A photon climbs a tower of height m near the Earth's surface

PHYSICS · Multi-part · Fermi University Physics · Relativity & Gravitation · The equivalence principle and gravity as geometry · The equivalence principle and gravity as geometry

The Geometry of Gravity, Ch The equivalence principle under test

Problem

A photon climbs a tower of height ℓ=22.6\ell = 22.6 m near the Earth's surface, where g⊕=9.81 m s−2g_\oplus = 9.81\ \mathrm{m\,s^{-2}}. ((a)) If local position invariance holds exactly (αLPI=0\alpha_{\rm LPI}=0), compute the fractional frequency shift Δν/ν\Delta\nu/\nu of the photon over the climb, using Δν/ν=ΔΦ/c2\Delta\nu/\nu = \Delta\Phi/c^2. ((b)) Now suppose local position invariance is violated with αLPI=1.0×10−4\alpha_{\rm LPI} = 1.0\times10^{-4}, so that Δν/ν=(1+αLPI)ΔΦ/c2\Delta\nu/\nu = (1+\alpha_{\rm LPI})\Delta\Phi/c^2. Compute the extra term this adds to the shift found in part (a). ((c)) The gravitational-redshift argument was built by an analogy with an observer undergoing proper acceleration a=g⊕a=g_\oplus in flat spacetime, whose worldline and clock-rate ratio between two points separated by ξ=ℓ\xi=\ell follow from the line element ds2=−(1+aξ/c2)2c2 dη2+dξ2ds^2=-(1+a\xi/c^2)^2c^2\,d\eta^2+d\xi^2. Does this line element itself need to be changed to accommodate αLPI≠0\alpha_{\rm LPI}\ne0, or does the violation enter only in the identification of the accelerated-frame result with a real gravitational clock-rate ratio? Justify your answer in one or two sentences.

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