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Charged Black Holes

  1. Jul 17, 2006 #1
    Let's assume we have a black hole with a net charge Q. If this charge is a substantial value, wouldn't this imply different event horizons for particles of different charge values? Wouldn't there be a black hole charge to mass ratio where the event horizon completely disappears for particles with the opposite charge sign?

    If so, doesn't this imply you could "feed" a black hole until its Q is such that the event horizon for certain particles disappears, and one could probe the singularity directly?

    Example: I have a black hole of 10 solar masses, and a charge of +10^12 coulombs. Electrons fall into the black hole easily, while protons can travel much closer to the center before falling in.

    Thoughts?
     
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  3. Jul 17, 2006 #2

    pervect

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    It is possible to have a "naked singularity" if the charge / mass ratio of a black hole is too high.

    One reference:

    http://staff.science.uva.nl/~jpschaar/report/node35.html

    However, clasically speaking, such a naked singularity would have to be created naked - it could not be formed by adding charge to an existing black hole. (I'm pretty sure this is discussed in MTW, but I'd have to look up the details to referesh my memory).

    Basically, to force the required amount of charge into the BH would require adding even more mass to it, so that a BH that started out not being a naked singularity would not become a naked singularity when the charge was added.

    I don't know enough about the quantum mechancial aspect to address the quantum mechanical issues (such as naked singularities being formed by evaporation of charged black holes mentioned on the webpage). I would tend to doubt very much that this is possible, but I don't have any calculations to demonstrate it.
     
  4. Jul 17, 2006 #3
    Is this because as the event horizon shrinks, adding more charged particles to the singularity becomes increasingly difficult?
     
  5. Jul 17, 2006 #4

    pervect

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    The event horizon never winds up shrinking. That's one of the laws of classical black hole mechanics - the area of the event horizon never decreases. (Again, I'm not going to address the quantum issues).

    If it were possible to add only charge to a black hole without adding mass as well, you could make the event horizon decrease. But it's not possible to do so.

    For example:

    http://en.wikipedia.org/wiki/Reissner-Nordström_black_hole

    puts the outer horizon at
    M = M + sqrt(M^2 - Q^2)

    Note there is not a separate one for charged vs uncharged particles however. Increasing Q -> decreasing horizon area.

    But it is not possible to add only charge to a black hole without increasing its mass. It requires work to add more charge to a negatively charged black hole. Imagine the classical case - you have a charged sphere, and you want to add more charge to it. To bring the additional charge to the sphere requires energy. To shoot a particle into the sphere, it would have to have a high enough velocity to overcome the repulsive potential barrier.

    In the black hole case, this energy winds up increasing the mass of the black hole. The mass of the black hole always increases by the "energy at infinity" of the particles you are shooting into it. This energy will be gamma*rest mass, where gamma is the relativistic gamma factor associated with the velocity at infinity of the particles being shot into the hole.

    This added mass is always enough so that the horizon area increases, and never decreases.
     
    Last edited: Jul 17, 2006
  6. Jul 17, 2006 #5
    Ah, I was forgetting to include the kinetic energy of the particles when considering the mass added to the black hole.

    What if you positioned a spaceship right outside the event horizon, which caught charged particles speeding towards event horizon, slowed them down to almost 0 velocity, and then deposited the particles into the black hole. Wouldn't this increase the mass of the black hole only by the rest mass?
     
  7. Jul 17, 2006 #6

    Office_Shredder

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    Guillochon, think of it this way.

    You're trying to make a black hole with a positive charge capable of pushing protons away, despite the massive gravitational pull. Because of the nature of electrical vs. gravitational force, the electrical force is going to be capable of pushing the proton away no matter how close it gets. However, imagine if you go back one step. How did the last charged particle enter the black hole if the black hole is already capable of repelling charged particles?
     
  8. Jul 17, 2006 #7
    Well, that's why I now think that eliminating the event horizon would probably be impossible, since it would take infinite energy to completely eliminate it. However, I am not convinced that the black hole's event horizon cannot shrink at all.

    What if we start with a completely neutral black hole, and we feed it zero-velocity electrons? Shouldn't the event horizon shrink, since the charge/mass ratio of the electron is large?

    BTW, I always have difficulty interpreting equations where the fundamental constants are set to 1. I can't figure out how to get an actual number out of the equation in pervect's link because of the "convenient" units used.
     
  9. Jul 18, 2006 #8

    pervect

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    See the Wiki article on the second law of black hole mechanics:

    http://en.wikipedia.org/wiki/Laws_of_black_hole_mechanics

    The weak energy condition basically says that you don't have negative mass. If you had a particle with a negative rest mass, you could put it into a black hole and have it's horizon area decrease.

    Quantum mechanics muddies the water considerably, unfortunately, but clasically speaking the area of the event horizon does not decrease.

    There is some discussion of this in the textbook "Gravitation" on pg 910, the same conclusion is reached. The detailed equations are a bit messy, but the same conclusion is reached.


    There's some info in the Wikipedia

    http://en.wikipedia.org/wiki/Geometrized_unit_system

    but their treatment of charge is really not very clear.

