Shouldn't objects entering black holes be instantly shredded?

[skeptic2]

A naive calculation implies you should be traveling at the speed of light once you reach the EH, which is presumably impossible for a massive particle. The logical answer is you would be converted to massless particles at that point.

Please excuse my naivete Chronos, but why is it [presumably] impossible for a massive to reach the speed of light at the horizon? The free falling massive particle is not being accelerated in the usual sense but is at rest.

Why would being converted to massless particles be the logical answer?

 

[rustytxrx]

I think the problem is the Special theory of Relativity. Einstein predicts if a particle reached the speed of light the mass and energy would be infinite. There seems to be a problem with that.

 

[skeptic2]

I think the problem is the Special theory of Relativity. Einstein predicts if a particle reached the speed of light the mass and energy would be infinite. There seems to be a problem with that.

That is true if the particle is accelerated, but in the case of free fall, the particle is at rest. It is not being accelerated. If you were in a spaceship, free falling towards the event horizon, you would be weightless which is not possible if you were being accelerated.

 

[WannabeNewton]

Please excuse my naivete Chronos, but why is it [presumably] impossible for a massive to reach the speed of light at the horizon? The free falling massive particle is not being accelerated in the usual sense but is at rest.

A freely falling particle is only locally inertial; it is not "at rest" given the strict definition of the term: if the space-time is stationary, a notion of being "at rest" with respect to the space-time exists and this is defined as following an orbit of the time-like killing vector field of the space-time. This is what people mean when they say "what is the force needed at infinity in order to keep stationary a unit test mass near the event horizon of a static black hole". A particle stationary ("at rest") in the above sense has non-vanishing proper acceleration (whereas a freely falling particle has vanishing proper acceleration). In summary, don't confuse free fall with being "at rest" in the above sense (which is roughly what you intuitively think of as being at rest with respect to the space-time).

 

[Aryianna]

There is a physicist by the name of Amos Ori whose research is about this subject matter. I have met him many years ago at Caltech, and he did explain to me that it is possible to escape the event horizon. I had a copy of his paper, and he has worked with Kip Thorne on this subject matter. I'll look it up.

 

[George Jones]

There is a physicist by the name of Amos Ori whose research is about this subject matter. I have met him many years ago at Caltech, and he did explain to me that it is possible to escape the event horizon. I had a copy of his paper, and he has worked with Kip Thorne on this subject matter. I'll look it up.

Do you mean "escape the event horizon" or "survive the curvature singularity singularity at inner (Cauchy) horizon"?

There is a weak curvature singularity at the Cauchy inner horizon (IH) of a rotating black hole. Shredding is caused by tidal force, and tidal force is caused by spacetime curvature, so isn't shedding guaranteed by a curvature singularity?

Seminal work on this was done by Poisson and Israel, and this work was continued by Ori. See

http://physics.technion.ac.il/~school/Amos_Ori.pdf .

Ori writes "Consequence to the curvature tensor at the IH ... However, the IH-singularity is weak (namely, tidally non-destructive.)"

Roughly, if components of g (the metric) are continuous but "pointy" (like the absolute value function), then first derivatives of g have step diiscontinuities (like the Heaviside step function), and second derivatives of g (used in the curvature tensor) are like Dirac delta functions. If a curvature singularity blows up like a Dirac delta function, then integration produces only a finite contribution to the tidal deformation of an object, which, if the object is robust enough, it can withstand.

 

[Fredrik]

Do you mean "escape the event horizon" or "survive the curvature singularity singularity at inner (Cauchy) horizon"?

There is a weak curvature singularity at the Cauchy inner horizon (IH) of a rotating black hole. Shredding is caused by tidal force, and tidal force is caused by spacetime curvature, so isn't shedding guaranteed by a curvature singularity?

Chronos posted a link to a paper earlier in the thread (post #8) that says that you may get shredded at the horizon even when the tidal forces are negligible. The argument is obviously not based on GR only, but on some assumptions about how GR and QM work together.

I have no idea if that's the sort of thing Aryianna was referring to. Edit: I think you're right that this must be about missing the singularity, not about passing through the event horizon in one piece.

 

[rustytxrx]

cartoons anyone?

cartoon = 10 chalk boards of calculus :)

 

[Naty1]

Chronos posts:

[Before the firewall paper, most of us were probably smugly confident that passing through an event horizon would be unremarkable, or even unnoticeable providing the black hole was sufficiently massive. I am considerably less confident than I was before that Polchinski guy crashed into the apple cart. /QUOTE]

I had not heard about that...Interesting.

