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Convex (null) neighbourhoods

  1. Mar 10, 2015 #1
    In general relativistic spacetimes, convex neighbourhoods are guaranteed to exist. As a reminder: a convex neighbourhood ##U## is a neighbourhood ##U## such that for any two points ##p## and ##q## in U there exists a unique geodesic connecting ##p## and ##q## staying within ##U##.

    With that established, does it somehow follow that within a small enough neighbourhood there exist a unique null-geodesic connecting them?

    In this paper the author seem to deduce this in his Proposition 1 -- something that he uses to establish the existence of neighbourhoods for which one can assign coordinates to events by sending and receiving light signals.
     
    Last edited: Mar 10, 2015
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  3. Mar 10, 2015 #2

    jim mcnamara

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    Your link (for me) has problems - Springer.com is often behind a paywall....
     
  4. Mar 10, 2015 #3
    Does it work for you now? The title is "On the Radar Method in General Relativistic Spacetimes".
     
  5. Mar 10, 2015 #4

    Matterwave

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    This doesn't sound right to me, think physically about what you are saying. In SR, a subset of GR, two points which are time-like or space-like connected can't possibly be null connected right?

    And indeed in GR, barring the existence of conjugate points, the boundary of the chronological future of a point is exactly the causal future of that point minus the chronological future of that point. In other words, even in GR, for small neighbourhoods, the null connected points to a point lie on the boundary of the chronological future of that point.
     
  6. Mar 10, 2015 #5

    jim mcnamara

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    Thanks for fixing the link.
     
  7. Mar 10, 2015 #6

    George Jones

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    This is not what the paper states.

    As Matterwave suggests, consider special relativity as an example.

    Let ##\gamma## be a worldine, let ##p## be an event on the worldline, and let ##U## be a neighbourhood that contains ##p##. Consider any event ##q## in the neighbourhood ##U## that is not on the the worldline ##\gamma##. Then, there exists a unique future-directed null geodseic that starts at ##q## and intersects ##\gamma##, and there exists another unique future-directed null geodesic that starts at an event on ##\gamma## and runs to ##q##.

    If ##p## and ##q## are not lightlike related, these null geodesics do not intersect ##\gamma## at ##p##, i.e., these null geodesics do join ##p## and ##q##.

    The null geodesics might leave the neighbourhood ##U## before they intersect the worldline ##\gamma##, but everything can be contained in some larger neighbourhood ##V##.

    See Fig, 1 from the paper. In Fig. 1, ##p## and ##q## are not lightlike related, and the null geodesics intersect ##\gamma## at ##\gamma \left( t_2 \right)## and ##\gamma \left( t_1 \right)##, not at ##p##.
     
  8. Mar 11, 2015 #7
    I agree with you that what the paper states is that one can find two neighbourhoods ##U,V## where ##p\in U \subset V## such that for any ##q \in U\ \text{Im}(\gamma)## there exists a unique future pointing geodesic, as well as a unique past pointing geodesic -- that stays within ##V## -- connecting ##q## to ##\gamma##.

    But how do we prove this from convexity?

    I would presume that we first take ##V## to be a convex neighbourhood and let ##\gamma## be a worldline of some observer going through ##V##. By convexity of ##V##, we can then connect any point ##q## in ##V - \text{Image}(\gamma)## to a point ##r \in \text{Image}(\gamma)## by a unique geodesic that stays within ##V##. This geodesic might be spacelike, null, or timelike. Now, I imagine sliding the point ##r## along ##\gamma## until this unique geodesic becomes null: I guess the subset ##U \subset V## has the purpose of being those point within ##V## than can be connected by a null geodesic, and not just any geodesic, from ##\gamma##. However, I do not see any good arguments on why ##V## has to be such that the geodesic can be made into a null geodesic by sliding ##r## along ##\gamma##.

    Is there such a reason? And how would I prove the statement more formally?
     
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