Consider the GR metric in the Newtonian limit, i.e. a small perturbation $\varphi$ from the Minkowski metric $\eta_{\mu\nu}$: $$ds^2=-(1+2\varphi)dt^2+(1-2\varphi)\delta_{ij}dx^idx^j$$
My objective is to calculate the Einstein tensor in this limit. I have already found that $R_{tt}=\delta^{ij}\partial_i\partial_j\varphi$ and $R_{ij}=\delta_{ij}\delta^{mn}\partial_m\partial_n\varphi$, and I believe these to be correct. Now, then, I only need the Ricci scalar. Here is my attempt: $$R=R^\mu_{\ \ \mu}=g^{\mu\nu}R_{\mu\nu}=g^{tt}R_{tt}+g^{ij}R_{ij}$$ Now I believe the inverse metric has components $g^{tt}=(-1-2\varphi)$ and $g^{ij}=(1+2\varphi)\delta^{ij}$, so I find that $$R=-\delta^{ij}\partial_i\partial_j\varphi+\delta^{mn}\partial_m\partial_n\varphi=0.$$ where I neglected terms with both $\varphi$ and a derivative of $\varphi$ as both are small terms. I believe this is wrong -- in fact, I know that $G_{tt}=2\delta^{ij}\partial_i\partial_j\varphi$, which leads me to believe that the minus sign in front of the first term on the RHS above should in fact by a plus sign -- that would mean $-\frac{1}{2}2\delta^{ij}\partial_i\partial_j\varphi g_{tt}=\delta^{ij}\partial_i\partial_j\varphi$ as expected. However, I really can't see why that should be a negative sign.
