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A spherical surface with areal radius $R>3m$, where $m=GM/c^2$, has apparent angular radius $\psi$ for a static exterior observer at $r_0>R$. Its grazing null geodesic has impact parameter $b=R/\sqrt{1-2m/R}$. Projecting the photon tangent into the observer's local rest frame gives $\sin \psi =b\sqrt{1-2m/r_0}/r_0$, yielding the displayed result. The condition $R>3m$ ensures the grazing ray escapes, since the surface is outside the photon sphere. For $R=4m$, the distant image radius is $\sqrt{2}$ times the flat-space prediction.

In a spherically symmetric spacetime, the areal radius $r$ is defined by the area $A=4\pi r^2$ of each symmetry sphere. It is independent of which radial coordinate chart is used.

A stationary time-coordinate change $t^{\prime }=t+F(r)$ leaves both the areal radius and the time-translation Killing vector field unchanged. The conserved Killing energy and angular momentum of a null geodesic therefore yield the same radial equation in either chart, ${\dot{r}}^2+h^2(1-2\mu /r)/r^2=c^2{k}^2$. The effective potential and radial turning points agree, even when the coordinate time velocities and horizon regularity differ.

A Finkelstein diagram plots areal radius $r$ against $T=v-r$ in ingoing Eddington-Finkelstein coordinates. Its radial null directions show explicitly that both future light-ray families decrease $r$ inside a Schwarzschild black hole. It is not a compactified Penrose diagram.

At the boundary of an Oppenheimer-Snyder cloud, its areal radius is $R=a\sin {\chi }_0$. Matching to a Schwarzschild metric gives $M=4\pi \rho R^3/3=a_{\max }{\sin }^3{\chi }_0/2$. Thus ${\sin }^2{\chi }_0=2M/R_0$ and $a_{\max }=\sqrt{R_0^3/(2M)}$. The boundary is a radial timelike geodesic with specific Killing energy $\sqrt{1-2M/R_0}=\cos {\chi }_0$. Its radial equation is ${\dot{R}}^2=2M/R-2M/R_0$. The induced metric and the angular extrinsic curvature agree on both sides, so no surface layer is required.

The area of a symmetry sphere in the weak-field isotropic metric is $4\pi {\rho }^2(1+2GM/\rho )$. Equating it to $4\pi r^2$ defines the areal radius and gives the displayed change. To first order $dr=d\rho$, and replacing $1/\rho$ by $1/r$ changes the perturbation only at second order. This explains why the isotropic angular correction disappears in the linearized Schwarzschild metric without discarding spatial curvature.

A Schwarzschild static observer at areal radius $R$ emits an outward null geodesic at a small angle $\psi$ to the radial direction. Its impact parameter is $b=R\sin \psi /\sqrt{f}$, and the accumulated azimuth to infinity is ${\int }_R^{\infty }bdr/[r^2\sqrt{1-f(r)b^2/r^2}]=b/R+O(b^3/R^3)$. Thus weak curvature magnifies a narrow outward beam angle by $1/\sqrt{f(R)}$, rather than producing the full infinity-to-infinity deflection $4GM/(b{c}^2)$.

Spatial geometry. An embedding of the three-sphere isTaking differentials and collecting terms gives the induced metric tensorPut $\bar{r}=A\sin \chi$. Then $d\bar{r}=A\cos \chi d\chi$, soThe areal radius $\bar{r}$ is not a global radial coordinate on the entire three-sphere: its maximum occurs at the equator and two hemispheres share each smaller value. Use separate charts there; the coordinate singularity at $\bar{r}=A$ is not a physical singularity.

For Euclidean space and hyperbolic space of curvature $-1/A^2$, respectively,For the latter, $\bar{r}=A\sinh \chi$ gives $d{\ell }_{-}^2=A^2[d{\chi }^2+{\sinh }^2\chi d{\Omega }^2]$. In all three cases the induced two-metric at fixed $\bar{r}=\bar{R}$ is ${\bar{R}}^{2}d{\Omega }^2$. Its surface area is thereforeThis equality compares the same areal radius, not the same proper radial distance. For a proper radial distance $\ell$, the corresponding areas are $4\pi A^2{\sin }^2(\ell /A)$, $4\pi {\ell }^2$, and $4\pi A^2{\sinh }^2(\ell /A)$.

