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I'm trying to understand the notion of a naked singularity on a more mathematical level (intuitively, it's a singularity "one can see and poke with a stick", but I'm having troubles on how to actually show it).

Based on what (little) is written in Choquet-Bruhat's, a naked singularity is the one for which we can extend the outgoing time-like geodesics to infinity. Now, I was wondering, assuming I had a given solution, how would I "test" the nakedness of the singularity? A natural thing to do would be to write the solution in some null coordinates, but what then? How to do I actually combine it with the (rather abstract) definition of a naked singularity?

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Related question by the author: – Waffle's Crazy Peanut Apr 19 '13 at 13:06
@CrazyBuddy I disagree. This question is very different from the one I asked about extendability. This one concerns the usage of null coordinates – ConciseAndClear Apr 19 '13 at 13:09
well, the problem is, with your tag it now looks like my question "may" already have an answer, which significantly lowers its chances of being answered... – ConciseAndClear Apr 19 '13 at 13:13
As an example, look at the Penrose diagram for the Schwarzschild metric. You have a pair of null coordinates you can define if you like, along axes that lie diagonally at slopes of +1 and -1. It's clear just by looking at the diagram that you can't extend a timelike or lightlike geodesic forward in time from the singularity to $\mathscr{I}^+$. – Ben Crowell Apr 19 '13 at 16:44
@BenCrowell thanks. I'm still not clear on the naked singularity case though. I looked at the Pensrose diag for Schwarzschild… and, if I understand it right, if I try to extend a geodesic coming out of r=0 (above) then I hit the r=2M and that's why it's inextendable, right? But this is more of an intuitive explanation. My question originally was more laong the lines of: how does a naked singularity mathematically look in null coordinates? – ConciseAndClear Apr 20 '13 at 6:47

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