DLuckyE said:
So why assume that it collapses into a singularity instead of still having some kind of non infinite density?
The the Schwarzschild radius is the radius of the event horizon for a black hole, and all black holes are described by General Relativity. The singularity in General Relativity is intrinsic to the Einstein field equations as the Planck force, which is the ratio of electromagnetic energy per gravitational length:
Planck force:
F_P = \frac{c^4}{G}
The Planck force itself is mathematically intrinsic to General Relativity due to the integration of the initial General Relativity equations with Quantum Mechanics in absence of Quantum Gravitation.
Einstein field equation:
G_{\mu \nu} = \frac{8 \pi T_{\mu \nu}}{F_P} = 8 \pi \frac{G}{c^4} T_{\mu \nu}
This means that the maximum force that can be spherically symmetrically applied under General Relativity to the Einstein tensor and stress-energy tensor differential and is equivalent to the Planck force, which is a singularity:
F_P = \frac{c^4}{G} = 8 \pi \left( \frac{T_{\mu \nu}}{G_{\mu \nu}} \right)
Therefore, any mathematical model using an Equation of State described by General Relativity, absent any Quantum Gravity, must contain a Planck singularity.
If Planck pressure and Planck density are the maximum upper limits in the Universe, and both pressure and density functions both relativistically contribute to the total Equation of State for hydrostatic equilibrium inside a black hole, then the total integration for the differential pressure that describes the perfect fluid inside a black hole must be equivalent to the Planck pressure at the singularity core.
Relativistic Equation of State functions for hydrostatic equilibrium: (J = 0, Q = 0)
\frac{dP(r)}{dr} = - \frac{G}{r^2} \left( \rho(r) + \frac{P(r)}{c^2} \right) \left(M(r) + 4 \pi r^3 \frac{P(r)}{c^2} \right) \left( 1 - \frac{2 G M(r)}{c^2 r} \right)^{-1} \;
Note that this equation under General Relativity has a mathematical singularity as r \neq 0, because of metric geometry as r^{-1} and r^{-2}.
Integration of the relativistic differential pressure Equation of State function for hydrostatic equilibrium for core pressure:
P_c = \int_{r_P}^{R_s} \left( \frac{dP(r)}{dr} \right) dr = \frac{c^7}{4 \pi \hbar G^2}
Where r_P is the Planck radius and R_s is the Schwarzschild radius.
Relativistic black hole singularity core pressure is equivalent to Planck pressure:
\boxed{P_c = P_P = \frac{c^7}{4 \pi \hbar G^2}}
Relativistic black hole singularity core density is equivalent to Planck density:
\boxed{\rho_c = \rho_P = \frac{3 c^5}{4 \pi \hbar G^2}}
Note that the core singularity inside a black hole under General Relativity does not have infinite dimensions and are limited to the Planck radius r_P as the smallest spatial unit mathematically possible in any model in absence of Quantum Gravitation.
In physical cosmology, the Big Crunch is one possible scenario for the ultimate fate of the Universe, in which the metric expansion of space eventually reverses and the Universe recollapses, ultimately ending as a black hole singularity.
The Big Crunch cosmological theory is also based upon that premise that the Planck units are the maximum attainable limits in the Universe and cannot be exceeded without Quantum Gravity.
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Reference:
http://en.wikipedia.org/wiki/Planck_force"
http://en.wikipedia.org/wiki/Planck_density"
http://en.wikipedia.org/wiki/Planck_pressure"
http://en.wikipedia.org/wiki/Planck_length"
http://en.wikipedia.org/wiki/Tolman%E2%80%93Oppenheimer%E2%80%93Volkoff_equation"
http://en.wikipedia.org/wiki/Big_Crunch"