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For smooth radial infall through a future Schwarzschild event horizon, outgoing retarded time obeys $u\sim -4M\log ({\tau }_H-\tau )$. A distant observer sees the redshift grow as $e^{t_{\mathrm{rec}}/(4M)}$. Its exponent in emission Schwarzschild time is instead $1/(2M)$, because $t_{\mathrm{rec}}\sim 2t_{\mathrm{em}}$. At finite fixed observer radius the wavelength ratio includes $\sqrt{1-2M/r_B}$.

With $f=1-2\mu /r$, the time coordinate $t^{\prime }=t+(2\mu /c)\log |r/(2\mu )-1|$ gives $d{s}^2=f{c}^{2}d{t}^{\prime 2}-(4\mu c/r)d{t}^{\prime }dr-(1+2\mu /r)d{r}^2-r^{2}d{\Omega }^2$. Its radial metric tensor block has determinant $-c^2$, so it extends regularly across the future Schwarzschild event horizon. The advanced null time of ingoing Eddington-Finkelstein coordinates is $v=t^{\prime }+r/c$, not $t^{\prime }$ itself. Radial null directions satisfy $dr/(cd{t}^{\prime })=-1$ or $(1-2\mu /r)/(1+2\mu /r)$.

For an equatorial null geodesic in Schwarzschild spacetime, let $k=(1-2\mu /r)dt/d\lambda$ and $h=r^{2}d\varphi /d\lambda$ be the conserved quantities for an affine parameter $\lambda$. The null metric condition gives ${\dot{r}}^2/2+V_{\mathrm{eff}}=c^2{k}^2/2$. For $h\ne 0$, the effective potential is zero at the Schwarzschild event horizon, positive outside, and tends to zero at infinity. Its unique exterior maximum is at $r=3\mu$, with value $h^2/(54{\mu }^2)$ and second derivative $-h^2/(81{\mu }^4)$. This maximum gives the unstable photon sphere.

The impact parameter of an incoming ray is $b=|h|/(ck)$ for $k>0$. Capture requires its radial energy to exceed the maximum, namely $b<3\sqrt{3}\mu$. At equality the incoming ray approaches the circular orbit asymptotically; it does not cross the horizon. For $h=0$, the potential is identically zero and the ingoing radial ray is captured.

For radial motion outside the Schwarzschild event horizon, the equations needed to determine the acceleration components areThey are respectively the four-velocity norm, velocity-acceleration orthogonality, and the maximum proper acceleration norm. Together with part d they give the stated thrust equation; no integration of the trajectory is required. The positive radial sign corresponds to outward thrust.

Let $r_I=\sqrt{GM/g}$. A first transition from inward to outward motion requires $u^r=0$ at a minimum radius and $d{u}^r/d\tau \ge 0$ there. The largest possible radial derivative at such an event isAt $r=r_I>r_s$, this isThus the ship cannot reverse its inward motion there. It cannot reverse at any smaller exterior radius either: $GM/r^2$ grows while $\sqrt{1-r_s/r}$ decreases. If it first reaches $r_I$ while falling inward, it cannot return home using the stipulated bounded radial thrust. If $r_I\le r_s$, it would instead have reached the black-hole horizon before that radius, which already prevents escape.

The minimum radius for radial turnaround with bounded proper acceleration lies strictly outside the proposed radius. Define $r_c>r_s$ byThe right side is the proper acceleration required to hover, decreases monotonically from infinity to zero, and exceeds $GM/r^2$. A genuine turnaround needs $r_{\mathrm{turn}}>r_c>r_I$; equality gives no positive initial outward acceleration at rest. A sufficiently early burn is also needed to remove the pre-existing inward velocity. The failure at $r_I$ follows from the turning-point condition, not from treating the moving ship as static throughout its fall.

Take $M>0$ and write $f=1-2M/r$. Integrating $d{r}_{\ast }/dr=f^{-1}$ gives the Schwarzschild tortoise coordinateThe advanced and retarded Eddington-Finkelstein coordinates are respectively $v=t+r_{\ast }$ and $u=t-r_{\ast }$. Substituting $dt=dv-dr/f$ or $dt=du+dr/f$ givesBoth radial-time blocks have determinant $-1$ and smooth coefficients at $r=2M$. Consequently the Schwarzschild event horizon is a coordinate singularity of the original chart, not a degeneracy of the geometry. The Kretschmann scalar $48M^2/r^6$ is finite there, while it diverges at $r=0$. Angular coordinate degeneracies at the polar axes are removed separately by ordinary sphere charts.

