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So I've been doing a lot of research on spinning black holes and fumbled upon the concept of a "naked singularity" where a gravitational singularity exists without an event horizon. Ignoring the physics-breaking problems and questions it raises, how would this behave in comparison to a spinning black hole?

For example(s): Would it have an accretion disk? Would matter still get spaghettify if it got too close? Could you get in and out to observe it while once you pass the even horizon of a reg black hole, you can't get out?

tl;dr, would it act the same as a regular spinning black hole?

Qmechanic
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buiud
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    It depends a lot on the type of naked singularity - different GR solutions have very different singularities. If it has curvature that increases as the singularity is approached one gets gravitation-like effects, but there are AFAIK crease- and point-like singularities with no increasing curvature too. – Anders Sandberg Nov 03 '19 at 10:06
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    If the singularity is naked, then predictability is lost and you cannot say anything. So, you can have an accretion disk or you could have a green slime monster as the standard example goes. – MBN Nov 04 '19 at 11:34
  • related: https://physics.stackexchange.com/questions/455726/why-exactly-are-singularities-avoided-or-deleted-in-physics –  Nov 04 '19 at 13:51
  • You could get in and out to observe a naked singularity. It has no event horizon by definition, which means it's necessarily true that you can observe it – Jim Nov 04 '19 at 14:26
  • @safesphere : The big bang and crunch singularities are not naked. At least that is the standard terminology. – MBN Nov 05 '19 at 10:29
  • @safesphere inside a Schwarzschild black hole, there is always an event horizon between you and the singularity because every surface below the event horizon is another surface that light cannot come back from. That singularity never looks naked – Jim Nov 05 '19 at 20:30
  • @safesphere Yes, actually, it does mean exactly that we have a horizon at midnight. You can reword the problem so that it sounds ridiculous all you want but the inside of a black hole is a ridiculous place. An observer falling in would constantly see light from above them and never see light coming from below them after crossing the event horizon. It would always appear like there is a horizon below them. The singularity would never be observable. If you want to say this is like having a horizon at midnight, then fine, but the fact is that the singularity wouldn't be observable – Jim Nov 07 '19 at 16:27
  • @safesphere I'm not confusing anything. I've analyzed the solutions as well. I'm aware of the timelike nature of space beyond the event horizon. Whether you're inside or outside a BH, the singularity is obscured for the same reason; timelike space immediately surrounds it and you can't see light from the future. But while the outer, lightlike boundary between regions is the event horizon, every layer beneath it is an effective horizon as it shields the singularity from view. That's why it's called a "horizon". It makes anything past it unobservable, whether or not space is timelike. – Jim Nov 08 '19 at 13:15

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Assuming you are asking about the type of naked singularity you get if consider the Kerr solution with spin greater than 1.

In some respects this solution behaves just like a Kerr black hole. For example, there still is an innermost stable circular orbit. Hence it could still have an accretion disk with an inner edge. Tidal effects would also still diverge as you approach the singularity leading to "spaghettification".

However, there also some crucial differences: For example there is no photon sphere, meaning the object would look very different when view by the event horizon telescope. Also, most matter falling toward the singularity will miss the singularity and be flung back out.

(And of course, there is the singularity itself, for which we have no clue how it would interact with other objects or fields.)

TimRias
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  • not saying I don't believe you LOL; you're def more knowledgeable than I am and with the research I've done, I agree. Though I'm confused on the "most matter falling toward the singularity will miss the singularity an be flung back out." Could you explain how/why that works? – buiud Nov 07 '19 at 03:36