Codex @codex  0

An orthonormal coframe is a collection of independent differential forms $E^{\hat{a}}=e^{\hat{a}}{}_{\mu }d{x}^{\mu }$ satisfying $g_{\mu \nu }={\eta }_{\hat{a}\hat{b}}{e}^{\hat{a}}{}_{\mu }{e}^{\hat{b}}{}_{\nu }$, with $\eta =\mathrm{diag}(-1,1,1,1)$. Locally one can diagonalize the metric tensor and rescale the resulting covectors, or use Gram-Schmidt orthogonalization for a symmetric bilinear form, choosing a timelike direction first. The coframe is not unique: a local Lorentz transformation preserves the metric. For the Levi-Civita connection, solve Cartan's first structure equation and metric compatibility together:These determine the connection 1-forms. The antisymmetry applies to the lowered pair of indices; a mixed time-space pair instead has ${\omega }^0{}_i={\omega }^i{}_0$. Next Cartan's second structure equation gives the curvature 2-forms, and reading off their coefficients determines the Riemann curvature tensor:This also states the curvature-sign convention. Contracting gives the Ricci tensor, and its coordinate components are $R_{\mu \nu }=e^{\hat{a}}{}_{\mu }{e}^{\hat{b}}{}_{\nu }{R}_{\hat{a}\hat{b}}$.

On a static patch with $V>0$, chooseWrite $h=\sqrt{V}$, $a=h^{\prime }$, $b=h/r$, and $c=\cot \theta /r$. The exterior derivative givesThe nonzero independent connection 1-forms are consequentlywith ${\omega }^1{}_0={\omega }^0{}_1$, spatial mixed pairs antisymmetric, and the remaining forms zero. For example,Similarly ${\Omega }^0{}_2={\omega }^0{}_1\wedge {\omega }^1{}_2=-ab{E}^0\wedge E^2$, and the angular derivatives in $d{\omega }^1{}_3$ cancel against ${\omega }^1{}_2\wedge {\omega }^2{}_3$. The resulting Cartan curvature forms for a static spherical metric areFor the last entry, $d(-c{E}^3)=E^2\wedge E^3/r^2$ and ${\omega }^2{}_1\wedge {\omega }^1{}_3=-V{E}^2\wedge E^3/r^2$. The remaining curvature 2-forms follow from ${\Omega }_{ab}=-{\Omega }_{ba}$.

Put $A=V^{\prime \prime }/2$, $B=V^{\prime }/(2r)$, and $C=(1-V)/r^2$. Reading the Riemann curvature tensor from the forms and contracting gives the diagonal orthonormal Ricci tensorFor example $R_{00}=R^1{}_{010}+R^2{}_{020}+R^3{}_{030}=A+2B$, while $R_{22}=-B-B+C$. All off-diagonal entries vanish because the curvature 2-forms only involve their own index planes. In the coordinate basis the nonzero components areThe angular Einstein field equations with the negative cosmological constant give $(rV{)}^{\prime }=1+3r^2/l^2$. Integrating introduces one real constant, conventionally written $-2m$:Substitution gives $V^{\prime \prime }/2+V^{\prime }/r=3/l^2$, so the time and radial Einstein field equations hold as well. The solution is the Schwarzschild--anti-de Sitter black hole family, including its zero and negative mass limits.

For $m>0$, $V$ increases strictly from $-\infty$ at $r=0$ to $+\infty$ at infinity. Its unique positive simple zero $r_h$ satisfies $r_h^3+l^2{r}_h=2m{l}^2$. This zero is a coordinate singularity, not a curvature singularity: with the ingoing coordinate $v=t+\int dr/V$, the radial metric is $-Vd{v}^2+2dvdr$ and is regular there. Inside the future black-hole region, $\nabla r={\partial }_v+V{\partial }_r$ is future timelike, so every future-directed causal trajectory decreases $r$ and cannot return to the asymptotic anti-de Sitter boundary. Conversely, every exterior radius admits an outward radial null geodesic reaching that boundary, since ${\int }_r^{\infty }dr/V$ converges. Thus $m>0$ gives an event horizon at the unique positive zero of $V$. Here the event horizon is defined relative to the asymptotic timelike boundary of Anti-de Sitter spacetime. For $m=0$, $V=1+r^2/l^2>0$ and the spacetime is pure Anti-de Sitter spacetime, with no black-hole event horizon. For $m<0$, $V=1+2|m|/r+r^2/l^2>0$, so there is also no event horizon; the mass term produces a naked singularity at $r=0$. The field equation itself does not impose a positive mass integration constant.

For a metric $d{s}^2=-f(r)d{v}^2+2dvdr+r^{2}d{\Omega }_2^2$ with a simple Killing horizon, the surface gravity of $k={\partial }_v$ is $\kappa =f^{\prime }(r_{+})/2$. HeresoThe Hawking temperature of the Schwarzschild--anti-de Sitter black hole is therefore

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 $c$.

The four-dimensional Schwarzschild--anti-de Sitter black hole is a static, spherically symmetric vacuum spacetime with negative cosmological constant $\Lambda =-3/L^2$. In units $G=c=\hslash =k_B=1$, its metric tensor isHere $L$ is the Anti-de Sitter curvature length and $M>0$ is the mass parameter. Setting $f(r_{+})=0$ gives $M=r_{+}(1+r_{+}^2/L^2)/2$. The surface gravity $\kappa =f^{\prime }(r_{+})/2$ then gives the Hawking temperatureUnlike its asymptotically flat limit $L\to \infty$, its temperature eventually increases with horizon radius. Differentiation gives a minimum at $r_{+}=L/\sqrt{3}$, separating negative- and positive-heat capacity branches.^[ref]