# Are black hole event horizons filled with black holes? [closed]

An observer hovering close to an event horizon will observe huge energies, like blue shifted radiation falling in, or Hawking radiation going out. So does the observer observe that black holes are created, when high energy particles collide, and these black holes then absorb energy at fast rate?

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Currently, your question isn't that clear... If you clarify it, ping me in the comments and I'd be happy to reopen it :) –  Manishearth Dec 23 '12 at 8:50

## closed as not a real question by Sklivvz♦, Manishearth♦Dec 23 '12 at 8:50

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First of all, Hawking radiation is a bold speculation. It has never been observed and its theoretical background is discutable as so far we do not know quantum theory of gravity which appears to be needed to speculate precisely about such phenomena. However, most phisicists agree that its existence is well possible.

Even if it existed, it is predicted to be really, really weak for any black holes of reasonable size. So, there will be no new black holes formed near the horizon because of that.

When it comes to very small black holes (e.g. size of proton) current theroies are helpless and we can not make any predictions. Therefore the answer is: they may or they may not.

Now, any other known radiation is too weak to produce black holes near the horizon for practical astrophysical situations. This is because energy concentration required to produce a horizon is immensly immense :) . Remember $E=mc^2$ - you need a lot of matter to form a black hole and still $c^2$ is sooo huge. I am working on black hole formation in my thesis and radiation can of course produce it - but simulations suggest that energy concentration has to be unreasonably strong.

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Well, I think that an observer hovering at event horizon will consider just about any energy to be nearly infinite energy! The blue shift of falling energy is the reason for that, or even better reason: the red shift of the observer's energy. I assumed everybody knows this :) –  kartsa Jan 31 '12 at 11:58
Ok, so the answer is a bit different now. The redshift is a consequence of coordinate singularity - appearence of $(1-2M/r)$ in metric expressions. Coordinate system which is singular at the horizon cannot be used to analyze observers there. You need to chage coordinates to Kruskal ones - and if you do you will find that there is no infinite blue-shift. Just 'some' shift, but not that significant. See en.wikipedia.org/wiki/Kruskal–Szekeres_coordinates for details (unusually good page for wikipedia) –  Terminus Feb 1 '12 at 12:13
Hovering at the event horizon is an infinitely hard thing to do, and when you are doing that, your energy is zero. That's what I think! .... If Kruskal-Szekeres-coordinates tell a different story, then that's a problem of Kruskal-Szekeres-coordinateds. –  kartsa Feb 2 '12 at 8:34