satuon said:
Lol, just saw in the Similar Threads section someone has asked about "Stuffing a black hole full of electrons?" (link is
https://www.physicsforums.com/showthread.php?t=392989)
Here is another thread on a similar theme.
https://www.physicsforums.com/showthread.php?p=2024222
satuon said:
So I'm not the first one to think about it. In his thread they reach the conclusion that when the black hole is charged enough it will start to actually repel electrons. So I guess my idea about blowing it up won't work.
Your question has inspired me to look into this a bit deeper and it appears there are ways to overcome the objections and it seems possible to create a naked singularity. Hawking bet that naked singularities could not exit but lost when it was shown that "extremal black holes" are a valid solution. He then set a new bet (as I understand it) that an object that is not initially an extremal black hole (naked sigularity) could not be turned into an extremal black hole by practical means.
Stingray said:
If you add enough energy to the electrons to overcome the electrostatic repulsion, you'll be adding a significant amount of mass (as well as charge) when those electrons are absorbed. As far as anyone can tell right now, this extra mass will always be large enough that the inequality quoted by xepma remains satisfied. There are some possible loopholes to this, but it seems that you can't destroy a black hole by shooting little charges at it.
Lets look at this objection.
The inequality quoted by xepma can be written like this:
Q^2_* + (\frac{J_*}{M_*})^2 \leq M^2_*
where
Q^2_* = \frac{GQ^2}{4\pi\epsilon_o c^4 }\ , \quad J^2_* = \frac{J^2}{M^2c^2} \quad and \quad M^2_*= \frac{G^2M^2}{c^4}
where the variables are qualified by asterisks to distinguish them from the normal meaning of Q, J and M and the above equation is dimensionally correct.
For an exactly extremal black hole the following is satisfied:
Q^2_* + (\frac{J_*}{M_*})^2 = M^2_*
and for the case when J* = 0 it is fairly easy to obtain:
Q = rc^2 \sqrt{\frac{4\pi\epsilon_o}{G}} = \frac{rc^2}{\sqrt{GKe} }
The above equation means that the charge of an extremal Reissner–Nordström (charged non-rotaing) black hole is proportional to its radius which is equal to GM/c^2.
The force acting on a test charge Q2 by the black hole of charge Q1 is
F= \frac{Q_1Q_2}{r^2Ke}
A charged sphere acts as if all the charge is located at its centre as far as calculating the force is concerned so we can simply use the event horizon radius for r in the above equation. For a given charged non rotating black hole, the charge and the mass are proportional to the radius while the force is proportional to the inverse square of the radius. For an near extremal RN black hole, the force required to add the final electron can be made arbitrarily small by starting with an arbitrarily large RN black hole. The above suggests that an extremal black hole can be created without requiring infinitely large forces or charges.
It should also be remembered that an electron has a very very large charge to mass (Q*/M*) ratio equal to approx 2*10^21.
In the other thread George mentions that Thorne calculated the maximum practical angular momentum to mass ratio is about 0.998. I won't show the calculation but it is possible to calculate that if you start with a uncharged black hole just slightly larger than a plank mass with a J*/M* ratio of 0.998 you can end up with an extremal black hole by adding a single electron. The required mass is just over the Planck mass whether or not you take the spin angular momentum of the electron into account. It is not clear to me at the moment whether the intrinsic quantum spin of an electron would contribute to the total angular momentum of the black hole or not, but it works either way. We could of course use a Stern-Gerlach type device to add electrons of the desired spin orientation if required. Anyone any ideas on this?
I am sure there are a lot of subtle aspects to all this, but it would seem that there is no reason why an exactly radially falling (no angular orbital momentum) uncharged particle should not be able to enter a near extremal rotating uncharged black hole and if the black hole is uncharged, there is no reason why it should be any harder to add a radially infalling electron. On the face of it, it seems fairly easy to create extremal black holes.
