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How can one proof that there is a singularity inside a black hole ?
..without jumping in, of course...
this is one of the questions where i'm not that sure..
is there any mathematical proof ?
I know there are some alternative ideas, but i heard it was shown that there must be a singularity
So can someone please be so kind and shed some light onto this ? ;)
10 Answers
- Mercury 2010Lv 71 decade agoFavorite Answer
check some of my previous answers and you might find some info to help you.
http://physics.syr.edu/courses/modules/LIGHTCONE/s...
looks like this its an equation by Schwarzschild and endorsed by Einstein.
here ya go
The singularity at r = 1 in the Schwarzschild space-time metric
$ds^2= (1 - 1/r)dt^2-\frac {dr^2} {(1-1/r)}- r^2(d\theta ^2 + \sin^2 \theta d\phi^2)$
is investigated from a physical standpoint.
- Tim CLv 51 decade ago
technically it might not be a singularity, it could be just a bit smaller than the event horizon itself. the only "proof" we have of a singularity in a black hole is that at that extreme gravitational region, no known force can stop gravity, so the matter must collapse into a singularity, but we do not absolutely know this for sure, maybe there is some other force that we have not found yet.
- mathematicianLv 71 decade ago
The only way to do a mathematical proof is to assume that, say, general relativity holds. Unless you have the physics understood, you won't be able to mathematically prove anything.
What has been shown is that, under general relativity, certain situations inevitably lead to singularities (Hawking and Penrose proved these). The conditions around some collapsing stars are good enough for these theorems to apply. Of course, since we don't know how quantum mechanics affects general relativity, our conclusions do not take such things into consideration.
- 1 decade ago
without gettiing in to the philosphy of "proof"....the singularity is purely a mathematical construct having infinite density and zero "size". A quantum theory of gravity is needed before a more realistic model of black holes is formed. More likely there is a new state of matter at the centre of black holes...no singularity.
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- 1 decade ago
The term,"singularity," is ill defined if there can be more than one of them. There are many black holes, as a matter of fact they are quite common. If mathematical was available to prove otherwise it would have been published by now.
- Vincent GLv 71 decade ago
At that level, is is all theoretical. But what we know is that a black hole has to have a mass. We also know that this mass in within the event horizon, the distance from the center of mass where anything traveling at less than the speed of light could not stay in orbit.
From then, it can be deducted that all the mass is in an area where it would need to travel at a speed faster than the speed of light just to stay put; so anything inside has to collapse.
To what degree? Perhaps to zero dimension. Our physical laws do not apply there, so what is in there?
No one knows.
- 1 decade ago
Cosmologists prefer the term 'singularity' within their community. BH is a metaphor for something quite amazing that defies analysis and explanation.
The term 'singularity' sadly only implies that there is a relatively small space which has profoundly strong gravity.
- spir_i_tualLv 61 decade ago
There is a Black Hole in the center of each Galaxy, but the thing is, they're not holes at all, they're thee most powerful predator stars in the cosmos. The gravitational force produce by these stars is so strong, that the speed of photons are made heavy. These predator stars are black cause light is heavy like rocks to them.
Source(s): Don't you just love it? - science_joe_2000Lv 41 decade ago
There is a singularity, because there is a different universe in there - that is the functional definition of a singularity.
Source(s): www.joebrownscience.net Articles: Millennial Physics - Chapter 6 Advanced Gravity Black Holes