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Blow up a black hole by filling it with electrons?

  1. Jul 29, 2010 #1
    This may be a stupid question but what would happen to a black hole if you point a stream of electrons at it and keep it up indefinitely, until it's made almost entirely of electrons repulsing each other, with no protons to balance the charge? Or is the principle of charge conservation violated in a black hole?

    Wouldn't the repulsive electromagnetic force of the electrons eventually become greater than the force of gravity that's binding the black hole?
  2. jcsd
  3. Jul 29, 2010 #2
    Yes, this is why there is an upper bound on the charge due to the mass.


    [tex]Q^2 + (\frac{J}{M})^2 \leq M^2[/tex]

    Where Q is the charge of the black hole and J is the angular momentum. 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.
  4. Jul 29, 2010 #3
    Hey, thanks for the answer. So is a naked singularity going to happen, and if not, what mechanism will prevent the negative charge from accumulating? You seem to imply that something will happen that will prevent us from seeing the naked singularity.
  5. Jul 29, 2010 #4
    Disclaimer: Anything below may be inexact, based on misunderstanding or simply incorrect.

    Cosmic censorship is still just a hypothesis. There are known solutions that lead to naked singularities and simulations seem to create them as well. It's possible the solutions are "too exact" in the sense that collapse of a real star would be infinitely unlikely to become a naked singularity and similarly that some simplification in the simulations makes them naked singularities while they should really have an event horizon. There's no known mechanism that would prevent the scenario in your original post.
  6. Jul 29, 2010 #5
    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)

    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.

    But they seem to miss a point - if you put energy inside the electrons you can still hurl them and *overcome* the still mild repulsive force (which would still be only as strong to cancel out the attractive gravity force but not repel very strongly). So for a time the black hole could have a negative repulsive force greater than its gravity force. What would happen to electrons inside it then? May be it will begin to spit out electrons until its electric charge becomes less than or equal to its gravity.
  7. Jul 29, 2010 #6


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    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.
  8. Aug 3, 2010 #7
    Here is another thread on a similar theme. https://www.physicsforums.com/showthread.php?p=2024222
    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. Hawkings 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.
    Lets look at this objection.

    The inequality quoted by xepma can be written like this:

    [tex]Q^2_* + (\frac{J_*}{M_*})^2 \leq M^2_*[/tex]


    [tex]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}[/tex]

    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:

    [tex]Q^2_* + (\frac{J_*}{M_*})^2 = M^2_*[/tex]

    and for the case when J* = 0 it is fairly easy to obtain:

    [tex]Q = rc^2 \sqrt{\frac{4\pi\epsilon_o}{G}} = \frac{rc^2}{\sqrt{GKe} }[/tex]

    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

    [tex]F= \frac{Q_1Q_2}{r^2Ke}[/tex]

    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:
    and a similar entry states:
    I believe the above Wikipedia statements are not correct. First of all, an electron has [tex]2R_Q >>> R_S[/tex] 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:

    [tex]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}[/tex]

    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.

    I think it would be better to say "These are forbidden by the unproven cosmic censorship conjecture".
    Last edited: Aug 3, 2010
  9. Aug 3, 2010 #8


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    It has been known for a very long time that extremal (and superextremal) black holes are valid solutions of Einstein's equation. The issue with cosmic censorship has always been whether these things can arise "in practice." The meaning of that qualifier is imprecise and has changed several times as various counterexamples have been found. As far as anyone knows, however, there is no generic mechanism to create a naked singularity from nonsingular initial data. You can construct strange examples in precisely-tuned systems, but that's about it.

    You should not be using Newtonian equations here. This is a regime where order of magnitude estimates will not work. Regardless, the force here doesn't matter. Neither does the small rest mass of an electron. Electrons must be thrown into the black hole at relativistic speeds in order to overcome the electrostatic repulsion. Against my own advice regarding Newtonian equations, I'll estimate the energy of that electron as at least

    E = m_e + e Q/R,

    where [itex]m_e[/itex] is the rest mass, [itex]Q[/itex] the initial charge of the BH, and [itex]R = M + (M^2-Q^2)^{1/2}[/itex] its radius. For a near-extremal BH, [itex]Q/R \sim 1[/itex]. As you mentioned, [itex]e/m_e \gg 1[/itex], so such an electron is ultrarelativistic.

    After absorbing this electron, the BH's new charge-to-mass ratio is something like

    \frac{e+Q}{M+m_e + e Q/R} \simeq \frac{Q}{M} + \frac{e}{M} (1-Q^2/M R).

    As the BH gets more extremal, the charge-to-mass ratio increases less and less. A careful treatment of this problem was first given by Wald: Ann. Phys. 82, 548 (1974). He found that you can't overcharge the black hole by shooting in charged particles in the test body regime. Much more recently, http://xxx.lanl.gov/abs/gr-qc/9808043" gave an argument that it might be possible. She finds a very small region of parameter space where things might work, but her equations could (and probably are) destroyed by the inclusion of self-force effects. This is something people are working on checking right now.

    An electron is not known to be described by a Kerr-Newman solution to Einstein's equation. You can't make statements like this without that knowledge.

    This is wrong. Extremal black holes have genuine curvature singularities that can be observed by external observers. See GR textbooks (like Wald). There are three "apparent singularities" in the standard RN metric. They are the two you've given plus one at [itex]r=0[/itex]. This last one always remains, and is a genuine curvature singularity.
    Last edited by a moderator: Apr 25, 2017
  10. Aug 8, 2010 #9
    In theory, increasing the rotational velocity would cause a similar effect, would it not?
  11. Aug 8, 2010 #10


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    Yes. There are similar no-go results for trying to overspin a black hole by shooting in offset masses.
  12. Aug 8, 2010 #11
    Thanks for the quick answer Stingray. :smile:
  13. Sep 18, 2010 #12
    I found a good article on the subject at "www.physics.umd.edu/grt/taj/776b/chappell.pdf"[/URL]
    Last edited by a moderator: Apr 25, 2017
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