Rindler radar distance between two coaxial observers is not symmetric under interchange

(Radar distance from $O$ to $R$.) By [L3] applied with observer radius ${\rho }_0$ and reflector radius ${\rho }_m=2{\rho }_0$, emitted from $O$ at ${\theta }_e=0$, the reflection event on $R$ is at Rindler time $\log 2$ and the return event on $O$ at Rindler time $2\log 2$, so the round trip takes $2{\rho }_0\log 2\mathrm{/}c$ of $O$'s proper time and the radar distance assigned by $O$ is $D_1={\rho }_0\log 2$; by [L1] and [L4] the reflection event is $x^{-1}({\rho }_0\cdot 2\cdot (3\mathrm{/}4),{\rho }_0\cdot 2\cdot (5\mathrm{/}4),0,0)=x^{-1}(3{\rho }_0\mathrm{/}2,5{\rho }_0\mathrm{/}2,0,0)$ and the reception event is $x^{-1}({\rho }_0\sinh (2\log 2),{\rho }_0\cosh (2\log 2),0,0)=x^{-1}(15{\rho }_0\mathrm{/}8,17{\rho }_0\mathrm{/}8,0,0)$.

(Radar distance from $R$ to $O$.) Now take $R$ as observer, emitting at its Rindler time zero from $r_{-}=x^{-1}(0,2{\rho }_0,0,0)$. The future null line from $r_{-}$ toward $O$ is the constant-$U$ line with $U=2{\rho }_0$ (the constant-$V$ line leaves the radius increasing), and it meets $\rho ={\rho }_0$ at the unique Rindler time $\theta$ with ${\rho }_0{e}^{\theta }=2{\rho }_0$, i.e. $\theta =\log 2$, at the reflection event $q_m^{\mathrm{\prime }}=x^{-1}({\rho }_0\sinh \log 2,{\rho }_0\cosh \log 2,0,0)=x^{-1}(3{\rho }_0\mathrm{/}4,5{\rho }_0\mathrm{/}4,0,0)$ by [L4]. The return leg is the constant-$V$ line through $q_m^{\mathrm{\prime }}$, with $V=-{\rho }_0{e}^{-\log 2}=-{\rho }_0\mathrm{/}2$, and it reaches $R$ at the unique Rindler time $\theta$ with $-(2{\rho }_0)e^{-\theta }=-{\rho }_0\mathrm{/}2$, i.e. $e^{-\theta }=1\mathrm{/}4$ and $\theta =2\log 2$, at the event $x^{-1}(2{\rho }_0\sinh (2\log 2),2{\rho }_0\cosh (2\log 2),0,0)=x^{-1}(15{\rho }_0\mathrm{/}4,17{\rho }_0\mathrm{/}4,0,0)$ by [L4].

(The two assignments.) By [L2] and [L1], the round trip of step 1.2 lasts $(2\log 2)$ in Rindler time of $R$, whose proper time satisfies $d\tau ={\rho }_{m}d\theta \mathrm{/}c=(2{\rho }_0\mathrm{/}c)d\theta$, so the elapsed proper time of $R$ is $4{\rho }_0\log 2\mathrm{/}c$ and the radar distance assigned by $R$ is $D_2=c\cdot (1\mathrm{/}2)\cdot 4{\rho }_0\log 2\mathrm{/}c=2{\rho }_0\log 2$.

(Asymmetry and scope.) Steps 1.1 and 2.1 give $D_1={\rho }_0\log 2$ and $D_2=2{\rho }_0\log 2$, and $\log 2>\log 1=0$ with ${\rho }_0>0$ makes $D_1>0$, so $D_2=2D_1\mathrm{\ne }{D}_1$ and the radar distance between the two worldlines depends on which worldline is the observer; both reflection events lie in the causal future of their emission events by step 1.2 and [L1], and the protocol's symmetric half-split convention of [L2] was retained. This is a property of the two radar assignments, not a physical anisotropy; no measurement, statistical or empirical claim is made, and no choice principle was used.