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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 $a{e}^{-a\zeta }$.