Time dilation when falling into a black hole

[guss]

Let's say I start out a few thousand kilometers from a black hole, and I begin to move toward the black hole due to it's gravitational pull.

What type of time dilation would I experience as I fell into the black hole before the event horizon, and after the event horizon? By the time I die inside of the black hole, approximately how many years on Earth would have gone by (no years, a year, a few hundred years...)?

Thanks.

 

[Mark M]

Well, remember that you never experience time dilation. For you, time is always normal, and it's everyone else's clocks that are running at a different rate.

An observer on Earth will never see you cross the event horizon. As you get closer and closer to it, light from you gets more and more redshifted, and you appear to move slower and slower. Eventually, you'll just redshift out if the visible spectrum, and the astronomers on Earth will have seen you 'frozen' at the event horizon. So, no amount of time can pass on Earth until you hit the center, as that will never happen according to them.

As I mentioned in the beginning, time is always normal for you. So, if you jump into a black hole, you won't notice anything peculiar. However, you'll either be torn apart by tidal forces, or you'll hit the singularity and die after a very short period of time.

 

[guss]

Well, remember that you never experience time dilation. For you, time is always normal, and it's everyone else's clocks that are running at a different rate.

An observer on Earth will never see you cross the event horizon. As you get closer and closer to it, light from you gets more and more redshifted, and you appear to move slower and slower. Eventually, you'll just redshift out if the visible spectrum, and the astronomers on Earth will have seen you 'frozen' at the event horizon. So, no amount of time can pass on Earth until you hit the center, as that will never happen according to them.

As I mentioned in the beginning, time is always normal for you. So, if you jump into a black hole, you won't notice anything peculiar. However, you'll either be torn apart by tidal forces, or you'll hit the singularity and die after a very short period of time.

Okay, thanks. So I understand that time will always be normal for me, and that people on Earth won't be able to see me. But, what about the time on Earth relative to the person falling into the black hole's time? By the time the person died, (very) roughly how much time on Earth would have passed?

I have heard, since spacetime inside the black hole is infinitely warped, that an infinite or near infinite amount of time will pass on Earth before I die. But, I'm not sure if this is correct.

 

[phinds]

Okay, thanks. So I understand that time will always be normal for me, and that people on Earth won't be able to see me. But, what about the time on Earth relative to the person falling into the black hole's time? By the time the person died, (very) roughly how much time on Earth would have passed?

I have heard, since spacetime inside the black hole is infinitely warped, that an infinite or near infinite amount of time will pass on Earth before I die. But, I'm not sure if this is correct.

If people on Earth could SEE you die, it would take approximately forever, but in actual fact you would die in quite a small amount of Earth time and physicists would pronounce you dead. Amateurs with good telescopes would be shouting, NO ... he's still outside the event horizon and quite alive.

 

[guss]

If people on Earth could SEE you die, it would take approximately forever, but in actual fact you would die in quite a small amount of Earth time and physicists would pronounce you dead. Amateurs with good telescopes would be shouting, NO ... he's still outside the

Haha, thanks, good way to put it. Could you distinguish between your first two points a bit? I don't understand why physicists would pronounce you dead even though you are still technically alive inside of the event horizon due to the warped spacetime. Or are you? And if you are not technically alive, then why would you die quickly in Earth time in this infinitely warped time?

 

[phinds]

Haha, thanks, good way to put it. Could you distinguish between your first two points a bit? I don't understand why physicists would pronounce you dead even though you are still technically alive inside of the event horizon due to the warped spacetime. Or are you? And if you are not technically alive, then why would you die quickly in Earth time in this infinitely warped time?

You would NOT be technically alive. You would be technically and actually dead. The fact that there are some gravitationally red-shifted photons that say otherwise is on the order of an optical illusion. You need to read up on the relativity of simultaneity and how people in different frames of reference see things differently.

 

[A.T.]

in actual fact you would die in quite a small amount of Earth time

Nope. You would die in quite a small amount of your own time. It would take forever of Earth time.

