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The REAL way to escape a black hole?


questionposter

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So you can't escape a black hole using energy, so why not just teleport out? If a particle's probability extend's infinitely, then theoretically a particle could cross the event horizon while it's probability still occupies spacial coordinates outside of the event horizon and teleport outside the event horizon before getting sucked in again.

Edited by questionposter
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Although it does mix-n-match QF theory and GR.

 

It is my understanding that this isn't much of a problem apart from when dealing with situations like the centre of a black hole. I'm not well versed on this though, during my Masters I had a choice a theory module on this or one on analytical dynamics. I chose analytical dynamics...

 

Does gravity make any difference to the probability wave?

 

I can think of no reason why they would not alter the probabilities.

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Does gravity make any difference to the probability wave?

Or is that more mix n match?

 

My understanding is that gravity is perfectly fine, and routinely used as a backdrop for QM calculations.

Either Newtonian gravity (where viewed as a potential) or as the geometry for the coordinates involved (ie. QM with a general relativistic backdrop). My only encounters with this are in a rather abstract and completely non-applied/mathematical context where I didn't fully understand the mapping between what I was doing and anything related to my personal experience in the macroscopic world.

 

In both of these cases, the gravitational field is important for modifying the quantum system you are examining, but the quantum system is assumed to have no influence on the gravitational field (ie. the mass of the quantum system is negligible compared to the source of the gravity).

 

The problem comes in when you try to calculate the spacetime curvature generated by your quantum system and how the quantum system reacts to the gravitational field at the same time.

To simplify my (already incomplete) understanding, you need your geometry settled before you can put your spatial coordinates into your quantum equations, so if the way the coordinates interact depends on your wavefunction, you can't solve the equations anymore.

 

MigL, ignoring the change in the gravitational field due to the interactions involved during Hawking radiation is fine as they are miniscule compared to the overall gravitational field. As such, mixing and matching in this case is going to give you an answer that is very very close to the truth (ie. QM on a GR background, a hard, but frequently dealt with situation).

Edited by Schrödinger's hat
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My understanding is that gravity is perfectly fine, and routinely used as a backdrop for QM calculations.

Either Newtonian gravity (where viewed as a potential) or as the geometry for the coordinates involved (ie. QM with a general relativistic backdrop). My only encounters with this are in a rather abstract and completely non-applied/mathematical context where I didn't fully understand the mapping between what I was doing and anything related to my personal experience in the macroscopic world.

 

In both of these cases, the gravitational field is important for modifying the quantum system you are examining, but the quantum system is assumed to have no influence on the gravitational field (ie. the mass of the quantum system is negligible compared to the source of the gravity).

 

The problem comes in when you try to calculate the spacetime curvature generated by your quantum system and how the quantum system reacts to the gravitational field at the same time.

To simplify my (already incomplete) understanding, you need your geometry settled before you can put your spatial coordinates into your quantum equations, so if the way the coordinates interact depends on your wavefunction, you can't solve the equations anymore.

 

MigL, ignoring the change in the gravitational field due to the interactions involved during Hawking radiation is fine as they are miniscule compared to the overall gravitational field. As such, mixing and matching in this case is going to give you an answer that is very very close to the truth (ie. QM on a GR background, a hard, but frequently dealt with situation).

 

So in other words, your saying because the fabric of space is distorted, the probability will tend to follow that curvature? What if half of the particle is in the black hole and the other half is outside of it? I guess maybe the half that's inside would like like a cone being stretched towards the event horizon, and the outer half would still look relatively like a sphere?

Edited by questionposter
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So in other words, your saying because the fabric of space is distorted, the probability will tend to follow that curvature? What if half of the particle is in the black hole and the other half is outside of it? I guess maybe the half that's inside would like like a cone being stretched towards the event horizon, and the outer half would still look relatively like a sphere?

 

On a local scale the event horizon of a black hole looks no different from normal space.

For a big enough black hole ie. galactic centre, even a person/house sized object would have difficulty detecting the tidal forces (ie. they wouldn't be able to figure out they're near/entering the event horizon without looking out the window or doing some quite precise experiments).

 

The tidal effect (or curvature in the region if you want to view it that way) influences the solutions to the wave equation, but it's not quite as simple as the idea of squishing/stretching the wavefunction as if it were macroscopic squishy ball in euclidean space with a force acting on it (although this may be a reasonable heuristic, I do not know enough of the details of QM in a curved background to know).

Edited by Schrödinger's hat
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