= Solution
The <Einstein tensor> and its contractions are
$$
G_{\mu\nu}=R_{\mu\nu}-\frac12g_{\mu\nu}R,\qquad
R_{\mu\nu}=R^\alpha{}_{\mu\alpha\nu},\qquad
R=g^{\mu\nu}R_{\mu\nu}.
$$
To match the supplied <Ricci tensor> components, use the <Riemann curvature tensor> convention
$$
R^\rho{}_{\sigma\mu\nu}
=\partial_\nu\Gamma^\rho_{\mu\sigma}
-\partial_\mu\Gamma^\rho_{\nu\sigma}
+\Gamma^\rho_{\nu\lambda}\Gamma^\lambda_{\mu\sigma}
-\Gamma^\rho_{\mu\lambda}\Gamma^\lambda_{\nu\sigma}.
$$
This 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 \b[$R_{\mu\nu}=\Lambda g_{\mu\nu}$].
Differentiate the angular field equation to obtain
$$
-2A'-rA''=2\Lambda r,\qquad
\frac{A'}r+\frac{A''}2=-\Lambda.
$$
Consequently $R_{00}=\Lambda A=\Lambda g_{00}$ 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 $(rA)'=1-\Lambda r^2$ gives $A=1+C/r-\Lambda r^2/3$. Match $g_{tt}\simeq1+2\Phi$ with the <Newtonian gravitational potential> $\Phi=-GM/r$ in the weak-field overlap region to identify $C=-2GM$. Hence the <Schwarzschild-de Sitter spacetime> has
$$
\boxed{A(r)=1-\frac{2GM}{r}-\frac{\Lambda r^2}{3}\quad(c=1).}
$$
Negative $\Lambda$ 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 $L=(A\dot t^2-\dot r^2/A-r^2\dot\phi^2)/2$. The <cyclic coordinates> give $A\dot t=k$ and $r^2\dot\phi=h$; the angular <canonical momentum> is $-h$ in this signature. The <four-velocity> normalization $2L=1$ then gives
$$
\boxed{\dot r^2=k^2-A(r)\left(1+\frac{h^2}{r^2}\right).}
$$
Here $k$ 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 $\Lambda\ne0$.
For radial motion $h=0$, differentiation gives $2\dot r\ddot r=-A'\dot r$. The radial <Euler-Lagrange equation> extends the resulting expression through a <turning point> where division by $\dot r$ would fail. Thus
$$
\boxed{\ddot r=-\frac{A'}2=-\frac{GM}{r^2}+\frac{\Lambda r}{3}\quad(c=1).}
$$
This 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> gives
$$
\boxed{\frac{d^2r}{dt^2}\simeq-\frac{GM}{r^2}+\frac{\Lambda c^2r}{3},\qquad
\Phi=-\frac{GM}{r}-\frac{\Lambda c^2r^2}{6}.}
$$
The extra term describes the gravitational effect of a <cosmological constant>, not drag or a contact force. The approximation requires $GM/(rc^2)\ll1$, $|\Lambda|r^2\ll1$ and <speed> much less than $c$.
Back to article page
