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Equation 116 · Part 3 · The Clock That Comes Back Wrong by Exactly Its Mass

Symbol Δnu

ΔE=hΔν≈2.846×10−19 J\Delta E=h\Delta\nu\approx2.846\times10^{-19}\,\mathrm J
Δν\Delta\nu

What this part means

Δnu is one of the signed contributions combined to compute the quantity on the left.

Its job in the formula

Δnu is one of the signed contributions combined to compute the quantity on the left.

The passage around this formula

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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Learn the underlying idea

A variable is a named place for a value. Its letter is a local label: x can mean position in one formula and a data point in another.

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Sources cited in the article section

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