A vertical shaft of height is cut in the Earth

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

Relativity and Gravitation, Ch Gravity as curved spacetime: redshift and local inertial frames

Problem

A vertical shaft of height h=480 mh = 480\ \mathrm{m} is cut in the Earth. Over this height treat the gravitational field as uniform with g=9.81 m s−2g = 9.81\ \mathrm{m\,s^{-2}}; take c=2.998×108 m s−1c = 2.998\times10^{8}\ \mathrm{m\,s^{-1}}. An idealised machine runs the following cycle with a body of mass m=1.50 kgm = 1.50\ \mathrm{kg}: the body is lowered from the top to the bottom on a light inextensible cable so slowly that its kinetic energy is negligible, and a generator captures the work done; at the bottom a converter anchored to the Earth turns the body entirely into light, the Earth taking up the recoil momentum with negligible energy; the light is beamed up the shaft; at the top a second anchored converter turns the arriving light entirely into a body at rest. Work to first order in gh/c2gh/c^2, and use the fact that a pulse's energy is proportional to its frequency. ((a)) Compute the work WW captured by the generator during the lowering. ((b)) Show that the cycle can create no energy only if the light arrives at the top with energy Etop=mc2(1−gh/c2)E_{\rm top} = mc^2\left(1 - gh/c^2\right), and deduce Δν/ν=−gh/c2\Delta\nu/\nu = -gh/c^2 for the climbing light. ((c)) Evaluate the fractional frequency shift, and state how much energy the machine would deliver for nothing on each cycle if the light arrived with its full energy mc2mc^2.

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