Homogeneity, isotropy and redshift. The FRW metric has constant-time spatial slices of constant curvature $k/a^2(t)$, with $k=+1,0,-1$ selecting the three geometries above. Their isometries act transitively on points and isotropically about every point, implementing the cosmological principle. Multiplication by one common time-dependent scale factor preserves these properties. With the source's $k=\pm 1$ normalization, $r$ is dimensionless and $a$ has units of length; using a dimensionless scale factor instead rescales the spatial curvature parameter. The origin of the spherical coordinates is arbitrary; it does not identify a physical centre. Cosmic time is the proper time of the comoving observers, and the absence of time-space cross terms makes their worldlines orthogonal to the homogeneous slices.

For a radial null geodesic between comoving endpoints,The next wave crest travels between the same endpoints. Subtracting the two path integrals, for wave periods short compared with the expansion time, gives $\delta t_0/a_0=\delta t_E/a_E$. Thus the cosmological redshift isBoth observers' clocks are cosmic proper-time clocks. Peculiar velocities and local gravitational-potential shifts have been left out of this homogeneous calculation.

Angular size and brightness. For a galaxy small compared with its distance, the angular diameter distance is $D_A=a_{E}R$ and its angular diameter isHere the emitted proper diameter is measured transverse to the line of sight; radial proper distance is a different integral when $k\ne 0$. Ignoring absorption, lensing caustics and intrinsic evolution, photon-number conservation spreads an isotropic-equivalent luminosity across the sphere $4\pi a_0^2{R}^2$. Each photon loses energy by $1+z$, and its arrival interval is stretched by the same factor. The received bolometric flux isThe disc's proper projected area is ${\mathcal{A}}_d=\pi d^2/4$, and its observed solid angle is ${\Omega }_d\simeq {\mathcal{A}}_d/D_A^2$. Therefore its surface brightness isThis is cosmological surface-brightness dimming. The normalization uses $L$ as isotropic-equivalent luminosity along the viewing direction; another disc emission law changes that normalization but not the redshift power. This is a bolometric result. In a specified observed frequency band the emitted spectrum is sampled at a shifted frequency and requires a spectral correction.

The distinction in telescope aperture and resolved surface brightness matters here. A larger telescope collects more photons and improves angular resolution, but cannot restore the incoming radiance lost by cosmological dimming. For aperture diameter $D$ and a diffraction resolution element of solid angle ${\Omega }_{\mathrm{res}}\sim (\lambda /D{)}^2$, the collected power from uniform resolved brightness scales asindependent of $D$. This is consistent with conservation of etendue. It explains why increasing aperture while keeping every gain in resolution does not increase counts per native resolution element from a smooth extended source. It does not make larger telescopes useless: at a fixed angular bin the counts grow as $D^2$, so a photon-noise-limited or uniform-sky-noise-limited signal-to-noise ratio grows roughly as $D\sqrt{t_{\exp }}$. They also resolve smaller structures and improve point-source sensitivity. Aperture, exposure and angular binning help measure a dim galaxy, without changing its intrinsic $(1+z{)}^{-4}$ brightness penalty.

The observer is static at $r_0$, with four-velocity $U=f_0^{-1/2}{\partial }_t$, where $f_0=1-2m/r_0$. Angles are measured in the observer's rest frame, using the positive spatial metric tensoron the equatorial plane. One cannot directly normalize a photon tangent with the full Lorentzian metric to define an angle, because that tangent is null.

At the observer the spatial direction of an outward-moving grazing ray is proportional to $w=A{\partial }_r+{\partial }_{\varphi }$, with $A=(dr/d\varphi {)}_{r_0}$. Its angle $\psi =\alpha /2$ to the outward radial direction equals the corresponding sky angle to the inward center of the image. Taking the spatial inner product with $e_r={\partial }_r$,ConsequentlyThe limiting value $A=0$ is understood by continuity.

