Time, Gravity

[Umbran]

Won't that cause a huge stress???

Not in and of itself, as I have said already. You keep returning to the escape velocities, but that is likewise not the relevant feature.

An additional issue, which I think is what you are getting at, is what to call the tether stress, and is it appropriate to think of that as being a "tidal" stress.

Yes. Stress results when forces are applied unevenly across a solid object. When you're talking about gravity applying a greater force on one end of an object than another, and thus causing stress, that's pretty much the textbook definition of a tidal force.

 

[tomBitonti]

Not in and of itself, as I have said already. You keep returning to the escape velocities, but that is likewise not the relevant feature.

But ... is the setup that the lower end of the tether is close to the event horizon, and the upper end of the tether is in a stable orbit?

Putting the lower end just slightly above the event horizon and the upper end slightly above a minimal stable orbit, is there no curvature between those points? The points must be at a distance that gets larger as the black hole gets larger. The local curvature does get smaller (flatter) as the black hole gets larger, but this is looking at two points at a distance which gets larger as the black hole gets larger.

Thx!

TomB

 

[freyar]

The thing that is kind of twisting my brain: the cable connecting the anchor and the orbiting spacecraft essentially stretches through time as much as it stretches through space between the two ends. One end is pretty much stuck at one point in time (the past), and the other end is continuing through the future. What happens if we send an astronaut down the cable? Does he go back in time the closer he gets to the anchor (which is stuck in the time when it was dropped)?

Bullgrit

You have to be careful about how you think about time (especially concepts like past, present, and future) in situations like this. The anchor (closer to the horizon) is not really "stuck in the past." Let's suppose our spaceship, the anchor, and your astronaut all carry incredibly precise atomic clocks which are all synchronized when they're all sitting together in that stable orbit. Now we lower the anchor. If you're watching from the spaceship, you see that the anchor's clock slows down compared to our clock, so that time seems to be passing very slowly at the anchor. However, if I'm on the anchor, I feel time passing normally but notice that the spaceship's clock appears to be running very fast. It's very important that neither is the "right" time; both of us are correct for our own time. If the astronaut starts out hanging out with you, then travels down the tether to visit me, then climbs back up, the astronaut's clock (wristwatch?) will not agree with either your clock or mine, but it will correctly register how much time has elapsed for the astronaut. Note that everyone is always moving forward in time. You never see my clock or the astronaut's watch stop completely. We just move forward in time at different rates.

What sticks in my brain is a question of whether a point infalling to a black hole is considered to actually cross the event horizon. Won't any signals emitted by an infalling object be red-shifted to be as faint as the hawking radiation of the black hole? Can external observers model the history of the infalling object as a fall which is gradually turned into an outflow of hawking radiation?

Thx!

TomB

This is sort of the million dollar question, and I now have to answer you very carefully. Let's start with Einstein's general relativity, which is where we really _know_ the answer. In GR (classical physics), there is no such thing as Hawking radiation, and we just see light from the infalling object get more and more redshifted. From our point of view far away, that's all that happens: the object falls toward the black hole and gradually goes dark as light from it redshifts --- we also see its clock slow down, and we never see it actually enter the black hole. From the infalling object's point of view, time just passes normally, and it never actually notices entering the black hole.

OK, now let's imagine some sort of quantum gravity. Hawking has famously given a strong argument that black holes should radiate like they are objects at a fixed temperature. The redshifting light from the object dropping in will, in fact, get dimmer than the Hawking radiation, but it can't be the Hawking radiation because it generally won't look like it has that fixed temperature. A big puzzle in quantum gravity is how what goes into a black hole turns into Hawking radiation. Some people claim that, if you collapse stuff to make a black hole, you don't actually end up with Hawking radiation, essentially saying that Hawking radiation is an artifact of imagining an eternal black hole. The puzzle is even stranger when you consider that someone falling into the black hole doesn't (classically) notice that they've fallen inside. They shouldn't see themselves turn in to Hawking radiation! Rather recently, another group of researchers claims that quantum mechanically, people falling into a black hole should actually burn up outside. They're probably wrong, but it's quite provocative.

Anyway, there is more to say, but it's past my bedtime...

 

[Umbran]

But ... is the setup that the lower end of the tether is close to the event horizon, and the upper end of the tether is in a stable orbit?

Yes.

