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In the exterior orthonormal coframe $E^0=\sqrt{V}dt$, $E^1=dr/\sqrt{V}$, $E^2=rd\theta$, $E^3=r\sin \theta d\varphi$, use ${\Omega }^a{}_b=\frac{1}{2}{R}^a{}_{bcd}{E}^c\wedge E^d$. The independent mixed-index curvature 2-forms are ${\Omega }^0{}_1=-V^{\prime \prime }{E}^0\wedge E^1/2$, ${\Omega }^0{}_A=-V^{\prime }{E}^0\wedge E^A/(2r)$, ${\Omega }^1{}_A=-V^{\prime }{E}^1\wedge E^A/(2r)$ for $A=2,3$, and ${\Omega }^2{}_3=(1-V)E^2\wedge E^3/r^2$. The orthonormal Ricci tensor is diagonal with entries $V^{\prime \prime }/2+V^{\prime }/r$, $-V^{\prime \prime }/2-V^{\prime }/r$, and twice $(1-V-r{V}^{\prime })/r^2$. The Einstein field equations with negative cosmological constant then give $V=1-2M/r+r^2/l^2$.

// Target: general-relativity.bigb

For the orthonormal coframe of a planar warped spacetime, Cartan's first structure equation gives ${\omega }^i{}_3=a{e}^i$ and ${\omega }^3{}_i=-s_{i}a{e}^i$, where $i=0,1,2$ and $(s_0,s_1,s_2)=(-1,1,1)$. All other connection 1-forms vanish. Cartan's second structure equation gives ${\Theta }^i{}_3=-q{e}^i\wedge e^3$ and ${\Theta }^i{}_j=-s_j{a}^2{e}^i\wedge e^j$ for $i\ne j$. Lowering the first index makes both the connection 1-forms and the curvature 2-forms antisymmetric. Contraction yields $R_{00}=q+2a^2$, $R_{11}=R_{22}=-q-2a^2$, and $R_{33}=-3q$.

For $g=-d{t}^2+A(u{)}^{2}d{x}^2+B(u{)}^{2}d{y}^2+d{z}^2$, $u=t-z$, with nowhere-zero smooth $A,B$, take $e^0=dt$, $e^1=Adx$, $e^2=Bdy$, $e^3=dz$, $\ell =du$. In signature $(-,+,+,+)$, Cartan's first structure equation and metric compatibility give the lower-index connection 1-forms ${\omega }_{01}={\omega }_{13}=-(A^{\prime }/A)e^1$ and ${\omega }_{02}={\omega }_{23}=-(B^{\prime }/B)e^2$. Cartan's second structure equation gives ${\Theta }_{01}={\Theta }_{13}=-(A^{\prime \prime }/A)\ell \wedge e^1$ and ${\Theta }_{02}={\Theta }_{23}=-(B^{\prime \prime }/B)\ell \wedge e^2$. All other independent curvature 2-forms vanish: in particular the $12$ component cancels between the time and longitudinal terms. Contracting $R^a{}_{bad}$ yields $\mathrm{Ric}=-(A^{\prime \prime }/A+B^{\prime \prime }/B)du\otimes du$. Thus the Vacuum Einstein equations are $A^{\prime \prime }/A+B^{\prime \prime }/B=0$, while nonzero $A^{\prime \prime }/A=-B^{\prime \prime }/B$ gives opposite transverse tidal curvatures and plus polarization relative to these axes.

The curvature operator is diagonal in an orthonormal coframe $\{e^a\}$ when every curvature 2-form satisfies ${\mathcal{R}}^{ab}=K_{ab}{e}^a\wedge e^b$. Its matrix on the corresponding basis of 2-forms is then diagonal.

In an orthonormal coframe in spacetime, antisymmetry applies after lowering the first index with the Lorentzian frame metric. Thus mixed time–space connection 1-forms are equal under index interchange, while mixed spatial forms are opposite. The same rule holds for curvature 2-forms. Confusing lowered antisymmetry with ordinary mixed-index antisymmetry changes the curvature signs.

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.