For a grazing ray the areal radius has a turning point at the stellar surface $R$, so $\dot{r}=0$ there. The radial equation givesSince $R=4m>3m$, the surface lies outside the photon sphere. The function $f/r^2$ decreases for $r>3m$, so this tangential ray can escape to every $r_0>R$ and correctly defines the image's rim. For the specified surface,Also $dr/d\varphi =-r^{2}du/d\varphi$, henceIts sign depends on the orientation of the ray and of $\varphi$, and does not affect the angular diameter. Substitution givesThis apparent angular radius of a spherical star in Schwarzschild spacetime has useful limiting checks. As $r_0\downarrow R$, the star occupies a hemisphere, $\alpha \to \pi$. Far away, $\alpha \simeq 2b/r_0=8\sqrt{2}m/r_0$: Schwarzschild light deflection increases the apparent radius by a factor $\sqrt{2}$ relative to the flat-space value $R/r_0$.

Approximate Earth by an isolated, nonrotating, spherically symmetric body. By Birkhoff's theorem, the exterior vacuum metric tensor is the Schwarzschild metric. With $x^0=ct$ it isHere $r$ is the areal radius, $\theta$ and $\varphi$ are spherical angles, $t$ is Schwarzschild time normalized to the proper time of a static observer at infinity, $M$ is Earth's mass, $G$ the gravitational constant, and $c$ the speed of light. The nonzero diagonal metric tensor components are $g_{00}=-f$, $g_{rr}=f^{-1}$, $g_{\theta \theta }=r^2$, $g_{\varphi \varphi }=r^2{\sin }^2\theta$.

The apple is a test particle; neglect its effect on the metric tensor, air resistance, Earth's rotation and nonspherical structure, and external tidal forces. Near Earth's surface, $GM/(R_E{c}^2)\simeq 7\times 10^{-10}$, $h/R_E\ll 1$, and the falling speed remains much smaller than $c$. Thus the Newtonian limit of this metric tensor is appropriate, while the exact radial geodesic is available as a check. A static observer under the tree has proper time $T=\sqrt{f(R_E)}t$, up to an additive constant. Such an observer is supported by the ground and has nonzero proper acceleration; the released apple instead has zero proper acceleration.

The intended orbit is a freely falling timelike circular Schwarzschild geodesic. Use the Schwarzschild metric with $t$ rather than $ct$ as the zeroth coordinate, and put the orbit at $\theta =\pi /2$. Thus $g_{tt}=-f{c}^2$, $g_{rr}=1/f$, $g_{\varphi \varphi }=r^2$, with $f=1-2GM/(r{c}^2)$. The radial geodesic equation at constant $r$ isConsequently ${\Omega }^2=(u^{\varphi }/u^t{)}^2=GM/r^3$. The four-velocity normalization givesand henceHere $t$ is Schwarzschild time, $\tau$ the orbiting clock's proper time, $G$ the gravitational constant, $M$ the central mass, $r$ the constant areal radius, and $c$ the speed of light. The dimensionless $u^0$ follows the question's time-valued zeroth-coordinate convention; with $x^0=ct$ the zeroth four-velocity component would be $cdt/d\tau$.

The condition is $r>3GM/c^2$ for a massive circular geodesic; radial stability further requires $r>6GM/c^2$, with equality marginal. The phrase “any circular orbit” is false without free fall: a powered circular trajectory has $dt/d\tau =[f-r^2{\Omega }^2/c^2{]}^{-1/2}$ with independently chosen $\Omega$. For example, at $r=10GM/c^2$ choose $\Omega =\frac{1}{2}\sqrt{GM/r^3}$; this valid accelerated circular trajectory has $dt/d\tau =\sqrt{40/31}$, instead of the displayed $\sqrt{10/7}$.

For a timelike circular Schwarzschild geodesic, the coordinate orbital period and onboard proper time per orbit areUsing $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$ and $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$, the Schwarzschild radius and orbital areal radius are $R_s=2GM/c^2\simeq 1.21789\times 10^{10}\mathrm{m}$ and $R=200R_s\simeq 2.43578\times 10^{12}\mathrm{m}$. ThereforeThe crew accumulates about $3836$ fewer seconds, or $63.9$ fewer minutes, per orbit. Both gravitational time dilation and kinematic time dilation contribute: a static clock at $R$ has $d{T}_{\mathrm{static}}=\sqrt{1-R_s/R}dt$, while the orbiting clock additionally has $d\tau =d{T}_{\mathrm{static}}/\gamma$, where $\gamma =\sqrt{398/397}$. A distant observer measures the coordinate period; comparing arrivals emitted at the same orbital phase also removes the repeated light propagation delay.

