 Solution

The Einstein tensor and its contractions areTo match the supplied Ricci tensor components, use the Riemann curvature tensor conventionThis convention also fixes the sign used for geodesic deviation above. In vacuum, tracing the Einstein field equations gives $-R+4\Lambda =0$. Substitution therefore yields $R_{\mu \nu }=\Lambda g_{\mu \nu }$.

Differentiate the angular field equation to obtainConsequently $R_{00}=\Lambda A=\Lambda g_{00}$ and $R_{11}=-\Lambda /A=\Lambda g_{11}$. The other angular component follows from $R_{33}={\sin }^2\theta R_{22}$. Thus the single angular equation implies all the displayed vacuum equations wherever this coordinate chart is regular.

Integrating $(rA{)}^{\prime }=1-\Lambda r^2$ gives $A=1+C/r-\Lambda r^2/3$. Match $g_{tt}\simeq 1+2\Phi$ with the Newtonian gravitational potential $\Phi =-GM/r$ in the weak-field overlap region to identify $C=-2GM$. Hence the Schwarzschild-de Sitter spacetime hasNegative $\Lambda$ instead gives the Schwarzschild--anti-de Sitter black hole case. This matching fixes the normalization locally; a spacetime with nonzero $\Lambda$ is not asymptotically flat.

For timelike geodesics in the orbital plane, take proper time as parameter and use $L=(A{\dot{t}}^2-{\dot{r}}^2/A-r^2{\dot{\varphi }}^2)/2$. The cyclic coordinates give $A\dot{t}=k$ and $r^2\dot{\varphi }=h$; the angular canonical momentum is $-h$ in this signature. The four-velocity normalization $2L=1$ then givesHere $k$ is conserved Killing energy per unit rest mass in the chosen time normalization, and $h$ is conserved angular momentum per unit rest mass. There is no general interpretation of $k$ as energy measured at an asymptotically flat infinity when $\Lambda \ne 0$.

For radial motion $h=0$, differentiation gives $2\dot{r}\ddot{r}=-A^{\prime }\dot{r}$. The radial Euler-Lagrange equation extends the resulting expression through a turning point where division by $\dot{r}$ would fail. ThusThis is the proper-time second derivative of the areal radius, not a nonzero four-acceleration of the freely falling particle. Positive $\Lambda$ contributes an outward term; negative $\Lambda$ reinforces inward attraction. Restoring units in the slow-motion, weak-field Newtonian limit givesThe extra term describes the gravitational effect of a cosmological constant, not drag or a contact force. The approximation requires $GM/(r{c}^2)\ll 1$, $|\Lambda |r^2\ll 1$ and speed much less than $c$.