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In 3 dimensions, gravity can usually be approximated using Newton's equation for gravity, $g=G\frac{m}{r^2}$. There have been answers here saying the acceleration of gravity in $n$ dimensions would be, but they are based on Newton's gravity equation. What does general relativity say about it, and what would the Newtonian approximation look like?

zucculent
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  • In relativity, gravity is not a force at all. So not sure what you are looking for here. – Marius Ladegård Meyer Nov 03 '21 at 18:32
  • Ah you caught a typo. Thanks. – zucculent Nov 03 '21 at 18:33
  • An object following a geodesic (i.e. in free fall, only influenced by the "force" of gravity) has zero four-acceleration by definition. That is the essence of the principle of equivalence. Again, not sure what you are looking for here. The equation of a geodesic? Those are what you can get when you solve Einsteins field equations. – Marius Ladegård Meyer Nov 03 '21 at 18:41
  • When I said "acceleration" of gravity, I meant the acceleration an observer on the ground would observe in an object in freefall. Does my new edit clear things up? – zucculent Nov 03 '21 at 18:44
  • Am I correct that the expression you are looking for in the 3+1 dimensions case is present in this Q&A? And you are wondering what the corresponding expression is for other numbers of dimensions? – Marius Ladegård Meyer Nov 03 '21 at 18:57
  • @MariusLadegårdMeyer Maybe? I'm not sure what you mean. – zucculent Nov 03 '21 at 19:17
  • I don't think GR can say anything about higher-D spaces, like 4+1. In 5D, the $T$ has 15 independent components, up 5 from 10 in 3+1D. I have no idea what physical sense can be ascribed to them. Do you? An additional vector field and a scalar field of unknown dynamics? Besides, solving the EFE fully in 5D even with a simplest nontrivial metric is horrendously hard. Of note is the Kaluza-Klein theory, q.v., a 5D theory but with a set of "cylindrical constraints," IIRC, of all $\partial/\partial x_5$ vanishing, describes classical GR+EM; still, it generates an extra non-physical scalar field. – kkm -still wary of SE promises Nov 03 '21 at 20:12

2 Answers2

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Unsurprisingly, GR recovers Newton. With $1$ time dimension and $n$ space dimensions, the Schwarzschild metric is $ds^2=-fdt^2+dr^2/f+r^2d\Omega_{n-1}^2$ with $f(\infty)=1,\,f^\prime\propto m/r^{n-1}$. The geodesic deviation equation $\ddot{x}^a=-\Gamma^a_{bc}\dot{x}^b\dot{x}^c$ includes the nonrelativistic special case$$\ddot{x}^r\approx -\Gamma^r_{tt}=\frac12g^{rr}g_{tt,\,r}=-\frac12ff^\prime\approx-\frac12f^\prime\propto-\frac{m}{r^{n-1}}.$$

J.G.
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Here is a (perhaps oversimplified) way of looking at this.

For any description of gravity which can be well-approximated with a 1/r^2 dependency, the presence of extra spatial dimensions can be inferred by "leakage" of gravity into the extra dimensions which manifests as a deviation from strict 1/r^2 dependence in the (3+1) dimensions we inhabit. So one test of the presence of extra spatial dimensions is to look for deviations from 1/r^2 behavior caused by leakage into the extra dimensions, and to date none have been detected.

niels nielsen
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