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Let's assume a universe with only some ($n$) single point masses $m_i$ in it. The point masses have initial positions in space-time, $x_{i0}$. The spacetime between them is curved due to general relativity. (n-body-problem)

The curvature of the spacetime between those single point masses must be a solution of Einstein's field equations. However, they are difficult to solve and there aren't solutions for the n-body-problem yet (to the best of my knowledge).

The curvature of spacetime is fully described by the metric tensor. (Or did I get something wrong here?) To derive the metric tensor, Einstein's field equations would have to be solved.

I'm wondering whether it is possible to approximate the true, curved, metric tensor at every point using different cartesian metric tensors. That would be the same principle as doing infinitesimal small steps when calculating the derivative of a function f(x). It's like using the flat tangent space-time at every point to describe the curved space-time.

Does this work? Has it been done somewhere in the literature? Are there circumstances for which this approximation doesn't work anymore?

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What you describe is the basic idea behind Regge calculus. In Regge calculus, spacetime is approximated by a triangulation of (flat) simplices. All curvature is pushed to live on the codimension-2 edges (2-faces) between the simplices. The curvature on each 2-face can be described by the holonomy of closed curves around the 2-face. The Einstein equation reduces to conditions on the holonomies of the 2-faces meeting at each 1-face.

Any solution to the Einstein equation can be approximated in this way.

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  • $\begingroup$ Is it possible to use cubes as well? $\endgroup$
    – MartyMcFly
    Commented Oct 24, 2023 at 7:19
  • $\begingroup$ I would like to use cubes (4D cubes) instead: A cartesian grid. For simplicity. Could that be possible as well? $\endgroup$
    – MartyMcFly
    Commented Oct 24, 2023 at 10:13
  • $\begingroup$ @MartyMcFly If you mean literal cubes (all sides, and faces the same size, and parallel), then I doubt that would be possible. If you just mean using rectangular grids, then that is what some numerical relativity codes are doing, e.g. the Einstein Toolkit (einsteintoolkit.org) $\endgroup$
    – TimRias
    Commented Oct 24, 2023 at 11:55
  • $\begingroup$ Helpful link, thank you! Why you doubt that would be possible with literal cubes? If I force the grid having all the same length, in my opinion the result is just: less precision in some direction. (??) Another question is regarding time: I would time just give the inverse of the volume of the cubes. Is that right as an approximation?? $\endgroup$
    – MartyMcFly
    Commented Oct 24, 2023 at 12:22

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