
#1
Jan2705, 04:46 PM

P: 15

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
Jan2705, 04:55 PM

P: 482

quantum effects probably prevent a singularity from reaching zero/infinite parameters




#3
Jan2705, 04:57 PM

Sci Advisor
P: 8,470





#4
Jan2705, 05:32 PM

P: 991

Black Hole: Infinite Density, Infinite Buoyancy?'nonrotating' Classical UniverseSchwarzschild Singularity Density solution for a onedimesional 'pointlike' object: [tex]\boxed{\rho_u = \frac{M_u}{2} \sqrt{ \frac{c^3}{\hbar G}}}[/tex] BH Singularity Density infinities do NOT exist. Reference: http://www.physicsforums.com/showpos...2&postcount=24 



#5
Jan2705, 05:38 PM

Emeritus
Sci Advisor
PF Gold
P: 2,018

There is no "inside" to a BH singularity (point).




#6
Jan2705, 05:43 PM

Sci Advisor
P: 8,470





#7
Jan2805, 01:44 AM

Sci Advisor
PF Gold
P: 9,182

I would argue the Planck density is the limit in the physical universe.




#8
Feb105, 11:58 AM

P: 991

[tex]\boxed{\rho_u = \frac{M_u}{2} \sqrt{ \frac{c^3}{\hbar G}}}[/tex] The solution for this 'nonrotating' classical schwarzschild singularity density for a one dimensional 'pointlike' 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 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 'nonrotating' Classical Schwarzschild Singularity Density for a two dimensional 'pointlike' 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] [tex] \begin{picture}(100,100)(0,0) \put(0,0){\circle{3}} \put(0,0){\line(1,0){100}} \put(0,0){\line(0,1){100}} \put(0,33){\circle{3}} \put(34,55){\circle{3}} \put(67,77){\textcolor{red}{\circle{3}}} \put(67,77){\textcolor{blue}{\circle{6}}} \put(100,100){\circle{3}} \put(100,5){{n}} \put(5,100){{ln p}} \end{picture} [/tex] Based upon the current logarithmic slope in the chart, at what density value does the slope cross the yintercept? Reference: http://www.physicsforums.com/showpos...2&postcount=24 


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