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An analytic extension embeds a spacetime isometrically as a proper open region in a larger analytic Lorentzian manifold with a nondegenerate metric tensor. A divergence of coordinate components does not obstruct extension if a new chart makes all metric coefficients analytic and the determinant nonzero. For a static spherical metric with an analytic lapse $f$ and a positive simple horizon radius, ingoing coordinates replace $-fd{t}^2+d{r}^2/f$ by $-fd{v}^2+2dvdr$. This removes the coordinate singularity at $f=0$; transverse metric coefficients must also remain regular. An extension across such a horizon is distinct from extending through a curvature singularity.

A coordinate singularity is a failure of a coordinate description which disappears in a regular chart. The horizon singularity of the Schwarzschild metric in $(t,r)$ coordinates is removable; its $r=0$ curvature singularity is not.

Declination $\delta$ is the signed angle from the reference celestial equator to a direction, measured along its hour circle: $-90^{\circ }\le \delta \le +90^{\circ }$. Positive declination means celestial north, rather than necessarily north of the observer's zenith. A direction with $\delta =0$ lies on the celestial equator; $\delta =+90^{\circ }$ is the north reference celestial pole. Declination is an angle, not a height or a measure of Euclidean distance. ^[ref].

The connection to an observation follows from a dot product. In the elementary spherical-Earth, unrefracted model, an observer at geographical latitude $\varphi$ has zenith direction $\mathbf{z}=(\cos \varphi \cos \Theta ,\cos \varphi \sin \Theta ,\sin \varphi )$ in the same reference axes. Since $\mathbf{n}\cdot \mathbf{z}=\sin a$ for altitude $a$, the cosine addition formula givesAt upper meridian transit, $H=0$, so $a_{\mathrm{upper}}=90^{\circ }-|\varphi -\delta |$. For the synthetic site $\varphi =52^{\circ }$ and synthetic direction $\delta =20^{\circ }$, this yields $a_{\mathrm{upper}}=58^{\circ }$. A direction with $\delta =70^{\circ }$ instead reaches $72^{\circ }$, north of the zenith; positive declination does not imply a crossing on the southern half of the celestial meridian. A negative computed upper altitude means the direction never rises above the geometric astronomical horizon at that site.

For an angular readout, define signed zenith distance $z_s$ on the upper celestial meridian, positive toward the south and negative toward the north. Then $\delta =\varphi -z_s$. A synthetic southward readout $z_s=32^{\circ }$ at $\varphi =52^{\circ }$ recovers $\delta =20^{\circ }$. An unsigned distance $|z_s|$ alone would leave two possibilities, $\varphi \pm |z_s|$, when both are in range. A calibrated optical telescope must therefore record which side of the zenith it points toward. In this linear relation, a $1^{\prime }$ error in the assumed geographical latitude produces a $1^{\prime }$ error in inferred declination; averaging repeated readings cannot remove that shared offset. Near the astronomical horizon, uncorrected atmospheric refraction also invalidates the geometric readout.

The unit vector construction explains a subtle feature of sky maps. At fixed declination, changing right ascension traces a parallel whose radius on the unit sphere is $\cos \delta$. Its arc length is $\cos \delta |d\alpha |$, with angles in radians; at $\delta =60^{\circ }$, a $15^{\circ }$ right ascension increment spans a $7.5^{\circ }$ arc along that parallel. This arc is not generally the shortest great circle separation. At either celestial pole, all right ascension values represent the same direction: the coordinate singularity belongs to the coordinate grid. The unit vector remains regular there.

In spherical coordinates on a sphere of radius $R$, the position is $X=R(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$. Its coordinate tangent vectors satisfy $X_{\theta }\cdot X_{\varphi }=0$, $|X_{\theta }{|}^2=R^2$, and $|X_{\varphi }{|}^2=R^2{\sin }^2\theta$. Expanding $|dX{|}^2$ proves the displayed line element, the first fundamental form of the sphere. Along a parametrized path it gives the arc length by integrating $R\sqrt{{\dot{\theta }}^2+{\sin }^2\theta {\dot{\varphi }}^2}$. The vanishing azimuthal coefficient at a pole is a coordinate singularity, not a singularity of the sphere.

