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Why is matter drawn into a black hole not condensed into a single point within the singularity?

When we speak of black holes and their associated singularity, why is matter drawn into a black hole condensed into a single point within the singularity?

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marked as duplicate by Qmechanic, Manishearth Dec 11 '12 at 11:46


As I was told my assumption was incorrect I wanted to see how many people agree with this explanation and more importantly what they believe this is true. –  Argus May 20 '12 at 13:23
The question(v1) asks precisely the opposite of this question. However, the answers are likely to be essentially just a repetition. –  Qmechanic May 20 '12 at 14:00
true but in this case if the light is condensed to a single point then its speed never reaches zero as described from an outside observers point of view. light with no velocity must have mass and as i do not want to believe that i am trying to cite specific reasons observations do not effect what is physically occuring. –  Argus May 20 '12 at 14:38

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I seem to be answering a lot of questions about black holes at the moment!

In my answer to Why is matter drawn into a black hole not condensed into a single point within the singularity? I explained how matter falling into a black hole reaches the event horizon, and in my answer to Why is a black hole black? I showed how matter that reaches the event horizon can never escape it. So combining these two shows that once you've fallen into a black hole you can never escape. If you are prepared to restrict yourself to radially infalling matter then my answer to Why is a black hole black? shows that matter must hit the singularity because once inside the event horizon even light has a negative radial velocity i.e. even a light beam can't escape the singularity.

However that isn't the same as showing that once you've fallen into the black hole you are inevitably drawn into the singularity, because you might argue you could orbit the singularity. The proof that anyone falling into a black hole must hit the singularity was given by Hawking and Penrose. Unfortunately my mission to make GR understandable hits a problem here because I don't understand the Hawking Penrose theorem so I certainly can't explain it to anyone else.

To make life more interesting, if the black hole is electrically charged (Reissner–Nordström metric) or spinning (Kerr metric) it is possible to find timelike paths that do not hit the metric but instead lead back out of the black hole into another region of spacetime. I even ranted on about this as well, see Entering a black hole, jumping into another universe---with questions.

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agreed as must hit the singularity is a very specific statememnt and the answer is no not 100% of infalling matter strikes the singularity as hawking radiation escapes the BH. as light travels inward it is slowed in time from the outside observers perspective. light without momentum has mass. is the mass what creates an un-observable state beyond the Actual Horizon –  Argus May 20 '12 at 14:41
Maybe I wasn't clear. For a Schwarzschild black hole 100% of all infalling matter strikes the singularity. The Hawking radiation is not an exception to this as the Hawking radiation we see coming from the black hole is not escaping from inside the event horizon. Light doesn't have mass under any circumstances. –  John Rennie May 20 '12 at 14:46
sorry not has mass but Is mass. as currently most high level theoris appear to place "Energy" in a specific state as appearing "observably" to "Be" massive ie m theory saying that resonations in the String are observed as physical objects. the peaks of the waves are observable and the movement between peaks is not observable. in "MY" terms strictly visually a jump rope "one person holding each end and shaking up and down respectively" how fast do they have to shake before all you see is the top and bottom or peak of the movement. –  Argus May 20 '12 at 17:29

I think the explanation of colapse is due to Schwarzschild; you can find it in Wald's book, the chapter on Schwarzschild's solution. Very briefly, solving the Einstein's equation inside the massive star yields an infinite pressure at the center (above some mass threshold) with the only imaginable consequence being the colapse into a point. This all is solid only for the simple case of a spherically symmetric black hole, but I guess the qualitative feature is the same for all massive enough bodies which become a blackhole. And the matter drawn inside the horizon can not be differentiated from the blackhole anymore; it's part of the blackhole and sooner or later gravity destroys any structure it may have, near the infinite pressure center.

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as this is a brief answer can you elaborate as to the importance of an outside observer to this event and does the outside observation effect what is takeing place inside the singularity. –  Argus May 20 '12 at 14:29
Argus, in a classical context, the outside observer is of no significance; in a quantum context, on the other hand, I'm of no significance, because I'm very ignorant about quantum backholes. Sorry! –  Arash May 20 '12 at 14:40
