= Light reception by a uniformly accelerated observer
{title2=$\tau_{\rm rec}=-(c/a)\log(1-ad/c^2)$}
Let an observer start at rest at $T=0$, $X=R=c^2/a$ and accelerate towards increasing $X$ with constant <proper acceleration> $a>0$. A signal emitted simultaneously from $X=R-d$ travels along $X=R-d+cT$. The observer follows <hyperbolic motion>, $X=R\cosh(a\tau/c)$ and $cT=R\sinh(a\tau/c)$. Their intersection obeys
$$
e^{-a\tau/c}=1-\frac{ad}{c^2},\qquad
\tau_{\rm rec}=\frac ca\log\frac1{1-ad/c^2}.
$$
A finite future reception exists only for $0\leq d<c^2/a$. At equality the ray lies on the limiting null horizon and never reaches the observer; for larger separation it remains outside the observer's receivable past. The proper reception time tends to $d/c$ as $a\to0$. The simultaneous emission and initial-rest conditions are essential; changing either changes the timing problem.
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