= Solution
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> is
$$
X_s(\tau)=R\cosh(a\tau/c),\qquad
cT_s(\tau)=R\sinh(a\tau/c).
$$
A photon emitted from $X=R-d$ at $T=0$ towards the departing ship follows $X_\gamma=R-d+cT$. At reception,
$$
R[\cosh(a\tau/c)-\sinh(a\tau/c)]=R-d,
\qquad e^{-a\tau/c}=1-\frac{ad}{c^2}.
$$
Therefore <light reception by a uniformly accelerated observer> occurs after
$$
\boxed{\tau_{\rm rec}=\frac ca\log\frac1{1-ad/c^2},\qquad 0\leq d<\frac{c^2}a.}
$$
The 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 $d\tau=\alpha R\,dt/c$ 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 $ad/c^2\ll1$, $\tau_{\rm rec}=d/c+ad^2/(2c^3)+\cdots$, 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.
Back to article page