The adjacent paper in ARXIV is from Leonard Susskind who so far disagrees...
should be fun to see how it plays out...


http://arxiv.org/abs/1301.4505

Black Hole Complementarity and the Harlow-Hayden Conjecture


Conclusion

[Amps is the Amheiri-Marolf-Polchinski-Sully firewall argument referenced by Chronos.]

The AMPS paradox is currently forcing a rethinking of how, and where, information is
stored in quantum gravity. The possible answers range from more or less conventional
localization (proximity and firewalls) to the radical delocalization of A = RB:
The argument in favor of the proximity postulate assumes the possibility of an Alice
experiment", which in some respects resembles time travel to the past. From this per-
spective, the firewall would function as a chronology protection agent, but at the cost of
the destruction of the interior of the black hole. The Harlow-Hayden conjecture opens an
entirely new perspective on chronology protection. It is based on the extreme fine-grained
character of information that Alice needs to distill before returning to the present."
Fine-grained information is something that has never been of much use in the past,
given how hard it is to extract. But there is clearly a whole world of fine-grained data
stored in the massive entanglements of scrambled pure states. That world is normally
inaccessible to us, but if BHC is correct, it is accessible to an observer who passes through
the horizon of a black hole. Thus complementarity implies a duality:
Ordinary coarse-grained information in an infalling frame is dual to the fine-grained
information in the exterior description...

It's obviously premature to declare the paradox resolved, but the validity of the HH
conjecture would allow the strong complementarity of Bousso and Harlow [14] to be con-
sistent, without the need for firewalls. For these reasons I believe that BHC, as originally
envisioned by Preskill, 't Hooft, and Susskind-Thorlacius-Uglum, is still alive and kicking.

 

[Aryianna]

Do you mean "escape the event horizon" or "survive the curvature singularity singularity at inner (Cauchy) horizon"?

There is a weak curvature singularity at the Cauchy inner horizon (IH) of a rotating black hole. Shredding is caused by tidal force, and tidal force is caused by spacetime curvature, so isn't shedding guaranteed by a curvature singularity?

Seminal work on this was done by Poisson and Israel, and this work was continued by Ori. See

http://physics.technion.ac.il/~school/Amos_Ori.pdf .

Ori writes "Consequence to the curvature tensor at the IH ... However, the IH-singularity is weak (namely, tidally non-destructive.)"

Roughly, if components of g (the metric) are continuous but "pointy" (like the absolute value function), then first derivatives of g have step diiscontinuities (like the Heaviside step function), and second derivatives of g (used in the curvature tensor) are like Dirac delta functions. If a curvature singularity blows up like a Dirac delta function, then integration produces only a finite contribution to the tidal deformation of an object, which, if the object is robust enough, it can withstand.

George, sorry for the very late reply (four months later), but yes I was referring to the probability of traveling through the singularity, not the event horizon. The interesting study is where the traveler ends up upon entering the singularity. My apologies for not being able to contribute or explain myself quantitatively. It has been years since I have been active in Physics.

 

[Low-Q]

I have learned that if a spaceship approach a black hole, the gravity that is stronger at the front end than the rear end will stretch the spaceship into pieces. Is this theory taking into account that the stronger gravity at the front end also slows down time more than the less gravity at the space ships rear end? Will the spaceship actually "feel" that it is stretched into pieces? Will it be stretched at all - relatively speaking? And will it ever reach the event horizon as the time finally stops (from what I have learned about time and gravity)?

Vidar

 

[Algr]

As you pass the event horizon there are still a wide variety of speeds that an object might be moving. A particle doesn't need to move away from the black hole in order to interact with other particles, you can have a state where some particles move faster than others.

It's rather like the path of the moon, relative to the sun. When viewed from the Earth, the Moon seems to move in an ellipse. But from the sun's POV, the moon never moves in a retrograde direction. (As moons of the outer planets do)

 

[Nugatory]

I have learned that if a spaceship approach a black hole, the gravity that is stronger at the front end than the rear end will stretch the spaceship into pieces. Is this theory taking into account that the stronger gravity at the front end also slows down time more than the less gravity at the space ships rear end? Will the spaceship actually "feel" that it is stretched into pieces? Will it be stretched at all - relatively speaking?

yes, yes, and yes - if the tidal forces are strong enough.
Somewhat counterintuitively, the tidal forces are less intense for larger black holes; we're looking at the difference between the force at the nose and the tail, not the absolute strength of the force. For a sufficiently large black hole, not only will you not be stretched, you may not even notice when you pass through the event horizon.

And will it ever reach the event horizon as the time finally stops (from what I have learned about time and gravity)?

Time doesn't stop, or even slow down, for the observer free-falling into the black hole. An observer hovering far away from the black hole may never see the free-faller reach the event horizon, but plenty of other observers will.