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I am the first in the world

theprestige said:
In thinking this over, I have a question, which I am hoping one of the other thread participants can answer for me:

If I am in the same frame as an observer falling through the event horizon, but my motion keeps me outside the event horizon for the duration of my observations, will I observe the horizon-crosser's clock to keep ticking at the same rate that she observes?
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ben m said:
As you yourself are falling in towards the event horizon, you will always say that the event horizon is in front of you and receding away from you. If you started out with a partner a bit ahead of you, you'll see that partner approach (but never cross) the receding horizon, and (as with an outside observer) light from that partner will be seen as redshifted.

You have to be careful with the idea of falling in with an observer in the "same frame". Since there's a strong gravity gradient, an observer at a fixed distance from you cannot be in an inertial frame. Another observer in an inertial frame who starts out co-moving with you, will accelerate away from you.
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On that subject (sort of), here are some very nice simulations of what it would look like to fall into a black hole:

http://jila.colorado.edu/~ajsh/insidebh/schw.html

They illustrate the effect you (ben m) mentioned above - you can't tell when you have crossed the horizon.
 
ben m said:
As you yourself are falling in towards the event horizon, you will always say that the event horizon is in front of you and receding away from you. If you started out with a partner a bit ahead of you, you'll see that partner approach (but never cross) the receding horizon, and (as with an outside observer) light from that partner will be seen as redshifted.

You have to be careful with the idea of falling in with an observer in the "same frame". Since there's a strong gravity gradient, an observer at a fixed distance from you cannot be in an inertial frame. Another observer in an inertial frame who starts out co-moving with you, will accelerate away from you.
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Thanks, I totally forgot that the gravity difference would totally prevent me from actually being in the same frame as the horizon-crosser!

ctamblyn said:
On that subject (sort of), here are some very nice simulations of what it would look like to fall into a black hole:

http://jila.colorado.edu/~ajsh/insidebh/schw.html

They illustrate the effect you (ben m) mentioned above - you can't tell when you have crossed the horizon.
Click to expand...

Thanks, those are definitely neat pictures!
 
steenkh said:
I was thinking more of the coordinates of the Earth where the North Pole is a strange place where North is not defined, but if we choose a coordinate system with North at a different place, the place where the North Pole was is not special at all any longer.
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That, in fact, is precisely the right analogy. If you're familiar with polar coordinates (more or less the same thing), even better. In Farsight's favorite coordinates, the radial distance from the horizon r is analogous to latitude from the north pole, and time is longitude (mathematically, the only difference is a minus sign).

The coordinate singularity at the horizon where time stops is analogous (indeed, almost identical) to exactly what you say, that there is no north - and more importantly, no east or west - at the north pole.
 
theprestige said:
In thinking this over, I have a question, which I am hoping one of the other thread participants can answer for me:

If I am in the same frame as an observer falling through the event horizon, but my motion keeps me outside the event horizon for the duration of my observations, will I observe the horizon-crosser's clock to keep ticking at the same rate that she observes?

ETA: And is there a better way to formulate this question?
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Let me reformulate slightly: Alice and Bob step out of the hatch of their spaceship near a black hole. The ship is on autopilot, and continually blasting its rockets just enough to stay a fixed distance from the horizon. Alice steps out first, followed a few seconds later by Bob. Neither has any means of maneuvering in space.

What they will see is the rocket accelerate away from them, picking up speed as it moves away. They themselves will stay almost exactly a fixed distance apart, at relative rest. This continues as they approach and then cross the horizon. They will continue to see the rocket the entire time (although it moves further and further away), and they will remain almost at rest with respect to each other.

In other words, what they will see is almost identical to what they would see if they stepped out of an accelerating rocket in flat, empty space - unsurprisingly, since that's more or less a consequence of the equivalence principle.

The difference is that some time after crossing the horizon they will start to feel some unpleasant stretchings as their bodies are acted on by tidal forces. Shortly thereafter, they will be spaghettified and ripped apart as they approach the singularity (ouch).

This ignores any other objects falling in, stars behind the hole getting lensed, etc., and it also assumes the hole is fairly big.
 
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Farsight said:
Come on sol, you know that gravitational time dilation is not symmetrical. All observers will agree that a clock near the event horizon is going slower than a clock in free space.
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No, they will not. You yourself posted a plot that shows that in sufficiently low earth orbit the orbiting clocks run slow, not fast.

In general, whether a clock that waited near a horizon reads more or less elapsed time than its twin that spent its time far from the hole depends on the trajectories each one followed.
 
sol invictus said:
Let me reformulate slightly: Alice and Bob step out of the hatch of their spaceship near a black hole. The ship is on autopilot, and continually blasting its rockets just enough to stay a fixed distance from the horizon. Alice steps out first, followed a few seconds later by Bob. Neither has any means of maneuvering in space.

What they will see is the rocket accelerate away from them, picking up speed as it moves away. They themselves will stay almost exactly a fixed distance apart, at relative rest. This continues as they approach and then cross the horizon. They will continue to see the rocket the entire time (although it moves further and further away), and they will remain almost at rest with respect to each other.

