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Gauss - Bonnet Gravity -> Curvature variations

  1. Nov 21, 2014 #1
    So I am working on the next quadratic Lagrangian:

    $$ L = \alpha R_{\mu\nu}R^{\mu\nu} + \beta R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma} + \gamma R² $$

    I have already derived $$ \delta (R_{\mu\nu}R^{\mu\nu}) = [- \frac{1}{2}g_{\mu\nu}R_{\alpha\beta}R^{\alpha\beta} + 2R^{\alpha\beta}R_{\mu\alpha\nu\beta}]\delta g^{\mu\nu} $$ right?

    How can I compute the other variations in order to compute the equations of motion?
  2. jcsd
  3. Nov 22, 2014 #2
    Do not ask for results, just a tip to start deriving. Not sure if the most appropiate here would be using the Leibniz rule.
  4. Nov 22, 2014 #3


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    you can use the Leibniz rule...
  5. Nov 22, 2014 #4
    But lots of Christoffel symbols appear and I do not know to simplificate the equations. I will try again.

    What about R², should I set it as $$ \delta R R + R\delta R$$


    $$ \delta (g^{\mu\nu}R_{\mu\nu}) R + g^{\mu\nu}R_{\mu\nu} \delta R $$ ?

    Another way?
  6. Nov 22, 2014 #5


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    I hope this can help.

    As for the second question... you have written the same thing twice...but yes, the second way, by replacing the other ricci scalar can help... All in all, what you will mainly need is the variation of the Riemann tensor, and all the rest can come by contractions...

    [itex]\delta R^2 = R ~\delta R + \delta R~ R= g^{\mu \nu} R_{\mu \nu} \delta (g^{\alpha \beta}R_{\alpha \beta} ) + \delta (g^{\alpha \beta}R_{\alpha \beta} ) g^{\mu \nu} R_{\mu \nu} = 2 g^{\mu \nu} R_{\mu \nu} \delta (g^{\alpha \beta}R_{\alpha \beta} )=[/itex]
    [itex]=2 g^{\mu \nu} R_{\mu \nu} (g^{\alpha \beta} \delta R_{\alpha \beta} + R_{\alpha \beta} \delta g^{\alpha \beta}) [/itex]
    Last edited: Nov 22, 2014
  7. Nov 22, 2014 #6
    I do not see clear what it says (bolds):

    In the second line we used the previously obtained result for the variation of the Ricci curvature and the metric compatibility of the covariant derivative, d5e4060b536bfbdcfc59f5a4120f2615.png .

    The last term, 8ee39a0d3cfd3c964d90b9bfae336c06.png , multiplied by b7a06bf4acbc66ea40626975edce6e49.png becomes a total derivative, since

    and thus by Stokes' theorem only yields a boundary term when integrated. Hence when the variation of the metric 984b69a6d964372ae6e86cbf49d9b176.png vanishes at infinity, this term does not contribute to the variation of the action. And we thus obtain,

  8. Nov 22, 2014 #7


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    I don't understand what is not clear? Maybe if I call it Gauss theorem?
    [itex] \int_V d^4 x \partial_\mu A^\mu = \oint_{\partial V} d \sigma_{\mu}A^{\mu} [/itex]

    So here if you set a boundary condition that [itex] \delta g^{\mu \nu}=0[/itex] at the limits of your integral, that integral can vanish...so the last term in [itex]\delta R[/itex] won't contribute...
    Last edited: Nov 22, 2014
  9. Nov 22, 2014 #8
    Oh right! So we are anticipating the action integral, right?.
  10. Nov 22, 2014 #9


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    Yes... If you don't anticipate the action integral, but keep carrying that term, you'll be carrying a lagrangian that comes with a total derivative...
    [itex]L'= L + \partial_\mu A^\mu[/itex]
    you can work with both [itex]L',L[/itex] since they are equivalent...so it's better (convenient) to work with [itex]L[/itex].
  11. Nov 22, 2014 #10
    I guess $$ \delta (R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}) $$ should vanish due to the indices location (the variation of all up or all down indices should be zero? I notice this because of the Christoffels symbols obtained...). Otherwise I tried to do:

    $$ \delta (g_{\mu\alpha}R^{\alpha}_{\nu\rho\sigma})R^{\mu\nu\rho\sigma} + R_{\mu\nu\rho\sigma}\delta(g^{\nu\beta}g^{\rho\gamma}g^{\sigma\delta}R^{\mu}_{\beta\gamma\delta} = (\delta g_{\mu\alpha}R^{\alpha}_{\nu\rho\sigma})R^{\mu\nu\rho\sigma} + R_{\mu\nu\rho\sigma}(\delta g^{\mu\alpha}R_{\alpha}^{\nu\rho\sigma}) $$