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 $G=c=1$ until converting the clock readings. By the Birkhoff theorem, the spherical vacuum exterior is a portion of the Schwarzschild spacetime with fixed mass $M$:The exterior is not flat and is not radiating merely because the surface moves. When the surface crosses $r=2M$, this exterior must be represented in regular ingoing Eddington-Finkelstein coordinates, rather than treating the Schwarzschild-coordinate divergence as a physical barrier.

The Oppenheimer-Snyder model has a homogeneous pressureless matter interior. Since its initial expansion vanishes but its density is positive, the Friedmann equation requires positive spatial curvature. Choose a comoving radial angle $\chi$ and the boundary's proper time $\tau$, with the stopwatch set to zero at release. The interior FLRW metric and density areThe density law follows from the dust stress-energy tensor and its covariant conservation. The Friedmann equation and the initial rest condition giveIts contracting branch has the parameterizationIndeed $d\tau /d\eta =a$ and $\dot{a}=-\sin \eta /(1+\cos \eta )$ satisfy the displayed equation; $\eta =0$ is the turning point and $\eta =\pi$ is the final curvature singularity.

For homogeneous dust-ball matching, the boundary has areal radius $R=a\sin {\chi }_0$. Matching its mass to the exterior givesThis $M$ is the gravitational mass parameter, not the integral of rest density over the curved proper-volume element. The matching can also be checked directly. The outward angular extrinsic curvature inside is $K_{\theta \theta }=R\cos {\chi }_0$; outside it is $R\sqrt{1-2M/R+{\dot{R}}^2}$. The boundary motion hasso these curvatures agree. The induced metric is $-d{\tau }^2+R^{2}d{\Omega }^2$ on both sides and $K_{\tau \tau }=0$ because the pressure-free boundary is geodesic. No artificial surface stress layer is required.

The observer moving with this boundary meets the event horizon when $R=2M$. From $R=R_0(1+\cos \eta )/2$, the proper-time horizon crossing in homogeneous dust collapse occurs atThe singularity is reached at ${\tau }_s=\pi a_{\max }/2$, so the remaining clock time isThese are finite proper times, even though the external static time diverges at horizon crossing.

Convert the supplied rounded units consistently: $M=10^{38}$ Planck masses and $R_0=6.25\times 10^{39}$ Planck lengths. Thus $2M/R_0=0.032$, ${\chi }_0\simeq 0.17985$, and $a_{\max }\simeq 3.4939\times 10^{40}$ in Planck units. Multiplying geometric times by $5\times 10^{-44}$ seconds givesAt the accuracy justified by the rounded constants, these are about three milliseconds to horizon crossing and about seven microseconds more to the singularity. For $R_0\gg 2M$, expanding ${\eta }_H=\pi -2{\chi }_0$ gives $\Delta \tau \simeq (2/3)a_{\max }{\chi }_0^3\simeq 4M/3$, independently confirming the microsecond scale. The exact expression above retains the finite release-radius correction.

Strictly, this boundary clock reading is when the observer crosses the horizon, not when the horizon first exists anywhere. The event horizon inside an Oppenheimer-Snyder cloud is an outgoing radial null ray. Since $d\tau =ad\eta$, the interior metric is conformal to $-d{\eta }^2+d{\chi }^2+{\sin }^2\chi d{\Omega }^2$, and that ray has $d\chi /d\eta =1$. Tracing it back from $({\eta }_H,{\chi }_0)$ givesFor this cloud it starts at the centre after release, at comoving time ${\tau }_{\mathrm{birth}}\simeq 2.722\times 10^{-3}$ seconds. This answers the literal global-formation interpretation as well; it is not an event on the outer observer's worldline. The distinction matters because an event horizon is defined by the entire future causal escape problem, not by a locally detectable signal of formation.

