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ADM decomposition for Gauss-Bonnet like gravity

+ 3 like - 0 dislike
136 views

Suppose we have a Gauss-Bonnet like theory which its Lagrangian can be written as, 
\[\mathcal{L}=aR^2+bR_{\mu\nu}R^{\mu\nu}+cR_{\mu\nu\lambda\sigma} R^{\mu\nu\lambda\sigma}\]
I want to write above after the ADM decomposition. I know that Riemann tensor can be written in term of the ADM variable via Gauss, Codazzi and Ricci equations. Namely via Eq 2.11 in this paper. Essentially what I am looking after is to write the above Lagrangian as 4.25 in the same paper. So far I tried to use the definitions of the terms appearing in 4.25 but I was not so successful. (The definitions are given in 2.13). Any one has any idea about this? 

asked Dec 20, 2015 in Theoretical Physics by William [ no revision ]
recategorized Dec 21, 2015 by dimension10

Nice paper. I do not understand what is the problem, the Lagrangian in the paper is identical up to coefficients and all the relations for the computation are just handed over. Of course, to get the same results, you must use the same constraint/divergence-addition procedure ($(2.1)\to (2.18) \to (3.2)$, also maybe read Faddeev & Jackiw to gain insight on the de-constrainization).

Thanks Void. Yes I know the procedure but I can not get exactly as it is in 4.25. That is why I asked for some deeper expertise. However thanks for the paper. 

@Wiliam The relations in $(4.25)$ are of course easily obtained by considering the definitions in $(2.13)$ and $h^{\mu \nu} \equiv g^{\mu \nu} + n^\mu n^\nu$.

@Void, Thanks, I think however the relation you have use shall be corrected to \(h^{\mu\nu}=g^{\mu\nu}-n^\mu n ^\nu\)

@Wiliam In $-+++$ signature (which is used in the paper) you can check explicitly that the sign is correct, e.g. by consulting $(2.8),\,(2.9)$.

Oh yes sure, you're right. 

1 Answer

+ 0 like - 0 dislike

Use \[h^{\mu\nu}=g^{\mu\nu}+n^\mu n^\nu\]

answered Dec 21, 2015 by Bernard [ revision history ]
edited Dec 21, 2015

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