For the three cases, distinguish a curvature singularity from a coordinate singularity. The Reissner-Nordstrom Kretschmann scalar isIt diverges at $r=0$ when either $M$ or $Q$ is nonzero. Any positive root of $F$ instead admits Eddington-Finkelstein coordinates: set $v=t+\int dr/F$ to obtainThe determinant of its $(v,r)$ block is $-1$, including at $F=0$. Hence positive roots of $F$ are removable coordinate singularities of the static chart. The apparent degeneracy at the poles of the angular chart is also a coordinate singularity.

For $0<Q^2<M^2$ there are two distinct positive roots:Both roots are coordinate singularities of the static chart. In the standard extension, $r_{+}$ is the outer event horizon and $r_{-}$ is an inner Cauchy horizon. The curvature singularity is $r=0$. Between the horizons $F<0$, so the radial direction is timelike; for $0<r<r_{-}$, $F>0$ again and the central curvature singularity is timelike. These are the three radial regions of the global structure of a charged spherical black hole.^[ref]

For completeness, if $Q=0$ and $M>0$, the Schwarzschild spacetime limit has just the positive horizon $r=2M$ and a spacelike curvature singularity at $r=0$; the formal root $r_{-}=0$ of $r^{2}F$ is not a regular inner horizon. If $M=Q=0$, the metric tensor is flat and the radial origin is regular in Cartesian coordinates.

If $M^2=Q^2$ with $M>0$, thenThe curvature singularity remains at $r=0$. At $r=M$ the static chart has a coordinate singularity, removed by the Eddington-Finkelstein coordinates above. The Kretschmann scalar there equals $8/M^4$, which is finite. The coincident horizons form a degenerate Killing horizon, and its future outer branch is the event horizon of the extremal black hole. A double zero of $F$ does not remove the horizon.

A nondegenerate Lorentzian metric requires $M\ne 0$; take $M>0$ as the radius convention. The original coordinates describe $\lambda >0$, with $\lambda =0$ a candidate boundary rather than an ordinary point of that chart. The change of coordinate givesSubstituting and cancelling the two terms proportional to $d{\lambda }^2/{\lambda }^2$ producesThe $(v,\lambda )$ block of this metric tensor has determinant $-M^2$ and inverse metricAll coefficients are smooth through $\lambda =0$, and the block has one negative and one positive eigenvalue. Together with the positive round two-sphere factor this is a regular Lorentzian metric. It extends to a neighborhood with either sign of $\lambda$. The usual polar-coordinate problem at the poles of the two-sphere is removed by separate angular charts. Therefore $\lambda =0$ is only a coordinate singularity of the original chart. This is the degenerate-horizon patch of Bertotti-Robinson spacetime.

Put $d{\Omega }_2^2=d{\theta }^2+{\sin }^2\theta d{\phi }^2$, the standard metric tensor on the unit sphere. From $t=v+m/\lambda$ one has $dt=dv-m{\lambda }^{-2}d\lambda$. The apparently divergent terms cancel:The $(v,\lambda )$ metric tensor block has determinant $-m^2\ne 0$, independently of $\lambda$. Its coefficients are real analytic, so this formula defines a regular Lorentzian metric onThe polar-coordinate degeneracies at the poles are handled by ordinary two-sphere charts. Thus $\lambda =0$ is a coordinate singularity of the original chart and is crossed by this spacetime extension.

