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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.

The inward radial geodesic congruence with unit Killing energy has $U=f^{-1}{\partial }_t-\sqrt{2M/r}{\partial }_r$, where $f=1-2M/r$. Its metric dual is $U^{♭}=-d\tau$, so $\tau$ is a synchronized proper time for the congruence. Locally $\tau =t+2\sqrt{2Mr}+2M\log |(\sqrt{r}-\sqrt{2M})/(\sqrt{r}+\sqrt{2M})|$. This yields Painlevé–Gullstrand coordinates with flat spatial slices and a regular future horizon.

Each Killing vector field supplies a conserved quantity for freely moving test particles: it turns a geometric symmetry into a first integral of the geodesic equation. Let $\gamma (\lambda )$ be a geodesic with affine parameter $\lambda$, tangent vector $U^{\mu }=d{x}^{\mu }/d\lambda$, and ${\nabla }_{U}U=0$. Wherever a Killing vector field $K$ is defined along the geodesic,The statement applies to Riemannian manifolds and to timelike geodesics, null geodesics and spacelike geodesics in a Lorentzian manifold. It requires an affine parameter; for a massive test particle, proper time is one choice. See Tong, section 4.3.2, equation (4.32).

For a worked example in the Euclidean plane, take $\gamma (t)=(1+t,2-t)$ with affine parameter $t$ and velocity $U=(1,-1)$. The Euclidean Killing vector fields for translation and rotation giveThe rotational field along the geodesic is $K_{\mathrm{rot}}(\gamma (t))=(t-2,1+t)$: it changes from $(-2,1)$ at $t=0$ to $(-1,2)$ at $t=1$, yet its pairing with $U$ stays $-3$. With toy mass $m=1$ and $t$ interpreted as time, the first constant is the $x$ component of linear momentum, and the second is the signed $z$ component of angular momentum about the origin. Their numerical values use toy units. The geodesic is straight; the rotation flow orbits are circles.

This conservation law follows from metric compatibility and the Killing equation:The geodesic equation kills the acceleration term, and the symmetric product $U^{\mu }{U}^{\nu }$ selects the vanishing symmetric derivative. This tensor contraction is the whole proof outline; see Carroll, equation (5.43).

In a stationary spacetime, a chosen time-translation Killing vector field $K$ gives Killing energy. With metric signature $(-+++)$, $c=1$, and a massive test particle's four-momentum $p^{\mu }=m{U}^{\mu }$, the physical normalization isHere $\Phi$ generates the axial rotational symmetry when one exists. In coordinates with $K={\partial }_t$ and $\Phi ={\partial }_{\varphi }$, these are $E=-p_t$ and $L_z=p_{\varphi }$. Both are conserved quantities along the geodesic, even if the metric has a time-angle cross term. A future timelike $K$ makes $E$ positive for a future massive particle; the scale of $E$ depends on the normalization of $K$. Not every spacetime has a global timelike Killing vector field, and a stationary field can become spacelike in an ergoregion of a stationary spacetime. Conservation on the field's domain does not imply a universal global positive energy; see Tong, sections 4.3.1-4.3.2 on the timelike-domain qualification and conserved charges.

In a static spacetime with $g_{tt}=-N^2$, a photon has conserved Killing energy $E=-p_t$. A static observer with four-velocity $N^{-1}{\partial }_t$ measures frequency proportional to $E/N$. Hence light emitted at one lapse and received at another obeys the displayed ratio. For $N=1+\Phi /c^2+O({\Phi }^2/c^4)$ it gives the weak-field shift $({\nu }_{\mathrm{r}}-{\nu }_{\mathrm{e}})/{\nu }_{\mathrm{e}}=({\Phi }_{\mathrm{e}}-{\Phi }_{\mathrm{r}})/c^2$. This local clock effect does not require light-ray bending.

An upward photon has lower received frequency when compared with the receiver's local clock: ${\Phi }_{\mathrm{r}}>{\Phi }_{\mathrm{e}}$ makes the shift negative. The corresponding upper clock runs faster in a comparison using the same static time coordinate. These are reciprocal descriptions of gravitational redshift and gravitational time dilation, not opposite predictions. A locally measured speed of light remains $c$.

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.

For a geodesic tangent $u$ and a stationary Killing vector field $K$, the Killing energy is $E=-g(K,u)$. It is conserved along the geodesic. When $K$ is future timelike and $u$ is future causal, $E>0$ for a nonzero tangent; it need not be positive where $K$ becomes spacelike.

For a radial timelike geodesic of specific Killing energy $k=fdt/d\tau$, the static observer measures $v_{\mathrm{loc}}=(dr/dt)/f=\dot{r}/k$. Its relative Lorentz factor is $k/\sqrt{f}$, giving $v_{\mathrm{loc}}^2=c^2(1-f/k^2)$.