The ingoing Eddington-Finkelstein coordinates extend the right exterior through its future black hole horizon. Radial null geodesics obey either $dv=0$ or $dr/dv=f/2$. Inside $r=2M$, the latter also moves towards smaller $r$ as $v$ increases, and the future-directed ingoing family has decreasing $r$. Thus neither family can escape to the exterior. The future horizon is the boundary separating points that can send signals to future null infinity from those that cannot. The retarded chart extends through the past horizon instead. Reversing the time orientation gives the white hole: future-directed signals can leave it, but exterior signals cannot enter it. These are distinct branches in the eternal extension; the white-hole branch need not occur in a spacetime formed by collapse.

For a simultaneous extension through both branches, start in the right exterior and define the Kruskal–Szekeres coordinatesTheir product and the extended metric areIndeed $du=-4MdU/U$ and $dv=4MdV/V$ transform $-fdudv$ into the displayed expression. The function of $r$ on the right of the product relation has derivative $-r{e}^{r/(2M)}/(4M^2)$, nonzero at $r=2M$, so $r$ is a smooth function of $UV$ across zero. The coefficient of $dUdV$ there is $-16M^2/e$, finite and nonzero. Allowing either sign of $U,V$ therefore gives the regular maximal extension with $UV<1$ and $r>0$.

With $T_K=(V+U)/2$ and $X_K=(V-U)/2$, the radial metric is a positive multiple of $-d{T}_K^2+d{X}_K^2$. The two exterior regions have $(U,V)=(-,+)$ and $(+,-)$. The future interior, or black hole region, has $(+,+)$; the past interior, or white hole region, has $(-,-)$. The Killing horizons are $U=0$ and $V=0$, meeting on the bifurcation surface at $U=V=0$.

For the Penrose diagram, compactify the null variables before forming time and space:This Schwarzschild conformal compactification preserves radial null directions because $dUdV={\sec }^2\bar{U}{\sec }^2\bar{V}(d{T}^2-d{X}^2)$. The ranges $|\bar{U}|,|\bar{V}|<\pi /2$ bound the diagram, and $UV<1$ cuts off its future and past tips. On $UV=1$ with $U,V>0$, $\bar{U}+\bar{V}=\pi /2$, so the future curvature singularity is $T=\pi /4$. The past curvature singularity is similarly $T=-\pi /4$. Both have $|X|<\pi /4$ and are spacelike. The horizon lines remain $T=\pm X$.

Each point of this radial Penrose diagram represents a two-sphere, except at limiting boundaries. Each exterior has its own future and past null infinity, ${\mathcal{I}}^{+}$ and ${\mathcal{I}}^{-}$. For example in the right exterior, $V\to \infty$ at fixed $U<0$ gives ${\mathcal{I}}^{+}$, whereas $U\to -\infty$ at fixed $V>0$ gives ${\mathcal{I}}^{-}$. Its spacelike infinity is $(T,X)=(0,\pi /2)$; its future and past timelike infinity are the vertices $(\pi /4,\pi /4)$ and $(-\pi /4,\pi /4)$. The left exterior has the reflected counterparts. These conformal boundary points are ideal endpoints, not finite-distance spacetime events; the physical affine or proper parameters of escaping geodesics diverge.

The horizons are regular null surfaces, while the horizontal $r=0$ boundaries are genuine curvature singularities. The central point represents the regular bifurcation surface, not a singularity. Future-directed radial causal curves have nondecreasing $U,V$, so a curve cannot go from one exterior to the other: changing the requisite signs would require one null variable to decrease. Crossing a future horizon instead leads towards the future singularity. The maximal extension has two causally disconnected exteriors, a future black-hole region and a past white-hole region.

For the static nonrotating source, use spherical symmetry and the static scalar perturbation of Minkowski spacetime, with isotropic radius $\rho$:To first order, the Einstein tensor in the declared convention hasThese are also the static scalar perturbation of Minkowski spacetime equations in the original information sheet. The source has $T_{00}=M{\delta }^{(3)}(\mathbf{x})$ and zero spatial stress. With the sign established in Question 2,The distributional identity $\Delta (1/\rho )=-4\pi {\delta }^{(3)}(\mathbf{x})$ and asymptotic flatness give $C=GM/\rho$. For $F=A+C$, the spatial field equation has trace $-2\Delta F=0$ and therefore ${\partial }_i{\partial }_{j}F=0$. Its decaying spherical solution is $F=0$, so $A=-GM/\rho$. This determines both potentials; the spatial correction cannot be inferred from the Newtonian gravitational potential alone.

