Throughout the literature Wormholes are typically constructed by "Minkowski" or "Schwarzschild Surgery" (see e.g. Visser, Lorentzian Wormholes...), i.e. under quite simple and/or highly symmetric circumstances.

In the former case, regions are excised from a single manifold and their boundaries identified, and in the latter, two manifolds are joined.

In the Minkowski case, the joining is trivial (modulo orientational considerations) since the metric is constant (-1,1,1,1), and in the Schwarzschild case shell Jump Conditions (ref, Israel, Darmois, Lichnerowicz, O'Brien & Synge - of which I have only been able to track down a copy of Darmois online) are applied.


If one were to attempt similar surgery on a more general spacetime foliated by Cauchy hypersurfaces, adopting the approach that two regions are excised and identified on each slice, what conditions (metric, derivatives, etc.) should be imposed on the spacelike identifications within surfaces and what conditions on the timelike identifications between surfaces?


1 Answer 1


Well general relativity doesn't really have any standard condition on the smoothness of the metric tensor, but generally the metric is assumed to be at least $\mathcal C^2$, so that the Levi Civitta connection will be $\mathcal C^1$ and the Riemann tensor $\mathcal C^0$. Minkowski surgery usually involves a bit of a weaker condition in the thin shell formalism, with the metric being $\mathcal C^0$, the connexion with some step discontinuities and the Riemann tensor some Dirac distributions. Since the computations done on the Riemann tensor are generally linear, this works out alright.

So the surgery should have the metric at least have the same value along the stitching, and perhaps better if you could identify up to second derivatives as well.

  • $\begingroup$ [Back from travels...] Thanks, and though that is all true (I mean, sounds right to me) I was trying to work out what happens to two Eulerian geodesics crossing a join - each losing one half in the cut process and becoming a complete geodesic again by joining up . My intuition, based on geodesic uniqueness is that one should end up with a single smooth geodesic... would that be an equivalent condition to the $C^{n}$ conditions you described? $\endgroup$ May 13, 2016 at 14:35
  • $\begingroup$ The uniqueness of geodesics is based on the Picard–Lindelöf theorem, which states that for a Lipschitz continuous differential equation, a first-order initial value problem will have a unique solution. This is not the case if the metric is $\mathcal C^0$, as the connection will fail to be continuous. I don't think we can guarantee geodesic uniqueness in those circumstances. You might get some odd behaviours trying to join geodesics at the joint. $\endgroup$
    – Slereah
    May 13, 2016 at 15:02
  • $\begingroup$ OK, but if as you suggested the metric is at least $C^{2}$, does that theorem then say, yes, the geodesic edit works as I suggested? $\endgroup$ May 13, 2016 at 15:07
  • $\begingroup$ If the metric is $\mathcal C^2$, then you have no problem concerning the uniqueness of the geodesic, I'm not sure it is required to be smooth, though. Just consider the equation $\ddot x + \vert x \vert \dot x^2 = 0$. The connection is continuous, but I don't think you'll get a smooth geodesic out of it. $\endgroup$
    – Slereah
    May 13, 2016 at 15:28

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