For the observer at infinity, use the boundary's conserved Killing energy $E$ in the exterior:As $\tau \uparrow {\tau }_H$, $\dot{R}\to -E$ and $R-2M\simeq E({\tau }_H-\tau )$. Consequentlywhere ${\tau }_{\ast }>0$ fixes a harmless dimensionless logarithm. For actual outgoing light signals the relevant retarded time is $u=t-r_{\ast }(R)$. Since $r_{\ast }\simeq 2M\log (R-2M)+O(1)$,Therefore she never receives a signal showing the surface cross the horizon at any finite asymptotic time. Radiation emitted ever closer to crossing arrives ever later and is increasingly redshifted and diluted; photons emitted at or inside the event horizon cannot reach her. The formal crossing time at infinity is infinite, not the finite boundary stopwatch reading. Nor can the earlier global birth of the horizon be observed directly as a local flash from its central birth event.

On the sphere, both cosmic time and the comoving coordinate $r$ are constant. The positive spatial induced metric obtained from the FLRW metric is therefore $d{\ell }^2=a^2{r}^2(d{\theta }^2+{\sin }^2\theta d{\varphi }^2)$. Its area element is the square root of its determinant:Integrating over the spherical coordinates givesThus $ar$ is the areal radius. In a curved spatial slice it need not equal the proper radial distance $a{\int }_0^r(1-k{r}^{\prime 2}{)}^{-1/2}d{r}^{\prime }$; the angular part of the FLRW metric, rather than a Euclidean radial-distance assumption, establishes the surface area.

Work in geometrized units $G=c=1$, with the stationary Killing vector field $K={\partial }_t$ normalized by $K^2\to -1$ at infinity. Put $q=2m{\sinh }^2\alpha$, $f=F/\sqrt{H}$, and $R^2=r^2\sqrt{H}$; $R$ is the areal radius. The static spherical Komar mass in reciprocal radial gauge givesIndeed the future and outward unit normals to a round sphere are $n=f^{-1/2}{\partial }_t$ and $s=f^{1/2}{\partial }_r$, and $s^a{n}^b{\nabla }_a{K}_b=-f^{\prime }/2$. The Komar mass integral $-(4\pi {)}^{-1}\int s^a{n}^b{\nabla }_a{K}_{b}dA$ is therefore $R^2{f}^{\prime }/2$. Since $F^{\prime }=2m/r^2$ and $H^{\prime }=-q/r^2$,Taking the asymptotic limit yieldsThis agrees with $g_{tt}=-1+(2m+q/2)/r+O(r^{-2})$. The finite-radius Komar integral need not equal its asymptotic value, because this solution is not vacuum. Restoring $G$ divides the mass expression by $G$ when $m$ has dimensions of length.

A future trapped surface is a smooth compact spacelike two-surface without boundary whose two future-directed orthogonal null expansions are strictly negative. In four-dimensional general relativity, the Penrose singularity theorem states: a time-oriented globally hyperbolic spacetime with a noncompact Cauchy hypersurface, a trapped surface, and the null convergence condition $R_{ab}{k}^a{k}^b\ge 0$ for every null vector is future null-geodesically incomplete. With the Einstein field equations, the null energy condition implies this null convergence condition; a cosmological constant drops out of the null contraction.

The Kruskal spacetime is an example. In its black hole interior, use future null normals in ingoing Eddington-Finkelstein coordinates:For a round sphere of areal radius $r$, its area is $4\pi r^2$ and its null expansions areBoth are negative for $0<r<2M$. The vacuum Einstein field equations give $R_{ab}=0$, and a two-ended Kruskal Cauchy hypersurface is noncompact. The future radial null geodesics reaching $r=0$ in finite affine parameter provide precisely the incompleteness predicted by the Penrose singularity theorem. A trapped surface at $r<2M$ is strictly trapped; the horizon sphere has one zero null expansion instead.

Take overdots in this question to mean differentiation with respect to the orbiting particle's proper time $\tau$. Write $\mu =GM/c^2$, $x=r/\mu$ and $\ell =hc/(GM)$, with $h$ the specific angular momentum. Then the dimensionless timelike effective potential isThe last term is the relativistic correction to the Newtonian centrifugal-plus-gravitational potential. At large $r$ the two agree. Newtonian angular momentum gives an infinite inner centrifugal barrier for nonzero $h$; the relativistic correction lowers that barrier and permits plunge. The exterior problem starts at $r=2\mu$, where $U=-c^2/2$; extrapolating the potential towards $r=0$ is not a static exterior description.

