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?
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$\begingroup$ In relativity, gravity is not a force at all. So not sure what you are looking for here. $\endgroup$– Marius Ladegård MeyerNov 3, 2021 at 18:32
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$\begingroup$ Ah you caught a typo. Thanks. $\endgroup$– zucculentNov 3, 2021 at 18:33
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$\begingroup$ 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. $\endgroup$– Marius Ladegård MeyerNov 3, 2021 at 18:41
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$\begingroup$ 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? $\endgroup$– zucculentNov 3, 2021 at 18:44
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$\begingroup$ 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? $\endgroup$– Marius Ladegård MeyerNov 3, 2021 at 18:57
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2 Answers
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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$$\ddot{x}^r\approx -\Gamma^r_{tt}=\frac12g^{rr}g_{tt,\,r}=-\frac12ff^\prime\approx-\frac12f^\prime\propto-\frac{m}{r^{n-1}}.$$
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$\begingroup$ I think what you're saying is that it ends up behaving like the Newtonian model in simple cases. Is this correct? $\endgroup$ Nov 4, 2021 at 2:38
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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.
answered Nov 4, 2021 at 2:21