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Variation of quadratic term in modified Einstein-Hilbert actions

+ 2 like - 0 dislike
106 views

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
$$
S_\chi=\int\,d^4x\,\sqrt{-g}\frac{1}{2}\gamma\chi^2,\qquad(\star)
$$
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
$$
\delta S_\chi^2=\int\,d^4x\,\Big[\sqrt{-g}\gamma\chi\delta g^{\mu\nu}\nabla_\mu\partial_\nu\phi\Big]=-\int\,d^4x\,\gamma\Big[\nabla_\mu(\sqrt{-g}\chi)\partial_\nu\phi\delta g^{\mu\nu}\Big]
$$
or
$$
\delta S_\chi^2=-\int\,d^4x\,\gamma\sqrt{-g}\partial_\mu\chi\partial_\nu\phi\delta g^{\mu\nu}
$$
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 
$$
G_{\mu\nu}=\cdots+\gamma\partial_\mu\phi\partial_\nu\chi.
$$
My question is: the reason that i find in literature that $\delta S_\chi^2$ contribute as
$$
\gamma\Big(\partial_\alpha\phi\partial^\alpha\chi\delta^\mu_\nu-(\partial_\nu\phi\partial^\mu\chi+\partial_\nu\chi\partial^\mu\phi)\Big)
$$
is because the missing step is to decompose the rank two tensor into its antisymmetric, trace and symmetric trace free parts
$$
\partial_\mu\phi\partial_\nu\chi=K_{\mu\nu}=K_{[\mu\nu]}+\frac{1}{n}\delta_{\mu\nu}\delta^{\alpha\beta}K_{\alpha\beta}+(K_{(\mu\nu)}-\frac{1}{n}\delta_{\mu\nu}\delta^{\alpha\beta}K_{\alpha\beta})
$$
or I'am missing something?

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

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