Work on a regular patch with a smooth nonzero conformal factor. Its sign is constant there, so replace it by $|\Omega |$ if necessary and assume $\Omega >0$. Put $u=\log \Omega$, and let all partial derivatives, index raising on $u_i,{\Omega }_i$, and the operators $\Delta ,|\nabla \cdot {|}^2$ in the following formulas use the flat Cartesian metric $\delta$. The orthonormal coframe $E^i=\Omega d{x}^i$ hasThe proposed connection 1-forms reduce toLowering the first frame index with $\delta$ makes these antisymmetric, which establishes metric compatibility. Moreover,Thus Cartan's first structure equation has zero torsion. Uniqueness of the metric-compatible torsion-free connection identifies these forms as the Levi-Civita connection; this also explains why merely guessing a skew connection would not have sufficed.

To calculate the Riemann curvature tensor, use Cartan's second structure equation. Direct differentiation and multiplication giveIntroduce the Schouten tensor in Cartesian coordinate components,The preceding calculation groups into the curvature 2-formsKeep the convention $R_{ijkl}=g({\partial }_i,R({\partial }_k,{\partial }_l){\partial }_j)$, with $R(X,Y)=[{\nabla }_X,{\nabla }_Y]-{\nabla }_{[X,Y]}$, as in the first solution. Converting frame indices to coordinate indices gives the curvature of a conformally Euclidean metric:These are coordinate components. Components in the dual orthonormal frame are ${\Omega }^{-4}{R}_{ijkl}$, hence the same expression with the overall factor ${\Omega }^{-2}$. This distinction is important when contracting the Riemann curvature tensor.

Contract with $g^{ik}={\Omega }^{-2}{\delta }^{ik}$. In four dimensions the Ricci tensor becomes $R_{jl}=2S_{jl}+{\delta }_{jl}{\sum }_k{S}_{kk}$, and thereforeTaking the remaining metric trace yields the Ricci scalarThe cancellation of the gradient-squared terms in the last expression is special to dimension four.

For the Tolman wormhole, compute on $r>0$:Substitution into the Ricci tensor and Ricci scalar givesFor example, the coordinate radial eigenvalue is $-12\lambda /(r^2+\lambda {)}^2$, while each of the three tangential eigenvalues is $4\lambda /(r^2+\lambda {)}^2$. Their sum vanishes, but none vanishes when $\lambda \ne 0$. The Vacuum Einstein equations with zero cosmological constant require $R_{ij}-\frac{1}{2}R{g}_{ij}=0$ and hence $R_{ij}=0$. The metric is scalar-flat but not a vacuum solution for any nonzero $\lambda$; only $\lambda =0$ gives flat vacuum space. A nonzero cosmological constant does not rescue it: its trace would force $R=4\Lambda$, so $R=0$ would first force $\Lambda =0$.

The smooth wormhole interpretation uses $\lambda >0$; $r=0$ is another asymptotically flat end, not an included origin, as inversion $\rho =\lambda /r$ shows. For $\lambda <0$, the conformal factor vanishes at $r=\sqrt{-\lambda }$, where the metric is degenerate. The local non-vacuum calculation remains valid on each regular region away from that sphere, but such a metric is not a smooth wormhole across it.

Treat the orthonormal coframe in spacetime and the Lorentz connection 1-form as independent variables, with nondegenerate coframe and ${\omega }_{ab}=-{\omega }_{ba}$. Fix the curvature conventionThe second expression is Cartan's first structure equation; the first is Cartan's second structure equation. Use compactly supported variations, or boundary data that make the variation's surface term vanish.

For a connection variation, $\delta R^{ab}=D\delta {\omega }^{ab}$, where $D$ is the covariant exterior derivative. Since $D{ϵ}_{abc}=0$, its graded product rule givesFor a coframe variation, relabeling the dummy indices makes the three variations of the cubic term identical. HenceThe independent variations therefore give ${ϵ}_{abc}{T}^c=0$ and ${ϵ}_{abc}(R^{ab}+3\lambda E^a\wedge E^b)=0$. In three dimensions the alternating symbol identifies an antisymmetric index pair with a single index, so these equations are equivalent toThese are the equations of first-order three-dimensional gravity. The first selects the Levi-Civita connection; the second fixes constant sectional curvature $K=-3\lambda$. The factor of three follows from varying the coefficient exactly as printed.

On a static coordinate patch where $f>0$, choose the Lorentz frame metric ${\eta }_{ab}=\mathrm{diag}(-1,1,1)$ and coframePut $A=f^{\prime }/(2\sqrt{f})$ and $B=\sqrt{f}/r$. Exterior differentiation gives $d{E}^0=-A{E}^0\wedge E^1$, $d{E}^1=0$, and $d{E}^2=B{E}^1\wedge E^2$. Solving the equation of vanishing torsion forms givesHere antisymmetry is in the two raised or two lowered frame indices; mixed time-space components have equal, rather than opposite, signs. This is the Lorentzian connection-form antisymmetry needed to retain the curvature signs.

