# Cosmology Question-What Happens When The Stars All Burn Out?

**URL:** <https://boards.straightdope.com/t/cosmology-question-what-happens-when-the-stars-all-burn-out/106020>\
**Category:** Factual Questions\
**Created:** [April 27, 2002, 11:45pm UTC](https://boards.straightdope.com/t/cosmology-question-what-happens-when-the-stars-all-burn-out/106020 "2002-04-27T23:45:42Z")\
**Posts on this page:** 1\
**Page:** 2

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**Author:** ![CanTak](https://avatars.discourse-cdn.com/v4/letter/c/ecae2f/32.png) [@CanTak](https://boards.straightdope.com/u/CanTak)\
**Post date:** [May 1, 2002, 10:58pm UTC](https://boards.straightdope.com/t/cosmology-question-what-happens-when-the-stars-all-burn-out/106020/21 "2002-05-01T22:58:05Z")

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> [@](#):
>
> \*Originally posted by Chronos \*  
> \*\*I’ve never heard of a ring black hole. The closest I can think of is a Kerr (rotating) black hole, where the central singularity is a ring, but there’s still only one event horizon, and it’s still shaped like a (squashed) sphere. Kerr black holes will evaporate just like Schwartzschild ones; in fact, black holes will lose angular momentum (and charge) much faster than they lose mass, so the final stages of a Kerr hole will be Schwartzschild anyway. \*\*

Not always, I was specifically talking about a Kerr black hole, yes, however they can have two event horizons. If the rotation is fast enough, these can cancel each other out leaving a naked singularity that is completely in view. It’s also important to realize that ONLY in a Schwartzchild model is the singularity zero radius and infinite density. A Kerr black hole would have a ring that is close to zero thickness but not quite, and the density is not infinite. So the question still posed is: What happens when this specific type of black hole cannot evaporate (since it cannot pull in one particle of a two particle virtual pair anymore without gravity around it) and it cannot deteriorate as nothing can leave its surface. There would be no gravity at this black hole except exactly at its surface and below, hence any decay would remain exactly where it was before.

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