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Black Hole: Infinite Density, Infinite Buoyancy?

  1. Jan 27, 2005 #1
    If a black hole has an infinite density, then one would think that anything would float inside of it. And since it's infinitely dense, the object(s) being pulled in would have an infinite buoyancy, causing it to be shot back out of the black hole at a seemingly infinite speed. So why don't these two forces cancle each other out?
  2. jcsd
  3. Jan 27, 2005 #2
    quantum effects probably prevent a singularity from reaching zero/infinite parameters
  4. Jan 27, 2005 #3


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    Only the singularity at the center would have infinite density according to general relativity, inside the event horizon is empty space (and infalling matter) just like outside the event horizon.
  5. Jan 27, 2005 #4
  6. Jan 27, 2005 #5


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    There is no "inside" to a BH singularity (point).
  7. Jan 27, 2005 #6


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    Is this equation something you derived yourself based on your own ideas, or did you get it from a textbook or something written by a professional physicist? The derivation you gave on that thread seemed to involve both QM and GR...would you agree that according to classical GR alone, the singularity has infinite density?
  8. Jan 28, 2005 #7


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    I would argue the Planck density is the limit in the physical universe.
  9. Feb 1, 2005 #8
    Planck Philosophy...

    [tex]\boxed{\rho_u = \frac{M_u}{2} \sqrt{ \frac{c^3}{\hbar G}}}[/tex]

    The solution for this 'non-rotating' classical schwarzschild singularity density for a one dimensional 'point-like' object was derived by me based on research on these physical models.

    Note that the Schwarzschild solution is only a solution for Schwarzschild BHs with zero angular momentum , this is a highly improbable state.

    Neutron star spin increases with increased density, therefore an object generating in the core of a neutron star or supernova without spin is...impossible. Only BHs with angular momentum can exist in the Universe, a rotating Kerr BH.

    The Classical General Relativity model is based upon four total dimensional space-time [tex]n_t = 3 + 1 = 4[/tex] (3 space + 1 time). Solutions for models for that contain dimensions of less than four are not valid solutions in the Universe.

    The classical solution stated for 1 dimension is actually 2 dimensions [tex]n_t = 1 + 1 = 2[/tex] (1 space + 1 time), because solutions with with a total dimensional range of less than 4 [tex]n_t < 4[/tex] cannot exist in the Universe, all solutions for total dimensional ranges between 0 and 3 are not real valid solutions because they cannot exist in a four total dimensional General Relativity Universe.

    Classical General Relativity models based upon 0 to less than 2 total dimensions are typical of producing solutions with 'infinities', and is only a division by zero in an 'undefined' model.

    This solution is based upon 2 dimensional space, the singularity described 'exists' in only 2 space dimensions (and 1 time) [tex]n_t = 2 + 1 = 2[/tex] (2 space + 1 time). [tex]n_t = n_s + n_t[/tex].

    Classical Schwarzschild Singularity Dimension Number:
    [tex]n_s = 2[/tex] - dimension #
    [tex]dV_s = \pi r_p^2[/tex] - volume
    [tex]L = 0[/tex] - angular momentum

    Solution for 'non-rotating' Classical Schwarzschild Singularity Density for a two dimensional 'point-like' object: (flat disc)
    [tex]\rho_s = \frac{dM_s}{dV_s} = \frac{dM_s}{\pi r_p^2} = \frac{M_u c^3}{\pi \hbar G}[/tex]

    [tex]\boxed{\rho_u = \frac{M_u c^3}{\pi \hbar G}}[/tex]

    \put(5,100){{ln p}}

    In a four dimensional space-time physical Universe, the average Planck density is a solution and a physical 'limit' in the Universe.

    Based upon the current logarithmic slope in the chart, at what density value does the slope cross the y-intercept?

    Last edited: Feb 1, 2005
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