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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?

Waffle's Crazy Peanut
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ConciseAndClear
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    Related question by the author: http://physics.stackexchange.com/q/61592/11062 – 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
  • Hi user. I can see that. That's why I didn't say that this question is an exact duplicate of the previous one ;-) – Waffle's Crazy Peanut Apr 19 '13 at 13:11
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    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
  • Please don't take it pessimistically. It's not like that. The "getting answers" thing is based on how good the question is, and how good it has been defined. Then, users definitely try to answer the question. Moreover, you can't say that the question can't be answered. I've seen many duplicate questions getting answers. Yours is not like that ;-) – Waffle's Crazy Peanut Apr 19 '13 at 13:18
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    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 $\mathscr{I}^+$. –  Apr 19 '13 at 16:44
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    @BenCrowell thanks. I'm still not clear on the naked singularity case though. I looked at the Pensrose diag for Schwarzschild http://upload.wikimedia.org/wikipedia/commons/5/5b/Diagramme_Penrose_Kruskal.png 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 06:47
  • @user2206544: No, it's not because it hits the event horizon. It's because a lightlike geodesic from the singularity can only run diagonally down and to the right, to $\mathscr{I}^-$, not to $\mathscr{I}^+$. –  Apr 21 '13 at 19:25

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