= Solution
Throughout this solution $c=1$ and $a>0$.
\b[Null <metric tensor> and boost.] Inverting the coordinate definitions gives $t=(p+q)/2$, $x=(q-p)/2$, and hence
$$
\boxed{ds^2=dp\,dq,\qquad
(g_{AB})_{p,q}=\begin{pmatrix}0&1/2\\1/2&0\end{pmatrix}}.
$$
For a frame moving at speed $v$ along positive $x$, use $t'=\gamma(t-vx)$, $x'=\gamma(x-vt)$. Their difference and sum give
$$
\boxed{p'=\alpha p,\qquad q'=q/\alpha,\qquad
\alpha=\gamma(1+v)=\sqrt{\frac{1+v}{1-v}}}.
$$
The reciprocal <scale factors> preserve $dp\,dq$. They are $e^{\pm\eta_v}$ in <rapidity> notation.
\b[<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 $\ddot t^2-\ddot x^2=-a^2$; therefore
$$
\boxed{\dot p\dot q=1,\qquad\ddot p\ddot q=-a^2}.
$$
For 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 $\eta=\eta_0+a\tau$.
\b[<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 family
$$
p=-\frac1a+\frac{e^{-\eta_0}}a(1-e^{-a\tau}),\qquad
q=\frac1a+\frac{e^{\eta_0}}a(e^{a\tau}-1).
$$
The 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,
$$
\boxed{p(\tau)=-\frac1ae^{-a\tau},\qquad
q(\tau)=\frac1ae^{a\tau}},
$$
and consequently
$$
\boxed{t(\tau)=\frac1a\sinh a\tau,\qquad
x(\tau)=\frac1a\cosh a\tau}.
$$
With 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.
\b[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 $p\geq0$ 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>.
\b[<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> gives
$$
\boxed{\nu(\tau)=\nu_0(\dot t-\dot x)=\nu_0\dot p=\nu_0e^{-a\tau}}.
$$
The 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 requires
$$
\boxed{t_e=L-\frac1ae^{-a\tau}<L}.
$$
Thus \b[$t_e=L$ 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.
\b[<Rindler coordinates>.] Substitution of the selected trajectory into the <coordinate transformation> gives
$$
\boxed{\xi(\tau)=\tau,\qquad\zeta(\tau)=0}.
$$
Differentiating the transformation also gives $ds^2=e^{2a\zeta}(d\xi^2-d\zeta^2)$. 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> $a e^{-a\zeta}$.
\Image[/past-exam-of-the-relativity-course-of-the-university-of-cambridge/2013/paper-1-rindler-signals.png]
{title=Original accelerated worldline, reception horizon and stellar photon rays, with a separate Rindler coordinate grid}
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