A static spherical star of mass has a weak external field

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 static spherical star of mass MM has a weak external field, GM/rc2≪1GM/rc^2 \ll 1 everywhere of interest. Outside it the gravitational potential is Φ(r)=−GM/r\Phi(r) = -GM/r, zero at infinity, and the local field strength is g(r)=GM/r2g(r) = GM/r^2. Light is emitted by a source at rest at radius r1r_1 and received by an observer at rest at radius r2>r1r_2 > r_1. Work throughout to first order in GM/rc2GM/rc^2. ((a)) Starting from the local result Δν/ν=−g Δr/c2\Delta\nu/\nu = -g\,\Delta r/c^2 for a short radial climb Δr\Delta r, stack short climbs into an integral and show that νrecνem=1−GMc2(1r1−1r2).\frac{\nu_{\rm rec}}{\nu_{\rm em}} = 1 - \frac{GM}{c^2}\left(\frac{1}{r_1} - \frac{1}{r_2}\right). ((b)) Evaluate the redshift zz for light leaving the Sun's photosphere and received at the Earth's orbit. Take M⊙=1.989×1030 kgM_\odot = 1.989\times10^{30}\ \mathrm{kg}, R⊙=6.960×108 mR_\odot = 6.960\times10^{8}\ \mathrm{m}, r2=1.496×1011 mr_2 = 1.496\times10^{11}\ \mathrm{m}, G=6.674×10−11 N m2 kg−2G = 6.674\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}} and c=2.998×108 m s−1c = 2.998\times10^{8}\ \mathrm{m\,s^{-1}}; ignore the Earth's own field and its orbital motion. ((c)) Obtain the same formula a second way, from energy conservation: a mass mm is lowered slowly from r2r_2 to r1r_1 with the work captured by a generator, is then converted entirely into a light pulse which is beamed up to r2r_2 and there converted entirely back into a mass at rest. Show that the cycle can create no energy only if the pulse's energy obeys the ratio derived in (a), and hence, since a pulse's energy is proportional to its frequency, that the frequency ratio is the same.

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