= Doppler shift for a uniformly accelerated receiver
{c}
{title2=$\nu_{\rm obs}=\nu_0e^{-a\tau}=a\nu_0(L-t_0),\quad c=1$}
In units $c=1$, a receiver on the <hyperbolic motion> worldline $t=a^{-1}\sinh(a\tau)$, $x=a^{-1}\cosh(a\tau)$ has retarded <null coordinate> $p=t-x=-a^{-1}e^{-a\tau}$. A right-moving light ray emitted at $(t_0,L)$ has $p=t_0-L$, so it meets the receiver only if $t_0<L$. The <photon> wave covector is proportional to $(1,-1)$, and contraction with the receiver's <four-velocity> gives its measured <frequency> as $\nu_0(\dot t-\dot x)=\nu_0e^{-a\tau}$. The intersection condition gives the second displayed equality. Emission at $t_0=L$ lies on the <Rindler horizon> and never reaches the receiver at finite <proper time>; $L$ is the supremum of receivable emission times, not an attained latest time.
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