Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 311 / 3

Use the intended positive-mass family: write $b=r_{+}>0$ and $a=r_{-}=2P^2/b\ge 0$, so $R^2=r(r-a)$, $V=1-b/r$ and $ab=2P^2$. Lowering the index of the Killing vector field gives $k^{♭}=-Vdt$, henceFor the stipulated orientation and Lorentzian signature, $\sqrt{-g}=R^2\sin \theta$ and the Hodge star operator obeysThe minus sign follows from $g^{tt}{g}^{rr}=-1$. Thus the Komar mass of the magnetic dilaton black hole isIts asymptotic mass is $M=r_{+}/2$.

For any Killing vector field, its antisymmetric derivative satisfies ${\nabla }_a(d{k}^{♭}{)}^{ab}=-2R^b{}_c{k}^c$. Applying Stokes theorem between two spheres relates the difference of their Komar mass to this Ricci source. It vanishes in vacuum but need not vanish for the nonzero Ricci tensor of Einstein-Maxwell-dilaton theory. Here, explicitly, $d{M}_{\mathrm{Komar}}/dr=P^2/r^2$. The potential $A=P\cos \theta d\phi$ is only a local gauge potential near the polar axes; the mass computation uses its globally defined field strength and the metric, so does not depend on this gauge-patch issue.

By spherical symmetry, rotate a geodesic into the equatorial plane $\theta =\pi /2$. The geodesic conserved quantities from Killing vectors arewhere the dot denotes a suitable affine parameter. Normalize the tangent by $g(\dot{x},\dot{x})=-ϵ$, with $ϵ=1$ for timelike proper time, $ϵ=-1$ for spacelike proper length, and $ϵ=0$ for null geodesics. ThenTherefore the geodesic potential of the magnetic dilaton black hole isTogether with the two conserved first integrals, this reduces the radial motion to ${\dot{r}}^2/2+\tilde{V}=0$. To cover turning points as well, the radial geodesic equation itself givesThus a constant-radius geodesic must satisfy both $\tilde{V}=0$ and ${\tilde{V}}^{\prime }=0$. The intermediate static-coordinate formulas apply where $V\ne 0$; the radial equation extends through a regular horizon in regular coordinates.

A radial null geodesic has $L=0$ and $ϵ=0$, so $\dot{r}=\pm E$ and $\dot{t}=E/V$. In the exterior, $E>0$ for a future-directed ray. Define the analogue of the Schwarzschild tortoise coordinate byThen an outgoing ray satisfies $\dot{u}=\dot{t}-\dot{r}/V=0$, whereas an ingoing ray satisfies $\dot{v}=\dot{t}+\dot{r}/V=0$. Thus $u=t-r_{\ast }$ is constant on outgoing rays and $v=t+r_{\ast }$ on ingoing rays. The radial two-dimensional metric becomes $-Vdudv$.

The ingoing coordinate is regular at the future horizon: substituting $t=v-r_{\ast }$ gives $d{s}^2=-Vd{v}^2+2dvdr+R^{2}d{\Omega }_2^2$. Inside the horizon, the words outgoing and ingoing refer to the continued null families; both future-directed families in the future interior decrease $r$.

Given an asymptotically flat spacetime with a conformal completion and future null infinity ${\mathcal{I}}^{+}$, its black hole region isThe event horizon is its boundary. For the magnetically charged dilaton black hole, first assume $0\le a=r_{-}<b=r_{+}$. The physical asymptotic component has $r>a$, and the ingoing chart extends regularly across $r=b$. Its future ingoing null direction is $-{\partial }_r$. For $a<r<b$,Hence $\nabla r$ is future timelike. Every nonzero future causal tangent $X$ satisfies $X(r)=g(\nabla r,X)<0$. A future causal curve from this interior can neither cross back to $r=b$ nor reach the asymptotic end $r\to \infty$, so it cannot reach ${\mathcal{I}}^{+}$. It lies in $\mathcal{B}$.

This proof uses the future black-hole extension. The radial interval alone does not distinguish it from the past white hole interior of the maximal extension; that interior has the same interval and can send signals to infinity. Thus the statement needs this future-component qualification if the maximal spacetime is intended.

The Kretschmann scalar confirms that $r=a$ is a genuine curvature singularity: for $0<a<b$, its numerator at $r=a$ is $3a^4(a-b{)}^2$, andFor $a=0$ it instead reduces to $12b^2/r^6$. Since $V<0$ on approaching the singularity from the interior, the singularity is spacelike. At $r=b>a$ the curvature is finite.

For the maximal extension, take $U=-e^{-u/(2b)}$, $W=e^{v/(2b)}$ in the right exterior and continue analytically across the horizons. ThenThe singularity is $UW=(1-a/b)e^{a/b}>0$ and the horizons are $U=0$ and $W=0$. Rescaling the null coordinates by the square root of this positive constant before arctangent compactification produces the usual qualitative Penrose diagram of Schwarzschild spacetime: two exteriors, a future black-hole region, a past white-hole region, and future and past spacelike singularities. The dashed lines below are the horizon branches; only the future interior is the black-hole region.

If $a>b$, the physical domain remains $r>a$, but now $V>0$ throughout it. The apparent zero $r=b$ is beyond the curvature singularity and is not an accessible horizon. The singularity is timelike, and outgoing radial null rays with $\dot{r}=E>0$ escape to infinity from arbitrarily near it. ThusThe excluded equality $r_{-}=r_{+}$ is a singular limiting case, not a regular horizon. These facts describe the causal structure of the magnetic dilaton black hole.

In the regular ingoing chart, the Killing vector field is $k={\partial }_v$, with $k^2=-V$. The normal covector to $r=b=r_{+}$ is $dr$; raising its index gives $\nabla r={\partial }_v+V{\partial }_r=k$ at $r=b$. Thus $k$ is null and normal to that hypersurface, which is a Killing horizon. It is regular because $R^2(b)=b(b-a)>0$ when $a=r_{-}<b$.

The defining identity for surface gravity on a Killing horizon is ${\nabla }_a(k^2)=-2\kappa k_a$. Here $k_a=-V(dv{)}_a+(dr{)}_a$, so at the horizon this becomes $-V^{\prime }(b)(dr{)}_a=-2\kappa (dr{)}_a$. ThereforeThis uses the asymptotic normalization $k^2\to -1$. A constant rescaling of $k$ rescales the surface gravity. As a check, the horizon area is $A_H=4\pi r_{+}(r_{+}-r_{-})$ and $\kappa A_H/(4\pi )=(r_{+}-r_{-})/2=M_{\mathrm{Komar}}(r_{+})$. The formula is being applied only to the regular strict-inequality case.