Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 58 / 1

A Killing horizon is a null hypersurface with a Killing vector field $\xi$ as its normal and generator. Write $\xi =d/d\alpha$ along its orbits. These are null geodesics, but their Killing parameter need not be affine:The coefficient is the surface gravity for the chosen normalization; nondegeneracy means $\kappa \ne 0$. If $\lambda$ is affine and $q=d\lambda /d\alpha$, then $\xi =qd/d\lambda$ and $d(\log q)/d\alpha =\kappa$. For constant $\kappa$, $\lambda ={\lambda }_0+C{e}^{\kappa \alpha }$.

The Killing equation yields ${\partial }_{\mu }({\xi }^2)=2{\xi }^{\nu }{\nabla }_{\mu }{\xi }_{\nu }=-2{\xi }^{\nu }{\nabla }_{\nu }{\xi }_{\mu }$. On the horizon, $\boxed{{\partial }_{\mu }{\xi }^2=-2\kappa {\xi }_{\mu }}$. Multiplying $\xi$ by a positive constant multiplies $\kappa$ by that constant; physical temperature therefore needs a stated time normalization.

A static observer has four-velocity $u=F^{-1/2}{\partial }_t$. Its only nonzero acceleration component is $a^r={\Gamma }^r{}_{tt}/F=F^{\prime }/2=-r/R^2$. Using $g_{rr}=F^{-1}$, the magnitude isIt vanishes only at $r=0$, where $F=1$ and $d\tau =\sqrt{F}dt=dt$. Thus $t$ is proper time for the central inertial static observer. Other static observers need inward acceleration to resist de Sitter separation and have a redshifted proper time.

For the static patch of de Sitter spacetime, $r_{\ast }=\int dr/F=(R/2)\log |(R+r)/(R-r)|$. With $v=t+r_{\ast }$ or $u=t-r_{\ast }$, respectively,Both radial blocks have determinant $-1$ and smooth coefficients at $r=R$. The normal has norm $g^{rr}=F$, so that surface is null. There $k={\partial }_v$ has covector $dr$, or $k={\partial }_u$ has covector $-dr$, making it normal as well as tangent: it is a Killing horizon.

For the signed surface gravity of a de Sitter horizon, since ${\partial }_r{k}^2=-F^{\prime }(R)=2/R$, part a gives $\kappa =-1/R$ in the ingoing chart on the past branch, and $\kappa =+1/R$ in the outgoing chart on the future branch, with $k$ future-directed in the static patch. The physical magnitude is $\boxed{|\kappa |=1/R}$, normalized by central proper time. A negative signed $F^{\prime }(R)/2$ must not become a negative temperature.

In $c=k_B=\hslash =1$ units, the Hawking temperature is $T_H=|\kappa |/(2\pi )$ for the stated time normalization. The inertial central observer sees the de Sitter horizon temperature $T_0=1/(2\pi R)$. The Tolman temperature law states $T\sqrt{-k^2}=\text{constant}$ in static thermal equilibrium. HenceWith part b, $T^2=(a^2+R^{-2})/(4{\pi }^2)$ and $T/(a/(2\pi ))=R/r\to 1$ at the horizon. This tends to the Unruh effect temperature for the increasingly accelerated observer.

Let $s$ be proper distance inward from the horizon. Then $R-r\simeq s^2/(2R)$, $F\simeq s^2/R^2$, givingThe radial factor has Rindler coordinates: $T_{\mathrm{M}}=s\sinh (t/R)$ and $X_{\mathrm{M}}=s\cosh (t/R)$ make it Minkowskian. Fixed $s$ has acceleration $1/s$ and temperature $1/(2\pi s)$, agreeing with the leading behavior. At the center $a=0$ but the finite de Sitter temperature remains; $T=a/(2\pi )$ is only the near-horizon limit here.

With $r=R\sin \chi$, the static metric isThe radial coefficient is regular at $\chi =\pi /2$, but the time coefficient vanishes and these coordinates cease to be a chart. Part c supplies smooth horizon-crossing coordinates; the local Rindler coordinates also prove the degeneracy is a coordinate one. Extending $\chi$ while keeping the same static $t$ alone is not a regular coordinate extension.

The complete analytic extension is the hyperboloid $-Y_0^2+{\sum }_{i=1}^4{Y}_i^2=R^2$. Static coordinates are $Y_0=\sqrt{R^2-r^2}\sinh (t/R)$, $Y_4=\sqrt{R^2-r^2}\cosh (t/R)$, and $(Y_1,Y_2,Y_3)=r\mathbf{n}$. Global coordinates giveWith $\tan \eta =\sinh (T/R)$ this isThe Penrose diagram is a rectangle, with spacelike past/future infinity, regular pole lines $\psi =0,\pi$, and radial null rays at $45$ degrees. Observer horizons divide static patches from inaccessible regions; they are not curvature boundaries.