 

[A.T.]

The fact that there are some gravitationally red-shifted photons that say otherwise is on the order of an optical illusion. You need to read up on the relativity of simultaneity and how people in different frames of reference see things differently.

Relativistic effects have nothing to do with "optical illusions" or just seeing things differently. They are what's left after you factor out the signal delay and optical effects.

 

[Nugatory]

Nope. You would die in quite a small amount of your own time. It would take forever of Earth time.

If the cause of death is being ripped apart by tidal forces (which doesn't necessarily have to happen at all) before you've fallen through the event horizon... That will happen in a finite and possibly a quite short amount of Earth time.

If we're going to declare the moment of death to be when you actually are observed to pass through the event horizon... yes, that will take forever in Earth time.

 

[phinds]

Nope. You would die in quite a small amount of your own time. It would take forever of Earth time.

DOH ! Once again I have proven how stupid/careless I sometimes am.

 

[Darwin123]

Well, remember that you never experience time dilation. For you, time is always normal, and it's everyone else's clocks that are running at a different rate.

An observer on Earth will never see you cross the event horizon. As you get closer and closer to it, light from you gets more and more redshifted, and you appear to move slower and slower. Eventually, you'll just redshift out if the visible spectrum, and the astronomers on Earth will have seen you 'frozen' at the event horizon. So, no amount of time can pass on Earth until you hit the center, as that will never happen according to them.

As I mentioned in the beginning, time is always normal for you. So, if you jump into a black hole, you won't notice anything peculiar. However, you'll either be torn apart by tidal forces, or you'll hit the singularity and die after a very short period of time.

Suppose that there is a large mass of material that is compacting. The center is a black hole that is expanding. Suppose that you, a small inhomogeneity, are falling into the black hole.
Also assume that this mass of material is spinning. So the black hole that forms is also spinning. You are falling toward the equator. However, centrifugal forces are slowing your approach to the black hole.
Wouldn't you see an ellipsoidal black hole that is expanding near the equator?
Couldn't the black hole expand to envelope you before you are torn apart or crushed?

The original Kerr calculations where done for a static black hole with no spin. In this model, the only expansion is spherically symmetrical. There are no inhomogeneities. However, a spaceship falling into a black hole is a small inhomogeneity. I am not sure the calculations for a spherically symmetric black hole apply to a black hole system with both inhomogeneity and anisotropy.

 

[guss]

So, wait, I'm confused now. Would it take forever/a long time in Earth time for you to technically die?

 

[phinds]

So, wait, I'm confused now. Would it take forever/a long time in Earth time for you to technically die?

Sorry for mis-stating previously.

In YOUR time, you would die rather quickly, and by the time you hit the singularity (not long after dying) the Earth would have aged an ENORMOUS amount.

 

[ExecNight]

So, wait, I'm confused now. Would it take forever/a long time in Earth time for you to technically die?

When you run into a wall with 500 MPH, the amount of time people wait before cleaning your remains from the wall is quite irrelevant.

 

[guss]

Sorry for mis-stating previously.

In YOUR time, you would die rather quickly, and by the time you hit the singularity (not long after dying) the Earth would have aged an ENORMOUS amount.

No worries. So, would it be correct to say that if you fell towards and into a black hole, you would outlive Earth itself, and maybe the human race?

And, I assume this is because as you fall towards the event horizon, time slows and slows relative to Earth time, then as you are almost touching the event horizon time practically stops?

 

[PAllen]

Nope. You would die in quite a small amount of your own time. It would take forever of Earth time.

It depends on who is correlating with Earth - which simultaneity is used. Barring tidal shredding, a point particle sees signals continuing to arrive from Earth between passing the horizon and reaching the singularity. They can add signal delay, and arrive at a specific Earth time for when they reached the singularity.