This Wikipedia entry
http://en.wikipedia.org/wiki/Reissner-Nordström_black_hole states:
Black holes with 2R_Q > R_S are believed not to exist in nature because they would contain a naked singularity; their appearance would contradict Roger Penrose's cosmic censorship hypothesis which is generally believed to be true.
and a similar entry states:
It is not difficult to construct spacetimes which have naked singularities, but which are not "physically reasonable;" the canonical example of such a spacetime is perhaps the "superextremal" M < | Q | Reissner-Nordstrom solution, which contains a singularity at r = 0 that is not surrounded by a horizon. A formal statement needs some set of hypotheses which exclude these situations.
I believe the above Wikipedia statements are not correct. First of all, an electron has 2R_Q >>> R_S and is technically a super-extremal black hole, so super-extremal black holes do exist in nature and are in fact extremely common. What is incorrect about the above statements is the assumption that a super-extremal black hole contains a naked singularity. If fact a super-extremal BH contains no singularities, no matter how compact the mass is.
The location Rz of the singularities in a generic Kerr-Newman black hole is given by:
Rz = \frac{GM}{c^2} \ \pm \sqrt{\frac{G^2M^2}{c^4} - \frac{GQ^2}{Kec^4} - \frac{J^2}{M^2c^2}} = (1/2) ( R_S \ \pm \sqrt{R^2_S -4R^2_Q -4R^2_Q}} = M_* \ \pm \sqrt{M^2_* -Q^2_* - J^2_*/M^2_*} = M_* \ \pm \sqrt{M^2_* -Q^2_* - a^2}
For a black hole with J=0 and Q=0 the singularities are located at 0 and 2GM/c^2 which is the Schwarzschild solution. The physical singularity is at the centre and the coordinate singularity at Rs protects the inner physical singularity from outside view. As charge or angular momentum is added to the BH the inner physical singularity moves outwards and the outer coordinate singularity moves inwards until in the exact extremal case both singularities are located at GM/c^2. The physical singularity is now exposed to the outside world. Any further addition of charge or angular momentum makes the quantity inside the square root negative and the both singularities become imaginary (this is the super-extremal case). In other words the singularities cease to exist and the BH is technically no longer a BH. An electron or any other super-extremal object does not contain any real singularities and Wikipedia seems to have got this wrong. The equations show that this is true no matter how small the physical radius of the super-extremal black hole is. Even if an electron had a radius of less than the Plank length or even if it was literally a "point particle" it would not have a singularity and presumably would not have any other properties that define it as a black hole. I have seen some arguments that the location of the inner singularity is not clearly defined because the solutions are "external". This is a misunderstanding. The solutions are only external to the physical mass, not external to any event horizon. If we assume an extremal BH with all its mass confined to a point at the centre with zero volume, then the equation for Rz tells us where all the singularities are and it tells us that is there is no singularity at the centre of an extremal or super-extremal BH.
Now if singularities can be naked in real world situations and more importantly, if something that was once a black hole can be turned into something that is not a black hole (recovering all the particles and energy and information that went into it) then it has important consequences for our understanding of physics and I am surprised that more effort has not been put into determining a definitive answer to this question.
Determining a definitive answer requires defining the surface gravity of a black hole in a well defined way. (See
http://en.wikipedia.org/wiki/Surface_gravity) One problem with the concept of surface gravity and black holes is that surface gravity is defined in a finite way while the coordinate force of gravity is infinite. If the "real" force of gravity at the surface of a black hole is finite, then the usual understanding that nothing can possibly remain stationary at the event horizon can not be true. We can not have it both ways. We can not say the force is infinite at the event horizon to suit our purpose when interpreting what happens inside a black hole on the one hand and then explain why extremal black holes can not exist because the real force of gravity at the surface of a black hole is finite. We have to jump one way or other and live with the consequences.
xepma said:
If this bound is violated you simply do not have a black hole in the usual sense. Violation of this inequality leads to singularities without a horizon -- naked singularities. These are forbidden by the cosmic censorship.
I think it would be better to say "These are forbidden by the
unproven cosmic censorship
conjecture".
