# Einstein field equations for infinite cylinder

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.

• A mass distribution can’t have a zero energy-momentum tensor. Do you mean zero outside the cylinder? – G. Smith Jan 17 at 13:58
• @G.Smith Yes, sorry. I was in a rush when I made this post. – Ryan Parikh Jan 17 at 14:07
• 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. – user4552 Jan 17 at 14:26
• @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. – Ryan Parikh Jan 17 at 14:37
• 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? – Alex Trounev Jan 17 at 14:39

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 $$r\to \infty$$.