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2 black holes, 2*M -> 1 b.h., 2*M - f*2*M?

  1. Dec 7, 2009 #1
    2 black holes, 2*M --> 1 b.h., 2*M - f*2*M?

    For equal sized black holes of mass M, what fraction, f, of the energy typically gets lost as gravitational radiation when roughly equal sized black holes merge?

    If f were plotted verses mass would the function have a maximum?

    Thanks for any help!
     
  2. jcsd
  3. Dec 7, 2009 #2

    atyy

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    Science Advisor

    Re: 2 black holes, 2*M --> 1 b.h., 2*M - f*2*M?

    For starting point, try page 4 of

    http://arxiv.org/abs/0708.4202
    Gravitational waves from black-hole mergers
    John G. Baker, William D. Boggs, Joan M. Centrella, Bernard J. Kelly, Sean T. McWilliams, James R. van Meter
     
  4. Dec 7, 2009 #3

    xantox

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    Re: 2 black holes, 2*M --> 1 b.h., 2*M - f*2*M?

    For two identical static black holes, the rigorous theoretical upper limit is the Hawking bound of (2-√2)m (eg 29% of the initial total mass). Numerical simulations typically show much lower amounts of radiation on the order of 0.1..4% (and up to 14% in the scattering regime).
     
    Last edited: Dec 7, 2009
  5. Dec 7, 2009 #4
    Re: 2 black holes, 2*M --> 1 b.h., 2*M - f*2*M?

    If I follow the paper above, for star sized black holes the fraction lost is about 1/20 the mass of the initial black hole pair. Are there any simple arguments one could give to explain how this fraction changes (or does not change) with black hole mass.

    Thanks for your help!
     
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