Curvature components and the vanishing of the Ricci and Einstein tensors of the Schwarzschild exterior metric

(Exhaustive pair-index table.) By [L4], a lowered component with a repeated index inside either pair vanishes, and the two skew-symmetries and pair exchange reduce all other components to a symmetric $6\times 6$ array indexed by the six choices $01,02,03,12,13,23$. Substitution of every pair-pair choice into [L3] gives this complete array; entries below the diagonal equal their transposes, and reversing the order within either pair changes the sign: [L2, L3, L4] $\begin{array}{crrrrrr}& 01& 02& 03& 12& 13& 23\\ 01& f^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}2& 0& 0& 0& 0& 0\\ 02& & rf{f}^{\mathrm{\prime }}\mathrm{/}2& 0& 0& 0& 0\\ 03& & & rf{f}^{\mathrm{\prime }}{\sin }^2\vartheta \mathrm{/}2& 0& 0& 0\\ 12& & & & -r{f}^{\mathrm{\prime }}\mathrm{/}(2f)& 0& 0\\ 13& & & & & -r{f}^{\mathrm{\prime }}{\sin }^2\vartheta \mathrm{/}(2f)& 0\\ 23& & & & & & r^2(1-f){\sin }^2\vartheta \end{array}$ Thus all fifteen unequal-pair entries are explicitly zero; the six diagonal entries are computed in steps 1.2, 1.3 and 2.1. This exhausts the $21$ independent pair-pair entries and, by [L4], every curvature component.

(The $01$ and $23$ components.) For $R_{0101}=g_{00}{R}^0{}_{101}$: by [L3], [L2] and ${\mathrm{\partial }}_0{\mathrm{\Gamma }}^0{}_{11}=0$, $R^0{}_{101}=-{\mathrm{\partial }}_1{\mathrm{\Gamma }}^0{}_{01}+{\mathrm{\Gamma }}^0{}_{01}{\mathrm{\Gamma }}^1{}_{11}-{\mathrm{\Gamma }}^0{}_{10}{\mathrm{\Gamma }}^0{}_{01}=-{\mathrm{\partial }}_1(f^{\mathrm{\prime }}\mathrm{/}(2f))-\frac{(f^{\mathrm{\prime }}{)}^2}{4f^2}-\frac{(f^{\mathrm{\prime }}{)}^2}{4f^2}=-\frac{f^{\mathrm{\prime }\mathrm{\prime }}}{2f}+\frac{(f^{\mathrm{\prime }}{)}^2}{2f^2}-\frac{(f^{\mathrm{\prime }}{)}^2}{2f^2}=-\frac{f^{\mathrm{\prime }\mathrm{\prime }}}{2f}$, and multiplication by $g_{00}=-f$ gives $R_{0101}=f^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}2$. For $R_{2323}=g_{22}{R}^2{}_{323}$: $R^2{}_{323}={\mathrm{\partial }}_2{\mathrm{\Gamma }}^2{}_{33}+{\mathrm{\Gamma }}^2{}_{21}{\mathrm{\Gamma }}^1{}_{33}-{\mathrm{\Gamma }}^2{}_{33}{\mathrm{\Gamma }}^3{}_{23}=({\sin }^2\vartheta -{\cos }^2\vartheta )-f{\sin }^2\vartheta +{\cos }^2\vartheta =(1-f){\sin }^2\vartheta$, using [L2]; hence $R_{2323}=r^2(1-f){\sin }^2\vartheta$.

(The $02$ and $12$ components.) For $R_{0202}=g_{00}{R}^0{}_{202}$: $R^0{}_{202}={\mathrm{\Gamma }}^0{}_{01}{\mathrm{\Gamma }}^1{}_{22}=-(r{f}^{\mathrm{\prime }})\mathrm{/}2$ by [L2] and [L3] (all remaining terms contain a vanishing symbol or a ${\mathrm{\partial }}_0$-derivative), so $R_{0202}=(-f)(-r{f}^{\mathrm{\prime }}\mathrm{/}2)=rf{f}^{\mathrm{\prime }}\mathrm{/}2$. For $R_{1212}=g_{11}{R}^1{}_{212}$: $R^1{}_{212}={\mathrm{\partial }}_1{\mathrm{\Gamma }}^1{}_{22}+{\mathrm{\Gamma }}^1{}_{11}{\mathrm{\Gamma }}^1{}_{22}-{\mathrm{\Gamma }}^1{}_{22}{\mathrm{\Gamma }}^2{}_{12}=-(f+r{f}^{\mathrm{\prime }})+\frac{r{f}^{\mathrm{\prime }}}{2}+f=-\frac{r{f}^{\mathrm{\prime }}}{2}$, using [L2], hence $R_{1212}=f^{-1}(-r{f}^{\mathrm{\prime }}\mathrm{/}2)=-r{f}^{\mathrm{\prime }}\mathrm{/}(2f)$.

