Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 311 / 1

Use $G=c=1$, the Einstein field equations with zero cosmological constant, and metric signature $(-,+,+,+)$. Write $m=m_{a}d{x}^a$ for the differential one-form dual to the axial Killing vector field. To fix the sign of the volume formula, use the component Hodge star operator convention $(\star j{)}_{abc}={ϵ}_{abcd}{j}^d$ and $(\star F{)}_{ab}=\frac{1}{2}{ϵ}_{abcd}{F}^{cd}$. The Killing equation gives $(dm{)}_{ab}=2{\nabla }_a{m}_b$, and its contracted curvature identity givesTo see the curvature step, tracing the Killing equation gives ${\nabla }_b{m}^b=0$. The second covariant derivative of a Killing vector then gives ${\nabla }^b{\nabla }_b{m}_a=-R_{ab}{m}^b$, while commuting the covariant derivatives gives ${\nabla }^b{\nabla }_a{m}_b=R_{ab}{m}^b$. Subtracting these two terms is precisely ${\nabla }^b(dm{)}_{ab}$ above. The component Hodge star operator converts this divergence to $d\star dm$ with the displayed minus sign.

In a vacuum spacetime region $R_{ab}=0$, so $d\star dm=0$. If $S_1$ and $S_2$ are enclosing spacelike submanifolds in the same homology class bounding a vacuum three-dimensional region $W$, Stokes theorem yieldsThus the Komar angular momentum is independent of the enclosing vacuum spacelike two-manifold, provided the surfaces have the same orientation and enclose the same sources and inner boundaries. The vacuum region need not be stationary: an axial Killing vector field suffices.

For a regular filling hypersurface $\Sigma$ with $\partial \Sigma =S$, the Einstein field equations giveThe pullback of $\star m$ to $\Sigma$ vanishes because $m^a$ is tangent to $\Sigma$: the dual three-form measures the normal component, which is zero. ThereforeThis is the stress-energy current from a Killing vector integrated over the slice, with the signs fixed by the stated Hodge star operator convention.

There is a necessary boundary qualification omitted from the printed formulation. If the slice has inner boundaries $S_i$, orient them so $\partial \Sigma =S-{\bigcup }_i{S}_i$. The actual Komar angular momentum with inner boundaries identity isFor a vacuum Kerr black hole, $T_{ab}=0$ on the exterior slice but $J\ne 0$; its horizon supplies precisely the inner boundary contribution. Thus the matter-only formula is false for arbitrary exterior slices. It is valid when a nonsingular filling with no inner boundaries exists, or when all omitted inner charges vanish. The Komar angular momentum independence likewise concerns homologous surfaces in the same vacuum region, not an unrestricted comparison of differently enclosed objects.

A gravitational wave can propagate through a region with vanishing stress-energy tensor; its gravitational field is not a matter current $T_{ab}{m}^b$. Nevertheless, exact axisymmetry gives an axial Killing vector field, and the vacuum identity $d\star dm=0$ makes the Komar angular momentum equal on enclosing spacelike submanifolds in the same homology class. Applying the same identity to the spacetime tube between surrounding spacelike submanifolds at different times gives zero net flux of this charge through the tube.

Consequently gravitational waves in exact axisymmetry carry no net angular momentum about the symmetry axis. They can still carry energy and decrease the total mass. In a mode description, rotationally invariant radiative data have zero azimuthal mode number; exact axisymmetry therefore excludes the axial angular momentum flux present in nonaxisymmetric merger radiation. This does not mean there are no gravitational waves, or that radiation cannot transport angular momentum when the exact symmetry is absent. The relevant conserved charge includes any horizon contributions identified in the preceding Komar angular momentum with inner boundaries identity.

The common axial Killing vector field fixes the same sign convention for both spins. Initial rest and alignment of the separation with the symmetry axis give no orbital angular momentum. At large separation, the total axial charge is therefore $J_1+J_2$. Since axisymmetric radiation carries no axial angular momentum, all of this charge remains in the remnant after the radiation has escaped and the exterior has settled:Here the spins are signed quantities about the same axis, not their magnitudes. The conclusion assumes the exact axisymmetry stipulated throughout the merger and no remaining exterior matter carrying angular momentum; the final stationary vacuum remnant is a Kerr black hole under the usual uniqueness hypotheses.

Assume the classical black-hole area theorem applies: the null energy condition and the required global regularity or strong asymptotic predictability hypotheses hold. The disconnected initial horizon sections then give $A_f\ge A_1+A_2$. DefineFor any regular Kerr black hole, $|J_i|\le M_i^2$, so $X\ge |J_1|+|J_2|\ge |J_f|$. The irreducible mass identity obtained by inverting the Kerr area relation isIndeed, $Y=M_f^2+\sqrt{M_f^4-J_f^2}$ implies $(Y-M_f^2{)}^2=M_f^4-J_f^2$, which gives this expression. At fixed $J_f$, its derivative with respect to $Y$ is $\frac{1}{2}(1-J_f^2/Y^2)\ge 0$ on the physical branch. Therefore the area bound for an axisymmetric Kerr merger isThis uses the ideal limit of infinite initial separation, in which the initial total energy is $M_1+M_2$. At finite separation it is the initial total gravitational energy, including binding energy, that replaces this sum. The inequality is an upper limit, not a prediction or a guarantee of saturation; real mergers increase the horizon area.

Changing only a spin sign leaves $X$ unchanged, whereas the square of the final angular momentum changes. For fixed nonzero $|J_1|,|J_2|$, same-sign spins give $(|J_1|+|J_2|{)}^2$, and opposite signs give $(|J_1|-|J_2|{)}^2$. The remnant must retain more rotational energy in the same-sign case, so its minimum allowed mass is larger and the radiated-energy upper limit is smaller. Cancellation of opposite spins permits more of the initial rotational energy to be radiated while respecting the same initial total area.

For example, equal extremal initial holes have $M_1=M_2=M$ and $|J_1|=|J_2|=M^2$. Same-sign spins allow at most $(2-\sqrt{2})M$, a fraction $1-1/\sqrt{2}$ of the initial energy; opposite signs allow at most $M$, a fraction $1/2$. These are geometric bounds under exact axisymmetry, not achievable-efficiency claims.