The physical spherical coordinates cover $t\in \mathbb{R}$, $r\in [0,\infty )$, $\theta \in [0,\pi ]$, and $\varphi \in [0,2\pi )$ with periodic identification. The center and the polar axes are the usual coordinate singularities, rather than physical curvature singularities; a regular polar chart restricts $r>0$ and $0<\theta <\pi$. For the null coordinates,Thus the physical metric tensor isThe conformal transformation followed by $p=\arctan u$, $q=\arctan v$ gives $du=(1+u^2)dp$, $dv=(1+v^2)dq$, andConsequently $d{s}^2=4dpdq-{\sin }^2(q-p)d{\Sigma }^2$. Finally $T=p+q$, $R=q-p$ converts $4dpdq$ to $d{T}^2-d{R}^2$, yieldingThe ranges must retain their joint constraints:Equivalently, $-\pi <T-R<\pi$ and $-\pi <T+R<\pi$, with $R\ge 0$. The angles retain their original ranges and identifications. Extending $R$ to negative values would double-count the radial polar geometry. The conformal factor isThe unphysical metric tensor extends onto the Einstein static universe, while physical Minkowski spacetime occupies the triangular region in its radial Penrose diagram. The edge $R=0$ is the regular center and belongs to the physical region away from its endpoints.

The conformal boundary consists of the following infinities. Future null infinity has $q=\pi /2$, hence $T+R=\pi$ with $0<R<\pi$, and corresponds to finite $u$ with $v\to +\infty$. Past null infinity has $p=-\pi /2$, hence $T-R=-\pi$, and corresponds to finite $v$ with $u\to -\infty$. Future timelike infinity is $(T,R)=(\pi ,0)$, past timelike infinity is $(-\pi ,0)$, and spacelike infinity is $(0,\pi )$. Although angles label points on null infinity, their two-spheres collapse at these three endpoint locations in the compactified picture.

For the geodesic endpoints in Minkowski conformal compactification, one can check the limits directly. A timelike inertial trajectory is $\mathbf{x}=\mathbf{b}+\mathbf{w}t$ with $|\mathbf{w}|<1$. At $t\to +\infty$ both $t-r$ and $t+r$ tend to $+\infty$, so $p,q\to \pi /2$ and the trajectory ends at future timelike infinity. At $t\to -\infty$ both tend to $-\infty$, giving past timelike infinity. This includes a stationary inertial worldline with bounded $r$.

For a null inertial trajectory, $|\mathbf{w}|=1$. As $t\to +\infty$, $r=t+\mathbf{w}\cdot \mathbf{b}+o(1)$, so $u$ has a finite limit while $v\to +\infty$; its future endpoint lies on future null infinity. As $t\to -\infty$, $r=-t-\mathbf{w}\cdot \mathbf{b}+o(1)$, giving a finite limiting $v$ and $u\to -\infty$, hence a past endpoint on past null infinity. Radial null geodesic paths have $T\pm R$ constant. A ray passing through the center changes its angular direction; in the angle-suppressed radial diagram it reflects at $R=0$ rather than terminating there.

A spacelike geodesic is also a straight line, now parametrized so that $|d\mathbf{x}/d\lambda |>|dt/d\lambda |$. At either end, $r>|t|$ asymptotically, $u\to -\infty$, $v\to +\infty$, and the endpoint is spacelike infinity. A radial spacelike line at $t=0$ has two opposite-angle halves which project onto the same horizontal interval in the diagram.

The conformal compactification endpoints occur at finite $T$ but at infinite physical proper time or affine parameter; Minkowski spacetime is geodesically complete. The conformal preservation of null geodesic paths does not preserve an affine parameter: if $\lambda$ is physically affine, an unphysical affine parameter satisfies $d\tilde{\lambda }/d\lambda ={\Omega }^2$ up to a constant. Along a complete null ray this can have a finite integral at infinity. Timelike and spacelike physical geodesic paths are not generally geodesics of the rescaled metric tensor, so their drawn curves must not be interpreted as freely falling unphysical observers.

Finally, every physical event can send an outgoing null ray to future null infinity, so $J^{-}({\mathcal{I}}^{+})$ is all of Minkowski spacetime. Its black-hole region and intrinsic event horizon are empty. The null edges in the diagram are infinity, not horizons concealing part of the physical manifold. Complete inertial observers likewise have no observer event horizon. Accelerated observers restricted to a Rindler wedge can have a Rindler horizon, but that is observer-dependent and does not contradict the horizon-free Minkowski spacetime. The conformal diagram compactifies infinity; it does not introduce a physical horizon.

