I understand that General Relativity explains gravity by framing it as a consequence of the curvature of spacetime rather than as a force. Does this theoretically guarantee that gravity must be an inverse square law? Would it be possible in a different hypothetical Universe to have gravity which is a different power law, e.g. inverse cube, while still obeying the rules of General Relativity?
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3$\begingroup$ General relativity predicts in $d$-dimensional space with $1$ time dimension the non-relativistic limit is an approximately $r^{1-d}$ force, but also includes slight corrections to this, which is why Mercury's perihelion precession is a famous test of GR. See e.g. the $d=3$ case here (where it's framed in terms of the potential energy, not the force). $\endgroup$– J.G.Commented Aug 23, 2022 at 7:52
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1$\begingroup$ Possible duplicates: physics.stackexchange.com/q/211930/2451 , physics.stackexchange.com/q/68067/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Aug 23, 2022 at 7:53
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$\begingroup$ @Qmechanic I thought that the question was about the mathematics of GR, i.e. could the mathematics allow another dependence. The way a wave equation can have many solutions but only the ones following the postulates are useful in QM? $\endgroup$– anna vCommented Aug 23, 2022 at 8:56
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$\begingroup$ Metric tensor signature is an assumption of GR. Indeed, if one says that the metric signature is ++-- (two "times" and two "distances"), nothing checks out = no Newton's law. no agreement with experiment. $\endgroup$– DanielCCommented Aug 23, 2022 at 9:02
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