Why singularities do not exist

[Dmitry67]

I posted it in SR/GR but probably it really belongs here:

I am looking at the way how Hawking radiation is derived:
http://en.wikipedia.org/wiki/Hawking_radiation

Where can I find more info about it? (for amaters like me)

Formula for the Unruh temperature gives an infinite temperature at the horizon.
At the same time AFAIK the apparent horizon recedes in front of the falling observer.
Also, is my assumption correct that observer sees radiation from an apparent horizon, not from an absolute one?
The Hawking radiation looks very intensive but extremely red-shifted by the gravitation. Like almost infinity devided by almost infinity.

So I wonder how intensive the Hawking radiation is for the freely falling observer.

I can explain my motivation. Non red-shifted Hawking radiation is very intense. So when observer approaches the singularity, singularity is always hidden behind the apparent horizon. However, the horizon it covered with a cloud of hawking particles. These particles (in a frame of a falling observer) are emitted from the horizon but then fall back into the singularity together with the observer.

But the singularity of mass M is represented with the mass of the matter inside the apparent horizon S and the 'cloud' C, M=S+C

While C is almost negligible far from the singularity, it is quite possible that C>S close to it. But then singularity itself does not exist because the cloud 'blurs' the singularity and flattens space-time.

 

[Finbar]

I posted it in SR/GR but probably it really belongs here:



I can explain my motivation. Non red-shifted Hawking radiation is very intense. So when observer approaches the singularity, singularity is always hidden behind the apparent horizon. However, the horizon it covered with a cloud of hawking particles. These particles (in a frame of a falling observer) are emitted from the horizon but then fall back into the singularity together with the observer.

But the singularity of mass M is represented with the mass of the matter inside the apparent horizon S and the 'cloud' C, M=S+C

While C is almost negligible far from the singularity, it is quite possible that C>S close to it. But then singularity itself does not exist because the cloud 'blurs' the singularity and flattens space-time.

So your saying that for the falling observer the photons in the cloud are not observed as they fall into the singularity? But they remain outside the apparent horizon? I'm confused!

 

[Dmitry67]

No, no.

At first, as it is well known, the very notion of the 'particle' is observer-dependent, for some observers particles can be real, for others they are virtual, for some observers space is vacuum, for others it is not.

For example, Hawking radiation observed for the outside observer is insivible for the freely falling observer.

I speculate that however freely falling observer observer ANOTHER type of hawking radiation - for HIS apparent horizon (different observers do not agree on the position of the apparent horizon).

As observer falls into black hole, apparent horizon recedes in front of him. So for the falling observer horizon is 'smaller' and emits much more radiation then for an observer 'at infinity'

 

[George Jones]

For example, Hawking radiation observed for the outside observer is insivible for the freely falling observer.

This isn't true.