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The interior of a curved cylinder is an Einstein manifold (the Ricci Curvature Tensor is proportional to the Metric $R_{\mu\nu}=kg_{\mu\nu}$) since it has a constant curvature.

Using the metric $$ {g_{\mu \nu }} = \left[ {\matrix{ { - 1} & 0 & 0 & 0 \cr 0 & {{{{C^2}} \over {{C^2} - 2{\pi ^2}{x^2}}}} & 0 & 0 \cr 0 & 0 & {{{{C^2}} \over {{C^2} - 2{\pi ^2}{y^2}}}} & 0 \cr 0 & 0 & 0 & {{{{C^2}} \over {{C^2} - 2{\pi ^2}{z^2}}}} \cr } } \right] $$ and where $C$ is the circumference of the cylinder.

However, I was unable to derive the constant of proportionality $k$.

May I know how one may find this constant?

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Well the handy mathematica notebook I have says $R_{\mu\nu}=0$, and actually, just having a look it seems like you can redefine your $x,y,z$ to make the metric constant. So your space is actually flat. –  Michael Brown Apr 27 '13 at 3:55
i.e., define $x' = \frac{C}{\sqrt{2}\pi} \sin^{-1}\left(\sqrt{2}\pi \frac{x}{C} \right)$ and similarly for $y,z$. Then your metric is Minkowski in terms of $t,x',y',z'$. :) –  Michael Brown Apr 27 '13 at 4:00
I was just writing that up when I noticed @Michael Brown already did it hours ago. He's right in that this is Minkowski (or rather, locally so); being a "cylinder" probably comes from identifying the coordinates so that the manifold 'wraps around'. 'Cylinder' is therefore bad naming; it should be called a flat $3$-torus. –  Stan Liou Apr 27 '13 at 9:14

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