Using Cartan's second structure equation, the nonzero curvature 2-forms areFor example, $d(A{E}^0)=-(\sqrt{f}{A}^{\prime }+A^2)E^0\wedge E^1=-f^{\prime \prime }{E}^0\wedge E^1/2$, while the $02$ form is the product ${\omega }^0{}_1\wedge {\omega }^{12}=-AB{E}^0\wedge E^2$. Thus the field equations require $f^{\prime \prime }/2=3\lambda$ and $f^{\prime }/(2r)=3\lambda$. Integrating the latter gives $f=C+3\lambda r^2$; the former is then satisfied automatically. The origin condition fixes $C=1$, yieldingAs a sign and normalization check, the Ricci tensor is $R_{\mu \nu }=-6\lambda g_{\mu \nu }$ and the scalar curvature is $R=-18\lambda$. The coframe calculation applies initially in the region with $f>0$; if $\lambda <0$, $f$ vanishes at $r=1/\sqrt{-3\lambda }$, delimiting that static chart without changing the derived function.

The displayed $e^a$ are one-forms, so they form an orthonormal coframe, dual to the orthonormal frame $E_0=A^{-1}{\partial }_t$, $E_1=A^{-1}{\partial }_x$, $E_2=A^{-1}{\partial }_y$, $E_3={\partial }_z$. Use ${\eta }_{ab}=\mathrm{diag}(-1,1,1,1)$ and abbreviateFor $i=0,1,2$, $d{e}^i=a{e}^3\wedge e^i$, while $d{e}^3=0$. The Levi-Civita connection is fixed by Cartan's first structure equation together with Lorentzian connection-form antisymmetry, ${\omega }_{ab}=-{\omega }_{ba}$, where ${\omega }_{ab}={\eta }_{ac}{\omega }^c{}_b$. Its nonzero connection 1-forms areIn particular, the time-index pair has the same sign with one index raised; incorrectly imposing ${\omega }^a{}_b=-{\omega }^b{}_a$ would lose the Lorentzian sign. All diagonal and all other entries vanish.

Apply Cartan's second structure equation. For $i=0,1,2$,For distinct $i,j$ in $\{0,1,2\}$, putting $s_0=-1$ and $s_1=s_2=1$ gives ${\Theta }^i{}_j={\omega }^i{}_3\wedge {\omega }^3{}_j=-s_j{a}^2{e}^i\wedge e^j$. Thus a complete independent set of curvature 2-forms isAll diagonal entries vanish. The remaining entries follow from ${\Theta }_{ab}=-{\Theta }_{ba}$: ${\Theta }^j{}_i=-s_i{s}_j{\Theta }^i{}_j$ for $i,j\le 2$, and ${\Theta }^3{}_i=-s_i{\Theta }^i{}_3$. These relations and the displayed six entries specify all sixteen curvature 2-forms. This computation is Cartan curvature of a planar warped spacetime.

Extract the Riemann curvature tensor from ${\Theta }^a{}_b=\frac{1}{2}{R}^a{}_{bcd}{e}^c\wedge e^d$ using the curvature 2-forms just computed. Contraction $R_{bd}=R^a{}_{bad}$ gives the diagonal Ricci tensor in the orthonormal coframe:For example, the $00$ component receives $a^2$ from each of the $1$ and $2$ directions and $q$ from direction $3$. The signs in the spatial components reflect the same Lorentzian index lowering used for the connection 1-forms.

For a null vector $k$ in this frame, $(k^0{)}^2=(k^1{)}^2+(k^2{)}^2+(k^3{)}^2$. ThereforeThe Einstein field equations imply $8\pi T_{ab}{k}^a{k}^b=R_{ab}{k}^a{k}^b$, since the metric term vanishes for a null vector; an included cosmological-constant term would also vanish. Choose $k^a=(1,0,0,1)$ and apply the null energy condition. It follows thatConversely, this inequality makes the same contraction nonnegative for every null vector, so it is exactly the null energy condition for a planar warped spacetime when its matter stress-energy tensor is defined by the Einstein field equations. Null vectors tangent to the planar slices saturate the condition. As checks, constant $A$ gives flat spacetime and $A=e^{bz}$ gives constant negative sectional curvature with $a^{\prime }=0$, also saturating it.