As for an Earth observer, the only strictly physical statements you can make are what light and signals they will receive. It is philosophy whether you call this an illusion or not. Specifically, simultaneity is not a physical observable and there is nothing inherently real about SC simultaneity for the outside observer. They could choose a different simultaneity convention (a foliation into spacelike surfaces parameterized by proper time along their world line), and also conclude a fixed time when singularity was reached. From which they would be inclined to think of a trapped light interpretation.

 

[Mark M]

So, wait, I'm confused now. Would it take forever/a long time in Earth time for you to technically die?

According to the observers on earth, you'll never pass the event horizon. This is because the time dilation factor goes to infinity as you approach the Schwarzschild radius. For a black hole, the SR is the event horizon.

 

[PAllen]

According to the observers on earth, you'll never pass the event horizon. This is because the time dilation factor goes to infinity as you approach the Schwarzschild radius. For a black hole, the SR is the event horizon.

Observer's on Earth will never see you reach the event horizon. How this is interpreted, is a choice, not physics. See my prior post.

Analogy: If you are accelerating in a rocket at 1 g forever, you are not required to believe that time on Earth slowed to a stop corresponding to your Rindler horizon. You can choose a different simultaneity convention and conclude no such thing. However, it is undisputed that no signals from Earth past a certain time will reach you as long as you continue accelerating. But you can stop accelerating and then see what you missed. Similarly, you can 'stop accelerating away from the SC horizon' - i.e. fall into it - and then see all the history of infaller's you missed.

 

[PeterDonis]

The original Kerr calculations where done for a static black hole with no spin.

I think you mean the original Schwarzschild calculations, or perhaps the original Oppenheimer-Snyder calculations of a spherically symmetric collapse of matter to form a black hole. The Kerr solution is for a spinning black hole.

a spaceship falling into a black hole is a small inhomogeneity. I am not sure the calculations for a spherically symmetric black hole apply to a black hole system with both inhomogeneity and anisotropy.

We don't have an *analytical* solution that includes inhomogeneity and anisotropy. But there have been plenty of numerical solutions done of such cases. Basically what happens is that any anisotropy/inhomogeneity in the BH's horizon, due to something like a spaceship falling in, quickly gets radiated away as gravitational waves, and the BH settles down to a new spherically symmetric state with a slightly larger mass. None of this affects what other people have been saying about the proper time elapsed for an infalling observer, or the relationship of that to the time elapsed on Earth.

 

[pervect]

Let's say I start out a few thousand kilometers from a black hole, and I begin to move toward the black hole due to it's gravitational pull.

What type of time dilation would I experience as I fell into the black hole before the event horizon, and after the event horizon? By the time I die inside of the black hole, approximately how many years on Earth would have gone by (no years, a year, a few hundred years...)?

Thanks.

You need to define a little bit more about how you measure time dilation. What comes to my mind is observing the frequency of a laser beam directed toward you from far away, "at infinity"m, to speak loosely.

The laser beam is a constant frequency as measured at its source, far away from the black hole, but as you fall into it, the frequency you directly measure will change, due to doppler shift. We can imagine that the laser beam also is amplitude modulated with some timestamp information, , so we know what time any light signal was sent.

As you fall into the black hole, you'll see the beam redshift. You'll attribute this redshift to tidal forces, as there won't be any other sort of forces affecting your trajectory. You will see that as the laser beam redshifts, and that as it redshifts the encoded "timestamp" information slows down at exactly the same rate as the carrier frequency of the beam itself is reduced.

This is what you will directly measure, in terms of doppler shift.

I calculated once that from a fall from infinity, the redshift factor at the event horizon would be 2:1 at the event horizon. I haven't done any more detailed calculations for the exact scenario you describe.

There isn't any good answer to " By the time I die inside of the black hole, approximately how many years on Earth would have gone by ", because simultaneity is relative. What you can answer in principle is the timstamp of the last signal you'll ever receive just before you reach the singularity. You can also answer, perhaps more easily, what the reading of a clock you carry with you will be just before you reach the singularity as well - i.e. the proper time it takes you to reach the singularity.

