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I think I read somewhere that knowing the curvature on a 3d slice of space-time is enough to determine the curvature of all 4D space-time everywhere?

Sort of like analytic continuation?

Is this true? How so?

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  • $\begingroup$ This seems a bit counterintuitive to me. I may be wrong, though. Is the spacetime without $T_{\mu\nu}$? $\endgroup$ – Sayan Mandal Oct 10 '17 at 1:52
  • $\begingroup$ "determine all of space-time" is not the same as "determine the curvature of all space-time". $\endgroup$ – Prahar Oct 10 '17 at 3:58
  • $\begingroup$ Related: physics.stackexchange.com/q/277872/2451 $\endgroup$ – Qmechanic Oct 12 '17 at 11:25
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Vacuum solutions of Einstein's equation tend to be analytic, (this is a selection effect. Analytic solutions are easier to work with). Consequently, if you have an analytic solution of the vacuum Einstein equations on some patch, you can use analytic extension it to the entire spacetime. Since the equations themselves are analytic, it is then guaranteed that this analytically extended solution is indeed a vacuum solution of the Einstein equation everywhere.

For example, this how you obtain the maximally extended version of the Schwarzschild metric (i.e. the Kruskal Szekeres metric).

However, this process won't work if you introduce a non analytic source (e.g. the universe), because then neither equation nor solution will be analytic.

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I believe you are talking about Cauchy surfaces. Spacetimes which admit Cauchy surfaces are called globally hyperbolic. The existence of these surfaces is an important feature to ensure the spacetime has nice notions of causality.

I would not relate this to an "analytic continuation," rather it's more a differential equation with initial values. Einstein's equations are the differential equation, the energy distribution at some time slice $t=0$ serves as an initial value, which in turn defines the metric (and curvature) thereafter.

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  • $\begingroup$ No, the curvature is not the energy distribution. And a patch of spacetime may actually not have a complete Cauchy surface. A Cauchy surface has to be a 3D hypersurface, simplistically a spacelike hypersurface or region covering all of space at one time. For instance if the patch of spacetime is a spatial 3D spherical region, where the curvature is known at one time for for some time, you don't have a Cauchy surface. You can't assume analyticity either. $\endgroup$ – Bob Bee Oct 10 '17 at 5:56
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Knowing the curvature of a patch of space-time determine all of space-time

is a wrong statement. It is enough to analyze the curvature for the onedimensional case to see this. Say your universe is made of a lot of masses, all in one line. From your point of observation you’ll somehow are able to measure the gravitational potential in some distances and from the this gradient you are able to conclude about the distribution of a mass in front of you and backside of you. This two masses are fictional because each of them is in reality the influence of different masses in different distances to you. Moving towards this fictional mass and repeating your measurement of the gravitational potential you’ll find out that the curvature calculations from the first point are diverge more and more.

I want to give an intuitive example. You see a slim rode which is supported on two points in front of you and the left and the right end are hidden behind curtains. Now you are able to measure the bending of the rode. Can you conclude from this about the distribution of masses behind the curtains? Not, you can’t. What you can do is to say that because the left rode end is more bended on this side the product of mass and distance is greater than on the other side. But about the real distribution of masses on the rode you can’t say anything and the curvature of the rode is well known to you only near your point of observation.

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