$$t_G/\beta = 4$$

Equation 126 · Evolutionary Physics

The Bit Comes Back Before the Bearing

This equation states an equality: the expressions on both sides have the same value under the article’s assumptions.

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 $\Delta$ $t_{LG}$ $\approx$…

Meanings in this article

• $t_G$: arbitrary on its own — each is anchored to a cited, checkable piece of theory, and the crossover tolerance $\delta^\ast$ is computed, not chosen after the fact to produce a headline.

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