    Basically, you can say in geometric units that

    c=1
    G=1
    [itex]4 \pi \epsilon_0[/itex] = 1

    (The last expression is motivated by making the Coloumb force law constant equal to 1. The Columb force law constant is equal to k = [itex] 1/4 \pi \epsilon_0[/itex]

    One usually converts all quanties to units of distance (or distance^n).

    Conversion factors to do this are given below. Note that they have all been defined to be 1 or are the products of constants defined to be 1, that's why they are acceptable conversion factors. (For the case of charge, we also have to take a square root). Note that most textbooks will be using cgs units for E&M, not the MKS units below.

    c: converts time to distance
    G/c^2: converts mass to distance
    [tex]\sqrt{\frac{G}{4 \pi \epsilon_0 c^4 }[/tex] : converts charge to distance

    Numerical values (from google calculator)

    c = 299 792 458 m / s
    G/c^2 = 7.42471382 × 10-28 m / kg
    [tex]\sqrt{\frac{G}{4 \pi \epsilon_0 c^4 }[/tex] = 8.61667791 × 10-18 m / amp-sec
     
    Last edited: Jul 18, 2006
  10. Jul 18, 2006 #9
    Mass of the electron converted to distance: 6.76e-58
    Charge of the electron converted to distance: 1.38e-38

    This means we can change the above equation to calculate the event horizons to the following for a 10 solar mass black hole (x is number of electrons we add. We assume that the black hole starts out electrically neutral. 2e21 is the ratio of the charge of the electron to the mass):

    [tex]
    1.48\times10^4 \pm \sqrt{(1.48\times10^4 + \frac{1.38\times10^{-36} x}{2\times10^{21}})^2 - (1.38\times10^{-36} x)^2}
    [/tex]

    Now add 10^40 electrons. The outer radius drops from 30km to 20km, while the inner radius goes from 0km to 10km. Unless this equation has some sort of caveat, this means the event horizon is shrinking for negatively charged particles, while increasing for positive ones. The weak energy condition claims this cannot occur though, so what am I missing here?

    EDIT: Several edits to make the equation look right in tex, still getting used to it.
     
    Last edited: Jul 18, 2006
  11. Jul 18, 2006 #10

    pervect

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    Your calculation is not correct.

    You have to add the "energy at infinity" to the mass of the black hole, not the rest mass of the electrons.

    The "energy at infinty" includes the potential energy of the electron, eA. In addition, you also have to deal with the metric coefficients, if you insist on dropping them "from a spaceship", a complicaiton that you are not addressing at all (and that I don't know how to explain fully without using tensors). Frankly I would suggest that you avoid doing that at this stage, and simply address the problem from the POV of an observer at infinity.

    The exact formula for "energy at infinity" for a charge e is just

    [tex]
    E = -p_t - eA_t
    [/tex]

    p is the covariant energy-momentum 4-vector, A is the covariant vector potential, and e is the charge. (See for instance MTW pg 899, or whatever GR textbook you can lay your hands on that talks about this issue).

    To avoid dealing with the metric coefficients and covariant vs contravariant vectors, which you are probably not familiar with, imagine an electron with the same energy escaping to infinity (because of a course diversion) rather than falling into the black hole.

    The energy at infinity, E, will in that case be just the energy of said electron - assuming that the electron does not radiate EM radiation on its outward journey due to its accleration.

    The apparent minus sign in the above formula is due to the "covariant vs contravariant" issue, and the sign convention of the metric: at infinity pt = - pt because g00 = -1. Thus a minus pt corresponds to a positive pt for the Minkowski metric -dt^2 + dx^2 + dy^2 + dz^2 "at infinity".
     
  12. Jul 18, 2006 #11
    The "dropping from a spaceship" part is a way to ensure that the electrons cross the event horizon with almost no momentum. With a mere undergraduate understanding of GR, you're right in that I would probably not get the tensor explanation at this point.

    Is the reason we treat the electron as falling from infinity because of time dialation near the singularity? The electron basically accelerates for eternity until it would have the speed it would have falling from an infinite distance?
     
  13. Jul 18, 2006 #12

    pervect

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    You get all kinds of weirdness when you deal with objects near the event horizon.

    For instance, relative to a hovering Schwarzschild observer, any infalling object appears to move arbitrarily close to 'c', the closer it gets to the event horizon.

    Relative to an observer at infinity, however, the infalling object appears to stop and have zero velocity.

    With proper bookeeping, this can all be accounted for, but one needs tensors to even be able to understand the notation used to keep the books :-(.

    There is also a coordinate singularity at the horizon in the usual Schwarzschild coordinate system that makes the bookeeping tricky even with tensors. (This can be avoided by using a different non-singular coordinate system but is another pitfall).

    The easy way around these obstacles is to look at what happens at infinity, where things are simple, and space-time is flat.

    There are deeper reasons as well - having a flat space-time at infinity turns out to be very important to even defining the concept of mass in GR.

    For a quick discussion of some of these deeper issues, you might take a look at

    http://www.math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html

     
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