Unfortunately, while the exterior geometry of the black hole is well understood, the interior geometry of a black hole is much less well understood. It hasn't been measured for reasons which I hope are obvious, and there is good reason to believe that the usual Schwarzschild geometry is not stable - and there are other puzzling issues that arise in particular with rotating black holes. And you'll be unlikely to find any other sort of black hole in reality, almost any black hole you'll find will have some angular momentum.

So while we can answer the above questions (what's the proper time, and the timestamp of the last signal you ever see, before you reach the singularity) for a Schwarzschild blacak hole, we can't be too terribly confident what the actual answer would be with a real black hole.

http://casa.colorado.edu/~ajsh/schw.shtml
might be a useful online reference, the author of the webpage has written a number of peere reviewed papers on black holes.

 

[PeterDonis]

So, would it be correct to say that if you fell towards and into a black hole, you would outlive Earth itself, and maybe the human race?

Once you fall into the black hole, you can't send any signals back out to Earth. So the people on Earth have no way of knowing what's happened to you--the last light signal you send just before you cross the hole's horizon will reach Earth very, very far in its future, and that's the last information Earth will receive from you. No one on Earth will ever see you fall inside the horizon, so in that sense all of your experience inside the horizon is "in the future" of every event on Earth, and you could be said to "outlive Earth" in this sense. That's one way of answering your question.

However, you can still *receive* light signals from Earth; so another way of answering your question is to ask, at what point on Earth's worldline will the *last* light signal you receive, just before you hit the singularity in the center of the BH, be emitted?

It turns out that that, if you freely fall into the black hole, the last light signal from Earth that you will receive, just before you hit the singularity, will be from not very far in Earth's future, compared to the time you left--so there will still be a lot of Earth's history to go, that you will never be able to see. The exact numbers will depend on the specifics of the scenario, but in general, you will *not* outlive Earth in this sense. (Whether you will outlive the human race probably depends more on how long you think the human race is likely to last, but that's a separate issue. :)

 

[guss]

You need to define a little bit more about how you measure time dilation. What comes to my mind is observing the frequency of a laser beam directed toward you from far away, "at infinity"m, to speak loosely.

The laser beam is a constant frequency as measured at its source, far away from the black hole, but as you fall into it, the frequency you directly measure will change, due to doppler shift. We can imagine that the laser beam also is amplitude modulated with some timestamp information, , so we know what time any light signal was sent.

As you fall into the black hole, you'll see the beam redshift. You'll attribute this redshift to tidal forces, as there won't be any other sort of forces affecting your trajectory. You will see that as the laser beam redshifts, and that as it redshifts the encoded "timestamp" information slows down at exactly the same rate as the carrier frequency of the beam itself is reduced.

This is what you will directly measure, in terms of doppler shift.

I calculated once that from a fall from infinity, the redshift factor at the event horizon would be 2:1 at the event horizon. I haven't done any more detailed calculations for the exact scenario you describe.

There isn't any good answer to " By the time I die inside of the black hole, approximately how many years on Earth would have gone by ", because simultaneity is relative. What you can answer in principle is the timstamp of the last signal you'll ever receive just before you reach the singularity. You can also answer, perhaps more easily, what the reading of a clock you carry with you will be just before you reach the singularity as well - i.e. the proper time it takes you to reach the singularity.

Unfortunately, while the exterior geometry of the black hole is well understood, the interior geometry of a black hole is much less well understood. It hasn't been measured for reasons which I hope are obvious, and there is good reason to believe that the usual Schwarzschild geometry is not stable - and there are other puzzling issues that arise in particular with rotating black holes. And you'll be unlikely to find any other sort of black hole in reality, almost any black hole you'll find will have some angular momentum.

So while we can answer the above questions (what's the proper time, and the timestamp of the last signal you ever see, before you reach the singularity) for a Schwarzschild blacak hole, we can't be too terribly confident what the actual answer would be with a real black hole.

http://casa.colorado.edu/~ajsh/schw.shtml
might be a useful online reference, the author of the webpage has written a number of peere reviewed papers on black holes.

