Ricci scalar in Newtonian Limit

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.

1 Answer

The spatial trace is wrong: you should have

$$g^{ij} R_{ij} \approx \delta^{ij} \delta_{ij} \delta^{mn}\partial_m \partial_n \varphi = \delta_{ii} \delta^{mn}\partial_m \partial_n \varphi = 3 \delta^{mn}\partial_m \partial_n \varphi.$$

The point is that the Kronecker delta is contracted with itself, which gives a factor of 3, not 1.

• oh my god i always forget that!! happens at least once a problem set. thanks. Commented Apr 23, 2020 at 21:18