# What would the Newtonian-esque approximation for gravity be in other numbers of dimensions, according to relativity?

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?

• In relativity, gravity is not a force at all. So not sure what you are looking for here. Nov 3, 2021 at 18:32
• Ah you caught a typo. Thanks. Nov 3, 2021 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. Nov 3, 2021 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? Nov 3, 2021 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? Nov 3, 2021 at 18:57

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$$\ddot{x}^r\approx -\Gamma^r_{tt}=\frac12g^{rr}g_{tt,\,r}=-\frac12ff^\prime\approx-\frac12f^\prime\propto-\frac{m}{r^{n-1}}.$$