Okay, thanks. I found this at http://casa.colorado.edu/~ajsh/schwp.html#slow:

This time dilation factor tends to zero as r approaches the Schwarzschild radius rs, which means that someone at the Schwarzschild radius will appear to freeze to a stop, as seen by anyone outside the Schwarzschild radius.

which further supports that time on Earth will move much faster.

Once you fall into the black hole, you can't send any signals back out to Earth. So the people on Earth have no way of knowing what's happened to you--the last light signal you send just before you cross the hole's horizon will reach Earth very, very far in its future, and that's the last information Earth will receive from you. No one on Earth will ever see you fall inside the horizon, so in that sense all of your experience inside the horizon is "in the future" of every event on Earth, and you could be said to "outlive Earth" in this sense. That's one way of answering your question.

However, you can still *receive* light signals from Earth; so another way of answering your question is to ask, at what point on Earth's worldline will the *last* light signal you receive, just before you hit the singularity in the center of the BH, be emitted?

It turns out that that, if you freely fall into the black hole, the last light signal from Earth that you will receive, just before you hit the singularity, will be from not very far in Earth's future, compared to the time you left--so there will still be a lot of Earth's history to go, that you will never be able to see. The exact numbers will depend on the specifics of the scenario, but in general, you will *not* outlive Earth in this sense. (Whether you will outlive the human race probably depends more on how long you think the human race is likely to last, but that's a separate issue. :)

Thanks for all the info. But, now I am sort of confused again. The bolded part contradicts what was previously established in this thread - that you would appear to freeze to someone on Earth (if they knew where you were) as you neared the event horizon due to the extreme gravitational time dilation. I'm sure I'm missing something...

 

[PAllen]

Thanks for all the info. But, now I am sort of confused again. The bolded part contradicts what was previously established in this thread - that you would appear to freeze to someone on Earth (if they knew where you were) as you neared the event horizon due to the extreme gravitational time dilation. I'm sure I'm missing something...

No, there is no contradiction. Please read my posts. From earth, the infaller would red shift and effectively disappear. But for the infaller, Earth would appear pretty normal (not infinitely fast), and there would be a modest, finite time passing on Earth (as seen by the infaller) between the infaller's crossing of the event horizon and the infaller's reaching the singularity.

What the Earth visually sees is at odds with what the infaller sees. I think it is not meaningful to attach much significance to what the the Earth visually sees. I put this in the same category of 'stopped light' in exotic materials similar to what Lene Hau has achieved in labs on Earth (google her).

 

[guss]

No, there is no contradiction. Please read my posts. From earth, the infaller would red shift and effectively disappear. But for the infaller, Earth would appear pretty normal (not infinitely fast), and there would be a modest, finite time passing on Earth (as seen by the infaller) between the infaller's crossing of the event horizon and the infaller's reaching the singularity.

What the Earth visually sees is at odds with what the infaller sees. I think it is not meaningful to attach much significance to what the the Earth visually sees. I put this in the same category of 'stopped light' in exotic materials similar to what Lene Hau has achieved in labs on Earth (google her).

Okay, thanks. I did read your replies, and I think it is just a little over my head. So, if I am correct, there are four (very) different things here:

• What the infaller sees looking at Earth - nothing bizarre, Earth appears mostly normal
  
• What Earth sees looking at the infaller - nothing (redshifted out of visible spectrum)/not significant
  
• What is actually happening to the observer relative to Earth - observer is frozen
  
• What is actually happening to Earth relative to the observer - lots of time goes by

I hope those last two make sense, what I am trying to do is ignore the optics, but I fear what I wrote makes no physical sense and is not at all meaningful.

Is this correct? If I'm not close I think I'll just come back to this in a while after I learn more about this stuff.

 

[PeterDonis]

The bolded part contradicts what was previously established in this thread - that you would appear to freeze to someone on Earth (if they knew where you were) as you neared the event horizon due to the extreme gravitational time dilation. I'm sure I'm missing something...