Treat the printed $e^{\mu }$ as an orthonormal coframe, with frame metric $\eta =\mathrm{diag}(-1,1,1,1)$. Work on a patch $z\ne 0$ and $F=1-\alpha z^3>0$, so the given real coframe exists. Define $f=\sqrt{F}$ and $A=z{f}^{\prime }-f$. The exterior derivatives areThe connection 1-forms satisfying Cartan's first structure equation and metric compatibility areAll other forms vanish. With a Lorentzian frame, it is the lowered forms ${\omega }_{ab}={\eta }_{ac}{\omega }^c{}_b$ that are antisymmetric; the two mixed time–space forms are equal. These displayed forms solve the torsion-free structure equation, and uniqueness of the Levi-Civita connection identifies them as the required connection.

Apply Cartan's second structure equation. For example,The remaining derivatives give ${\Theta }^1{}_3=Af{e}^1\wedge e^3$ and ${\Theta }^1{}_2=-f^2{e}^1\wedge e^2$, with analogous forms for index 2. Put $q=\alpha z^3$ and $B=1+q/2$. From $f^2=1-q$,Therefore all six independent curvature 2-forms areThe other six are fixed by ${\Theta }^b{}_a=-{\eta }_{aa}{\eta }_{bb}{\Theta }^a{}_b$ for $a\ne b$, with no sum: equal for time–space pairs and opposite for spatial pairs. All diagonal forms are zero. The Lorentzian connection-form antisymmetry is essential to these signs.

Use the convention ${\Theta }^a{}_b=\frac{1}{2}{R}^a{}_{bcd}{e}^c\wedge e^d$ and $R_{ab}=R^c{}_{acb}$. Each independent two-form has only one coordinate-plane wedge, so the Ricci tensor is diagonal. Contracting the six coefficients givesBut $2B+F=2(1+q/2)+(1-q)=3$, soIn coordinate-independent form, $R_{\mu \nu }=-3g_{\mu \nu }$. Thus the vacuum Einstein field equations $G_{\mu \nu }+\Lambda g_{\mu \nu }=0$ hold withThe curvature radius is one in the metric's normalization. The parameter $\alpha$ changes the individual curvature 2-forms but cancels from their Ricci contraction. For $\alpha =0$ this is Anti-de Sitter spacetime; for nonzero $\alpha$ it is the planar Einstein metric with cubic radial function, with a nontrivial Weyl tensor rather than universally constant sectional curvature. The result is local on regular coordinate patches; a zero of $F$ invalidates this particular static coframe, not the tensor equation in a suitable regular extension.

Keep $\ell =e^0-e^3$ and set $P=p^{\prime }+p^2=A^{\prime \prime }/A$, $Q=q^{\prime }+q^2=B^{\prime \prime }/B$. Differentiation gives $d(p{e}^1)=P\ell \wedge e^1$ and $d(q{e}^2)=Q\ell \wedge e^2$. In Cartan's second structure equation, the quadratic terms for these components vanish. The apparently possible transverse component cancels:Thus the potentially nonzero mixed-index curvature 2-forms areAll other components vanish. Equivalently, the independent lower-index forms are ${\Theta }_{01}={\Theta }_{13}=-P\ell \wedge e^1$ and ${\Theta }_{02}={\Theta }_{23}=-Q\ell \wedge e^2$, with the rest supplied by lower-index antisymmetry. These signs use ${\Theta }^a{}_b=\frac{1}{2}{R}^a{}_{bcd}{e}^c\wedge e^d$ and the curvature convention in question 2(c). This is the curvature of a diagonal plane wave in Rosen coordinates.

Let $E_a$ denote the frame dual to the coframe. For the convention just stated, the Ricci tensor is obtained from the curvature 2-forms through the Ricci one-forms ${\mathcal{R}}_b={\sum }_a{\iota }_{E_a}{\Theta }^a{}_b=R_{bd}{e}^d$. In particular ${\iota }_{E_1}(\ell \wedge e^1)=-\ell$ and likewise for index two. It follows thatThusThe scalar curvature vanishes even before imposing vacuum because $du$ is null, but the full tensor vanishes exactly whenThis is the Vacuum Einstein equations in this family. The nonzero-coframe hypothesis makes division by $A$ and $B$ legitimate. A nonzero $P=-Q$ can leave the Riemann curvature tensor nonzero even in vacuum.