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Geodesic completeness proofs

  1. Apr 7, 2010 #1
    I'm reading an article (http://arxiv.org/abs/gr-qc/0403075) which proves that a certain spacetime is geodesically complete. It does this by proving that the first derivatives fo all coordinates have finite bounds. My question is why this is enough.

    Is it just a simple ODE result? We know that the geodesic equation locally has a solution given an initial "position" and "velocity", by the basic existence result in ODEs. Hence if we show that this "velocity" is bounded everywhere then then the geodesic can extend indefinitely, because the geodesic equation can be solved everywhere. Is this right?
  2. jcsd
  3. Apr 7, 2010 #2


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    On p. 5, where they say this, they refer to their reference 5, which is a book on ODEs, so it sounds like they are claiming that it is a simple ODE result.

    This sounds convincing to me. The general impression I get is that the test they're applying is a very strict one, so it's fairly clear that if a spacetime passes the test, it's geodesically complete. On the other hand, I think there might be many spacetimes that were geodesically complete but would fail this test.

    If you had a coordinate singularity, I think your spacetime could be geodesically complete but you wouldn't be able to prove it by this test.

    If there is a non-coordinate singularity, but it can't be reached in a finite amount of time, then I believe that is generally considered to be a case where the spacetime is still geodesically complete. I don't know what their test would do in this case -- it might depend on the method they used to prove the limit on the bounds of the derivatives.

    You can also get cases like the following. Suppose you take a Minkowski space, described in the usual coordinates, and restrict it to the subset of events with t<0. Then the space would be geodesically incomplete, because every geodesic would terminate at t=0. However, the space can be extended past t=0, so the incompleteness isn't due to a singularity.
  4. Apr 8, 2010 #3
    Thanks! I think that all sounds right. I'm reassured now.
  5. Apr 8, 2010 #4
    No it couldn't! Every spacetime with a coordinate singularity is geodesically incomplete. This is definitely becase there are values of coordinates for which the geodesic equations have at least an infinite term. As an example of this, see http://en.wikipedia.org/wiki/Rindler_coordinates" [Broken].

    Last edited by a moderator: May 4, 2017
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