Use geometrized units $G=c=1$ and positive mass $M$. The Schwarzschild metric initially describes an exterior region, not the whole black hole:Here $r$ is the areal radius and $t$ is Schwarzschild time, normalized at infinity. We must distinguish a removable coordinate singularity from a genuine curvature singularity. The Kretschmann scalar isIt is finite at $r=2M$ and diverges at $r=0$. Thus $r=2M$ can be crossed in a better chart, whereas $r=0$ cannot be made a regular point by changing the coordinate chart.

First integrate the Schwarzschild tortoise coordinate:The radial metric becomes $f(-d{t}^2+d{r}_{\ast }^2)$. Set $v=t+r_{\ast }$; the ingoing Eddington-Finkelstein coordinates giveThe radial block has determinant minus one and smooth coefficients at $r=2M$, so it extends through the future Schwarzschild event horizon. Its ingoing radial null geodesics have constant $v$, while the outgoing radial null geodesics obey $dr/dv=f/2$. In the future interior both future-directed radial null families decrease $r$; $r=0$ is therefore an unavoidable future boundary for causal motion there. The outgoing chart $u=t-r_{\ast }$ similarly gives $d{s}^2=-fd{u}^2-2dudr+r^{2}d{\Omega }^2$, extending through the past horizon. A single one of these null charts does not display both horizons or the entire extension.

To regularize both horizons together, introduce Kruskal–Szekeres coordinates in the original exterior:ThenFor $r>0$ the right side defining $UV$ decreases monotonically from one to minus infinity. In fact its derivative is $-r{e}^{r/(2M)}/(4M^2)$. Thus it determines a unique smooth $r=r(UV)$ wherever $UV<1$, including $UV=0$. At $r=2M$ the radial metric coefficient is finite and nonzero, proving actual regularity of both horizons and their intersection, rather than merely relabelling the old coordinate divergence. Allowing all real $U,V$ with $UV<1$ produces the simply connected maximal analytic extension, the Kruskal spacetime.

There are four regions of the Kruskal spacetime. The original exterior is $U<0,V>0$; a second asymptotically flat exterior is $U>0,V<0$. The future black hole has $U>0,V>0$ and $0<r<2M$; the past white hole has $U<0,V<0$ and the same range of areal radius. The horizon branches are $U=0$ and $V=0$. Their intersection $U=V=0$ is a regular bifurcation surface, a two-sphere of radius $2M$, not the centre of the spacetime. Each point of the radial diagram represents a symmetry two-sphere.

Put $T_K=(V+U)/2$ and $X_K=(V-U)/2$. Radial null geodesics have slopes $d{T}_K=\pm d{X}_K$, and the curvature singularities haveThe upper branch is the future spacelike Schwarzschild singularity; the lower branch is the white hole past singularity. The Killing vector field extends as ${\partial }_t=(V{\partial }_V-U{\partial }_U)/(4M)$. It is future directed in the original exterior and past directed in the other exterior for the chosen global time orientation; inside, this stationary vector is spacelike. In the future interior, $r$ is a time coordinate decreasing toward the singularity. For example, a radial infaller of unit specific Killing energy obeys $dr/d\tau =-\sqrt{2M/r}$ and reaches $r=0$ from the horizon in finite proper time $4M/3$. The curvature divergence and this geodesic endpoint show why the extension stops there.

For the global causal picture, compactify the null coordinates using $p=\arctan U$, $q=\arctan V$ and let $\mathcal{T}=2(p+q)/\pi$, $\mathcal{X}=2(q-p)/\pi$. The conformal compactification rescales the metric to remove the coordinate factors from the radial metric while preserving its null directions. Since $UV=1$ gives $p+q=\pm \pi /2$ on the two singular branches, they become the horizontal lines $\mathcal{T}=\pm 1$. The resulting Penrose diagram displays the two exteriors, each with its own past null infinity, future null infinity and spatial infinity, plus the future and past interiors. The singularities are spacelike boundaries; the horizons are interior null surfaces. Future event horizons separate events that can escape to an exterior's future null infinity from events in the black-hole interior.