In other words, what they will see is almost identical to what they would see if they stepped out of an accelerating rocket in flat, empty space - unsurprisingly, since that's more or less a consequence of the equivalence principle.

The difference is that some time after crossing the horizon they will start to feel some unpleasant stretchings as their bodies are acted on by tidal forces. Shortly thereafter, they will be spaghettified and ripped apart as they approach the singularity (ouch).

This ignores any other objects falling in, stars behind the hole getting lensed, etc., and it also assumes the hole is fairly big.
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Thanks, sol!

If Bob were to observe Alice's clock as it crossed the horizon a few seconds ahead of his own, what would he observe?
 
ben m said:
As you yourself are falling in towards the event horizon, you will always say that the event horizon is in front of you and receding away from you. If you started out with a partner a bit ahead of you, you'll see that partner approach (but never cross) the receding horizon, and (as with an outside observer) light from that partner will be seen as redshifted.
Click to expand...

I don't think that's correct, unless I'm misunderstanding the situation you're analyzing. Do you agree with post #252?

Since there's a strong gravity gradient, an observer at a fixed distance from you cannot be in an inertial frame. Another observer in an inertial frame who starts out co-moving with you, will accelerate away from you.
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At least for a large black hole there aren't any strong "gravity gradients". The curvatures near the horizon are all of order the inverse horizon radius.
 
Perhaps he was thinking of the horizon you see (the antihorizon), which is never crossed, and is different from the true horizon, as per the Penrose diagram. Or for a realistic black hole, the stellar surface that seems to always be ahead ahead of you, getting far redshifted. That sort of (but not quite) fits.
 
sol invictus said:
I don't think that's correct, unless I'm misunderstanding the situation you're analyzing. Do you agree with post #252?
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Vorpal said:
Perhaps he was thinking of the horizon you see (the antihorizon), which is never crossed, and is different from the true horizon, as per the Penrose diagram. Or for a realistic black hole, the stellar surface that seems to always be ahead ahead of you, getting far redshifted. That sort of (but not quite) fits.
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Sorry, yes, Vorpal has the diagnosis. I meant to say that you see your partner approach, but never cross, the antihorizon. Isn't this correct?
 
ben m said:
Sorry, yes, Vorpal has the diagnosis. I meant to say that you see your partner approach, but never cross, the antihorizon. Isn't this correct?
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I'll confess I'd never heard of the "antihorizon" before now, despite knowing a fair amount about black holes. If it's labelled correctly in this diagram, there's a reason for that.

First, it only exists for eternal black holes - real black holes that formed at a finite time in the past do not have it. Here's a spacetime diagram showing that.

Second, even for an eternal black hole it really should be divided into two parts. The lower right part should be thought of as part of the event horizon. The upper left part is something else entirely. One way to see that is to draw the "stretched horizon" - a smooth, curved timelike line that is everywhere to the right of the horizon and lower right part of the antihorizon. That's physically what you should think of as the horizon of the hole.
 
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A White-Hole (WH) and a Black-Hole (BH) enter a bar. The WH and BH order drinks. As the bartender sends the WH its drink the hole through which the WH would take it's drink gets farther and farther away. On the other hand, as the bartender extends a hand to give the BH a drink (finish joke anyway you want...).
 
sol invictus said:
I'll confess I'd never heard of the "antihorizon" before now, despite knowing a fair amount about black holes. If it's labelled correctly in this diagram, there's a reason for that.
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Hmm. That can't be the same thing I'm thinking of.

The infalling observer, despite having crossed the event horizon, does not suddenly get a view of a naked singularity. What do you call the horizon that protects them from this view? I guess http://casa.colorado.edu/~ajsh/singularity.html calls it the "Schwarzschild surface".
 
ben m said:
The infalling observer, despite having crossed the event horizon, does not suddenly get a view of a naked singularity. What do you call the horizon that protects them from this view?
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There is only one horizon - the event horizon. Observers that cross it don't see the singularity because it (the singularity) is in their future. The singularity is a spacelike volume, quite like a big crunch, not a point in space that's extended in time.

I guess http://casa.colorado.edu/~ajsh/singularity.html calls it the "Schwarzschild surface".
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That's just the event horizon.

ETA - for eternal charged or spinning black holes, there is an inner horizon and the singularity is timelike. However, that's again a mathematical fiction. The inner horizon exists only for eternal solutions in a perfect vacuum. It is unstable to forming a spacelike singularity just like that of a Schwarzschild hole.
 
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sol invictus said:
There is only one horizon - the event horizon. Observers that cross it don't see the singularity because it (the singularity) is in their future. The singularity is a spacelike volume, quite like a big crunch, not a point in space that's extended in time.
Click to expand...

Aha! Now the Penrose diagram you linked to makes sense. Thanks.
 
Life and information

“I define Evolution, accumulating and processing information for biological systems” Adrian Ferent
Evolution is faster for those capable to accumulate and process more information.
I calculated precisely the lower bound, the lowest number of qubits after the Big Bang:
min Iuniverse = 15.392 × 10^61 qubits.
The funny thing will be if we will find out that an advanced civilization is using and controlling the information of our planet and of course we do not know about it.
An abstract of my Life and information theory on my blog: http://adrianferent.blogspot.com/
 
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