Constants and radial motion. Write $m=GM/c^2$ and $f(r)=1-2m/r$. A Schwarzschild black hole has spherical symmetry, so its conserved angular momentum fixes an orbital plane through the centre. A rotation of the angular coordinates puts any nonradial orbit in the equatorial plane. A radial orbit can also be placed in such a plane. This is a choice of coordinates, not a restriction on the possible orbital inclinations.

The quadratic geodesic Lagrangian, with a dot denoting differentiation with respect to the satellite's proper time, isThe last equality is four-velocity normalization, imposed after deriving the Euler-Lagrange equations. The cyclic coordinates $t,\varphi$ give independent conserved specific Killing energy and angular momentum:Here $e$ is the conserved energy per unit rest energy and $\ell$ has dimensions of angular momentum per unit mass. Substituting them in the normalization condition givesThe two signs of $\dot{r}$ distinguish outgoing and ingoing branches; the expression must be nonnegative on the allowed radial interval.

For a nonradial orbit, put $u=1/r$. Since $\dot{r}=-\ell du/d\varphi$, the radial equation becomesDifferentiating and extending continuously through isolated radial turning points givesA circular solution can equivalently be obtained directly from the radial geodesic equation, avoiding division by $du/d\varphi =0$ everywhere on that special orbit.

Circular launch and the station's clock. For $u=1/R$ the orbit equation givesThe static astronaut measures circumferential proper length $Rd\varphi$ and station proper time $d{\tau }_s=\sqrt{f(R)}dt$. Thus the Schwarzschild circular-orbit local speed and return time areThis is a local speed measured with the astronaut's clock, not the satellite's proper velocity $\gamma v$. After one azimuthal revolution the circular satellite meets the fixed station again. For comparison, the satellite's own elapsed proper time would instead be $2\pi \sqrt{R^3/(GM)}\sqrt{1-3m/R}$.

A static station requires $R>2m$; a timelike circular satellite requires $R>3m$, since the local circular speed must be below $c$. The circular orbit is stable only for $R>6m$, marginal at $6m$, and unstable for $3m<R<6m$. Thus a request for a robust near-circular trajectory implicitly needs more than merely placing the station outside the horizon.

Slightly changing the launch speed. In the stable, bound regime, a small tangential increase makes the launch point a pericentre; a small decrease makes it an apocentre, provided the perturbed orbit remains outside its plunge separatrix. In either case the orbit is a precessing rosette, not an exactly closed Newtonian ellipse. For an explicit first-order description write $A=A_c+\delta A$ and $u=1/R+\delta u$. ThenThe tangential launch gives $\delta u(0)=\delta u^{\prime }(0)=0$. At the station, $\ell =R\gamma v$, so $\delta \ell /{\ell }_c={\gamma }_c^2\Delta v/v_{\mathrm{circ}}$ and $\delta A=-2A_c{\gamma }_c^2\Delta v/v_{\mathrm{circ}}$. ThereforeThis perturbative expression also requires the resulting radial excursion to be small compared with $R$; close to the innermost stable orbit, $|\Delta v|\ll v_{\mathrm{circ}}$ alone does not ensure that.

The Schwarzschild launch-orbit recurrence needs care when deciding whether a collision is inevitable. The satellite returns to radius $R$ once per radial cycle, after an azimuthal advance approximately ${\Phi }_r=2\pi /{\kappa }_{\varphi }$. It returns to the station's exact point only if an integer number of radial cycles also gives an integer number of azimuthal revolutions. Rational ${\Phi }_r/(2\pi )$ gives an exact return; an irrational ratio gives arbitrarily close approaches but no exact post-launch return for ideal point objects. At $R=12m$, for example, the linearized ratio is $\sqrt{2}$, illustrating why precession alone does not prove an exact collision. For generic bound orbits the dense recurrence eventually threatens any station and satellite of nonzero size; a resonant closed rosette returns exactly. A small speed adjustment is therefore not a robust permanent collision-avoidance manoeuvre for a bound satellite.

There is no unconditional collision conclusion without specifying the regime. An unstable circular orbit cannot support the small bounded oscillations used above. In particular, near $R=3.5m$ the circular specific energy is already $e_c>1$; a small positive speed increment can send the satellite outward on an unbound escape orbit, with no subsequent return to $R$. Such an escape is a different way to avoid a future encounter. Radiation reaction and manoeuvres have been excluded throughout the ideal test-particle calculation.