Putting the lower end just slightly above the event horizon and the upper end slightly above a minimal stable orbit, is there no curvature between those points?

Yes, there is curvature. But there's also curvature between the ground and geostationary orbit on Earth. There's curvature around anything that has mass! The question isn't if there is curvature, but how curved is the space between the two points.

The thing you must remember is that, in terms of curvature, the event horizon is nothing special. Space gets very curved near the singularity at the center of a black hole. Now for smaller black holes, the event horizon is itself not all that far from the singularity, so we note great curvature, and thus great tidal stresses, near the Event Horizon. That's not a feature of the Horizon, but of the singularity beyond. But, for some (the very large ones), the event horizon is quite a ways off from the singularity, out where space is mostly flat, so that we no longer feel much in the way of tidal stresses.

 

[Umbran]

This is sort of the million dollar question, and I now have to answer you very carefully. Let's start with Einstein's general relativity, which is where we really _know_ the answer. In GR (classical physics), there is no such thing as Hawking radiation, and we just see light from the infalling object get more and more redshifted. From our point of view far away, that's all that happens: the object falls toward the black hole and gradually goes dark as light from it redshifts --- we also see its clock slow down, and we never see it actually enter the black hole. From the infalling object's point of view, time just passes normally, and it never actually notices entering the black hole.

Perhaps, to support the point - we keep thinking of how the anchor's clock slows down, asymptotically approaching a stop. Turn that around, and what the anchor sees is the outside universe speeding up, running through it's history to the end of the universe, while locally everything seems fine.

The redshifting light from the object dropping in will, in fact, get dimmer than the Hawking radiation, but it can't be the Hawking radiation because it generally won't look like it has that fixed temperature.

Correct - a thing cannot be converted into Hawking radiation until after it crosses the Event Horizon. GR, that never happens, from our external point of view. Infalling objects are caught in a Zeno's Paradox, always approaching the Event Horizon, but never crossing it. Quantum mechanics gives us the solution to that problem, as eventually matter can tunnel across the boundary.

Rather recently, another group of researchers claims that quantum mechanically, people falling into a black hole should actually burn up outside. They're probably wrong, but it's quite provocative.

They're probably wrong, but... There's also the issue of other radiation falling into the hole. We, at a distance, see radiation from the object red-shift ultimately to invisibility. But, that means that when we turn it around, and think of someone on the Anchor looking out at the universe, he or she sees the rest of the universe blue-shifted. That poor person on the anchor is getting sleeted with the hardest X-rays we can imagine! Even if there are no other effects ripping them apart, that's gonna sting!

 

[freyar]

Yes, there is curvature. But there's also curvature between the ground and geostationary orbit on Earth. There's curvature around anything that has mass! The question isn't if there is curvature, but how curved is the space between the two points.

The thing you must remember is that, in terms of curvature, the event horizon is nothing special. Space gets very curved near the singularity at the center of a black hole. Now for smaller black holes, the event horizon is itself not all that far from the singularity, so we note great curvature, and thus great tidal stresses, near the Event Horizon. That's not a feature of the Horizon, but of the singularity beyond. But, for some (the very large ones), the event horizon is quite a ways off from the singularity, out where space is mostly flat, so that we no longer feel much in the way of tidal stresses.

I just want to emphasize what Umbran said here. In classical general relativity, the event horizon is not special at all.

Perhaps, to support the point - we keep thinking of how the anchor's clock slows down, asymptotically approaching a stop. Turn that around, and what the anchor sees is the outside universe speeding up, running through it's history to the end of the universe, while locally everything seems fine.

Right, once again, the event horizon (or near it) is not special in classical general relativity.

Correct - a thing cannot be converted into Hawking radiation until after it crosses the Event Horizon. GR, that never happens, from our external point of view. Infalling objects are caught in a Zeno's Paradox, always approaching the Event Horizon, but never crossing it. Quantum mechanics gives us the solution to that problem, as eventually matter can tunnel across the boundary.

That's not actually right in classical relativity. You're right, from our point of view, an infalling object doesn't make it into the black hole. We see it slow down and never quite enter the horizon as we reach the infinite future. However, from the infalling object's point of view, as you said, time passes perfectly normally, and it falls in just fine. In fact, the infalling object doesn't see a horizon at all. There is no need for quantum mechanics whatsoever. Tunnelling is sometimes used as a way to explain where Hawking radiation comes from, but that's about stuff coming out of the black hole, not falling in.

