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Stuffing a black hole full of electrons?

  1. Apr 6, 2010 #1
    What would happen to black hole, if I just kept stuffing it with electrons (and nothing else)?

    It occurs to me that as I keep stuffing them in, the black hole's charge would go up and up, but since the repulsive force of electrons is greater then the attractive force of gravity, could I start sending electrons across the event horizon, or destroy the black hole or something?
     
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  3. Apr 7, 2010 #2

    bcrowell

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    With a rotating black hole, I believe that there is a maximum angular momentum, and what keeps you from exceeding that maximum is that if you try to throw in another particle that would add to the angular momentum, the particle's geodesic actually wouldn't cross the event horizon.

    A charged black hole is described by the Kerr-Newman metric: http://en.wikipedia.org/wiki/Kerr–Newman_metric The Kerr-Newman metric is only a solution of the Einstein field equations if the charge is less than or equal to a certain value.

    My guess would be that something similar happens in the charged case: if you try to keep charging it up once it's reached the maximum charge, you won't be able to do it, because an electron thrown toward the black hole will just rebound due to the electrical repulsion.
     
  4. Apr 7, 2010 #3
    The maximum charge a black hole can attain is when Q=M where-

    [tex]Q=C\frac{\sqrt{G\ k_e}}{c^2}[/tex]

    and

    [tex]M=\frac{Gm}{c^2}[/tex]

    where C is charge in coulombs which are equal to 1 Ampere seconds, also equal to 6.241e18 electrons, G is the gravitational constant, m is mass in kgs and ke is the Coulomb force constant.

    Once Q=M, the inner and outer horizons meet (according to [itex]r_{\pm}=M\pm\sqrt{(M^2-Q^2)}}[/itex]), the http://en.wikipedia.org/wiki/Surface_gravity#The_Kerr-Newman_solution" drops to zero and the black hole should begin to repel (though [itex]\kappa=0[/itex] is only theoretical as it implies a maximal BH which defies BH thermodynamics).

    If spin is involved then the following applies-

    [tex]Q_{max}\equiv M\sqrt{1-\frac{a^2}{M^2}}[/tex]

    where a=J/mc where J is angular momentum.


    You might say that the max number of electrons you could add to a static black hole is-

    [tex]e_{max}=\frac{\sqrt{G}m}{\sqrt{k_e}}\ C[/tex]

    where C=6.241e18 electrons, which for a 10 sol mass static BH is 1.0697e40 electrons, the mass contribution being negligible relative to the mass of the BH (9.7e9 kg), though realistically, charged BHs would neutralize fairly rapidly.
     
    Last edited by a moderator: Apr 25, 2017
  5. Sep 25, 2012 #4
    Thank you stevebd1. Mind a few more questions? How about rotation of the universe?
    http://news.discovery.com/space/do-we-live-in-a-spinning-universe-110708.html
    http://physics.stackexchange.com/questions/1048/what-if-the-universe-is-rotating-as-a-whole

    "We can therefore look at the cosmic microwave background and see whether its anisotropy contains a preferred axis.[Collins 1973] Such observations impose a limit that is tighter than provided by solar-system measurements (perhaps 10^-9 rad/yr[Su 2009] or 10^-15 rad/yr[Barrow 1985]), but such limits are model-dependent."

    We have to be a little careful with with this bound since V = ωr. At 10 billion light years (if i did the math correctly) would result in a tangential velocity 10 times the speed of light...

    By the way, I've always felt inflation is a bit contrived, perhaps a secondary effect of another process. If one were to use the observed rotational characteristics.. http://arxiv.org/abs/1104.2815...."A [Broken] preference for spiral galaxies in one sector of the sky to be left-handed or right-handed spirals would indicate a parity violating asymmetry in the overall universe and a preferred axis."
    1. Extrapolate to include the observable universe to calculate the universes"s angular momentum
    2. since momentum is conserved, the initial big bang particle must have had the same angular momentum.
    3. what is the resulting spin rate & tangential velocity of the initial particle?
    4. What impact would that have on the fabric of space time? How could that be estimated using frame dragging?
     
