# What is the nonrelativistic limit of the Einstein field equations in arbitrary dimensions?

A standard exercise in essentially any introductory textbook on general relativity is to work out the non-relativistic limit of the 3+1D Einstein field equations. This is most commonly done in order to nail down the proportionality constant between the Einstein tensor $$G$$ and stress-energy tensor $$T$$ that reproduces Newton's law of universal gravitation in the nonrelativistic regime.

That is, one typically starts with the ansatz $$G = \kappa T$$, where $$\kappa$$ is an unknown constant, and then works out that in the nonrelativistic limit, this reduces to $$\nabla^2 \Phi = \frac{1}{2} \kappa \rho, \tag{1}$$ which reproduces Newton's law of gravitation if we set $$\kappa = 8 \pi G$$.

What happens if we take the same nonrelativistic limit in arbitrary dimensions?

My guess is that we get the same limiting expression (1) in $$D \geq 4$$ spacetime dimensions, but we get something qualitatively different for $$D = 2$$ or $$D = 3$$. That's because in those lower dimensions, the Weyl tensor vanishes, so I don't think you can have long-distance gravitational effects that extend through vacuum. But I have no idea what you do get.

This seems like the kind of thing that someone would have already worked out somewhere, but I couldn't find it.

• Using classical physics, in $n$ spatial dimensions one can deduce that Newtonian force of gravity has the form $F\propto \frac{1}{r^{n-1}}$ So I assume that since Einsteinian gravity reduces to the force law $\frac{1}{r^2}$ in the Newtonian limit, it should reduce to the same ($\frac{1}{r^{n-1}}$) for arbitrary dimensions. Commented Jan 18, 2023 at 5:05
• @josephh That's a very natural assumption, but unfortunately it's incorrect. As John Baez says in a comment at math.columbia.edu/~woit/wordpress/?p=555, $(2+1)D$ Einstein gravity does not reduce to the Newtonian expression in the nonrelativistic limit. Hence my question. Commented Jan 18, 2023 at 5:28
• But aren't you interested $D\ge 4$? Or are you asking specifically about 2+1? Commented Jan 18, 2023 at 5:34
• @josephh As I said in both the title and the body of the question: arbitrary dimensions, so both greater than and less than 4. Commented Jan 18, 2023 at 5:52
• Do you want the number of time dimensions to be arbitrary, or $1$?
– J.G.
Commented Jan 18, 2023 at 7:34