Symbol delta

The passage around this formula

…be extended past the compass’s own total intrinsic Fisher information $F_Q^{(0)}$ = 4$\operatorname{Var}$$\hat J$/$\hbar^2$ , since no amount of radiation can reveal more about g than the original state ever carried. A tolerance $\delta$ tighter than 1/$\sqrt{F_Q^{(0)}}$ is not a slower recovery — it is a request for better precision than the compass itself possessed, which no observer, inside or outside any horizon, could ever meet. Within that ceiling, $t_G(\delta)$ grows toward the hole’s full evaporation lifetime as $\delta$…

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