    This seems wrong... :-/

    Assuming it must vanish then, after a few calcs I get the next equations of motion (avoiding the constants of the original lagrangian):

    $$ (\frac{3}{2}g_{\mu\nu}R + 2R_{\mu\nu}) = 0 $$

    This seems better if we choose in the Lagrangian of the statement: $$\alpha = \frac{1}{2} \ \space ; \space \ \gamma= -\frac{1}{8} $$ or multiples. So we obtain:

    $$ (R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R) = 0 $$

    Is this correct?
    Last edited: Nov 22, 2014
  12. Nov 22, 2014 #11


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    can you elaborate what you mean with the vanishing of the [itex]\delta(R_{abcd}R^{abcd})[/itex]?

    [itex] \delta R_{abcd}= \delta (g_{ka} R^{k}_{~bcd} )=R^{k}_{~bcd} \delta g_{ka} + g_{ka} \delta R^{k}_{~bcd} = R^{k}_{~bcd} \delta g_{ak} + g_{ak} \nabla_{[c}\delta \Gamma_{d]~ b}^k [/itex]


    [itex] \delta R^{abcd}= \delta (g ^{am} g^{bn} g^{cw} g^{dr} R_{mnwr}) [/itex]

    I am sorry, I just can't see how can you notice that they should vanish because of the indicies...
  13. Nov 22, 2014 #12
    My fault, I reached that conclusion before the trial. So what I calculated was correct?

    I am stucked then... Will keep trying. But after the result I reached assuming it vanishes, I think it really should vanish... I will try to forget that idea...

    Thank you very much ChrisVer for all your support!

    EDIT: Oh wait! Maybe it vanish because a factor of $$ g^{\alpha\beta} $$ will remain in the action so it will vanish? I mean I am calculating: $$ \frac{\delta S}{\delta g^{\mu\nu}} $$ so I can not remove the metric variations obtained for the Riemann tensor calc, right?
    Last edited: Nov 22, 2014
  14. Nov 22, 2014 #13


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  15. Nov 22, 2014 #14
    Not really, I am stucked in the computation of the riemann tensor square variation. :(
  16. Nov 22, 2014 #15
    I think still can not solve it. I get this:

    $$ \delta (R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}) = (\delta g_{\mu\alpha} R^{\alpha}_{\nu\rho\sigma})R^{\mu\nu\rho\sigma} + R_{\mu\nu\rho\sigma}(\delta g^{\mu\alpha} R_{\alpha}^{\nu\rho\sigma})$$

    Thus, the equations of motion:

    $$ 0 = \alpha RR_{\mu\nu} + \gamma ( R^{\rho\sigma} R_{\mu\nu\rho\sigma} - \frac{1}{4}g_{\mu\nu}R_{\rho\sigma}R^{\rho\sigma}) + \beta (R_{\nu\alpha\rho\sigma}R_{\mu}^{\alpha\rho\sigma}) $$

  17. Nov 22, 2014 #16
    I just read in the Zee's book "Quantum Field Theory in a Nutshell":

    "Also, use the Gauss-Bonnet theorem to get rid of the $$ R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}$$ term" pag. 459

    I am trying to find that fact but the most I found was a bit complex explanation in the Hawking - Ellis book. Can anyone give me a reasoning for beginners?

