Consider a cylindrical mass in empty space oriented along the x1 direction. In general relativity, do the space-time dimensions (x1, x2, x3, x4) curve about the mass in a completely symmetrical manner?
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2 Answers
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You're asking about the Ricci curvature tensor $R_{\mu\nu}$ provided the stress-energy tensor $T_{\mu\nu}$ (a local mass), i.e., a solution to Einstein's equations. The easy way to solve this is to realize since flat space is isotropic and homogeneous, only a spherical mass will produce a symmetric curvature (assuming the object is inertial). So the answer is no.
answered May 2, 2017 at 19:48
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$\begingroup$ The OP didn't rule out an infinitely long cylinder, and perhaps in that case you would have symmetry. They also didn't rule out the cylinder spinning, say on it's long axis. In that case neither the cylinder nor the sphere would produce symmetric curvature. right? $\endgroup$ May 2, 2017 at 20:18
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$\begingroup$ So then, for a stationary cylinder of finite length, can we conclude that the spacetime dimensions curve differently in different directions? $\endgroup$ May 2, 2017 at 20:24
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$\begingroup$ @docscience You are correct. But if it's infinitely long, you can reduce it to a 2-dimensional problem (where you find axial symmetry). If it's spinning then your life is hell because you need to deal with torsion terms, but you're right again - e.g. Kerr black holes. $\endgroup$ May 2, 2017 at 20:30
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$\begingroup$ @WillCunningham If the cylinder is moving at a constant velocity along the x axis, will the spacetime dimensions curve as it enters a region, and then will the spacetime dimensions return to being flat after the cylinder moves away from the region? $\endgroup$ May 2, 2017 at 20:33
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You need to be more precise about what you mean by do the space-time dimensions curve about it in a completely symmetrical manner?.
The cylinder has axial symmetry. Suppose we use cylindrical polar coordinates $r$, $\phi$ and $z$, then the axial symmetry means the stress-energy tensor is independent of the angle $\phi$. That means the metric will also be independent of $\phi$, which essentially means the spatial curvature is independent of the angle. If this is what you meant by completely symmetrical then the answer to your question is yes.
If the cylinder has an infinite length then the curvature will likewise be independent of $z$, so it will depend only on the distance from the axis.
answered May 3, 2017 at 12:04