Is a Black Hole the Most Dense Object in the Universe?

aise0603
Messages
6
Reaction score
0
Mass falling into a black hole will approach the center of the black hole at a given velocity. As it approaches the center of the black hole, from our reference frame, there is time dialation. In fact, at some given closeness, a unit of Planck time, in the mass's reference frame, is longer than the age of the universe.

Gravitational time dilation outside a non-rotating sphere
to = tf(1-2GM/(rc2))1/2

if to = Planck time
tf = the age of the universe
r = Planck distance

solve for M to find the smallest amount of mass that could possibly create a black hole

Ms = 1.09 X 10-8 kg




Density = Ms/Vs = 6.15718 X 1095kg/m3

Is that the most dense that anything could be in the universe?
 
Physics news on Phys.org
What you have calculated is so called Planck Density:
http://en.wikipedia.org/wiki/Planck_units

It means that at such density we can't use GR or QM as is
We need to use more advanced theory which does not exist yet.
 
But could it be that in a given reference frame, nothing can be more dense than this? Just like nothing can travel faster than the speed of light, in a given reference frame.
 
In what context do you want us to answer your questions?
As Dmitry mentioned, we have good reason to expect our current theories won't match nature very well in these regimes. And we definitely don't have experimental data from these regimes either. So no "real" answer can be given right now.

But if you are just asking what classical theory like GR says, then the answer is no: there is no density limit. Density is coordinate system dependent, so we can choose a coordinate system which makes it arbitrarily large if we wish. So there is no classical limit on density.

Even ignoring the coordinate system issues, what would you consider a point particle. For example, what is the density of an electron?
 
JustinLevy said:
In what context do you want us to answer your questions?
As Dmitry mentioned, we have good reason to expect our current theories won't match nature very well in these regimes. And we definitely don't have experimental data from these regimes either. So no "real" answer can be given right now.

But if you are just asking what classical theory like GR says, then the answer is no: there is no density limit. Density is coordinate system dependent, so we can choose a coordinate system which makes it arbitrarily large if we wish. So there is no classical limit on density.

Even ignoring the coordinate system issues, what would you consider a point particle. For example, what is the density of an electron?

Typical...
 
aise0603 said:
Typical...

Typical what? It was a good answer to your question.
 
I started reading a National Geographic article related to the Big Bang. It starts these statements: Gazing up at the stars at night, it’s easy to imagine that space goes on forever. But cosmologists know that the universe actually has limits. First, their best models indicate that space and time had a beginning, a subatomic point called a singularity. This point of intense heat and density rapidly ballooned outward. My first reaction was that this is a layman's approximation to...
Thread 'Dirac's integral for the energy-momentum of the gravitational field'
See Dirac's brief treatment of the energy-momentum pseudo-tensor in the attached picture. Dirac is presumably integrating eq. (31.2) over the 4D "hypercylinder" defined by ##T_1 \le x^0 \le T_2## and ##\mathbf{|x|} \le R##, where ##R## is sufficiently large to include all the matter-energy fields in the system. Then \begin{align} 0 &= \int_V \left[ ({t_\mu}^\nu + T_\mu^\nu)\sqrt{-g}\, \right]_{,\nu} d^4 x = \int_{\partial V} ({t_\mu}^\nu + T_\mu^\nu)\sqrt{-g} \, dS_\nu \nonumber\\ &= \left(...
In Philippe G. Ciarlet's book 'An introduction to differential geometry', He gives the integrability conditions of the differential equations like this: $$ \partial_{i} F_{lj}=L^p_{ij} F_{lp},\,\,\,F_{ij}(x_0)=F^0_{ij}. $$ The integrability conditions for the existence of a global solution ##F_{lj}## is: $$ R^i_{jkl}\equiv\partial_k L^i_{jl}-\partial_l L^i_{jk}+L^h_{jl} L^i_{hk}-L^h_{jk} L^i_{hl}=0 $$ Then from the equation: $$\nabla_b e_a= \Gamma^c_{ab} e_c$$ Using cartesian basis ## e_I...
Back
Top