The time-translation Killing vector field of the Schwarzschild metric gives a conserved Killing energy per unit rest mass,For unit rest mass, this is the total energy relative to infinity, including rest energy. At release $dr/d\tau =0$, so the four-velocity normalization gives $dt/d\tau =f(r_0{)}^{-1/2}$ andSubstitution into the radial four-velocity normalization yieldsChoosing the inward square root and evaluating at the ground provesThis identity is exact within the assumed Schwarzschild metric, despite its familiar Newtonian mechanics form. The quantity $dr/d\tau$ is not exactly the locally measured speed. A Schwarzschild static observer measures $v_{\mathrm{loc}}=(dr/dt)/f=(dr/d\tau )/\sqrt{f(r_0)}$. Thus the distinction disappears in the stated weak-field approximation, giving $v_{\mathrm{loc}}^2\simeq 2gh$ for $h\ll R_E$.

The inverse of the $(v,\lambda )$ metric tensor block isThe normal covector $d\lambda$ has squared norm ${\lambda }^2/m^2$. ThereforeMore explicitly, the induced metric tensor is $-{\lambda }_0^{2}d{v}^2+m^{2}d{\Omega }_2^2$; at zero its sole zero direction is ${\partial }_v$. The coefficients are independent of $v$, so $K={\partial }_v$ is a Killing vector field. On $\lambda =0$,Hence $K$ is both tangent and normal to this null hypersurface, proving it is a Killing horizon. The definition of surface gravity is ${\nabla }_{K}K=\kappa K$ on the Killing horizon, or equivalently $d(K^2)=-2\kappa K^{♭}$. Since $d(K^2)=-2\lambda d\lambda$ vanishes there while $K^{♭}\ne 0$,Thus this is a degenerate Killing horizon, and $v$ is an affine parameter on each of its null geodesics.

For a concrete horizon frequency comparison, fix the same angular position for a narrow high-frequency light beam and two freely falling radial observers. Work in geometric optics, with the beam's affinely transported wavevector tangent to the Killing horizon, $k=qK$. Here $q>0$ is constant because ${\nabla }_{K}K=0$. Choose the observers as translates of one timelike geodesic by the isometry generated by $K$, crossing at $v_1$ and $v_2$. Give them the same positive Killing energy per unit rest mass, $E=-K\cdot W$. At their crossings their normalized four-velocities havewhich satisfy $g(W,W)=2m{W}^v{W}^{\lambda }=-1$. They can be realized by local geodesics with precisely these initial data. The measured angular frequency isTo connect this measurement to surface gravity, if ${\nabla }_{K}K=\kappa K$ with constant $\kappa$, affine transport of $k=q(v)K$ would give $dq/dv=-\kappa q$. For observers related by the same Killing vector field translation, $-K\cdot W$ agrees at the crossings, giving ${\omega }_2/{\omega }_1=e^{-\kappa (v_2-v_1)}$. Here there is no accumulated redshift along the Killing horizon for these equally prepared observers. Arbitrary observers with different four-velocities can still measure different frequencies through their different Killing energies.

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 Planck units for the following formulas. The black-hole area theorem is a classical statement with hypotheses. Assume the Einstein field equations, the null energy condition and appropriate global future regularity/predictability. For a clean version of the focusing sketch, take the future horizon generators to remain on a regular horizon and to be future complete in affine parameter. The usual strong asymptotic predictability formulation supplies the global causal control needed to exclude the same focusing pathology; an arbitrary spacetime with naked future breakdown is not covered by the theorem.

Let ${\ell }^a$ be an affinely parametrized generator of a smooth portion of the event horizon, and let $\theta =d\log dA/d\lambda$ be its null expansion. The horizon is a null hypersurface, so its generators have zero null twist. In four dimensions the Null Raychaudhuri equation isThe Einstein field equations give $R_{ab}{\ell }^a{\ell }^b=8\pi T_{ab}{\ell }^a{\ell }^b\ge 0$, since the metric terms vanish when contracted with a null vector. The screen-space null shear term is also nonnegative, so ${\theta }^{\prime }\le -{\theta }^2/2$. If ${\theta }_0<0$ at ${\lambda }_0$, integrating this inequality while the congruence remains regular givesIt focuses within affine distance at most $2/|{\theta }_0|$. The vanishing area element is a focal point to an earlier spacelike horizon cut. Past such a point the null geodesic can no longer generate an achronal boundary, but the event horizon is precisely such a boundary. Future completeness and the global hypotheses exclude its escaping this contradiction by ending prematurely at a pathology. Therefore $\theta \ge 0$ everywhere the smooth horizon description applies.

Integrate $d(dA)/d\lambda =\theta dA$ between ordered horizon cuts. Every continuing generator contributes a nondecreasing area element. New generators can enter at past endpoints or merger crease sets, adding area; generators cannot leave a regular future horizon through future endpoints under these hypotheses. Thus, including the total area of disconnected components,This is the future-complete horizon focusing proof of the area theorem. A smooth elementary congruence proof requires the stated regularity; the general theorem also treats the horizon's nonsmooth joining set rather than assuming mergers have a globally smooth horizon.