The bolded part is not talking about light that travels from you to Earth. It is talking about light traveling from Earth to you--i.e., in the opposite direction (down into the hole, instead of up away from the hole). There's no contradiction; I'm talking about something different than what's been discussed in the rest of this thread.

 

[PeterDonis]

What the infaller sees looking at Earth - nothing bizarre, Earth appears mostly normal

Yes--with the additional proviso that the last thing the infaller sees of Earth, just before he hits the singularity, will be not too far in Earth's future.

What Earth sees looking at the infaller - nothing (redshifted out of visible spectrum)/not significant

Yes, with the qualifier that we are talking about when the infaller crosses the horizon--after that there is literally *nothing* getting back to Earth, not even redshifted out of the visible spectrum. Light can't escape from inside the horizon at all.

What is actually happening to the observer relative to Earth - observer is frozen

I would not put it this way; I would say that, once the observer falls inside the horizon, Earth simply has no way of knowing "what is actually happening" to the observer. It "seems" to the people on Earth that the observer is frozen; but it is possible for the people on Earth to calculate that this appearance is due to "optics", which you said you are trying to ignore. In other words, the people on Earth can calculate that, even though the infalling observer appears to be "frozen", he actually does pass through the horizon. (They can do this by calculating the proper time the infalling observer will experience from when he starts to fall until he reaches the horizon, and finding that it is finite. They can also calculate a similar result for the proper time he will experience before reaching the singularity, and find that that is finite too.)

What is actually happening to Earth relative to the observer - lots of time goes by

I don't see any interpretation of the physics that would make this true. Once the observer hits the singularity, he ceases to exist and can no longer receive signals from Earth. Since the last signal he receives before that is from not too far in Earth's future, the only interpretation of "what is actually happening to Earth relative to observer" would have not too much time passing on Earth.

 

[guss]

I don't see any interpretation of the physics that would make this true. Once the observer hits the singularity, he ceases to exist and can no longer receive signals from Earth. Since the last signal he receives before that is from not too far in Earth's future, the only interpretation of "what is actually happening to Earth relative to observer" would have not too much time passing on Earth.

Okay, thanks a lot, that makes sense. I feel like I have a basic understanding now. Then, how would you fit this in?:

Nope. You would die in quite a small amount of your own time. It would take forever of Earth time.

 

[pervect]

Okay, thanks. I found this at http://casa.colorado.edu/~ajsh/schwp.html#slow:



which further supports that time on Earth will move much faster.

You should also find

http://casa.colorado.edu/~ajsh/singularity.html#r=1

Answer to the quiz question 5: False. You do NOT see all the future history of the world played out. Once inside the horizon, you are doomed to hit the singularity in a finite time, and you witness only a finite (in practice rather short) time pass in the outside Universe.

Also check out the "redshift map" on the same page, which shows distant areas of the sky are redshifted, as per my description.

What has probably misled you into your incorrect conclusion is thinking in terms of "time passing faster. This is the wrong mental image. A more nearly correct mental image would tell you "it's wrong to try to compare time at different places".

Thanks for all the info. But, now I am sort of confused again. The bolded part contradicts what was previously established in this thread - that you would appear to freeze to someone on Earth (if they knew where you were) as you neared the event horizon due to the extreme gravitational time dilation. I'm sure I'm missing something...

Consider the Special Relativistic twin paradox. As one twin speeds up towards the speed of light, the other twin's time appears to slow down, in the limit they'd freeze. But the situation is symmetrical, the other twin can say the same thing!

Hanging onto absolute time, but believing that this absolute time "slows down" and "speeds up" is not compatible with even special relativity, (much less general relativity).

So, it's prefetly possible that E (the Earth) thinks that R's (the rocket's) clocks are slow, while R thinks that E's clocks are slow. This happens in SR with the twin paradox - and something vaguely similar ,though different in detail, happens when R falls into a black hole.

