Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 311 (Black Holes)

An isolated collapse emits transient matter and gravitational waves. The exterior is expected to settle to an asymptotically flat stationary spacetime. Because the star is uncharged, the black-hole uniqueness theorem identifies the regular final state with a Kerr black hole. Its higher multipole moments are fixed by its conserved charges: the black-hole no-hair theorem leaves only the total mass $M$ and angular momentum $J$. Thus the late-time spacetime is characterized by $M$ and $J$.

FactorIntroduce $v=t+r_{\ast }$ with $d{r}_{\ast }/dr=f^{-1}$. Since $dt=dv-dr/f$,These are the charged analogue of ingoing Eddington-Finkelstein coordinates. The $(v,r)$ block has determinant $-1$, so the metric and its inverse are nonsingular at $r=r_{+}$. For $M\ge Q$, $r_{+}$ is real and the metric extends analytically through it. The original potential becomes $A=-Qdv/r^2+Qdr/(r^{2}f)$; the radial term is locally exact, and a gauge transformation gives the horizon-regular potential $A=-Qdv/r^2$.

The stationary Killing vector field is $k={\partial }_t={\partial }_v$, with $k^2=-f$. The normal to $r=\text{constant}$ has squared norm $g^{rr}=f$, so $r=r_{+}$ is a null hypersurface; there $k$ is both tangent and normal. It is therefore a Killing horizon.

For a static metric of this form, the surface gravity is $\kappa =f^{\prime }(r_{+})/2$. Since $r_{+}^2{r}_{-}^2=Q^2$ and $r_{+}^2+r_{-}^2=2M$,For the extremal black hole $M=Q$, the horizons coincide and $\kappa =0$.

The zeros of $f$ give three causal structures.

Thus the cases are subextremal, extremal, and overextremal according as $M>Q$, $M=Q$, and $M<Q$.

For this charged scalar field, $\sqrt{-g}=r^3\sin \theta \cos \theta$. On $e^{-i\omega t}$, the gauge-covariant derivative of a charged scalar field supplies the shifted frequencyThe wave equation becomesSeparation of variables with constant $\lambda$ givesandThe angular equation is the axisymmetric spherical harmonic equation on the three-sphere; regular solutions have $\lambda =\ell (\ell +2)$ with $\ell =0,2,4,\dots$, the even values selected by the absence of $\varphi$ and $\psi$ dependence.

Take $n^a{n}_a=-1$. The first fundamental form, or induced metric, is the spatial projection tensorWith the convention used in the question, define the second fundamental form byLocally $n_a$ is proportional to the gradient of a defining function for the hypersurface, so $h_a{}^c{h}_b{}^d{\nabla }_{[c}{n}_{d]}=0$ and $K_{ab}=K_{ba}$. Both derivative indices are projected tangentially. If two extensions of $n$ agree on $\Sigma$, their difference vanishes there, as does its tangential derivative, so $K_{ab}$ is independent of the extension.

Let $D$ be the Levi-Civita connection of $h$. Apply $(D_c{D}_d-D_d{D}_c)$ to a tangent vector, replace each $D$ by the tangential projection of $\nabla$, and decompose $\nabla n$ using $K_{ab}$. Comparison with the intrinsic Riemann curvature tensor gives the Gauss equationChanging the sign convention for $K_{ab}$ does not alter these quadratic terms.

Contract the Gauss equation with $h^{ac}{h}^{bd}$. Since $h^{ab}=g^{ab}+n^a{n}^b$ and the antisymmetries of the Riemann curvature tensor eliminate the term containing four normals,The extrinsic terms contract to $K^2$ and $K_{ab}{K}^{ab}$. Hence the intrinsic Ricci scalar is

Contract the Einstein field equations $G_{ab}=8\pi T_{ab}$ twice with $n^a$. Since $n^a{n}_a=-1$ and $\rho =T_{ab}{n}^a{n}^b$,If $K_{ab}$ is proportional to $h_{ab}$ in three dimensions, tracing gives $K_{ab}=(K/3)h_{ab}$ and $K_{ab}{K}^{ab}=K^2/3$. The Scalar Gauss equation gives the Hamiltonian constraintTherefore

For any nonzero $g$-causal vector $v^a$,For timelike $v$, the first term is negative. For null $v$, a nonzero null vector cannot be orthogonal to a timelike vector, so $(t\cdot v{)}^2>0$. Thusin both cases. Every $g$-causal direction is strictly timelike for $\tilde{g}$, so its light cone is strictly wider.

Let $t^a$ exhibit the stable causality of $g$, so $g-t_a{t}_b$ has no closed timelike curves. SetThe vector $r^a$ remains timelike for $g^{\prime }$. Widening the cones of $g^{\prime }$ once more using $r$ giveswhich is causal by assumption. Hence

The relevant implications in the causal hierarchy areThe first follows from strict cone widening: failure of strong causality would give causal curves leaving and re-entering arbitrarily small neighborhoods, and the wider metric turns the limiting recurrence into a closed timelike curve. For the second, if distinct $p,q$ had $I^{+}(p)=I^{+}(q)$, arbitrarily short timelike segments near either point and this equality would produce a causal curve leaving and re-entering a sufficiently small causally convex neighborhood. ThereforeThe stated null geodesic completeness is stronger than needed: stable causality already gives the conclusion.

Write $I^{+}(U)=U=I^{-}(U)$. Since every chronological future and chronological past is open, $U$ is an open set. We show its complement is also open.

Take $p\notin U$ and choose $p_{-}\ll p\ll p_{+}$ in a convex normal neighborhood. The diamond $V=I^{+}(p_{-})\cap I^{-}(p_{+})$ is a neighborhood of $p$. If $x\in V\cap U$, then $p_{-}\ll x$ and $U=I^{-}(U)$ imply $p_{-}\in U$; then $p_{-}\ll p$ and $U=I^{+}(U)$ imply $p\in U$, a contradiction. Thus $V$ lies in the complement of $U$.

Hence $U$ is nonempty, open, and closed. Since $\mathcal{M}$ is a connected space,

For a real free scalar field on a globally hyperbolic spacetime, takeThe Klein-Gordon equation has a well-posed initial-value problem on every Cauchy hypersurface. Its complex solution space carries the conserved Klein-Gordon inner product. Choose modes withand expandThe canonical commutation relations produce the bosonic Fock space, and $a_i|0⟩=0$ defines the chosen vacuum.

The choice of positive-frequency solutions is extra structure. In a time-dependent geometry there is no preferred time translation, so different choices mix $u_i$ with $u_i^{\ast }$ and give the vacuum ambiguity in a nonstationary spacetime. In a stationary spacetime, a timelike Killing vector field defines positive frequency by $i{\mathcal{L}}_K{u}_i={\omega }_i{u}_i$, ${\omega }_i>0$, and hence the preferred vacuum state in a stationary spacetime.

If the spacetime is stationary in the remote past and future, define normalized in- and out-modes there. Their Bogoliubov transformation isThe expected out-particle number in the in-vacuum isThus nonzero beta coefficients measure particle creation by the intervening time dependence.

Classically, the laws of black-hole mechanics have the same form as the laws of thermodynamics, but this could have been only a formal analogy. Hawking radiation makes it physical: a stationary black hole emits approximately thermal radiation withso surface gravity is proportional to genuine temperature.

The correspondence is:

matches the first law of thermodynamics and fixes the Bekenstein-Hawking entropy

A real Hawking temperature therefore identifies horizon area as thermodynamic entropy and turns the geometric laws into thermodynamic laws.