Black holes and singularities

1. Oct 26, 2009

Pequinino

OK, this is a long one.

Black holes are a singularity, right?
As is, their dimensions are 0mX0mX0m?
That is why their gravity is so strong, because objects can get much closer and thus make the distance between them 0 and force of gravity infinite.

In order for this to be possible, there would have to pe a particle with mass, but no volume.
Since the majority of the volume of, say, an atom, is made of the empty space between the nucleus and the electrons, you can decrease the volume by a lot. Because the majority of the volume of a proton consists of the empty space between the quarks, you can further reduce the volume. But in order to remove all the volume of an object, that would mean that at some point, the entire object is made of empty space with some volumeless particle that has mass.
That would mean that everything is made up of singularities, and thus everything is made of black holes.

Is this valid?
If not, where did I go wrong?

2. Oct 26, 2009

Dmitry67

no, black holes and singularities are different things, even in most cases there is a singularity in a black hole.

black hole is an area covered by the event horizon.

one can imagine a naked singularity without a horizon. And a horizon without a black hole (coosmological horizons for example)

3. Oct 26, 2009

Dmitry67

What you describe is a known problem: for that very reason Gravity and Quantum Mechanics do not work together - because we get infinities for the point-like particles.

One of the solutions is a superstring theory, where particles are not point-like

4. Nov 12, 2009

Frank123

"black hole is an area covered by the event horizon". Wrong. A BH has no inside, as it takes forever to get there. See "Basic Assumptions And Black Holes", Physics Essays, 22, 559 (2009).
The geometry is very non-Euclidean.

5. Nov 12, 2009

Dmitry67

Oh no, not again.
After a long thread where it took several pages to bust a common misconception... there IS inside and it takes a limited time to fall into a black hole (by the proper time of free falling observer)

6. Nov 12, 2009

Dmitry67

7. Nov 12, 2009

Frank123

8. Nov 12, 2009

DaveC426913

There's a non sequitur between the above and the below.
Your logic assumes that what is at the centre is nothing more than a regular bit of matter with all the empty space removed. It does not allow for the possibility that there is something additional happening that causes a degeneration of the matter.

For example - and I'm not suggesting this is the case, just putting it out there as a way to refute your logic: perhaps the matter is converted to energy, then it can achieve a zero volume while still having a gravitational influence.

9. Nov 12, 2009

Dmitry67

Yes, there are errors because there were different people talking :)

I looked thru the article. He quotes "Misner et al" at looks like it is his only source.

The following parts are obviously false:

OMG, Gravitation in GR is not a force!

Oh, poor Misner. Yes, in some coordiante systems. But NOT for the poor observer!

Object crossed something far away is not a physically observable event.
It is like asking 'what is currently, at this very second, is happening at the Andromeda

So mistake on mistake...

10. Nov 12, 2009

Dmitry67

And finally: he mentioned Goedel several times. Well, this theorem is applicable to even very simple universes (like Conway Game of Life). However, it produces only 'non-local' results: some statements about the initial and final states are not provable.

However, every single step State(T+1)=F (State(T)) is deterministic and effectively calculable. It is very difficult to tell if Goedels theorem is applicable to our universe, and if yes, to what extent.

And even if it is applicable, then what's the problem? It is applicable to Game of Life - so what, do you have any problems with that 'toy universe'?

But he even does not try to ask these questions.

11. Nov 12, 2009

Naty1

A black hole IS a singularity. According to Roger Penrose, the big bang and black hole singularities have zero Weyl curvature while a big crunch singulairty appears different having an essentialy infinite Weyl curvature.

There is no mass as we know it; inside all particles have been destroyed and gravitational effects remain outside the event horizon along with a few characteristics (electric harge,spin,etc).
What you are neglecting is that the virtually infinite gravity obliterates all particles curshing them beyond recognition...what we observe outside the even horizon appeared during the contraction period.

There is a partial discussion here, https://www.physicsforums.com/showthread.php?p=2379694&highlight=weyl+curvature#post2379694
and I guess a white hole is also a singularity...How they all precisely relate I still don't know.

The simplest black hole has "mass" but neither charge nor angular momentum is the Schwarzschild black hole, I think it is the only vacuum solution that is spherically symmetric
The Reissner-NordstrÃ¶m metric describes a black hole with electric charge, while the Kerr metric yields a rotating black hole. The Kerr-Newman metric, describes both charge and angular momentum.
I'm also unsure about exactly the differences between these black hole types....

12. Nov 12, 2009

George Jones

Staff Emeritus
I disagree. A black hole is a region of spacetime, i.e., the complement of the region for which it's possible to escape to future null infinity. Penrose's singularity theorem proves that given certain conditions, any spacetime that contains a black hole is singular. But this is not the same as saying that a black hole is a singularity.

13. Nov 12, 2009

pervect

Staff Emeritus
Yes. This is mentioned by Misner et al in "Gravitation" for instance - in several spots, I"m sure, there's one particular example on pg 851 where MTW calculates the finite proper time for an observer on a ball of collapsing dust to reach the singularity at the center.

It's also instructive to consider the case of an accelerating observer and the associated "Rindler metric", which is a stationary metric in which time and space are orthogonal and time at the origin is given by the proper time of the accelerating observer.

This metric also has an event horizon, very similar to the black hole event horizon. But it would be a mistake to say that objects "never pass through the Rindler horizon" - or that things beneath the horizon "don't exist". A space-ship accelerating at 1g that is more than one light year from earth will place the Earth below their Rindler horizon. However, the Earth will still exist, and will still be able to receive signals from the spaceship. The space-ship, however, will never be able to receive signals from the Earth. Due to its hyperbolic motion, light signals will never be able to catch up with the spaceship (as long as it continues to accelerate) - if it is further than 1 light year away from the Earth, it has too much of a head-start.

But just because the signals from the Earth can't reach the observer on the spaceship is not sufficient grounds to claim that the Earth "doesn't exist" anymore....

14. Nov 15, 2009

wofsy

I do not think that singularities need to be points. The can have any shape as long as the light cones converge onto them.

15. Nov 16, 2009

ExecNight

Blackholes look more like a 3D Maelstrom, now the question is;

Is it mass and gravity that causes the hole in the universe to create this 3D maelstrom?
Can it be possible that blackholes are simply a result of a reaction in space-time, and it just takes too long for us to observe the space-time vibration caused by this reaction come to a rest.

So the most important and relevant question seems to me that, blackholes are irrelevant what is happening to the particles, atoms and how are they reacting with each other just the moment before the mass collapses. What speeds do they reach?

16. Nov 16, 2009

Ich

Depends on what you mean by "never", I think. Especially as in the Rindler case the EH is tied to a specific coordinate system, in which objects indeed never cross the horizon.
That said, I agree with you that there is no obvious reason to let a coordinate system tell us where the world ends - as long as there is finite curvature, it's natural to choose different coordinates and extend the decription of spacetime to the unobservable region. Falsifiability is then a bit hard, though.