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  Variation of quadratic term in modified Einstein-Hilbert actions

+ 2 like - 0 dislike

In the context of [mimetic gravity](http://www.google.com/search?as_epq=mimetic+gravity) at some point one try to add to an already modified Einstein-Hilbert action also a term like
where $\chi=g^{\mu\nu}\nabla_\mu\nabla_\nu\phi=g^{\mu\nu}\nabla_\mu\partial_\nu\phi$. Variation of this lead to the two contributes
\delta S_\chi^1+\delta S_\chi^2=\int\,d^4x\,\Big[\delta(\sqrt{-g})\frac{1}{2}\gamma\chi^2+\sqrt{-g}\gamma\chi\delta\chi\Big].
Looking only at the second term, after integration by parts, it become
\delta S_\chi^2=\int\,d^4x\,\Big[\sqrt{-g}\gamma\chi\delta g^{\mu\nu}\nabla_\mu\partial_\nu\phi\Big]=-\int\,d^4x\,\gamma\Big[\nabla_\mu(\sqrt{-g}\chi)\partial_\nu\phi\delta g^{\mu\nu}\Big]
\delta S_\chi^2=-\int\,d^4x\,\gamma\sqrt{-g}\partial_\mu\chi\partial_\nu\phi\delta g^{\mu\nu}
because the metric is covariantly conserved. So I found that the contribute to the right side of the Einstein equations of the term $\delta S_\chi^2$ that originate from ($\star$) is 
My question is: the reason that i find in literature that $\delta S_\chi^2$ contribute as
is because the missing step is to decompose the rank two tensor into its antisymmetric, trace and symmetric trace free parts
or I'am missing something?

asked Aug 6, 2015 in Theoretical Physics by yngabl (10 points) [ no revision ]

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