The time-symmetric spatial slice has a bridge between the exteriors, but it is not a traversable passage. Future-directed causal curves have nondecreasing $U$ and $V$. Moving from the right exterior to the left would require $V$ to decrease from positive to negative; the reverse passage would require $U$ to decrease. Both are impossible. This proves the non-traversability of the Schwarzschild bridge. The complete eternal vacuum extension has two exteriors, a black hole and a white hole; it has no causal route between its exteriors. A black hole made by stellar collapse is a different global spacetime: its matter-filled past replaces the white hole region, and it normally has only one exterior. The vacuum extension constructed here should not be mistaken for the collapse history of a single star.

With electric charge, spherical electrovacuum instead gives the Reissner-Nordstrom metric, in an electromagnetic normalization absorbing the charge conversion constants:For $0<|Q|<M$, there are two simple horizons. Outside $r_{+}$ the metric is static; between $r_{-}$ and $r_{+}$, $f<0$ and $r$ is timelike. Inside $r_{-}$, $f>0$ again, so $r$ is spacelike and the $r=0$ singularity is timelike. The outer horizon is the event horizon; the inner horizon is a Cauchy horizon. The same logarithmic tortoise-coordinate and exponential-null-coordinate method crosses each simple horizon locally, but one exponential scale does not regularize both simultaneously: their surface gravities have different magnitudes,Continuing the exact metric through successive inner horizons produces repeated blocks with further exteriors, and evolution beyond a Cauchy horizon is not uniquely fixed by data on the original Cauchy hypersurface. The timelike singularity can be avoided by some trajectories in this ideal extension, unlike the unavoidable spacelike singularity of the future Schwarzschild spacetime interior. However, the enormous inner-horizon blueshift makes this exact continuation unstable: perturbations can cause mass inflation. This is an important qualification on the global structure of a charged spherical black hole.

At $|Q|=M$ the two horizons coalesce, $f=(1-M/r{)}^2$ and surface gravity is zero. The horizon is a degenerate Killing horizon, the tortoise-coordinate divergence becomes a pole and the ordinary nondegenerate Kruskal spacetime construction must be replaced. The exterior develops an infinitely long spatial throat and the formal Hawking temperature vanishes. For $|Q|>M$, no real zero of $f$ remains: the electrovacuum solution has a naked singularity, not a black hole. Consequently arbitrarily allowing charge changes the causal structure, not just the horizon radius.

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.

Since $dt=dv-dr/f$, the inverse-$F$ radial terms cancel. The ingoing Eddington-Finkelstein coordinates giveAt $r=2m$, $H=1+{\sinh }^2\alpha ={\cosh }^2\alpha >0$. Thus all coefficients are analytic there, and the determinant of the $(v,r)$ metric block is $-1$, not zero. The angular area is also nonzero. This is an analytic extension of a spacetime through the apparent coordinate singularity, into the regular region $0<r<2m$.

Assume $M>0$ and use geometric units. The stationary Killing vector field $K={\partial }_t$ gives the conserved Killing energy $E=-g(K,U)=f\dot{t}$. For radial motion, normalization of the four-velocity gives $-f{\dot{t}}^2+f^{-1}{\dot{r}}^2=-1$. Combining these identities yieldsAt rest at infinity, $f\to 1$ and $\dot{r}\to 0$, so $E=1$. Choose the inward branch $\dot{r}=-\sqrt{2M/r}$. Its proper time from the Schwarzschild event horizon to the curvature singularity isIn ordinary units this is $4GM/(3c^3)$. The endpoint $r=0$ is not a regular centre of the spacetime. This is the Schwarzschild horizon-to-singularity proper time for fall from infinity, rather than a universal time for all infall energies.

For this geodesic congruence, set $w(r)=\sqrt{2M/r}$. Its four-velocity and metric dual areThe right side is an exact one-form because $w/f$ depends only on $r$. Thus $U_a=-{\partial }_a\tau$, with free-fall proper time in Schwarzschild spacetime determined byIndeed $U^a{\partial }_a\tau =-U^a{U}_a=1$, so this scalar advances by the actual proper time on every member of the infalling geodesic congruence, with a synchronized choice of additive offsets.

