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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.

Throughout this solution $c=1$ and $a>0$.

Null metric tensor and boost. Inverting the coordinate definitions gives $t=(p+q)/2$, $x=(q-p)/2$, and henceFor a frame moving at speed $v$ along positive $x$, use $t^{\prime }=\gamma (t-vx)$, $x^{\prime }=\gamma (x-vt)$. Their difference and sum giveThe reciprocal scale factors preserve $dpdq$. They are $e^{\pm {\eta }_v}$ in rapidity notation.

acceleration invariants. Proper-time normalization gives ${\dot{t}}^2-{\dot{x}}^2=1$, so $\dot{p}\dot{q}=1$. The four-acceleration has squared norm ${\ddot{t}}^2-{\ddot{x}}^2=-a^2$; thereforeFor future motion write $\dot{p}=e^{-\eta }$ and $\dot{q}=e^{\eta }$. Then the second equation is $-(\dot{\eta }{)}^2=-a^2$. acceleration toward positive $x$ selects $\eta ={\eta }_0+a\tau$.

hyperbolic motion from an initial event and rapidity. The stated initial event fixes $p(0)=-1/a$, $q(0)=1/a$, but not the initial rapidity ${\eta }_0$. Integrating with positive signed acceleration gives the general familyThe initial event alone does not fix the initial velocity or signed acceleration. The trajectory explicitly requested in the next paragraph selects the additional conditions ${\eta }_0=0$ and positive-$x$ acceleration. With that convention,and consequentlyWith arbitrary ${\eta }_0$ the equivalent expressions are $t=[\sinh ({\eta }_0+a\tau )-\sinh {\eta }_0]/a$ and $x=1/a+[\cosh ({\eta }_0+a\tau )-\cosh {\eta }_0]/a$. Thus position data alone do not select the displayed special hyperbola. The following calculations use that intended hyperbola, which is also explicitly given by the paper.

Diagram and horizon. The selected world line is the right-hand branch $x^2-t^2=a^{-2}$. Its future asymptote $x=t$ is a reception horizon: rightmoving signals have constant $p=t-x$, whereas the observer has $p<0$ and approaches zero only as $\tau \to \infty$. Events with $p\ge 0$ cannot send a future signal to this observer if the acceleration continues indefinitely. This is an observer-dependent causal restriction in flat spacetime, not a curvature singularity.

photon reception by a uniformly accelerated observer. A rightmoving photon from the stationary source has four-momentum $P^{\mu }=h{\nu }_0(1,1)$. Contracting with the observer's four-velocity givesThe same result follows by differentiating its emission-time relation. A photon emitted at $x=L$ and time $t_e$ has $p=t_e-L$, so intersection requiresThus $t_e=L$ is the supremum of receivable emission times, reached only in the limit of infinite receiver proper time. A photon emitted exactly then lies on the horizon and is never received at finite time; later photons cannot catch the ship. Restoring dimensions gives the limiting emission time $L/c$ for this chosen origin and trajectory.

Rindler coordinates. Substitution of the selected trajectory into the coordinate transformation givesDifferentiating the transformation also gives $d{s}^2=e^{2a\zeta }(d{\xi }^2-d{\zeta }^2)$. Curves of fixed $\zeta$ are hyperbolae, while curves of fixed $\xi$ are straight rays $t/x=\tanh (a\xi )$ within the right wedge. The coordinate $\xi$ is the selected ship's proper clock; at another fixed $\zeta$, proper time obeys $d{\tau }_{\zeta }=e^{a\zeta }d\xi$. The spatial coordinate is logarithmic in $\rho =\sqrt{x^2-t^2}=e^{a\zeta }/a$, rather than being proper distance itself. Fixed-$\zeta$ observers have proper acceleration $a{e}^{-a\zeta }$.

Differentiating the supplied coordinates givesSubtract the squares. The cross terms cancel and ${\cosh }^2-{\sinh }^2=1$, yieldingThe spacetime is flat Minkowski spacetime in Rindler coordinates. $x>0$ covers $X>|cT|$, the right Rindler wedge, while $x<0$ covers the left wedge $X<-|cT|$; neither is the whole spacetime.

The null combinations areFor $x=C{e}^{-\alpha t}$ the first is constant; for $x=C{e}^{\alpha t}$ the second is constant. Thus part (a)'s exponentials are simply the inertial light rays $X\pm cT=\text{constant}$. Their coordinate-time asymptotes reflect the chart's Rindler horizon, not a failure of light to move at $c$ in an inertial frame.

Let $a>0$ be the constant proper acceleration away from the emitter and set $R=c^2/a$. Take the signal emission to be simultaneous, in the initial inertial frame, with the ship being at rest and its clock reading $\tau =0$. Place that initial ship event at $T=0$, $X=R$. Its hyperbolic motion isA photon emitted from $X=R-d$ at $T=0$ towards the departing ship follows $X_{\gamma }=R-d+cT$. At reception,Therefore light reception by a uniformly accelerated observer occurs afterThe same result follows in Rindler coordinates: the ship stays at $x=R$, the emitter starts at $x=R-d$, and the right-moving null ray has $x=(R-d)e^{\alpha t}$. The onboard lapse $d\tau =\alpha Rdt/c$ converts its coordinate catch-up time to the expression above.

If $d=c^2/a$, the ray lies on the limiting null horizon and never intersects it at finite proper time; if $d>c^2/a$, it never reaches the ship. The reception time diverges as the emitter approaches this Rindler horizon. For $ad/c^2\ll 1$, ${\tau }_{\mathrm{rec}}=d/c+a{d}^2/(2c^3)+\cdots$, recovering the inertial light-travel time. A different emission time or initial relative velocity would require specifying that extra datum; the formula uses the initial-rest simultaneous-emission convention stated above.

Outside a Schwarzschild metric horizon $r_s$, write $r-r_s={\rho }^2/(4r_s)$. To leading order the radial metric isIt has Rindler coordinates with $\alpha =c/(2r_s)$. The proper-distance-like coordinate $\rho$ has null paths exponential in $\pm ct/(2r_s)$; the Schwarzschild coordinate difference $r-r_s\propto {\rho }^2$ is exponential in $\pm ct/r_s$.

Inside, writing $r_s-r={\rho }^2/(4r_s)$ instead gives $d{s}^2\simeq d{\rho }^2-[c\rho /(2r_s){]}^{2}d{t}^2$, where the radial coordinate is timelike. The negative-$x$ Rindler wedge is therefore not the black-hole interior: its time coordinate retains the same causal type as in the positive-$x$ wedge. The flat radial limit does not remove the genuine curvature of the full Schwarzschild spacetime.

For $d{s}^2=(\alpha x{)}^{2}d{t}^2-d{x}^2$ in a patch with $x\ne 0$, the null paths solve $dx/dt=\pm \alpha x$, giving $x=C{e}^{\pm \alpha t}$ with $C\ne 0$. They are geodesics after choosing an affine parameter. The Rindler coordinates $X=x\cosh (\alpha t)$, $cT=x\sinh (\alpha t)$ give $d{s}^2=c^{2}d{T}^2-d{X}^2$ and $X\pm cT=x{e}^{\pm \alpha t}$. Thus each exponential path is a straight inertial null line. A literal $x=0$ curve is outside this degenerate coordinate chart, not an extra null geodesic within it.

An observer at constant proper acceleration in the Minkowski vacuum detects a thermal response at this temperature. The associated Rindler coordinates separate the observer's accessible wedge from a horizon. This is an observer-dependent particle/detector response, not an assertion that inertial observers measure a thermal bath.