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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?

Qmechanic
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Victor Hakim
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    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). – J.G. Aug 23 '22 at 07:52
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    Possible duplicates: https://physics.stackexchange.com/q/211930/2451 , https://physics.stackexchange.com/q/68067/2451 and links therein. – Qmechanic Aug 23 '22 at 07:53
  • @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? – anna v Aug 23 '22 at 08:56
  • 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. – DanielC Aug 23 '22 at 09:02

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