Contraction in the orthonormal frame yields
$$
R_{00}=1,
\qquad R_{11}=R_{22}=0,
\qquad R=-1.
$$
Hence
$$
G_{00}=G_{11}=G_{22}=\frac12.
$$
For $u^\mu=(1,0,0)$, pressureless matter has $T_{00}=\rho$ and vanishing spatial components. The spatial Einstein equations $G_{ii}+\Lambda\eta_{ii}=0$ give $\Lambda=-1/2$, while the time component gives $1=8\pi\rho$. Thus, for $G_N=1$,
$$
\boxed{\Lambda=-\frac12,\qquad \rho=\frac1{8\pi}}.
$$
If the convention is $G_{ab}+\Lambda g_{ab}=T_{ab}$, the corresponding density is simply $\rho=1$.


Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 309 / 4 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-309/4/d/solution