Hawking radiation changes the situation because the quantum field state in a collapsing geometry is not a classical positive-energy fluid. For a large isolated Schwarzschild black hole,in natural units. Outgoing quantum modes at future null infinity have a nearly thermal occupation, modified by greybody factors; an accompanying negative Killing energy flux near the horizon reduces the hole's mass. On timescales short compared with the evaporation time but long compared with $M$, its geometry is approximately Schwarzschild spacetime with slowly decreasing $M(t)$. It is not an exactly stationary vacuum solution with a parameter changed by hand: the radiation's stress-energy tensor sources the time dependence through the semiclassical Einstein equation. The distant luminosity is carried by the outgoing radiation, while the horizon-area decrease accompanies the inward negative-energy contribution.

As the black hole loses mass, its horizon scale and Bekenstein-Hawking entropy shrink, its temperature rises and its luminosity increases. This is the negative heat capacity of a Schwarzschild black hole. Large astrophysical holes evaporate extremely slowly in isolation; small holes evaporate more rapidly, and increasingly massive particle species become accessible as the temperature rises. An incoming radiation bath or accretion can instead offset the loss, so the pure evaporation law assumes negligible incoming energy. A formal complete evaporation also raises the black hole information paradox; the low-curvature calculation alone does not determine how information or the final state is resolved.

For an elementary luminosity estimate, treat the horizon as emitting photon blackbody radiation. In natural units the Stefan–Boltzmann law has $\sigma ={\pi }^2/60$, givingThe horizon area used here is an estimate of emitting area, not the exact frequency-dependent absorption cross-section. More generally write $P=\alpha /M^2$, where $\alpha >0$ includes the greybody factors and the active particle species. For constant effective $\alpha$, energy balance and direct integration giveHere $t$ is time measured at infinity. The formal evaporation timescale is $M_0^3/(3\alpha )$; with the photon blackbody estimate it is $5120\pi M_0^3$. Restoring constants givesfor that specified estimate. If particle thresholds make $\alpha$ mass dependent, the precise leading law is instead $t={\int }_{M(t)}^{M_0}{m}^{2}dm/\alpha (m)$. The robust result is the inverse-square mass-loss rate for a fixed species regime and the cubic mass scaling of its lifetime. This is the semiclassical cubic mass law for Schwarzschild evaporation. Extrapolating its zero to a definite end state is unjustified once $M$ approaches the Planck mass and curvature requires quantum gravity.

There is no contradiction with the classical area theorem: the renormalized quantum stress-energy tensor can violate the null energy condition used in the focusing inequality. In the evaporation regime,which is permitted when that hypothesis fails. The thermodynamic replacement is the generalized second law, concerning $S_{\mathrm{outside}}+A/(4G\hslash )$ rather than horizon area alone. Outside radiation can carry entropy while the hole's Bekenstein-Hawking entropy decreases.

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.

Stationarity gives the conserved Killing energy per unit rest mass $E=(1+x^2)\dot{t}$, where dots here denote proper-time derivatives. It measures energy relative to the time-translation Killing field; a static local observer measures $E/\sqrt{1+x^2}$. Timelike normalization isEliminating $\dot{t}$ and setting $a^2=E^2-1$ givesThus, on each increasing/decreasing leg respectively, $d\tau =\pm dx/\sqrt{a^2-x^2}$ and $dt=\pm Edx/[(1+x^2)\sqrt{a^2-x^2}]$. The printed positive differentials describe the increasing leg, not both legs at once. Differentiation along a leg gives $\ddot{x}=-x$, extending smoothly through the turning points. Hence $x=a\sin (\tau -{\tau }_0)$ and the proper-time period is $2\pi$.

The requested velocity sketches are the branches $\dot{x}=\pm \sqrt{a^2-x^2}$ and $dx/dt=\pm (1+x^2)\sqrt{a^2-x^2}/E$. They share the turning points and opposite signs on the two legs; their speed ratio is $(1+x^2)/E$, smaller than one near zero and larger than one near the endpoints. The supplied integral is the coordinate time from $-a$ to $a$, equal to $\pi$. The return leg takes the same time, so the coordinate-time period is also $2\pi$.

For the perturbed metric the same Killing conservation gives $\dot{t}=E/(1+x^2)$, and normalization becomes $(1+f(x){)}^2{\dot{x}}^2=a^2-x^2$. Since $1+f>0$,Both weights without $f$ are even in $x$. Oddness of $f$ makes its integral over the symmetric excursion $[-a,a]$ vanish. Each increasing leg therefore still takes $\pi$ in either clock; the decreasing leg takes the same integrals with reversed orientation. Both periods remain exactly $2\pi$, although the proper-time waveform is no longer generally sinusoidal. This is the odd radial perturbations preserve timelike-geodesic periods mechanism. Smoothness of the metric and positivity of $1+f$ ensure regular turning and monotone coordinate time.

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.