Now use the isotropic-to-areal weak-field radius change. The angular coefficient defines $r^2=(1+2GM/\rho ){\rho }^2$, hence $r=\rho +GM$ to first order. Thus $dr=d\rho$ to the same accuracy, and replacing $1/\rho$ by $1/r$ inside a perturbation changes only second-order terms. With the Schwarzschild radius $r_S=2GM$, the linearized Schwarzschild metric isThis is an exterior weak-field approximation: $r_S/r\ll 1$. The point source is used distributionally to fix the mass, not to claim a regular or weak field at the origin.

For a null geodesic choose its orbital plane as $\theta =\pi /2$, using rotational symmetry, and let a dot denote an affine parameter derivative. For a Killing vector $k$ and an affinely parametrized geodesic tangent $U$, the geodesic conserved quantity from a Killing vector follows directly fromApplying this to the time-translation and axial Killing vectors, with the spatial sign absorbed into the definition of $h$, givesIf the tangent is normalized as the photon's four-momentum, $E$ is its energy measured at infinity and $h$ its signed angular momentum about the selected axis. Rescaling an arbitrary affine parameter rescales both; the ratio $b=|h|/E$ is the invariant impact parameter. Future-directed orbits have $E>0$. Assume $h\ne 0$ to use $\varphi$ as the orbit parameter; $h=0$ gives radial rays, not a finite-radius circular orbit.

The null vector constraint givesExpand consistently to first order in the weak-field approximation correction: $(1-u^2{)}^{-1}=1+O(u^2)$ and $(1+u{)}^{-1}=1-u+O(u^2)$. Also $du/d\varphi =-r_S\dot{r}/h$. Therefore the first-order orbital equation isThe cubic term is the first correction relative to the flat inverse-radius term; the equation is not a controlled exact equation for large $u$.

A constant solution of a first integral is not by itself sufficient to establish a circular geodesic. For a circular trajectory, the radial Euler-Lagrange equation also requires $A^{\prime }(r){\dot{t}}^2/2=r{\dot{\varphi }}^2$, where now $A(r)=1-r_S/r$ is the metric's time coefficient. Combining this with the null vector constraint gives $r{A}^{\prime }=2A$, or $u=2/3$. Equivalently, the formal potential $V(u)=u^2(1-u)$ must have a stationary point. Apart from the infinite-radius endpoint $u=0$, its only stationary point is $u_c=2/3$, with $V(u_c)=4/27$. Thus the formal circular solution requiresThe reduced radial equation is $u^{\prime \prime }=-u+\frac{3}{2}{u}^2$, with prime denoting $d/d\varphi$. It follows by differentiating the first integral on moving segments and agrees with the regular Euler-Lagrange equation at turning points. Perturbing $u=u_c+ϵ$ givesThere is an exponentially growing linear instability mode; equivalently $V^{\prime \prime }(u_c)=-2<0$. The formal circular orbit is unstable.

The weak-field photon-orbit validity limitation is essential: $u_c=2/3$ is not small. There is therefore no finite-radius circular photon orbit established within the controlled weak-field region. The formal extrapolation happens to reproduce the radius and impact parameter of the exact Schwarzschild photon sphere, but that physical result needs the exact geometry. The calculation also cannot establish a Schwarzschild event horizon from a first-order expansion at $u\sim 1$.

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.

Use geometrized units, and denote the beacon's proper time by $s$ to distinguish it from the short emitted duration $\tau$. Take $s=0$ at release. The period $M$, the duration $\tau$, and the wavelength $\lambda$ are measured by the beacon's own clock. Its radial timelike geodesic has conserved Killing energy per unit rest mass $E=fdt/ds$, where $f=1-2M/r$. Initially $dr/ds=0$ at $r=8M$, so $E=\sqrt{f(8M)}=\sqrt{3}/2$. Normalization of the four-velocity givesFor the Schwarzschild radial fall from rest at a finite radius, introduceIndeed $ds/d\eta =8M(1+\cos \eta )$ and $dr/d\eta =-4M\sin \eta$ reproduce the inward radial equation. At the Schwarzschild event horizon, $r=2M$ and $\cos {\eta }_H=-1/2$. HenceThis is the finite falling-clock time to the event horizon. By contrast, Schwarzschild time obeys $dt/ds=E/f$ and diverges at crossing. Near the horizon $r-2M\simeq E(s_H-s)$, so $dt/ds\simeq 2M/(s_H-s)$ and $t\simeq -2M\log ((s_H-s)/M)+\text{constant}$.