Substitution of $dt=d\tau -w{f}^{-1}dr$ in the Schwarzschild metric givesUsing $f=1-w^2$, the Painlevé–Gullstrand coordinates therefore haveAll coefficients are finite at $r=2M$; the determinant of the $\tau ,r$ block is $-f-w^2=-1$. Hence the horizon is a coordinate singularity of Schwarzschild time, not a degeneracy of the metric. The new spatial slices have $d{\ell }^2=d{r}^2+r^{2}d{\Sigma }^2$, the flat Euclidean metric, and $g^{ab}{\partial }_a\tau {\partial }_b\tau =-1$ makes their normals the freely falling four-velocities. By contrast, the curvature scalar $R_{abcd}{R}^{abcd}=48M^2/r^6$ diverges at $r=0$; this genuine singularity is not removed.

For ingoing Eddington-Finkelstein coordinates, use $v=t+r_{\ast }$ with $d{r}_{\ast }/dr=f^{-1}$, for example $r_{\ast }=r+2M\log |r/(2M)-1|$. ThenBoth constructions cross the future horizon. Their time functions have different meanings: $v=\text{constant}$ is an ingoing null hypersurface, whereas $\tau =\text{constant}$ is a flat spacelike hypersurface orthogonal to the infallers. Thus the former is adapted to ingoing null geodesics, and the latter to a timelike free-fall clock congruence.

For the Kruskal diagram, choose Kruskal–Szekeres coordinates with $U_K=-e^{-(t-r_{\ast })/(4M)}$, $V_K=e^{(t+r_{\ast })/(4M)}$ in the right exterior and extend them across its horizon. They obeyThe four regions of Kruskal spacetime are I, the right exterior $(U_K<0,V_K>0)$; II, the future black hole $(U_K>0,V_K>0)$; III, the other exterior $(U_K>0,V_K<0)$; and IV, the past white hole $(U_K<0,V_K<0)$. The null axes are horizons, their intersection is the bifurcation two-sphere, and the spacelike boundaries $U_K{V}_K=1$ are the future and past curvature singularities. A point in this radial diagram represents a symmetry two-sphere.

The domain of the ingoing Painleve-Gullstrand chart is region I together with region II and their common future horizon. More precisely, it is $V_K>0$, $U_K{V}_K<1$, or $r>0$ with finite $\tau$. To verify the boundary rather than infer it from the locally regular metric, put $\rho =\sqrt{r/(2M)}$ and writeThis follows by subtracting $r_{\ast }$ from the preceding radial primitive, and $S$ is finite at $r=2M$. Thus $V_K=\exp [(\tau -S(r))/(4M)]>0$ throughout the chart. Conversely every point with $V_K>0$ and $r>0$ gives a finite $\tau$. The past horizon $V_K=0$, the bifurcation sphere, region III and region IV are absent. Reversing the sign of the radial term produces the outgoing chart through the white hole instead.

Put $r_{+}=\ell \sqrt{M}$ and $f=(r^2-r_{+}^2)/{\ell }^2$. The apparent failure of the nonrotating BTZ black hole coordinates at $r=r_{+}$ is a coordinate singularity. Define the tortoise coordinate and ingoing time byThe Ingoing BTZ coordinates giveAll coefficients are analytic at $r_{+}>0$, and the determinant is $-r^2$, so the metric tensor extends nondegenerately through that radius. The Killing vector field $K={\partial }_v$ agrees with ${\partial }_t$ in the exterior. On $r=r_{+}$, $K^2=-f=0$ and $K_a=(dr{)}_a$. It is therefore both normal and tangent to this null hypersurface, proving that it is a Killing horizon.

