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For $r_{-}<r_{+}$, an ingoing chart with $v=t+r_{\ast }$ has metric $-Vd{v}^2+2dvdr+R^{2}d{\Omega }_2^2$. In its future interior, $\nabla r={\partial }_v+V{\partial }_r$ is future timelike, so every future causal curve decreases $r$ and cannot reach future null infinity. The singularity $r=r_{-}$ is spacelike. A Kruskal–Szekeres coordinates extension has the same qualitative Penrose diagram as Schwarzschild spacetime, with two exterior regions, a future black-hole interior and a past white hole interior. The past interior has the same radial interval but is not a black hole region; specifying the radial interval alone does not identify the future component. For $r_{-}>r_{+}$, the physical exterior $r>r_{-}$ has $V>0$ everywhere and outgoing null rays escape from arbitrarily near a timelike naked singularity.

The Kretschmann scalar is the scalar tensor contraction of two copies of the Riemann curvature tensor. In Schwarzschild spacetime, with metric coefficient $1-2M/r$ and $G=c=1$, it is $48M^2/r^6$. Divergence of this scalar along a curve approaching a proposed regular point proves that there is no twice continuously differentiable, nondegenerate metric extension through that point: such an extension would have continuous finite curvature components and hence a finite scalar contraction. This gives a coordinate-independent test of a curvature singularity. Boundedness of this single scalar does not prove regularity.

The Kruskal spacetime is the maximal analytic extension of the positive-mass vacuum Schwarzschild spacetime. It contains two exterior regions, a black hole, and a white hole. Kruskal–Szekeres coordinates remove both horizon coordinate singularities, while $r=0$ remains a curvature singularity.

The harmonic equation, including critical points. All derivatives in this argument use the three-dimensional Levi-Civita connection of $\gamma$. Write $U_i={\nabla }_{i}U$, $|\nabla U{|}^2={\gamma }^{ij}{U}_i{U}_j$ and ${\Delta }_{\gamma }U={\gamma }^{ij}{\nabla }_i{\nabla }_{j}U$. The curvature relation for the static vacuum conformal spatial metric gives $R=2|\nabla U{|}^2$. Substituting it into the contracted Bianchi identity yieldsThe Hessian of a scalar is symmetric, so the last terms cancel:Where $\nabla U\ne 0$, this implies ${\Delta }_{\gamma }U=0$. On the interior of the critical set $\{\nabla U=0\}$, $U$ is locally constant and its Laplacian is also zero. Every other critical point is a limit of noncritical points, so continuity of the Laplacian givesThis avoids incorrectly dividing by a gradient at its zeros.

Integration by parts and rigidity. The usual whole-space energy identity isWith standard static asymptotic falloff $U=O(R^{-1})$ and $\nabla U=O(R^{-2})$, the last term is $O(R^{-1})$ and vanishes. Positivity of the spatial Riemannian metric then makes $U$ constant, and its limiting value fixes that constant to zero.

There is also a direct integration by parts proof requiring only the stated vanishing of $U$ at infinity, rather than an assumed flux decay rate. For a regular value $\epsilon >0$, the region ${\Omega }_{\epsilon }=\{U>\epsilon \}$ has compact closure. Completeness and the absence of an inner boundary ensure no missing boundary pieces; standard asymptotic flatness and $U\to 0$ on all ends keep this positive level set away from infinity. Its outward unit normal is $n=-\nabla U/|\nabla U|$. Multiplying the harmonic equation by $U$ and integrating givesBoth terms are nonpositive, so both vanish. A nonempty component with $U>\epsilon$ cannot have zero gradient throughout and boundary value $\epsilon$. Thus ${\Omega }_{\epsilon }$ is empty. Choose arbitrarily small regular values and apply the same argument to $-U$; it follows thatThis is vanishing harmonic function by level-set integration. It also makes explicit why an inner boundary would invalidate the conclusion.

From zero Ricci curvature to Minkowski spacetime. In dimension three the Schouten tensor is $S_{ij}=R_{ij}-R{\gamma }_{ij}/4$, so it vanishes here. The three-dimensional curvature reconstruction then gives $R_{ijpq}=0$: $\gamma$ is a flat Riemannian manifold. This use of three-dimensional curvature from the Ricci tensor is essential; zero Ricci tensor would not by itself imply flatness in four dimensions.

A complete connected flat spatial manifold has Euclidean universal cover. Standard asymptotic flatness, with an ordinary Euclidean end, excludes a nontrivial free Euclidean quotient: a nontrivial fixed-point-free Euclidean isometry contains a translational or screw component, and its cyclic quotient has at most quadratic volume growth, incompatible with a three-dimensional Euclidean end. Extra identifications cannot restore cubic growth. Thus the complete spatial metric is globally Euclidean, not just locally flat. With $U=0$ and the standard global time coordinate the four-metric becomesTherefore horizonless static vacuum rigidity gives Minkowski spacetime as the sole solution under these hypotheses.