(The remaining $03$ and $13$ components.) From [L2]-[L3], $R^0{}_{303}={\mathrm{\Gamma }}^0{}_{01}{\mathrm{\Gamma }}^1{}_{33}=-(r{f}^{\mathrm{\prime }}{\sin }^2\vartheta )\mathrm{/}2$, so $R_{0303}=g_{00}{R}^0{}_{303}=rf{f}^{\mathrm{\prime }}{\sin }^2\vartheta \mathrm{/}2$. Also $R^1{}_{313}={\mathrm{\partial }}_1{\mathrm{\Gamma }}^1{}_{33}+{\mathrm{\Gamma }}^1{}_{11}{\mathrm{\Gamma }}^1{}_{33}-{\mathrm{\Gamma }}^1{}_{33}{\mathrm{\Gamma }}^3{}_{13}=-(r{f}^{\mathrm{\prime }}{\sin }^2\vartheta )\mathrm{/}2$, hence $R_{1313}=g_{11}{R}^1{}_{313}=-r{f}^{\mathrm{\prime }}{\sin }^2\vartheta \mathrm{/}(2f)$. These verify the final two diagonal entries of the table in step 1.1.

(Mixed and zero components.) The fifteen unequal-pair entries are exactly the off-diagonal entries of the exhaustive array in step 1.1. Every component with a repeated index inside a pair vanishes by [L4]; all remaining sign and order variants are determined by its skew-symmetries and pair exchange. Thus clause (a) is exhaustive.

(Contraction to the Ricci tensor.) By [L4] and clause (a), ${\mathrm{Ric}}_{00}=g^{11}{R}_{1010}+g^{22}{R}_{2020}+g^{33}{R}_{3030}=f\cdot \frac{f^{\mathrm{\prime }\mathrm{\prime }}}{2}+\frac{1}{r^2}\cdot \frac{rf{f}^{\mathrm{\prime }}}{2}+\frac{1}{r^2{\sin }^2\vartheta }\cdot \frac{rf{f}^{\mathrm{\prime }}{\sin }^2\vartheta }{2}=f(\frac{f^{\mathrm{\prime }\mathrm{\prime }}}{2}+\frac{f^{\mathrm{\prime }}}{r})$. Similarly ${\mathrm{Ric}}_{11}=g^{00}{R}_{0101}+g^{22}{R}_{2121}+g^{33}{R}_{3131}=-\frac{f^{\mathrm{\prime }\mathrm{\prime }}}{2f}-\frac{f^{\mathrm{\prime }}}{2rf}-\frac{f^{\mathrm{\prime }}}{2rf}=-\frac{1}{f}(\frac{f^{\mathrm{\prime }\mathrm{\prime }}}{2}+\frac{f^{\mathrm{\prime }}}{r})$, ${\mathrm{Ric}}_{22}=g^{00}{R}_{0202}+g^{11}{R}_{1212}+g^{33}{R}_{3232}=-\frac{r{f}^{\mathrm{\prime }}}{2}-\frac{r{f}^{\mathrm{\prime }}}{2}+(1-f)=1-f-r{f}^{\mathrm{\prime }}$, and ${\mathrm{Ric}}_{33}=g^{00}{R}_{0303}+g^{11}{R}_{1313}+g^{22}{R}_{2323}=-\frac{r{f}^{\mathrm{\prime }}{\sin }^2\vartheta }{2}-\frac{r{f}^{\mathrm{\prime }}{\sin }^2\vartheta }{2}+(1-f){\sin }^2\vartheta =(1-f-r{f}^{\mathrm{\prime }}){\sin }^2\vartheta$. For a mixed pair, say $(0,1)$, every term $g^{aa}{R}_{a0a1}$ of [L4] contains a component with unequal index pairs or a repeated index, which vanishes by step 3.1; hence ${\mathrm{Ric}}_{01}=0$, and all mixed components vanish likewise. This is clause (b).

(Specialisation and the Einstein tensor.) For $f=1-s\mathrm{/}r$ the identities of [L5] give $f^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}2+f^{\mathrm{\prime }}\mathrm{/}r=0$ and $1-f-r{f}^{\mathrm{\prime }}=0$. Clause (b) therefore gives $\mathrm{Ric}=0$ on the dense open regular angular locus $M^{\circ }=\mathbb{R}\times (s,\mathrm{\infty })\times (S^2\setminus \{\text{two poles}\})$. The Ricci tensor is smooth on all of $M$ by [L3]; in any chart about a pole its components are continuous and vanish on the dense subset from $M^{\circ }$, hence vanish at the pole too. Thus $\mathrm{Ric}=0$ globally. Its contractions then give $S=0$ and $G=0$ by [L4]. This proves (c).

(Nonvanishing curvature.) On $M^{\circ }$, the $01$ entry of step 1.1 and $f^{\mathrm{\prime }\mathrm{\prime }}=-2s\mathrm{/}{r}^3$ give the global base-field contraction $g(R({\mathrm{\partial }}_0,{\mathrm{\partial }}_r){\mathrm{\partial }}_r,{\mathrm{\partial }}_0)=-s\mathrm{/}{r}^3$. This contraction is continuous on $M$ and equals the smooth function $-s\mathrm{/}{r}^3$ on the dense subset $M^{\circ }$, so the identity holds everywhere. Since $s,r>0$, the contraction is nowhere zero and therefore the curvature tensor is nowhere zero. This proves (d).