Can Adding Spin to a Black Hole Create a Naked Singularity?

[JustinLevy]

"generic" naked singularity

I don't understand the arguments/discussion against naked singularities. The reason is that it seems obvious to me that given a black hole, there is a generic procedure to form a naked singularity. This reasoning is probably flawed (otherwise there wouldn't be debate about naked singularities), so if someone could point out the error, that would be quite helpful.

So the suggested procedure goes like this:
1) start with a black hole surrounded by vacuum
2) continue adding spin polarized electrons and protons until J/M > M in the black hole, yielding a naked singularity

This is possible because for an electron or proton, J/M > M. So if you continue adding spin polarized material into the black hole, one can exceed the "nakedness criteria".

The procedure is simple enough, that I can't see where there is a flaw. Any help?Very short discussion on Hawking's stance on naked singularities:
http://www.theory.caltech.edu/people/preskill/nyt_bet_story.html

 

[Passionflower]

1) start with a black hole surrounded by vacuum
2) continue adding spin polarized electrons and protons until J/M > M in the black hole, yielding a naked singularity

The procedure is simple enough, that I can't see where there is a flaw. Any help?

The catch is that your spacetime is obviously not a black hole surrounded by a vacuum.

It is alright to treat a bunch of test particles as 'vacuum' but obviously not so many that it becomes a significant mass-energy part of the total spacetime. Your spacetime is not a R-N metric.

 

[stevebd1]

While spin reduces Hawking radiation there are a number of other factors that extract energy from a rotating BH and probably reduce spin, these are the http://en.wikipedia.org/wiki/Penrose_process" . These would appear to stop a Kerr black hole from attaining a/M=1 and a naked singularity being formed.On a side note, you mention that particles have J/M>M which is presumably based on J=aM where a=J/mc and M=Gm/c², the equations possibly being slightly different for particles. Roughly speaking, for anything smaller than a sol mass, a is always greater than M, for instance J/M>M applies to the Earth, Jupiter and a white dwarf with a relatively rapid spin period of 30 seconds yet we know if any of these objects were to fall into a black hole, there contribution to BH's AM would be slight (unless they were thrown into orbit with great speed and then maybe their orbital angular momentum would contribute). The line/scale where a and M change places isn't clear but it appears to be around the 1 sol mass mark.

 

[Dmitry67]

JustinLevy, you are reding my mind!
I am also sure you can make naked singularities this way.

 

[JustinLevy]

JustinLevy, you are reading my mind!
I am also sure you can make naked singularities this way.

Haha, thanks, but I'm still quite sure there is a flaw here somewhere. I'm trying to dig it out so I can learn from it.

Considering how low my level of understanding is compared to Hawking, Preskill, and Thorne, it would be great hubris and arrogance to believe this procedure is correct for generically forming a black hole. So I'm sure there is a mistake in my reasoning.

The catch is that your spacetime is obviously not a black hole surrounded by a vacuum.

It is alright to treat a bunch of test particles as 'vacuum' but obviously not so many that it becomes a significant mass-energy part of the total spacetime. Your spacetime is not a R-N metric.

I don't see how this creates a general failure to the setup. One can just make sure they are sufficiently far away from the black hole such that the "perturbations" on the black hole due to external matter is negligible. Can't I always choose a distance far enough away that this is true? Or am I missing something important here?

After all, the astronomical black holes people are studying are obviously not "alone" in the universe, yet near the black holes we can quite well model things as moving according to a simple vacuum black hole metric, no?

While spin reduces Hawking radiation there are a number of other factors that extract energy from a rotating BH and probably reduce spin, these are the http://en.wikipedia.org/wiki/Penrose_process" . These would appear to stop a Kerr black hole from attaining a/M=1 and a naked singularity being formed.

While interesting, I'd like to leave considering quantum effects for later. So for now, let's consider whether in normal classical GR one can "generically" create a naked singularity. Maybe we can later return to whether quantum mechanics ruins this.

On a side note, you mention that particles have J/M>M which is presumably based on J=aM where a=J/mc and M=Gm/c², the equations possibly being slightly different for particles. Roughly speaking, for anything smaller than a sol mass, a is always greater than M, for instance J/M>M applies to the Earth, Jupiter and a white dwarf with a relatively rapid spin period of 30 seconds yet we know if any of these objects were to fall into a black hole, there contribution to BH's AM would be slight (unless they were thrown into orbit with great speed and then maybe their orbital angular momentum would contribute). The line/scale where a and M change places isn't clear but it appears to be around the 1 sol mass mark.

Now THIS is fascinating! That sounds very counter-intuitive to me, so this may be on the right track to correcting my understanding.

If a black hole has J=J1 and I throw in the Earth with J=J2, then won't the resulting black hole have J=J1+J2 due to conservation of angular momentum? Similarly, won't the black hole mass increase by the mass of the Earth + whatever kinetic energy the Earth had upon enterring?

I don't understand why throwing in a bunch of planets, if J/M > M, won't eventually cause J/M > M for the black hole? If you could expound upon this, it would be greatly appreciated.

 

[Dmitry67]

1. It is true that if you join 2 almost-extreme black holes, rotating in the same direction, the result is a black hole, 2 times more massive, but 2 times LESS extreme. (because it has 2J/2M, and mass 2M).

However, 'extremality' of elementary particles (in Planck units) is so HUGE that i think it is not a problem at all.

2. I think the problem could be that if you have non-extreme BH and you shine polarized light into it, making it extreme, it works, but it takes almot infinite time for you to see the result. So even if the result is achieved, it takes infinite time to see it. In another words, for the naked singularity to be observed, it must be naked from the very beginning.

 

[Dmitry67]

Hm, may be I was not right.


Say, we have non-rotating BH with mass=sun:

2*10^30kg = 10^38 planks masses

Lets say we inject electrons and positrons:

9*10^-31 kg = 4.5*10^-23 in natural units.

We need to add aprrox M^2=10^76 properly polarized electrons to make it superextreme
with a total mass of 10^53 - 10^15 sun masses...

Hm...

In order for that mechanism to work, the mass of the original BH must be
1/Me = 2.2*10^22 or 4.4*10^14kg

 

[stevebd1]

I don't understand why throwing in a bunch of planets, if J/M > M, won't eventually cause J/M > M for the black hole? If you could expound upon this, it would be greatly appreciated.

The a/M relationship relates to the AM and mass of a specific body, for the addition of objects with angular momentum, the equations for a and M might be written-

$$a_t=\frac{(j_{1}\pm j_{2})}{(m_{1}+m_{2})c}$$

$$M_t=\frac{G(m_{1}+m_{2})}{c^2}$$

where j and m are angular momentum and mass in SI units, sub 1 for the BH, sub 2 for the object entering the BH, + or - for whether the spin is complimentary to the BH or not. If we consider Jupiter entering a 1e+6 sol BH with a spin parameter of 0.95, even with the spin being complementary, the end result for a_t/M_t will be marginally less than the BH's original a/M of 0.95 due to Jupiter's mass being more significant than it's spin.https://www.physicsforums.com/showpost.php?p=2660592&postcount=3" of a BH can not become zero in a finite time (akin to reaching absolute zero) which would be the case with a maximal BH.