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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 $\sqrt{1-2M/r_B}$.

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 $-1$.

Capture from infinity. To reach the event horizon from infinity, the radial square must remain nonnegative throughout $r>2M$. Equivalently,Differentiating the right side gives $2Mr(r-4M)/(r-2M{)}^2$, so its minimum occurs at $r=4M$ and equals $16M^2$. Thusis the necessary capture bound. At equality the radial numerator is $2M(r-4M{)}^2$. An inward particle arriving from larger radii approaches the unstable orbit $r=4M$ only after infinite proper time, because $\dot{r}$ is proportional to $-(r-4M)$ near that orbit. Actual plunges from infinity require $|h|<4M$. This is Schwarzschild marginally bound capture.

The origin-at-infinity hypothesis is important and is not explicit in the PDF. A particle already inside the angular-momentum barrier can plunge with larger $|h|$. For example, $h=10M$ and initial radius $r_0=2.01M$ give positive radial numerator $7.0802M^3$, remaining positive as $r$ decreases to $2M$. This trajectory has $e=1$ and reaches the event horizon. Thus an unrestricted claim about every inward particle would be false; the bound is the intended capture-from-infinity statement.

Invariant collision energy. At a collision, the total four-momentum is $P^{\mu }=m(u_1^{\mu }+u_2^{\mu })$. The invariant center-of-mass energy uses the covariant metric:The PDF instead prints a raised metric multiplying raised velocities. That contraction is not a tensor scalar; the corrected expression above, or a raised metric with lowered momenta, is required. Since each four-velocity has norm $-1$,Let $D_i=2M{r}^2-h_i^2(r-2M)$. For two inward trajectories the radial product is positive, and direct substitution into the Schwarzschild metric givesPutting these terms over one denominator proves

Horizon limit and the upper bound. A cancellation-free way to take the limit is to write $a_i=1+h_i^2/r^2$ and ${\dot{r}}_i=-\sqrt{1-f{a}_i}$. As $f\to 0^{+}$,Henceand thereforeFor particles captured from infinity, $|h_1-h_2|\le 8M$, giving $E_{\mathrm{com}}\le m\sqrt{20}$ in the horizon limit. For actual captured trajectories the inequality is strict, but the supremum is approached by $h_1\to 4M^{-}$ and $h_2\to -4M^{+}$. Their azimuthal starting positions can be chosen so that the trajectories meet. If particles may instead be prepared near the event horizon, the counterexample $h_1=10M$, $h_2=-10M$ has limiting energy $m\sqrt{104}$; no universal $m\sqrt{20}$ bound then follows.

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.

The radial null geodesics satisfyThe ingoing family has $v=\text{constant}$, with $r$ decreasing toward the future; its tangent $-{\partial }_r$ is affinely parametrized because ${\Gamma }^a{}_{rr}=0$. The outgoing family hasAway from the Schwarzschild event horizon, integration gives $v=2r_{\ast }+\text{constant}$. These are null geodesics up to reparametrization: a null direction in the two-dimensional radial geometry is automatically pregeodesic, and the angular geodesic equations are satisfied by constant angles. On the Schwarzschild event horizon the outgoing family instead has $r=2M$, with tangent proportional to ${\partial }_v$.

In a Finkelstein diagram, plot $T=v-r$ vertically and $r$ horizontally. Ingoing rays obey $T+r=\text{constant}$; outgoing rays obeyOutside the Schwarzschild event horizon, outgoing rays increase $r$; on it they remain at $r=2M$; inside it they decrease $r$ even though $v$ increases. Thus both radial future null directions point toward smaller $r$ inside the black hole. Every future timelike direction lies between these null directions, so it also moves toward smaller $r$ there.

Arrows show future propagation. The vertical red line is the Schwarzschild event horizon; the black boundary at $r=0$ is the Schwarzschild singularity.

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

For radial free fall, let $E=f{\dot{t}}_A>0$ be Alice's conserved Killing energy. The inward branch of the timelike geodesic has $\dot{r}=-\sqrt{E^2-f}$, henceWith $\delta =r-2M$, the near-horizon geodesic equations giveThe constants in the logarithms are understood to make their arguments dimensionless. Therefore $u_A=t_A-r_{\ast }=2t_A+O(1)$. For an outgoing signal reaching the fixed-radius Bob, $t_B=u_A+r_{\ast }(r_B)$, so $t_B=2t_A+O(1)$. The redshift consequently behaves asThis is the answer when $t$ is Bob's Schwarzschild time, as appropriate to the stated observation. It is the Schwarzschild surface gravity $a=\kappa =1/(4M)$. If $t$ instead means the emission Schwarzschild time $t_A$, the same redshift is proportional to $e^{t_A/(2M)}$ and gives $M=1/(2a)$. The two answers use different clocks, not different dynamics. If the measured exponential uses Bob's proper time ${\tau }_B$, its rate is $1/(4M\sqrt{f_B})$. In SI units the reception-time result is $M=c^3/(4Ga)$.

The reception-time exponent also holds for smooth radial infall crossing the future Schwarzschild event horizon with finite nonzero $dr/d\tau$: regular Ingoing Eddington-Finkelstein coordinates give finite $v$ there, $u=v-2r_{\ast }\sim -4M\log ({\tau }_H-\tau )$. Thus no special value of $E$ is needed.

Let $F=1-2M/r$, define the tortoise coordinate by $d{r}_{\ast }/dr=F^{-1}$, and set the advanced time $v=t+r_{\ast }$. Since $dt=dv-F^{-1}dr$, substitution givesThese are Ingoing Eddington-Finkelstein coordinates; all coefficients are regular at the event horizon $r=2M$.

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.