They're probably wrong, but... There's also the issue of other radiation falling into the hole. We, at a distance, see radiation from the object red-shift ultimately to invisibility. But, that means that when we turn it around, and think of someone on the Anchor looking out at the universe, he or she sees the rest of the universe blue-shifted. That poor person on the anchor is getting sleeted with the hardest X-rays we can imagine! Even if there are no other effects ripping them apart, that's gonna sting!

That might be true in the real universe, but this argument is made in a universe devoid of anything but the infalling person, the black hole, and outgoing Hawking radiation. In this case, it's supposed to be the Hawking radiation that burns up the infalling person. The reason this claim is unusual sounding comes back to the point that the horizon isn't supposed to be special for an infalling object. In fact, in the usual calculation deriving Hawking radiation, the Hawking radiation doesn't come from the horizon at all but from generally all the space around the black hole. So it's a bit of a puzzle.

 

[tomBitonti]

Yes.
Yes, there is curvature. But there's also curvature between the ground and geostationary orbit on Earth. There's curvature around anything that has mass! The question isn't if there is curvature, but how curved is the space between the two points.

The thing you must remember is that, in terms of curvature, the event horizon is nothing special. Space gets very curved near the singularity at the center of a black hole. Now for smaller black holes, the event horizon is itself not all that far from the singularity, so we note great curvature, and thus great tidal stresses, near the Event Horizon. That's not a feature of the Horizon, but of the singularity beyond. But, for some (the very large ones), the event horizon is quite a ways off from the singularity, out where space is mostly flat, so that we no longer feel much in the way of tidal stresses.

Let me ask ... the upper tether endpoint is in a minimum stable orbit (which is way up where the escape velocity is almost 0.5c). The lower tether is as close as possible to the event horizon without falling through (where the escape velocity is almost 1.0c). What force will there be across the tether, and will it depend on the size of the black hole?

Thx!

TomB

 

[Umbran]

That's not actually right in classical relativity. You're right, from our point of view, an infalling object doesn't make it into the black hole. We see it slow down and never quite enter the horizon as we reach the infinite future. However, from the infalling object's point of view, as you said, time passes perfectly normally, and it falls in just fine.

Yes, but, by the watch of someone far away from the black hole, when does that happen?

The infalling person's clock keeps running, but at an ever-slowing rate as compared to the rest of the universe, asymptotically approaching (but never reaching) a stop. So their personal experience is that they fall and will eventually cross the horizon, the rest of the universe has gone off to heat death before that happens, hasn't it? It isn't "never" by his own clock, it is "never" by ours, no? If they had an amazing jet pack that could accelerate them to arbitrarily close to c, there is no time in our future that they could not emerge, isn't that correct?

Tunneling gets invoked so that they get across that boundary in what we, in the rest of the universe, would consider finite time.

 

[freyar]

Yes, but, by the watch of someone far away from the black hole, when does that happen?

The infalling person's clock keeps running, but at an ever-slowing rate as compared to the rest of the universe, asymptotically approaching (but never reaching) a stop. So their personal experience is that they fall and will eventually cross the horizon, the rest of the universe has gone off to heat death before that happens, hasn't it? It isn't "never" by his own clock, it is "never" by ours, no? If they had an amazing jet pack that could accelerate them to arbitrarily close to c, there is no time in our future that they could not emerge, isn't that correct?

Tunneling gets invoked so that they get across that boundary in what we, in the rest of the universe, would consider finite time.

What you say is true. The thing is that getting infalling stuff into the horizon in finite time isn't considered an issue, and I've never heard any of the "big thinkers" about black holes, Hawking radiation, etc, mention any worries about needing to tunnel into a black hole. So I just wouldn't want to say that tunnelling in is in any way a consensus view.

 

[freyar]

Let me ask ... the upper tether endpoint is in a minimum stable orbit (which is way up where the escape velocity is almost 0.5c). The lower tether is as close as possible to the event horizon without falling through (where the escape velocity is almost 1.0c). What force will there be across the tether, and will it depend on the size of the black hole?

There is definitely going to be a force through the tether, though the calculation isn't going to be very simple. As Umbran has mentioned, though, the force certainly will depend on the black hole size, as well as a number of other details such as spin.