What does this equation mean?

A third, more uncomfortable check is what the toy model of the previous section actually implies at the numbers already computed: $t_L$ $\approx$ 3.5\,$\mathrm{ms}$ , or $t_L$/$\beta$ $\approx$ 28.2 , for the solar-mass hole above. Taking $f_1$ = 1 as an illustrative, explicitly assumed per-quantum information content, a modest tolerance $\delta$ = 0.1\,$\mathrm{rad}$ (about six degrees) gives $t_G$/$\beta$ = 100 , so $t_G$ $\approx$ 12.4\,$\mathrm{ms}$ and $\Delta$ $t_{LG}$ $\approx$ +8.9\,$\mathrm{ms}$ : the compass is the slower debt, as the title claims. But a coarser tolerance, $\delta$ = 0.5\,$\mathrm{rad}$ (about twenty-nine degrees), gives $t_G$/$\beta$ = 4 , so $t_G$ $\approx$ 0.50\,$\mathrm{ms}$ and $\Delta$ $t_{LG}$ $\approx$ -3.0\,$\mathrm{ms}$ : under this same model, on this same hole, the compass now comes back first. The crossover, where 1/($f_1$$\delta^2$) = $\ln$ S/2$\pi$ , sits at $\delta^\ast$ $\approx$ 0.188\,$\mathrm{rad}$ , about eleven degrees, for this hole and this choice of $f_1$ . A fourth, trivial floor closes the set of checks: for $C_t$ smaller than the input systems themselves, or for t before scrambling has had time to run at all, neither $\epsilon_L(t)$ nor $\operatorname{Var}$$\hat g$ meets any nontrivial bound — nothing has left yet, which is simply causality, not a result.