What does the exterior metric look like for an infinitely long cylindrical mass distribution? I'm assuming the stress energy tensor, $T_{\mu\nu}=0$ outside the cylinder and that the cylinder has no angular momentum.
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$\begingroup$ A mass distribution can’t have a zero energy-momentum tensor. Do you mean zero outside the cylinder? $\endgroup$– G. SmithJan 17, 2020 at 13:58
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$\begingroup$ @G.Smith Yes, sorry. I was in a rush when I made this post. $\endgroup$– Ryan ParikhJan 17, 2020 at 14:07
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$\begingroup$ I don't think this is enough information to specify the problem. Even the analogous problem for a sphere is not straightforward and requires for self-consistency that you set up some kind of equation of state that allows the sphere to be in hydrodynamic equilibrium. $\endgroup$– user4552Jan 17, 2020 at 14:26
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$\begingroup$ @BenCrowell I'm not sure what you mean. You can assume the simplest case, for example, that the mass is uniformly distributed within the cylinder. $\endgroup$– Ryan ParikhJan 17, 2020 at 14:37
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$\begingroup$ The axisymmetric metric in the static case is investigated by Weil and Levy-Chevita. In the case of gravitational waves, the problem was studied by Rosen and Einstein. What is your case? $\endgroup$– Alex TrounevJan 17, 2020 at 14:39
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1 Answer
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The most general static vacuum solution of Einstein equations with a cylindrical symmetry is the Levi–Civita metric: $$ds^2=r^{8σ^2−4σ}(dr^2+dz^2) +D^2r^{2−4σ}dφ^2−r^{4σ}dt^2$$ where $σ$ and $D$ are constants and the coordinate $φ$ is assumed to be periodic with a period of $2\pi$ (if we drop periodicity requirement the solution could be interpreted as a metric outside of infinite wall). The metric generally has curvature singularity at $r=0$ and is flat in the limit $r\to \infty$.
More on this metric including possible sources for it could be found in this or that papers.
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$\begingroup$ I was trying to get the equations for an infinite cylinder, like how you can use Gauss' law to find the gravitational field outside an infinitely long cylinder in classical physics. In those cases the math simplifies and it can be used as an approximation for a finite cylinder. I plugged it into the Einstein Field Equations and ended up with imaginary numbers that I couldn't get rid of. I assume this means there is no solution for this in general relativity, you can only find the metric for finitely long cylinders. It is also possible that I made an error in the math. $\endgroup$ Mar 7, 2020 at 21:45
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$\begingroup$ @RyanParikh: I do not know what constraints you are trying to impose, so I could not comment on “imaginary numbers” problem, but there are perfect fluid sources for LC metric, see e.g. Haggag & Desokey, 1996, dx.doi.org/10.1088/0264-9381/13/12/012 $\endgroup$– A.V.S.Mar 8, 2020 at 15:28
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$\begingroup$ The issue is that I was trying to get a metric that is asymptotically flat. The field equations don't have a solution like that with cylindrical symmetry. I guess this can be seen as a feature of general relativity. $\endgroup$ Aug 29, 2020 at 23:40
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$\begingroup$ @RyanParikh: Nontrivial solution with cylindrical symmetry means that there is finite curvature (and also sources of stress–energy, etc) at arbitrarily large distances from the origin. This would directly contradict with “asymptotically flat” assumption. $\endgroup$– A.V.S.Aug 30, 2020 at 6:53