There are very important differences - in the black hole case, R actually vanishes from E's sight, though E never vanishes from R's sight.

 

[guss]

You should also find

http://casa.colorado.edu/~ajsh/singularity.html#r=1
Also check out the "redshift map" on the same page, which shows distant areas of the sky are redshifted, as per my description.

What has probably misled you into your incorrect conclusion is thinking in terms of "time passing faster. This is the wrong mental image. A more nearly correct mental image would tell you "it's wrong to try to compare time at different places".
Consider the Special Relativistic twin paradox. As one twin speeds up towards the speed of light, the other twin's time appears to slow down, in the limit they'd freeze. But the situation is symmetrical, the other twin can say the same thing!

Hanging onto absolute time, but believing that this absolute time "slows down" and "speeds up" is not compatible with even special relativity, (much less general relativity).

So, it's prefetly possible that E (the Earth) thinks that R's (the rocket's) clocks are slow, while R thinks that E's clocks are slow. This happens in SR with the twin paradox - and something vaguely similar ,though different in detail, happens when R falls into a black hole.

There are very important differences - in the black hole case, R actually vanishes from E's sight, though E never vanishes from R's sight.

Okay, thanks a lot, that makes sense too. I understand now that I have the wrong mental image.

I asked this to Peter but I'm sure you know the answer too. Where does this fit in?:

Nope. You would die in quite a small amount of your own time. It would take forever of Earth time.

I know now it is wrong to compare times at different places, so how can they be compared here (small amount of own time = lots of Earth time)?

 

[PAllen]

I know now it is wrong to compare times at different places, so how can they be compared here (small amount of own time = lots of Earth time)?

A.T.'s quote is not quite accurate. Accurate is:

Earth would never see you reach the horizon, therefore would certainly never see you die. However, they could easily conclude that all they are seeing is light delay, and that you have actually died at the singularity. Everything about 'what your state is now' is a matter of convention, not physics, even in special relativity. There are conventions where Earth concludes - that you never reach the horizon; and other conventions where they conclude you reach the singularity.

 

[PeterDonis]

Then, how would you fit this in?:

PAllen gave a good answer. The only thing I would add is that, using the convention that A.T. was using, it "takes forever" for you to reach the *horizon*--this convention simply can't comprehend *at all* anything that happens to you at or below the horizon. So assuming that you are still alive when you cross the horizon, it would take "more than forever", so to speak, for you to die from the viewpoint of the people on Earth. But as PAllen said, this is just a convention, and not the only possible one that the people on Earth could adopt.

 

[PAllen]

I thought it might be useful to make concrete the point about the conventionality of simultaneity in both SR and GR.

For non-inertial observers in SR, two common conventions are:

- Fermi-Normal coordinates (using the simultaneity of an instantaneously comoving inertial observer).
- Radar simultaneity - using round trip light signals to define simultaneity.

These become arbitrarily close to each other in a sufficiently small region of spacetime around the observer (as would be desired of any 'physically plausible' simultaneity). However, they are not only radically different 'far away', but they cover different regions of spacetime!. The Fermi-Normal convention typically covers only part of spacetime with a valid coordinate chart, while radar (in SR) will cover all of spacetime.

Similarly, I propose a physically plausible alternate simultaneity for SC geometry for a distant static observer (it closely matches SC coordinates over a small region of spacetime for the distant static observer). The convention is as follows:

- consider launching test bodies on radial geodesics at .9c from the distant static observer. For every SC (t,r) coordinate (both interior and exterior), there is an event on the the observe'rs world line that could reach that (t,r) with a test body probe. Compute the proper time along the test body world line, multiply by gamma(.9c), and declare this simultaneous with this amount past the launch event, on the observer's world line.

Having charted (t,r), the angular coordinates are the same as SC due spherical symmetry. This chart smoothly covers one exterior and one black hole interior section of the complete Kruskal geometry in physcially motivated way. You can now talk about 'when' any event inside the horizon occurs in relation to the distant observer, and even compute proper distances along these simultaneity surfaces to events inside the horizon.

