I've been trying to prove this equation:
$$
\delta G_{\alpha\beta}=-\frac{1}{2}\left(\square\bar{h}_{\alpha\beta}+2R{}_{\gamma\alpha\delta\beta}\bar{h}^{\gamma\delta}\right)+\frac{1}{2}\left(\bar{h}_{\alpha;\gamma\beta}^{\gamma}+\bar{h}_{\beta;\gamma\alpha}^{\gamma}-g_{\alpha\beta}\bar{h}^{\rho\delta}{}_{;\rho\delta}\right),
$$
for the first order Einstein tensor, where $\bar{h}_{\alpha\beta}:=h_{\alpha\beta}-\frac{1}{2}g_{\alpha\beta}\left[g_{\gamma\delta}h^{\gamma\delta}\right]$. I've tried it a dozen times and always get this expression:
$$
\delta G_{\alpha\beta}=-\frac{1}{2}\left(\square\bar{h}_{\alpha\beta}-2R{}_{\gamma\alpha\delta\beta}\bar{h}^{\gamma\delta}\right)+\frac{1}{2}\left(\bar{h}_{\alpha;\gamma\beta}^{\gamma}+\bar{h}_{\beta;\gamma\alpha}^{\gamma}-g_{\alpha\beta}\bar{h}^{\rho\delta}{}_{;\rho\delta}\right).
$$
Can anyone prove the right expression from:
$$
\delta G{}_{\alpha\beta}=\frac{1}{2}\left(-h_{\gamma;\beta\alpha}^{\gamma}+h_{\beta;\alpha\gamma}^{\gamma}+h_{\alpha;\beta\gamma}^{\gamma}-h_{\alpha\beta}{}^{;\gamma}{}_{;\gamma}\right)-\frac{1}{4}g{}_{\alpha\beta}\left[g{}^{\gamma\delta}\left(-h_{\rho;\delta\gamma}^{\rho}+h_{\delta;\gamma\rho}^{\rho}+h_{\gamma;\delta\rho}^{\rho}-h_{\delta\gamma}{}^{;\rho}{}_{;\rho}\right)\right].
$$
Best Answer
I won't do it with the messy formulas, because this is not the right way to check signs.
There is a right way to check signs--- by examples. Then you know for sure what the term means. The term whose sign you don't know is the only term without a derivative. So if you make a perturbation to g which is constant in the coordinate system you are using, it will be the only term.
The quickest way to check is to start in Riemann normal coordinates in the neighborhood of a single point. This is very useful for every local calculation--- it's a freefalling frame with the best possible conventional local form for the bending of the coordinate axes. In such a frame the metric near a point gives a simple and canonical expression for the curvature,
$$ g = \delta_{\mu\nu} + {1\over 6} R_{\mu\alpha\nu\beta} x^\alpha x^\beta $$
Now you can add a constant perturbation to g, say in the diagonal component:
$$ g_{\mu\nu} = \delta_{\mu\nu} + h_{\mu\nu} - {1\over 6} R_{\mu\alpha\nu\beta} x^\alpha x^\beta$$
Were $h_\mu\nu$ is a diagonal perturbation, which can be reconverted back to Riemann normal coordinates by rescaling each x by the corresponding factor of $(1+h/2)$, and this has the effect of multiplying R by (1-h) for each x it contracts. The first order change in R and then in G is with a minus sign, as you said it should be.
But I am sure you didn't really screw up. The two obvious places where a sign difference can creep in between your calculation in your conventions and your source:
I am sure that your disagreement is due to one of the two points above. What is most important for the sign issue is having enough quick checks (like Riemann normal coordinates, an explicit sphere metric, AdS metric, 2d metrics, etc) and only then will you stop getting confused over signs.