Allowing horizons. A black hole exterior can be static and asymptotically flat without being Minkowski. The Schwarzschild spacetime is the basic example. Its lapse function vanishes at the event horizon, so $U$ is unbounded below there; the horizon introduces an inner end or boundary in the spatial reduction, and the previous energy argument no longer has its hypotheses. In fact for Schwarzschild,so this conformal quotient terminates at $r=2M$ and is not the complete nonsingular quotient used above.

With the usual regularity, connected-horizon and global hypotheses of the static vacuum black-hole uniqueness theorem, the nontrivial black-hole exterior is Schwarzschild. Rotating Kerr black holes are stationary but not static and therefore are not alternatives in this question. This statement concerns a regular domain outside the horizon; the Schwarzschild interior contains a curvature singularity. If “globally static” and “nonsingular everywhere” are retained literally, Schwarzschild does not meet them. Allowing horizons means relaxing the horizonless global hypotheses to the corresponding static exterior problem.

Use geometrized units and metric signature $(-+++)$, with $M>0$. In the exterior of Schwarzschild spacetime, put $f=1-2M/r$. The Schwarzschild tortoise coordinate satisfiesThe retarded and advanced null coordinates $u=t-r_{\ast }$ and $v=t+r_{\ast }$ then give $d{s}^2=-fdudv+r^{2}d{\Omega }^2$. The logarithmic divergence of $r_{\ast }$ suggests exponentiating these null coordinates. In the right exterior define the Kruskal–Szekeres coordinatesTheir product eliminates $t$:Since $du=-4MdU/U$ and $dv=4MdV/V$, the Schwarzschild metric becomesHere $r$ is an implicitly defined function of $UV$. The derivative of the right side with respect to $r$ is $-r{e}^{r/(2M)}/(4M^2)$, which is nonzero at $r=2M$. The inverse function theorem therefore makes $r(UV)$ smooth across that surface, and the coefficient of $dUdV$ tends to $-16M^2/e$. Thus the Schwarzschild event horizon is a coordinate singularity of the original chart, while this Lorentzian metric remains regular there.

Extend the Kruskal–Szekeres coordinates to all real $U,V$ with $UV<1$. The signs give two exterior regions, $U<0<V$ and $V<0<U$, a future black hole region $U,V>0$, and a past white hole region $U,V<0$. The event horizons are $U=0$ or $V=0$, intersecting at the bifurcation surface. The boundary $UV=1$ has $r=0$ and is a genuine Schwarzschild singularity, as the Kretschmann scalar $48M^2/r^6$ diverges there.

Finally, $T=(V+U)/2$ and $X=(V-U)/2$ give $UV=T^2-X^2$ and a radial metric proportional to $-d{T}^2+d{X}^2$. Hence radial null geodesics have slopes $\pm 1$, the event horizons are $T=\pm X$, and the singular boundaries are $T^2-X^2=1$. This constructs the maximal Kruskal extension; a black hole produced by collapse need not contain the second exterior or the white hole of that eternal extension.

Given an asymptotically flat spacetime with a conformal completion and future null infinity ${\mathcal{I}}^{+}$, its black hole region isThe event horizon is its boundary. For the magnetically charged dilaton black hole, first assume $0\le a=r_{-}<b=r_{+}$. The physical asymptotic component has $r>a$, and the ingoing chart extends regularly across $r=b$. Its future ingoing null direction is $-{\partial }_r$. For $a<r<b$,Hence $\nabla r$ is future timelike. Every nonzero future causal tangent $X$ satisfies $X(r)=g(\nabla r,X)<0$. A future causal curve from this interior can neither cross back to $r=b$ nor reach the asymptotic end $r\to \infty$, so it cannot reach ${\mathcal{I}}^{+}$. It lies in $\mathcal{B}$.

This proof uses the future black-hole extension. The radial interval alone does not distinguish it from the past white hole interior of the maximal extension; that interior has the same interval and can send signals to infinity. Thus the statement needs this future-component qualification if the maximal spacetime is intended.

The Kretschmann scalar confirms that $r=a$ is a genuine curvature singularity: for $0<a<b$, its numerator at $r=a$ is $3a^4(a-b{)}^2$, andFor $a=0$ it instead reduces to $12b^2/r^6$. Since $V<0$ on approaching the singularity from the interior, the singularity is spacelike. At $r=b>a$ the curvature is finite.

For the maximal extension, take $U=-e^{-u/(2b)}$, $W=e^{v/(2b)}$ in the right exterior and continue analytically across the horizons. ThenThe singularity is $UW=(1-a/b)e^{a/b}>0$ and the horizons are $U=0$ and $W=0$. Rescaling the null coordinates by the square root of this positive constant before arctangent compactification produces the usual qualitative Penrose diagram of Schwarzschild spacetime: two exteriors, a future black-hole region, a past white-hole region, and future and past spacelike singularities. The dashed lines below are the horizon branches; only the future interior is the black-hole region.