    Thank you in advance.
  18. Nov 23, 2014 #17


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    http://relativity.livingreviews.org/open?pubNo=lrr-2010-3&page=articlesu33.html [Broken]

    maybe tis can be of help
    Last edited by a moderator: May 7, 2017
  19. Nov 23, 2014 #18
    It says the equations of motion gives second derivatives, but I did not obtain any second one...
  20. Nov 23, 2014 #19


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    $$R_{abcd} \delta (g^{ia}g^{jb}g^{kc}g^{wd}R_{ijkw})=R_{abcd} ( g^{jb}g^{kc}g^{wd}R_{ijkw} \delta g^{ia} + g^{ia}g^{kc}g^{wd}R_{ijkw} \delta g^{jb} +$$
    $$ +g^{jb}g^{ia}g^{wd}R_{ijkw} \delta g^{kc}+ g^{jb}g^{kc}g^{ia}R_{ijkw} \delta g^{wd} + g^{ia} g^{jb}g^{kc}g^{wd} \delta R_{ijkw})$$

    The first term is:

    $$R_{abcd} g^{jb}g^{kc}g^{wd}R_{ijkw} \delta g^{ia}$$

    The second term is giving:

    $$R_{abcd}g^{ia}g^{kc}g^{wd}R_{ijkw} \delta g^{jb}= R_{bacd} R_{jikw} g^{ia}g^{kc}g^{wd} \delta g^{jb}=R_{abcd} g^{jb}g^{kc}g^{wd}R_{ijkw} \delta g^{ia}$$

    where at first equation I changed the indicies position using the antisymmetry of the Riemann tensor and then at the second equation I just renamed null indices.

    The third term is again playing with indices:
    $$R_{abcd}g^{jb}g^{ia}g^{wd}R_{ijkw} \delta g^{kc} =R_{cdab}g^{wd}g^{kc}g^{jb}R_{kwij} \delta g^{ia}= R_{abcd}R_{ijkw} g^{wd}g^{kc}g^{jb} \delta g^{ia}$$

    The fourth:
    $$ g^{jb}g^{kc}g^{ia} R_{abcd}R_{ijkw} \delta g^{wd}= g^{kc}g^{jb}g^{wd} R_{dcba}R_{wkji} \delta g^{ia}= R_{abcd}R_{ijkw} g^{kc}g^{jb}g^{wd} \delta g^{ia} $$

    Using also that $$R_{abcd}g^{ai}g^{bj}g^{ck}g^{dw}= R^{ijkw}$$

    These terms in the above first equation will give:

    $$R_{abcd} \delta R^{abcd} = R_{abcd}R_{ijkw} (g^{jb}g^{kc}g^{wd}+g^{jb}g^{kc}g^{wd}+g^{wd}g^{kc}g^{jb}+g^{kc}g^{jb}g^{wd} ) \delta g^{ia} +R^{ijkw} \delta R_{ijkw} $$


    $$R_{abcd} \delta R^{abcd}= 4 R_{abcd}R_{ijkw} g^{jb}g^{kc}g^{wd} \delta g^{ia} + R^{ijkw} R^{s}_{~jkw} \delta g_{is} + R^{ijkw}g_{is} \nabla_{[k} \delta~\Gamma^{s}_{w]j}$$

    Now you can play again with indices and use that [itex] \delta g_{is}= - g_{ib} g_{sp} \delta g^{bp} [/itex]

    $$R_{abcd} \delta R^{abcd}= 4 R_{abcd}R_{ijkw} g^{jb}g^{kc}g^{wd} \delta g^{ia} - R^{rjkw} R_{ijkw} g_{ra} \delta g^{ia} + R^{ijkw}g_{is} \nabla_{[k} \delta~\Gamma^{s}_{w]j}$$

    $$=(4 R_{abcd}R_{ijkw} g^{jb}g^{kc}g^{wd} - R^{rjkw}R_{ijkw} g_{ra} ) \delta g^{ia}+ R^{ijkw}g_{is} \nabla_{[k} \delta~\Gamma^{s}_{w]j}$$

    To this you just add then the [itex] R^{abcd}\delta R_{abcd}[/itex] I gavee in a previous post and work out the result...

    Did you get this result by your calculations? or something similar....I did it really fast I may have done mistakes with the renaming of indicies and so on.....
    Last edited: Nov 23, 2014
  21. Nov 23, 2014 #20


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    To my last not about adding the previous result, you don't have to do it... because ere it appears the same quantity so you can just add a 2 in the appropriate terms.
    However do the calculations also by yourself to check whether I did any mistake or not.....
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