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Black Hole Paradox

  1. Dec 2, 2005 #1
    Consider the following thought experiment. Electrons are propelled into an ordinary (nonrotating) black hole until the black hole has a sufficient electrostatic charge so that the electrostatic repulsive force on the next electron is greater than the gravitational attractive force. Since both the electrostatic force and gravity vary inversely with the square of the distance from the black hole, this next electron will experience a net inverse square repulsive force. Now suppose that this next electron is propelled towards the black hole with sufficient initial velocity so that it crosses the event horizon of the black hole. While an observer on this electron would observe the electron cross the event horizon in a finite amount of time, an infinite amount of time will have elapsed in the outside universe. Once inside the event horizon, the electron cannot fall into the singularity due to the net repulsive force, but must at some point be propelled away from the singularity until it recrosses the event horizon. Meanwhile, an infinite amount of time will have gone by in the outside universe into which the electron has re-entered, while the observer on the electron will have experienced the elapse of only a finite amount of time. Is that possible?
    Last edited: Dec 2, 2005
  2. jcsd
  3. Dec 2, 2005 #2
    I'm not an expert, but don't your electrons have mass? as you pump more and more electrons into the black hole the the repulsive force may grow, but the gravitational force is also growing and will always be stronger. that's what makes it a black hole after all
  4. Dec 2, 2005 #3


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    In newtonian gravity, gravitational force is proportional to the inverse square of the distance, but obviously you can't assume the same in general relativity, since if this was true in GR then you could always escape a black hole's pull by applying a sufficiently large force, and there'd be no such thing as an 'event horizon'. GR doesn't express gravity in terms of a 'force' at all, although I guess in the Newtonian limit you should be able to approximate it that way.
  5. Dec 3, 2005 #4
    To be honest, this is a far more complicated problem than you probably realize. The first part of that starts with the fact that gravity is NOT a force in GR. You're using newtonian gravity in a GR regime, which does not work, at all. Not even as an approximation. Its Ptolemy's sphere's to Kepler's orbits (ok, its not that bad, but the analogy to their relative usefulness is...well, useful).

    That said, a particle cannot cross back over the event horizon, such a path is not allowed with a timelike (less that the speed of light) trajectory.

    Also, such a black hole would not remain charged for long, it would radiate electrons via hawking radiation (essentially absorbing positive charge from quantum vacuum fluctuations, leaving electrons produced by the fluctuations free to escape. It is believed that a charged black hole will not exist "in the wild" because of this (although, that does not affect the question at hand).
  6. Dec 3, 2005 #5


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    Your model is already too simplistic. What you need to do to get an accurate analysis for the path of a charged particle is to use the geodesic equation, modified to include the electrostatic force

    This is

    m \left( \frac{d^2 r}{d \tau^2} + \Gamma^r{}_{tt} \left(\frac{dt}{d\tau}\right)^2 + \Gamma^r{}_{rr} \left(\frac{dr}{d\tau}\right)^2 \right) = e E_r

    I've omitted a number of terms which turn out to be zero from the general geodesic eq.

    The charged BH is described by a Reissner-Nordstrom metric.

    [tex]ds^2 = f(r) dt^2 - \frac{dr^2}{f(r)} - r^2 d\theta^2 - r^2 sin^2 \theta d\phi^2 \hspace{.25 in}f(r) = (1-\frac{2M}{r}+ \frac{Q^2}{r^2})[/tex]

    This gives

    [tex]\Gamma^r{}_{tt} = \frac{(r^2-2 M r+Q^2)(M r - Q^2)}{r^5}[/tex]
    [tex]\Gamma^r{}_{rr} = -\frac{M r - Q^2}{r (r^2-2 M r+Q^2)}[/tex]

    E_r does turn out to be Q/r^2

    Howver, gravity is not of the simple form you propose - the static field is due to [itex]\Gamma^r{}_{tt}[/itex] via a much more complex expression than 1/r^2, and there is an additional quadratic term [itex]\Gamma^r{}_{rr}[/itex] that has no interpreation as a "force".

    Alternatively to using the geodesic equations, one could use some of the conserved quantities for this metric - this is computationally more convenient, but doesn't give as much insight.

    While your analysis above is much too simplistic, strange things do happen with charged black holes (much stranger than even those strange things that happen with uncharged black holes).

    An observer can never pass back through the event horizon of a black hole, but under certain circumstances, such as the idealized geometry I gave above, it is possible for an observer to enter such a charged black hole through one event horizon, and leave it through a different event horizion to "another universe".

    However, the above simple geometry, while valid outside the event horizon of a charged black hole, is probably NOT valid inside the event horizon of a black hole, so this is probably not realistic for an actual charged black hole.

    The best paper I know about on the topic of what really happesn to someone falling through a charged BH of the sort likely to form in nature is

    Last edited: Dec 3, 2005
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