One may include both horizon branches and their bifurcation surface explicitly. Set $\kappa =r_{+}/{\ell }^2$, $u=t-r_{\ast }$ and, initially in the exterior, $U=-e^{-\kappa u}$, $V=e^{\kappa v}$. ThenThis analytic extension of a spacetime is regular near $UV=0$, and $K=\kappa (V{\partial }_V-U{\partial }_U)$ extends analytically. The surfaces $U=0$ and $V=0$ form the extended Killing horizon at $r=r_{+}$; $K$ vanishes only at their regular bifurcation circle.

Write $f=1-r^2/a^2$ and $r_{\ast }=a\mathrm{artanh}(r/a)$. Since $d{r}_{\ast }/dr=f^{-1}$, $dt=du+dr/f$. Substitution gives the outgoing Outgoing Eddington-Finkelstein coordinatesThe radial block has determinant $-1$, independent of $f$, and all coefficients are smooth at $r=a$. This is a coordinate singularity of the static chart, not a spacetime singularity. The transformation was defined inside the horizon, but the new metric itself extends smoothly across it.

At $r=0$ the spherical angular coordinates degenerate as usual. Cartesian spatial coordinates make the original static metric regular there: $f=1+O(r^2)$ and the extra spatial radial correction is $x_i{x}_j/[a^2(1-r^2/a^2)]$. Thus the origin is not a spacetime singularity either.

Use geometrized units $G=c=1$ and metric signature $(-,+,+,+)$. Put $f=1-2M/r$ in the Schwarzschild metric. The Schwarzschild tortoise coordinate satisfies $d{r}_{\ast }/dr=f^{-1}$, soSubstituting $dt=dv-f^{-1}dr$ gives the ingoing Eddington-Finkelstein coordinates:The radial metric tensor has determinant $-1$ and inverse components $g^{vv}=0$, $g^{vr}=1$, $g^{rr}=f$. Thus it is nondegenerate and analytic at $r=2M$. The same expression defines a Lorentzian metric for every $r>0$, extending the exterior across the future Schwarzschild event horizon into the black hole. It does not include the other exterior or the white hole of the full Kruskal spacetime. At $r=0$, the Kretschmann scalar $R_{abcd}{R}^{abcd}=48M^2/r^6$ diverges, so this is a curvature singularity, not a removable coordinate singularity.

Write the position vector as $\mathbf{r}(\rho ,\varphi ,z)=(\rho \cos \varphi ,\rho \sin \varphi ,z)$. Its coordinate tangent vectors areTheir pairwise dot products vanish. Their lengths, the scale factors of orthogonal coordinates, areThus the cylindrical coordinate system is orthogonal wherever it is a nonsingular coordinate chart. At $\rho =0$, the angular tangent vector vanishes and the azimuth is undefined; this is the usual coordinate singularity on the axis, not a failure of the Euclidean geometry.

The radial unit vector in the spherical coordinate system isThe coordinate tangent in the polar direction has length $r$, so its corresponding unit vector isDifferentiating ${\mathbf{e}}_r$ at fixed $\varphi$ therefore givesThe factor is not $r$: the corresponding scale factors of orthogonal coordinates occur in the derivative of the position vector, whereas the question differentiates its unit direction. The formula is understood away from the coordinate singularities where a complete spherical basis is not uniquely specified.

Choose the vertical $z$ axis downward, and measure the polar angle $\theta$ from it. With azimuth $\varphi$ and fixed rod length $r$, the Cartesian coordinates areThe massless rod has no kinetic energy. Differentiating these coordinates and adding the squared components of velocity yieldsThe gravitational potential energy is $V=-mgz=-mgr\cos \theta$, up to an arbitrary constant. The Lagrangian of this spherical pendulum is thereforeApplying the Euler-Lagrange equation to the two generalized coordinates givesEquivalently, away from the coordinate singularities at the poles,Since $L$ is independent of $\varphi$, the azimuth is a cyclic coordinate. Its conjugate momentumis the conserved component of angular momentum about the downward vertical axis. The polar angle is not a cyclic coordinate. Using an upward $z$ axis instead reverses the sign of the cosine term in $L$ and of the corresponding gravity term; it describes the same motion with the opposite polar angle convention.