Gauss - Bonnet Gravity -> Curvature variations

[ChrisVer]

again you talk about \delta \Gamma while I don't know it...
In general it's good to bring \delta \Gamma in terms of \delta g in the last expression I gave in the previous page (with the factors of 2 you added) and take the \delta g out as a common factor...
Also get a factor from \sqrt{-g} although that's the easiest part in your Lagrangian, since it will come in complete analogy:
\delta_g S_1 = \int d^{4} x L \delta \sqrt{-g} = -\frac{1}{2}\int d^{4} x \sqrt{-g}L g_{\mu \nu} \delta g^{\mu \nu}

with L your Lagrangian..and so you don't have to care about it aver again...in your last overall Eom just add that term: -\frac{1}{2}\sqrt{-g}L g_{\mu \nu} \delta g^{\mu \nu}...

 

[Breo]

Ok I will (what ugly EoM :s).

So my last question about this: I do not understand the post #24 :P It would be possible that you are mixing the terms of the variation of square root of -g times L with the variation of L times square root of -g? Because the integral you write is for the latter, right? but you introduced it as the former.

 

[ChrisVer]

It's supposed to be ugly since it has so many higher order terms... and you have them all in your lagrangian (I wonder where did you find it).
Not that they don't exist (eg the Kretschmann scalar as I read has applications in quantum gravity), but the Lagrangian looks awful.

As for the post

Also , since you work with variations it's better to work with the action, rather than the lagrangian. The action will be for this R_{abcd}R^{abcd} only term [Kretschmann scalar]:
S = \int d^4 x \sqrt{-g} R_{abcd}R^{abcd}

So when you vary it with respect to the metric, you will get an additional term from the variation of the \sqrt{-g} which will also exist in the equations of motion... I would try to bring it to a form:

\delta S = \int d^4 x [A]_{ij} \delta g^{ij}

And thus the EoM will be \frac{\delta S}{\delta g^{\mu \nu}}=A_{ij} \delta^{i}_\mu \delta^j_\nu=A_{\mu \nu}=0

with A given by the terms coming from the variation of the Kretschmann scalar and the square root of the determinant of the metric...

In your case you have addiitional terms coming from the Ricci scalar^2 and the "Ricci curvature^2"...

I just sketched the idea of working this out... you will have:
S= \int d^4 x \sqrt{-g} L
take variation wrt the metric field:
\delta _g S = \int d^4 x \sqrt{-g} \delta_g L + \int d^4 x L \delta_g \sqrt{-g}
you will somehow try (I don't know whether that is possible or not) to bring \delta_g L= A_{ij} \delta g^{ij}~~(1)
and similar for \delta_g \sqrt{-g} = B_{ij} \delta g^{ij}~~(2) so that:

\delta _g S = \int d^4 x A_{ij} \sqrt{-g} \delta g^{ij}+ \int d^4 x L B_{ij} \delta g^{ij}
or
\delta _g S = \int d^4 x (A \sqrt{-g}+BL)_{ij} \delta g^{ij}= \int d^4 x (C)_{ij} \delta g^{ij}
The EoM will be then C_{ij}=0 where you determine C by A,B which you determine by the variations (1),(2).

In case you have the covarant derivatives of the variations of the metric, you can bring them out using:
\nabla_\alpha ( A B)=A \nabla_\alpha B + (\nabla_\alpha A) B

 

[Breo]

Ah! it is clear! I got lost because I expected to see the $\sqrt{-g}$ term explicitly out of $A_{ij}$ (in your last post $B_{ij}$).

I found this in my gravitation course notes. I am studying a masters right now.

I can not express how much I appreciate your efforts and help, ChrisVer :)

Thank you very very much!

 

[ChrisVer]

well I have a script right now where I derive the EoM for
R_{abcd}R^{abcd} -term...
If you want I can send it to you...however there may be mistakes...

 

[haushofer]

In this thesis,

http://thep.housing.rug.nl/sites/default/files/theses/Master thesis_Roel Andringa.pdf

some of these variations are derived in the later chapters :)