Every outward optical pulse emitted while $r>2M$ can reach a distant observer, and no pulse emitted after entry into the future black-hole region can do so. The infinite coordinate crossing time does not produce infinite emissions, because the pulse clock runs in proper time. If there is a pulse at release, the escape times are $s_n=nM$ for $n=0,\dots ,23$. Thus 24 pulses are received, counting the pulse at release; 23 are emitted subsequently during the fall. More generally the PDF does not specify the phase of the pulse clock. For a first emission at $s=\delta$ with $0\le \delta <M$, the finite pulse count for a falling Schwarzschild beacon isThe ceiling handles a pulse scheduled exactly at crossing correctly: it cannot escape. This count concerns pulses associated with the fall, not any pulses emitted before release, and assumes an ideal observer able to detect arbitrarily redshifted signals.

To find the stretching of an optical pulse, use the Schwarzschild tortoise coordinate $r_{\ast }=r+2M\log (r/(2M)-1)$. An outward radial null geodesic has constant Schwarzschild retarded null coordinate $u=t-r_{\ast }$. At an observer's fixed large radius $R$, its arrival time is $t_R=u+r_{\ast }(R)$; therefore only the varying emission value of $u$ contributes to the interval between arriving wavefronts. Differentiating along the falling source givesThe outgoing pulse redshift from a radial Schwarzschild infaller includes both gravitational redshift and the emitter's inward Doppler shift. A short pulse emitted at radius $r_e$ has, to leading order in its duration,For a finite receiver radius, replace $\mathcal{D}$ by $\sqrt{f(R)}\mathcal{D}$ to measure the receiver's proper time and frequency; this correction is negligible for $R\gg M$. The inverse scaling of frequency follows because the number of carrier cycles is unchanged when the pulse is stretched. Here $\nu =1/\lambda$ is ordinary frequency, not angular frequency, in units with light speed one.

Under the release-phase convention, the last emission is at $s_e=23M$. Solve $8({\eta }_e+\sin {\eta }_e)=23$ to obtain ${\eta }_e\simeq 1.94374184$, $r_e\simeq 2.54255983M$, and $\mathcal{D}(r_e)\simeq 7.49122492$. Because $\tau \ll M$ and this pulse begins $0.683364M$ before crossing, the local short-pulse approximation applies. Thus its approximate observed duration and frequency areWithout fixing that phase, these numerical coefficients are not unique: put $s_e=\delta +(N(\delta )-1)M$, find $r_e$ from the parameterization, and use the boxed general formulas. For a last pulse close to crossing, with $\Delta =s_H-s_e\ll M$ but $\tau \ll \Delta$, the horizon redshift of a Schwarzschild infaller gives the useful estimatesIf a phase instead places the pulse within one pulse-duration of crossing, $\Delta \lesssim \tau$, a single constant stretching factor is inappropriate. The received extent of a complete escaping pulse is exactly $u(s_e+\tau )-u(s_e)$; a pulse straddling crossing has an escaping tail extending to arbitrarily late arrival times, whose instantaneous frequency tends to zero. The finite count remains valid, while its last-pulse duration then has no finite constant multiple of $\tau$. This identifies precisely what the unspecified pulse phase can change.

Write $f(r)=1-2M/r$. Radial null paths satisfy $dt/dr=\pm f^{-1}$, whose integral uses the Schwarzschild tortoise coordinate $r_{\ast }=r+2M\log |r/(2M)-1|$. The incoming and outgoing families are thereforewith constants $v,u$. For an exterior outgoing ray, put $r/(2M)-1=\epsilon$. As $t\to -\infty$, $t=u+2M(1+\epsilon )+2M\log \epsilon$, so $\epsilon \sim C{e}^{t/(2M)}$ for a positive ray-dependent $C$. As $t\to \infty$, the logarithm is smaller than the linear term, giving $r\sim t$.

In ingoing Eddington-Finkelstein coordinates, $v=t+r_{\ast }$, so $dt=dv-dr/f$. Substitution cancels the singular $d{r}^2/f$ term and givesThis metric is regular at $r=2M$. In the requested chart $\hat{t}=v-r$, the incoming rays are $\hat{t}=v_0-r$ and the outgoing rays satisfyIncoming rays move inward everywhere. Outgoing rays move outward outside the horizon, remain at constant radius on it, and also move inward inside it.

The line $r=2M$ is the Schwarzschild event horizon, a null boundary of the future black-hole region. It is a coordinate singularity of Schwarzschild time, not a curvature singularity; the curvature singularity is at $r=0$. Signals from the future black-hole interior cannot cross it outward.