 

[PeterDonis]

Similarly, I propose a physically plausible alternate simultaneity for SC geometry for a distant static observer (it closely matches SC coordinates over a small region of spacetime for the distant static observer).

I have an even easier proposal, which basically amounts to using Painleve coordinate time as the standard instead of Schwarzschild coordinate time. Since at very large r, the two are almost the same, the idea is this: just use proper time along any infalling geodesic as the standard of simultaneity. That is, if it takes me one year of my proper time to fall from a faraway space station to the horizon of a BH, then the event of my crossing the horizon is deemed to be simultaneous with an event on the station's worldline one year, by the station clock, after I leave the station.

One key thing I would want to look at, for both these proposals, is how they match up with light travel times. That is, when we combine the suggested simultaneity convention with the travel times of light, do we get something reasonable? For example, if the faraway space station sends a laser message after me, at what time by the station's clock would it have to be sent in order to reach me just as I cross the horizon?

In the case of your suggested convention, I would want to formulate a similar comparison: the "time" by the station clock at which I am deemed to cross the horizon is determined by the time at which a 0.9c test probe would need to be sent after me in order to reach me just as I cross the horizon, plus the elapsed proper time for the probe times gamma. At what time, by the station clock, would a laser message have to be sent in order to reach me just as the test probe does?

(One thing that bothers me a bit about your proposal is that the comparison I just described depends partly on gamma--i.e., on the velocity of the test probe--as well as on when I leave the station. In my version, that's not the case.)

 

[PAllen]

I have an even easier proposal, which basically amounts to using Painleve coordinate time as the standard instead of Schwarzschild coordinate time. Since at very large r, the two are almost the same, the idea is this: just use proper time along any infalling geodesic as the standard of simultaneity. That is, if it takes me one year of my proper time to fall from a faraway space station to the horizon of a BH, then the event of my crossing the horizon is deemed to be simultaneous with an event on the station's worldline one year, by the station clock, after I leave the station.

One key thing I would want to look at, for both these proposals, is how they match up with light travel times. That is, when we combine the suggested simultaneity convention with the travel times of light, do we get something reasonable? For example, if the faraway space station sends a laser message after me, at what time by the station's clock would it have to be sent in order to reach me just as I cross the horizon?

In the case of your suggested convention, I would want to formulate a similar comparison: the "time" by the station clock at which I am deemed to cross the horizon is determined by the time at which a 0.9c test probe would need to be sent after me in order to reach me just as I cross the horizon, plus the elapsed proper time for the probe times gamma. At what time, by the station clock, would a laser message have to be sent in order to reach me just as the test probe does?

(One thing that bothers me a bit about your proposal is that the comparison I just described depends partly on gamma--i.e., on the velocity of the test probe--as well as on when I leave the station. In my version, that's not the case.)

Yours is effectively the gamma=1 case of mine, I think. I am really proposing a family of possible coordinate charts, that depend on two parameters: reference r value, and gamma. I think you do need a specific reference r value since proper time on a radial geodesic from infinity is infinite. My motivation for using a significant gamma was simply to make it 'look' more like half of the radar simultaneity convention. It would be interesting to compute how gamma affects the chart, but I am not motivated to do that. I suspect the effect may be subtle, because multiplication by gamma removes a large part of the difference.

Also, I believe these charts lose the 'manifest' static representation of SC coordinates. Some metric components would seem to depend on the new t. I suspect this is an inevitable tradeoff.

[edit: one can even conceive of taking the gamma=infinity limit of these charts. Not sure how that would work out.]

[edit2: Maybe you are thinking of the family of free fall from infinity geodesics. In which case this becomes the (gamma,r)=(1,infinity) member of my family of coordinates. ]

 

[guss]

Thanks a lot guys, I think I have a good understanding of it now (or at least a good understanding of what I currently understand and what I don't :p). That previous post by A.T. was sort of what was throwing me off the whole time, because I was misunderstanding what was being said.