R_{\alpha \rho \gamma \sigma}*R_\beta {} ^ \rho {} _\delta {} ^\sigma = R_{\alpha \rho \delta \sigma}R_\beta{}^\rho{} _\gamma {} ^\sigma - \frac {1}{8}g_{\alpha \beta}g_{\gamma \delta}R_{ \rho \sigma \lambda \mu }R^{ \rho \sigma \lambda \mu }##
     
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