    Last edited by a moderator: May 6, 2017
  6. Sep 25, 2012 #5
    Since the build up of charge inside the event horizon is not observable, I'm not sure what process would result in neutralization of the charge imbalance.

    Once Q=M, the inner and outer horizons meet (an interesting concept, thanks)...this implies an unmasking of the black hole since at least one of its 3 properties of a BH is now observable, charge. Seems we have created a naked BH?
     
  7. Sep 25, 2012 #6

    PAllen

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    The Kerr-Newman and Riessner-Nordstrom metrics are solutions to the EFE for any charge and mass and angular momentum, so far as I know. The issue is that super-extremal black holes have unphysical features - but these are not rejected by the EFE.

    For example:

    http://www.famaf.unc.edu.ar/~gdotti/18.pdf

    does not argue that super-extremal black holes are not EFE solutions, only that they are unstable and not plausible results of collapse (among other reasons due to not being able to capture more particles). Just making a technical point that the EFE themselves in no way prohibit super-extremal BH solutions if they magically appeared fully formed.
     
  8. Sep 25, 2012 #7

    bcrowell

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    This is a new topic -- actually two new topics -- so if you want to discuss them, please start new threads rather than reviving a trhead from 2 years ago on an unrelated subject.
     
    Last edited by a moderator: May 6, 2017
  9. Sep 25, 2012 #8

    bcrowell

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    Here are some papers and talks. I viewed the whole talk by Poisson, and through it was pretty informative.

    Gravitational Collapse and Cosmic Censorship, Wald, 1997 - http://arxiv.org/abs/gr-qc/9710068

    Veronika Hubeny, 1998 "Overcharging a Black Hole and Cosmic Censorship," http://arxiv.org/abs/gr-qc/9808043 -- claims yes; influential paper, seems to have shaped later debate; nonrotating

    Barausse et al., 2011, "Testing the Cosmic Censorship Conjecture with point particles: the effect of radiation reaction and the self-force," http://arxiv.org/abs/1106.1692 -- raises uncertainties about previous "yes" results, suggests that answer may actually be no because of effects neglected in those calculations

    Saa and Santarelli, 2011, "Destroying a near-extremal Kerr-Newman black hole," http://arxiv.org/abs/1105.3950 -- claims yes

    http://arxiv.org/abs/1104.3741 -- April 2011 review paper

    talk by Eric Poisson, http://pirsa.org/11050063/
     
  10. Sep 26, 2012 #9

    mfb

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    The total charge of a black hole influences its geometry, and the forces on charged objects. A significant total charge will attract the opposite charge more than the same charge, and that will balance the total charge quickly for realistic setups.
     
  11. Sep 26, 2012 #10
    There is no such charge build up.
     
  12. Sep 26, 2012 #11

    pervect

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    You can't get to the limit, which means you can't crack the hole open.

    My recollection of MTW"s treatment boils down to this. The inequality is satisfied because the energy carried by the electron in order for it to pass through the event horizon has a minimum value, so you must always add enough mass to satisfy the inequality.

    By energy I mean the "energy at infinity", a conserved quantity in the BH metric. Note that the energy at infinity adds to the mass of a black hole when an object falls into a black hole. I'm not sure if you're familiar with these ideas or not, they're also in MTW. Conservatively I would guess not.

    I should add that electric charge is a perfectly observable feature of the black hole - just integrate the normal electric field of a sphere outside the black hole. I'm not quite sure if you were thinking otherwise or not.

    The energy requirement can be informally said to be caused by the fact that the negatively charged event horizon repels incoming electrons - so unless they have enough energy to pass through the potential barrier, they won't make it.
     
    Last edited: Sep 26, 2012
  13. Sep 26, 2012 #12

    Ben Niehoff

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    You can prove mathematically that there is no physical process that can raise a sub-extremal black hole to extremality (let alone exceed it). This is directly analogous to the proof in thermodynamics that no physical process can lower the temperature of a system to absolute zero.

    In fact, an extremal black hole has zero Hawking temperature, thus completing the analogy.
     
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