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Equation 129 · Part 4 · The Bit Comes Back Before the Bearing

Symbol pi

1/(f1δ2)=ln⁡S/2π1/(f_1\delta^2) = \ln S/2\pi
π\pi

What this part means

pi is an input to the expression that computes the quantity on the left.

Its job in the formula

pi is an input to the expression that computes the quantity on the left.

The passage around this formula

A third, more uncomfortable check is what the toy model of the previous section actually implies at the numbers already computed: tLt_L ≈\approx 3.5\,ms\mathrm{ms} , or tLt_L/β\beta ≈\approx 28.2 , for the solar-mass hole above. Taking f1f_1 = 1 as an illustrative, explicitly assumed per-quantum information content, a modest tolerance δ\delta = 0.1\,rad\mathrm{rad} (about six degrees) gives tGt_G/β\beta = 100 , so tGt_G ≈\approx 12.4\,ms\mathrm{ms} and Δ\Delta tLGt_{LG} ≈\approx +8.9\,ms\mathrm{ms} : the compass is the slower debt, as the title claims. But a coarser tolerance, δ\delta = 0.5\,rad\mathrm{rad} (about twenty-nine degrees), gives tGt_G/β\beta = 4 , so tGt_G ≈\approx 0.50\,ms\mathrm{ms} and Δ\Delta tLGt_{LG} ≈\approx…

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Sources cited in the article section

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