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A geometrized luminosity is dimensionless because in geometrized units an energy and a time have the same dimension. Starting from SI luminosity, multiply by $G/c^5$. The corresponding SI power unit is $c^5/G$. Radiating a fraction $f$ of a mass $M$ at constant SI luminosity $L$ takes $fM{c}^2/L$, or $f(GM/c^3)/L_{\mathrm{geom}}$ in seconds.

Geometrized units set $G=c=1$, allowing mass, length, and time to be expressed in common units. Restore $GM/c^2$ for a mass parameter in a radial metric tensor.

For two equal point masses at $\pm z$ falling from rest at $\pm z_0$ under Newtonian gravity, in geometrized units one has $\ddot{z}=-m/(4z^2)$ and $\dot{z}=-\sqrt{m(1/z-1/z_0)/2}$. The only nonzero component of the second mass moment tensor is $I_{zz}=2m{z}^2$, and its third derivative is $-m^2\dot{z}/z^2$. The trace-free mass quadrupole moment and quadrupole formula give the displayed power. From infinity, integrating down to $z=z_c$ gives $E=2\sqrt{2}{m}^{9/2}/(105z_c^{7/2})$. Relative to the total initial mass $2m$ the fraction is $\sqrt{2}(m/z_c{)}^{7/2}/105$. The weak-field slow-motion approximation must be distinguished from its extrapolation to a compact endpoint.

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.

In geometrized units, a mass is converted to a length by multiplying its SI value by $G/c^2$, or to a time by multiplying by $G/c^3$. Using the supplied constants, the solar mass becomesThus the characteristic scales are kilometres and a few microseconds. With potential zero at infinity, the Newtonian gravitational potential at the surface, in units of $c^2$, isIts very small magnitude is the relevant weak-field measure. In SI potential units the same number corresponds to about $-1.91\times 10^{11}{\mathrm{m}}^2{\mathrm{s}}^{-2}$.

Power is energy per time. In geometrized units, energy has the dimension of length and time is converted to length using $c$, so the dimensionless geometrized luminosity isEquivalently, $c^5/G$ is the corresponding SI power unit. The lifetime for emitting the entire rest energy at the given constant luminosity isThe same result follows from $M_{\odot }^{(\mathrm{time})}/L_{\odot ,\mathrm{geom}}$. The stipulated full-mass radiation time is of order $10^{13}$ years; this is the constant-luminosity energy budget asked for, not a stellar-evolution calculation.

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 physical argument for the Penrose inequality combines weak cosmic censorship conjecture, the dominant energy condition, and relaxation to a stationary black hole. Work in geometrized units. Let $M_f$ and $A_f$ be the final Kerr black hole mass and horizon area. Positive energy radiated to infinity gives $E_{\mathrm{ADM}}\ge M_f$, where $E_{\mathrm{ADM}}$ is the initial ADM energy. For a Kerr black hole with $a_f=J_f/M_f$,If the initial apparent horizon obeys the necessary apparent-horizon area comparison with the enclosing event horizon, and Hawking's area theorem applies during the evolution, thenConsequently the anticipated answer, under those additional hypotheses, isThe bound is saturated by a nonrotating Schwarzschild black hole with no energy loss. Rotation or outgoing radiation makes the argument's inequalities stricter.

There is an essential qualification: inclusion inside an event horizon does not by itself compare areas. An arbitrary apparent horizon on general, non-time-symmetric initial data need not satisfy the displayed apparent-horizon area comparison; the unqualified version with its area is not universally true, even with the dominant energy condition. On time-symmetric data the relevant outermost minimal surface is an outer area-minimizing surface, as used in the Riemannian Penrose inequality, with nonnegative scalar curvature. In more general formulations an appropriate enclosing-area quantity is needed. The physical expectation is conditional on this area comparison, as well as on censorship, predictability, settling, and the energy assumptions; the mere presence of a trapped surface does not supply every step.

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.

The Penrose inequality compares asymptotic energy and an appropriate horizon or enclosing area in geometrized units. Its physical motivation uses censorship, positive radiated energy, settling to a Kerr black hole, and Hawking's area theorem. The area variable requires care: the area of an arbitrary apparent horizon on non-time-symmetric data does not give a universally valid inequality. The Riemannian Penrose inequality uses the relevant outermost minimal surface under nonnegative scalar curvature.

For a Schwarzschild black hole of mass $M$ in geometrized units, $f=1-2M/r$ gives $\kappa =f\prime (2M)/2=1/(4M)$. The Killing vector field is normalized at infinity; changing that normalization changes $\kappa$.