# Why is general relativity in (2+1) dimensions different from cylindrical systems in (3+1) dimensional GR?

The gravitational potential $$\Phi$$ of an infinite rod in newtonian gravity is $$\Phi \sim \ln(r)$$. This is the same as the gravitational potential of a point charge in two-dimensional Newtonian gravity (see https://en.wikipedia.org/wiki/Newtonian_potential). They are the same, because both systems exhibit cylindrical symmetry and Gauss Law yields a logarithmic potential in this case.

In general relativity the solution for a static cylindrical spacetime is the Levi-Civita spacetime, which in the Newtonian limit will also give a potential $$\Phi \sim \ln(r)$$ (see e.g. https://arxiv.org/abs/1901.06561).

But what I can't understand is that in (2+1) dimensional general relativity it is said the spacetime is flat outside of a mass point, so in the Newtonian limit $$\Phi \sim 0$$. This is claimed despite authors stating that a point particle in general relativity in (2+1) dimensions is equivalent to a (infinite) string in (3+1) dimensions.

"We discussed the global properties of the (locally flat) geometries generated by moving point particles in 2+1 dimensions, or equivalently by parallel moving cosmic strings in 3+1 dimensions." (Deser, Jackiw, t'Hooft (1992))

"There is also a close relation to cosmic strings in four dimensions since the space-time of an infinite straight string is effectively three-dimensional." (Deser, Jackiw (1988))

So why is there a difference between Newtonian Potential derived from Levi-Civita spacetime and the Newtonian potential derived from a (2+1) dimensional general relativity? Is general relativity in (2+1) dimensions not simply a cross section through a cylindrical symmetric spacetime in (3+1) dimensions? What is it then?

• About the last question: think about weyl curvature, which carries gravitational degrees of freedom vanishes identically in (2+1) dim, but is non-zero in (3+1) dim spacetimes. The curvatures appearing in EFE are the intrinsic curvatures of space time. If you take cross-section , then you will also have to take into account the extrinsic curvatures, depending on how you define the cross section.
– KP99
Dec 3, 2021 at 7:56

That's a good question! Let us consider vacuum solutions to the EFE with zero cosmological constant $$\Lambda=0$$:

1. The static axisymmetric/cylindrical spacetimes in 3+1D are classified by Levi-Civita solutions [1,2]. They represent a (possibly thick$$^1$$) cosmic string of total zero angular momentum $$J=0$$. There are effectively 2 parameters:

• $$\sigma$$, which has an interpretation of an energy-density per length of the string.

If there is no deficit angle, then the well-known Newtonian limit (with an attractive Newtonian $$ln$$ potential) exists and corresponds to the limit $$\sigma\to 0^+$$.

2. The static spherical symmetric spacetimes in 2+1D. The corresponding Schwarzschild-like$$^2$$ solution is locally on Minkowski-form$$^3$$, and it has only 1 topological parameter: A deficit angle, which grows with the mass of the point source.

In both cases, if there is a deficit angle, it extends all the way to $$r=\infty$$, i.e. there is no asymptotic region $$r=\infty$$ unaffected by the string/point source, and hence there is no Newtonian limit.

Perhaps surprisingly, in 2+1D the Newtonian $$ln$$ potential has no analog in GR, cf. e.g. this Phys.SE post.

References:

1. M.F.A. da Silva, L. Herrera, F.M. Paiva, & N.O. Santos, The Levi-Civita spacetime, arXiv:gr-qc/9607058 .

2. K. Bronnikov, N.O. Santos, & Anzhong Wang, Cylindrical Systems in General Relativity, arXiv:1901.06561 .

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$$^1$$ Angular motion of the thick cosmic string is not completely prohibited, which leads to 1 extra parameter, that saves the $$\ln$$ potential. This is the main answer to OP's title question.

$$^2$$ Not a black hole!

$$^3$$ This is related to the fact that GR in 2+1D is a topological field theory (TFT).