If $a>b$, the physical domain remains $r>a$, but now $V>0$ throughout it. The apparent zero $r=b$ is beyond the curvature singularity and is not an accessible horizon. The singularity is timelike, and outgoing radial null rays with $\dot{r}=E>0$ escape to infinity from arbitrarily near it. ThusThe excluded equality $r_{-}=r_{+}$ is a singular limiting case, not a regular horizon. These facts describe the causal structure of the magnetic dilaton black hole.

Work in the rest frame of the hole and neglect stellar velocities compared with $v$. In addition to $v\gg {\sigma }_{\ast }$, assume the galactic encounter speed is nonrelativistic, $v\ll c$. Treat a star as a test particle with specific angular momentum $\ell =bv$. At its pericentre $r_{\min }$, its radial velocity vanishes. Conservation of energy and conservation of angular momentum giveThe second term is gravitational focusing. For the rough Newtonian estimate in the question, assume capture when the pericentre reaches the Schwarzschild radius, $r_S=2G{M}_{\mathrm{BH}}/c^2$. It follows thatThe approximation keeps the focusing term because $v\ll c$.

Let ${\rho }_{\ast }=m{n}_{\ast }$ be the stellar mass density. Multiplying the geometric collision cross-section by the incoming mass flux gives the collisionless black-hole capture rateFor a supersonic gas with negligible sound speed, Bondi--Hoyle--Lyttleton accretion has capture radius $r_a=2G{M}_{\mathrm{BH}}/v^2$ and rateConsequentlyFor comparable densities and $v=200\mathrm{km}{\mathrm{s}}^{-1}$, the suppression is approximately $4\times 10^{-7}$. Gas can collide, dissipate orbital energy, and be captured from a region of size $GM/v^2$; collisionless particles generally leave on an unbound orbit unless their angular momentum is small enough for direct capture. Chandrasekhar dynamical friction can slow the hole by transferring energy to a stellar wake, but does not mean that the stars producing that wake are swallowed.

There is an important accuracy limitation in the requested numerical coefficient. Newtonian orbital mechanics evaluated at a relativistic event horizon cannot give the exact capture boundary. In Schwarzschild spacetime, a slowly incoming massive particle has dimensionless energy $\mathcal{E}\simeq 1$ and timelike geodesic effective potentialThe double turning point satisfying $V=1$ and $dV/dr=0$ occurs at $r=4r_g$ and $\ell =4G{M}_{\mathrm{BH}}/c$. The exact low-speed boundary is therefore $b_{\mathrm{cap}}\simeq 4G{M}_{\mathrm{BH}}/(cv)$, twice the rough value; it makes the rate ratio $4({\rho }_{\ast }/{\rho }_g)(v/c{)}^2$. The key suppression scaling is unchanged.

Direct capture of ordinary stars or dark matter is therefore an inefficient route to rapid growth of a supermassive black hole under these conditions. Tidal disruption events can capture stellar debris without swallowing an intact point-like star, and special dense environments can modify the supply. These qualifications do not make collisionless capture equivalent to dissipative gas accretion.

The radial equation for a circular geodesic, or equivalently $V^{\prime }(r)=0$, gives the Angular velocity of a circular Schwarzschild geodesicOne orbit therefore takes Schwarzschild coordinate time

Observer $O_3$ is static at infinity, where coordinate time equals proper time, so

For the static observer $O_2$ at radius $r$, $dr=d\theta =d\varphi =0$, and the Schwarzschild spacetime metric givesThus

Along $O_1$,where $d\varphi /dt=\Omega$ was used. Hence the Proper period of a circular Schwarzschild geodesic is

Because the shell lies well inside its Schwarzschild radius once it reaches $r=2E$, an event horizon forms and the final exterior is a black hole of mass $E$. In the ideal instantaneous-shell model the collapse ends at a spacetime singularity.

Outside the ingoing null shell the enclosed mass is $E$, so use the ingoing Ingoing Eddington-Finkelstein coordinatesThis is the Schwarzschild metric of mass $E$, regular across $r=2E$ and valid for $r>0$ on the exterior side of the shell.

The station's proper rest mass isA static local energy $d{E}_{\mathrm{loc}}$ at radius $r_{\mathrm{st}}$ contributes the redshifted energyto the mass measured at infinity. To first order in $M_{\mathrm{st}}/M$, the exterior mass parameter is thereforeTerms quadratic in $\rho$ describe the shell's self-gravity and are neglected.

The exterior of the shell is vacuum and spherically symmetric. The Birkhoff theorem therefore makes it a Schwarzschild spacetime with mass $M_{\mathrm{out}}$:This is the redshifted gravitational mass of a static thin shell.