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View Full Version : Retrieving Information from Black Holes (Please repond)


capnkissling
Oct19-09, 10:44 PM
I was reading an article about retrieving information from black holes when I came up with an idea.

Please bear with me, as I am not a scientist (or even close) and know little physics vocabulary.

A theory/hypothesis I read involves Hawking radiation retrieving information from the black hole. In this theory, a pair of virtual particles appear at the black hole's event horizon. The negative particle falls into the black hole slowly evaporating the black hole, while the positive particle flies off into space with information from the black hole. The question is how does the positive particle from the pair acquire information from deep in the black hole?

My idea is that the information sits on the edge of the event horizon, not at the singularity. In fact, anything that falls prey to the black hole will remain at the event horizon forever.

My reasoning involves time dilation. We know that time will move differently for an object experiencing acceleration, or in this case experiencing gravitation effects. The higher the gravitational pull on the object the faster time will move around the object. That is why an observer outside of the black hole will watch an object move slower and slower through time as it falls towards the event horizon. At the event horizon the gravity is so strong that the speed of light cannot escape it. Thinking logically, once the object reaches the event horizon, the gravitational pull is equal to the speed of light. That would mean that time is moving infinitely fast around this object. The object cannot go any further and is essentially frozen in time. Therefore, that object and any other information that falls into the black hole is frozen in time along the event horizon, not at the singularity.

Please respond, as I know little about quantum physics and relativity, I would love to hear your opinion. Thank you,
-Tom

Nabeshin
Oct19-09, 11:20 PM
My reasoning involves time dilation. We know that time will move differently for an object experiencing acceleration, or in this case experiencing gravitation effects. The higher the gravitational pull on the object the faster time will move around the object. That is why an observer outside of the black hole will watch an object move slower and slower through time as it falls towards the event horizon. At the event horizon the gravity is so strong that the speed of light cannot escape it. Thinking logically, once the object reaches the event horizon, the gravitational pull is equal to the speed of light. That would mean that time is moving infinitely fast around this object. The object cannot go any further and is essentially frozen in time. Therefore, that object and any other information that falls into the black hole is frozen in time along the event horizon, not at the singularity.


This is true for an observer far away from the black hole. These observers see objects fall closer and closer to the EH, and the light coming from them becomes increasingly redshifted and time becomes increasingly dilated.

However, for an freely falling observer, they would notice nothing special happening as they pass the event horizon. Indeed, they reach the singularity of a black hole in a finite amount of proper time.

In terms of information loss, I'm not sure how to reconcile the two.

A note: we don't allow personal theories to be expressed here at PF. But this is more legitimate question than advancement of some ridiculous theory, so I think it's okay.

clamtrox
Oct20-09, 01:20 AM
The question is how does the positive particle from the pair acquire information from deep in the black hole?

I always felt like the "spooky action from a distance" of quantum mechanics would be just the right kind of thing to transfer information from the negative energy particle falling into the hole into the positive energy particle that's flying off to infinity. I have no clue what the 'establishment' says about this though.

Then your problem would be reduced into understanding quantum mechanics, good luck with that! :)

capnkissling
Oct22-09, 10:56 AM
However, for an freely falling observer, they would notice nothing special happening as they pass the event horizon. Indeed, they reach the singularity of a black hole in a finite amount of proper time.

Unless I am mistaken, relativity states that the observer entering the black hole is not just free falling. They are feeling and having the effects of acceleration in the form of gravity. Objects outside of the massive field of gravity do not feel these effects. These objects should include space and time. Therefore, wouldn't time move around the falling observer infinitely quickly since it doesn't share the same frame of reference as the falling observer?

JesseM
Oct22-09, 11:28 AM
Unless I am mistaken, relativity states that the observer entering the black hole is not just free falling.
"Free falling" just means that an object isn't subject to any non-gravitational forces.
They are feeling and having the effects of acceleration in the form of gravity.
Because of the equivalence principle (http://www.aei.mpg.de/einsteinOnline/en/spotlights/equivalence_principle/index.html) in general relativity, a free-falling observer won't feel any acceleration, they'll feel weightless just like an observer moving at constant velocity in a gravity-free region.
Therefore, wouldn't time move around the falling observer infinitely quickly since it doesn't share the same frame of reference as the falling observer?
"Time" does not have a frame of reference, only physical objects do (and in general relativity the notion of a 'frame of reference' becomes more complicated, because a frame of reference is really just a spacetime coordinate system and all coordinate systems are equally valid in general relativity (http://www.aei.mpg.de/einsteinOnline/en/spotlights/background_independence/index.html), unlike in special relativity where there is a special class of inertial reference frames (http://id.mind.net/~zona/mstm/physics/mechanics/framesOfReference/inertialFrame.html) which have a preferred role, so that when we talk about the frame of an inertial object in SR we know we're talking about its inertial rest frame).

clamtrox
Oct22-09, 02:21 PM
Because of the equivalence principle in general relativity, a free-falling observer won't feel any acceleration, they'll feel weightless just like an observer moving at constant velocity in a gravity-free region.

This is not entirely correct; what the equivalence principle says that this is true in a sufficiently small region of space, which in the case of a highly curved space around a black hole can be smaller than an unlucky astrounaut falling into the hole. What you would then experience is in the jargon of astrophysicists known as "spaghettification".

JesseM
Oct22-09, 05:22 PM
This is not entirely correct; what the equivalence principle says that this is true in a sufficiently small region of space, which in the case of a highly curved space around a black hole can be smaller than an unlucky astrounaut falling into the hole. What you would then experience is in the jargon of astrophysicists known as "spaghettification".
Yes, the article I linked to discussed tidal forces, but capnkissling was just talking about regular G-forces due to acceleration so I oversimplified a bit...even if the astronaut gets ripped apart by tidal forces, it's still true that each sufficiently small region of his body is locally weightless as it falls. And for a large enough black hole, the effects of tidal forces wouldn't be noticeable until long after you'd crossed the event horizon.

clamtrox
Oct23-09, 01:35 AM
My idea is that the information sits on the edge of the event horizon, not at the singularity. In fact, anything that falls prey to the black hole will remain at the event horizon forever.


By the way, this makes some sense. For example, if you consider a BH with an electric charge, an observer far away from the hole does still observe an EM field generated by the charged particles, which in the observers frame are located close to the event horizon of the BH.