Finally parametrize any future-directed timelike observer in that interior by proper time $s$. Normalization in the given metric givesThe future direction in the ingoing chart has $\dot{r}<0$, so $-\dot{r}\ge \sqrt{2M/r-1}$. The proper-time bound inside a Schwarzschild black hole isThus a future-continued observer must reach the singular boundary in finite proper time, regardless of acceleration or angular motion. The black-hole future orientation is essential; a past-directed path or a future path in a white-hole interior is a different statement.

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.

Write $f=1-2M/r$ in the Schwarzschild metric. For the stationary Killing vector field $V={\partial }_t$, its dual is $V^{♭}=-fdt$ and the Papapetrou electromagnetic field isOutside the Schwarzschild event horizon, use the static orthonormal coframe $e^{\hat{0}}=\sqrt{f}dt$, $e^{\hat{r}}=dr/\sqrt{f}$, $e^{\hat{\theta }}=rd\theta$, $e^{\hat{\varphi }}=r\sin \theta d\varphi$. Then $F=(2M/r^2)e^{\hat{0}}\wedge e^{\hat{r}}$. With the electric field convention $E_{\hat{i}}=F_{\hat{i}\hat{0}}$ from the preceding part,Thus the stationary Killing vector field gives a Coulomb electric field, directed inward in this convention. If the charge is normalized by $E_{\hat{r}}=Q/r^2$, its charge is $Q=-2M$. A convention with $F\mapsto -F$ reverses that charge; the invariant content is a radial monopole electric field of magnitude $2M/r^2$.

For the axial Killing vector field $V={\partial }_{\varphi }$, $V^{♭}=r^2{\sin }^2\theta d\varphi$, givingIn the same static orthonormal coframe, this is a purely magnetic field withAt large $r$ these are the spherical components of $2{\mathbf{e}}_z$. This is therefore the regular magnetic test field asymptotic to a uniform axial field, distorted by the Schwarzschild black hole. Its net magnetic monopole flux is zero because the integral of $\cos \theta$ over a sphere vanishes. Both electromagnetic fields are regular at the future event horizon: in ingoing Eddington-Finkelstein coordinates, the electric field is $(2M/r^2)dv\wedge dr$, and the magnetic two-form already contains only regular spatial coordinates. Static observers themselves cease to exist on the event horizon.

In the final Schwarzschild spacetime, an outgoing ray with $r>2M_1$ escapes to future null infinity, whereas one with $r<2M_1$ cannot do so and ends at the Schwarzschild singularity. Their boundary is the outgoing null geodesic $r=2M_1$. Since the final region continues indefinitely, the boundary of the causal past of future null infinity there is exactly that Schwarzschild event horizon. Equivalently, integrating the outgoing equation backward from this final boundary selects the complete event horizon. ThusThe original Penrose diagram below shows the right-hand exterior and black-hole interior appropriate to ingoing coordinates. The green band is the ingoing accretion interval; the dashed blue curve is the spherical apparent horizon $r=2m(v)$. Outgoing null geodesics escaping to future null infinity lie to the right of the red event horizon; those to its left terminate on the future spacelike singularity. In the final stationary region the two horizons coincide. The red event horizon already lies outside $r=2M_0$ before the ingoing matter arrives, because its position is fixed by the future escape criterion.

The stress-energy tensor is already first order, so its energy flux can be evaluated on the unperturbed Schwarzschild event horizon $r=2M_0$, with the infinity-normalized Killing vector field $\xi ={\partial }_v$. On this background $g^{rv}=1$, and henceThe conserved stress-energy current from a Killing vector is $J^a=-T^a{}_b{\xi }^b$. Its radial component is negative, so the positive inward energy flux is $-J^r$. Using the directed null-surface measure $r_H^{2}dvd\Omega$ givesThe initial and final event horizon areas are $A_0=16\pi M_0^2$ and $A_1=16\pi (M_0+\Delta M{)}^2$. Thus $\Delta A=32\pi M_0\Delta M+O((\Delta M{)}^2)$. With Schwarzschild surface gravity ${\kappa }_0=1/(4M_0)$, the Physical-process first law of black-hole mechanics follows:The flux is Killing energy, including gravitational redshift, rather than the energy measured by a sequence of static observers arbitrarily close to the event horizon.

Use geometrized units $G=c=1$ and metric signature $(-+++)$; $\tau$ is proper time. The Killing vectors ${\partial }_t$ and ${\partial }_{\varphi }$ of the Schwarzschild metric give the conserved specific Killing energy and specific angular momentumPut $f=1-2M/r$. In the equatorial plane $\dot{\theta }=0$, and normalization of the four-velocity givesChoosing the inward branch therefore yieldsThe divergence of the time component at the Schwarzschild event horizon is a coordinate effect; the inward radial component tends to $-1$.