Circular orbits obeyThere are two extrema when $h^2>12(GM/c{)}^2$: an inner maximum representing an unstable circular orbit and an outer minimum representing a stable one. At equality they merge at $r=6\mu$; below it neither exists. More explicitly, evaluate the curvature at a circular orbit:Timelike circular orbits require $r>3\mu$. They are radially stable for $r>6\mu$, so the marginally stable inner limit is $r_{\mathrm{ISCO}}=6GM/c^2$. The Newtonian potential instead has one stable minimum at $r=h^2/(GM)$, without an innermost stable circular orbit.

For the Schwarzschild circular-orbit local speed in the question, $\gamma$ is the Lorentz factor relative to a local Schwarzschild static observer, not $dt/d\tau$. That observer's clock obeys $dT=\sqrt{1-2\mu /r}dt$. Since the particle passes with local speed $v$, $rd\varphi /d\tau =\gamma v$. Circular normalization givesThus $v=c/2$ at the innermost stable circular orbit. Using Schwarzschild time in place of the local clock would give a different speed.

For the conserved particle energy measured at infinity, substitute $h^2$ in the circular energy equation. Setting $y=\mu /r$ givesTake the positive root for a future-directed exterior orbit. It approaches $m{c}^2-GMm/(2r)$ at large radius. The marginally bound circular Schwarzschild orbit solves $k=1$, giving $y(4y-1)=0$ and hence $r_{\mathrm{mb}}=4GM/c^2$ at finite radius. It is unstable; strictly bound circular orbits have $r>4\mu$, with $E<m{c}^2$. The phrase smallest bound orbit here denotes this marginal circular limit, not a prohibition on a bound-energy trajectory plunging below it.

At the innermost stable circular orbit,A radiatively efficient thin disc with a zero-torque inner boundary condition can therefore release about 5.7% of the accreted rest energy before the plunge. energy advection, stress at the inner edge and photon capture can alter the escaping luminosity, so this is a model efficiency rather than a universal emitted fraction.

For the Kerr ISCO spin limits of an equatorial prograde disc around a rapidly rotating Kerr black hole, the innermost stable circular orbit moves inward and its specific energy decreases. In the ideal extremal limit its Boyer-Lindquist radius tends to $\mu$, $E/(m{c}^2)\to 1/\sqrt{3}$, and the corresponding binding-energy efficiency tends to 42.3%. Retrograde rotation has the opposite trend: the extremal limit is $r_{\mathrm{ISCO}}=9\mu$, $E/(m{c}^2)=5/(3\sqrt{3})$, giving about 3.8%. The prograde limiting coordinate radius does not identify a massive orbit with the null horizon generator.^[ref]

Finally $\dot{t}=k/(1-2\mu /r)=1/\sqrt{1-3\mu /r}$. Dividing $\dot{\varphi }=h/r^2$ by this givesIts algebraic form is exactly Newtonian, but $r$ is the Schwarzschild spacetime areal radius and $t$ Schwarzschild time, normalized at infinity. Local clock rates, orbit stability and binding energy remain relativistic. With $G{M}_{\odot }\simeq 1.3271\times 10^{20}{\mathrm{m}}^3{\mathrm{s}}^{-2}$,The marginally bound circular-orbit frequency is about $0.404\mathrm{kHz}$. These Schwarzschild ISCO orbital frequencies and marginally bound frequencies are rotation frequencies per distant stationary observer's time, as intended by using $d\varphi /dt$. A static observer at finite radius $r_o$ instead divides both frequencies by $\sqrt{1-2\mu /r_o}$, leaving their ratio unchanged. Identifying an observed quasi-periodic oscillation with the innermost stable circular orbit orbital frequency is an additional astrophysical assumption.

The Einstein tensor and its contractions areTo match the supplied Ricci tensor components, use the Riemann curvature tensor conventionThis convention also fixes the sign used for geodesic deviation above. In vacuum, tracing the Einstein field equations gives $-R+4\Lambda =0$. Substitution therefore yields $R_{\mu \nu }=\Lambda g_{\mu \nu }$.