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Equation 112 · The Clock That Comes Back Wrong by Exactly Its Mass

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T=1 sT=1\,\mathrm s

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Inputs and operations1mathrm s
Result or conditionT
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TT

Symbol T

T is part of the quantity the equation computes from the expression on the right.

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ss

Symbol s

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

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=

=

The expressions on both sides represent the same quantity under the stated assumptions.

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To make this concrete without simulation, take the weak-field proper-time rate dτ\tau/dt≈\approx1+gz/c2c^2 for a static height z in Earth’s field, so that holding two branches at a height difference Δ\Delta z for coordinate time T gives Δ\Deltaτ\tau≈\approx g\,Δ\Delta z\,T/c2c^2 . For Δ\Delta z=1\,m\mathrm m and T=1\,s\mathrm s , Δ\Deltaτ\tau≈\approx1.090×\times10^{-16}\,s\mathrm s . Pair this with a single-photon optical clock transition near 698\,nm\mathrm{nm} , comparable to the transition used in strontium-lattice-clock proposals for this kind of experiment [ 11 , 13 ] , giving Δ\Deltaν\nu≈\approx4.295×\times10^{14}\,Hz\mathrm{Hz} and Δ\Delta E=hΔ\Deltaν\nu≈\approx2.846×\times10^{-19}\,J\mathrm J , a mass excess Δ\Delta…
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To make this concrete without simulation, take the weak-field proper-time rate dτ\tau/dt≈\approx1+gz/c2c^2 for a static height z in Earth’s field, so that holding two branches at a height difference Δ\Delta z for coordinate time T gives Δ\Deltaτ\tau≈\approx g\,Δ\Delta z\,T/c2c^2 . For Δ\Delta z=1\,m\mathrm m and T=1\,s\mathrm s , Δ\Deltaτ\tau≈\approx1.090×\times10^{-16}\,s\mathrm s . Pair this with a single-photon optical clock transition near 698\,nm\mathrm{nm} , comparable to the transition used in strontium-lattice-clock proposals for this kind of experiment [ 11 , 13 ] , giving Δ\Deltaν\nu≈\approx4.295×\times10^{14}\,Hz\mathrm{Hz} and Δ\Delta E=hΔ\Deltaν\nu≈\approx2.846×\times10^{-19}\,J\mathrm J , a mass excess Δ\Delta E/c2c^2≈\approx3.167×\times10^{-36}\,kg\mathrm{kg} .

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