Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 52 / 1

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.

Start with the four-dimensional Minkowski metric $d{s}^2=-d{t}^2+d{r}^2+r^{2}d{\Omega }^2$, where $r\ge 0$. Choose an arbitrary length $L>0$, and use the retarded and advanced null coordinates $u=t-r$, $v=t+r$. For the Minkowski conformal compactification, setBoth $p,q$ lie between $-\pi /2$ and $\pi /2$. Since $v\ge u$, we have $R\ge 0$; the remaining inequalities are $|T|+R<\pi$. Moreover,Multiply by the square of the conformal factor ${\Omega }_c=2\cos p\cos q/L$. The resulting metric is regular on the appropriate boundary pieces and preserves the null directions. Suppressing the angular two-spheres gives a triangular Penrose diagram with radial null geodesics at $45$ degrees.

The line $R=0$ is the ordinary timelike centre $r=0$. The upper sloping edge $T+R=\pi$ is future null infinity, reached with $v\to +\infty$ and finite $u$; the lower sloping edge $T-R=-\pi$ is past null infinity, reached with $u\to -\infty$ and finite $v$. The vertices $(T,R)=(\pi ,0)$ and $(-\pi ,0)$ are future and past timelike infinity, denoted $i^{+}$ and $i^{-}$. The vertex $(0,\pi )$ is spacelike infinity, $i^0$. These are limiting endpoints in the conformal completion, rather than ordinary physical events. In particular, finite diagram coordinates at null infinity do not imply finite physical affine parameter.

The four-dimensional radial diagram is the triangle $R\ge 0$, $|T|+R\le \pi$. If one instead draws two-dimensional Minkowski spacetime with a signed Cartesian spatial coordinate, the diagram is the full diamond. The centre is a boundary of the radial quotient, not a boundary of the physical four-dimensional Minkowski spacetime.

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.