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Why is a black hole with q>e singular?

  1. Jul 12, 2003 #1
    Black holes have the properties of angular momentum, mass, and charge. Since electromagnetism is a much stronger inverse-square force than gravity, and like charges repel, wouldn't a black hole with charge q>e avoid singularity?
     
  2. jcsd
  3. Jul 12, 2003 #2

    jcsd

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    No, this is how a black hole is formed, i.e. the gravitational pull of the star becomes greater than the degenracy pressure caused by electromagnetism and the like.
     
  4. Jul 12, 2003 #3
    Even when a mass is collapsed beyond degeneracy to neutronium, any net charge its eventual "singularity" harbors still retains the property of predominant repulsion.
     
  5. Jul 14, 2003 #4
    Well Loren,..who says inverse square law holds true inside a singularity?:wink:

    Creator
     
  6. Jul 14, 2003 #5
    Creator,
    One must achieve the singularity first. In other words, can you derive a minimum net charge/mass relation for collapsing matter to attain a black hole?
     
    Last edited: Jul 14, 2003
  7. Jul 14, 2003 #6
    Beats me; I've not heard of such a calculation. However, even before going singular 1/r^2 likely becomes deficient.

    Creator
     
    Last edited: Jul 14, 2003
  8. Jul 14, 2003 #7
    What does q>e mean ?
     
  9. Jul 14, 2003 #8

    jcsd

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    There is no minimum mass needed for a black hole (though obviously one formed by steallr evolution has a minimum mass)and charge doesn't enter into it.
     
  10. Jul 14, 2003 #9
    bogdan, q>e means the net collapsing discrete charge, q, is at least 2 times the electon charge, e, in order for like charges to repel.

    jcsd, the minimum mass for a black hole is M*, the Planck mass. This quantity derives from the absolute radiative constants c, G and h. Only quanta or their composites weigh less.
     
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