Black Holes and Time Dilation

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When we approach a black hole, the effect of gravity is such that relative to us, time far from the black hole would approach infinity as the distance to the Event Horizon approaches zero. But what happens when we cross the Horizon? How do we measure time outside? Will infinite time have have passed outside once we crossed the Horizon?
 

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  • #2
Ibix
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You cannot define time that way once you reach the horizon. Once you get there you have a finite time to live - about 15μs per solar mass of the black hole by your watch - and most of the external universe is outside your past light cone. So you will see a little bit of the external universe before you die, but most definitely not all of it.
 
  • #3
Ibix
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To expand a bit on the above, there is no timelike Killing vector field inside the event horizon, so there is no natural way to define a simultaneity convention. And you can't communicate across the horizon, so there's no way to define one in practical terms. So there is no way to define "now" in the sentence "what's happening outside the horizon now?".

There is, for anything inside or outside the horizon, a last time it can send a signal that you could receive as you fall in to the hole. For anyone falling across the horizon there is a last time you can send a message and expect a response, and a last time you can send a message that will be received (although the recipient will have to wait an arbitrarily long time to receive it).
 
  • #4
Orodruin
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so there is no natural way to define a simultaneity convention
Just to point out that you do not need a time-like Killing field to define a simultaneity convention. For example, the standard simultaneity convention in a RW universe is not based on a time-like Killing field, but is still quite natural.

I feel that with age I am becoming more and more radical in my dislike of anything convention dependent in GR in general and simultaneity conventions in particular ...
 
  • #5
Ibix
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Just to point out that you do not need a time-like Killing field to define a simultaneity convention. For example, the standard simultaneity convention in a RW universe is not based on a time-like Killing field, but is still quite natural.
Fair enough. It's based on another symmetry which isn't available across the event horizon of a black hole, either.
I feel that with age I am becoming more and more radical in my dislike of anything convention dependent in GR in general and simultaneity conventions in particular ...
Unfortunately the question boils down to "what simultaneity convention can I use".
 
  • #6
Orodruin
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Unfortunately the question boils down to "what simultaneity convention can I use".
This I agree with. It does not mean I have to like introducing simultaneity conventions. :rolleyes:
 
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  • #7
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So for an observer inside a black hole to ask " how much time have past outside the black hole", is not a meaningful question?
 
  • #8
Orodruin
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So for an observer inside a black hole to ask " how much time have past outside the black hole", is not a meaningful question?
Even in special relativity, "how much time has passed for another observer" is not really a meaningful question as it is dependent on the simultaneity convention, although there are some standard choices there. (This is the reason time dilation is reciprocal, i.e., both observers will consider the other time-dilated.) Time-dilation is not a physical measurable (differential ageing is).
 
  • #9
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However, consider this thought experiment. For an observer outside, the other person will never enter the black hole, but "freeze" on the Horizon. But we also know that the hole will eventually disappear due to Hawking radiation. So the black hole will disappear before the person enter the black hole, viewed from outside. But from the perspective of the person entering the black hole, it will still exist.

How can the black hole exist and not exist at the same time?
 
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  • #10
Ibix
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How can the black hole exist and not exist at the same time?
It can't. Your question makes no sense.

It's perfectly possible for a distant observer to assign a finite time at which an infaller crosses the horizon. They just choose something other than Schwarzschild coordinates, which can't describe a horizon crossing. An infalling observer assigns a finite time to their horizon crossing naturally.

For an eternal black hole, a distant observer will have to wait an infinite time to be certain they've received the last photon from an infaller, true. But that doesn't mean they have to believe the infaller hasn't crossed the horizon. Just that they haven't seen it.

For an evaporating black hole, you can't possibly have to wait longer than the lifetime of the black hole. Once it's gone it's gone, and the infaller must have fallen in earlier. Again, Schwarzschild coordinates can't describe this.
 
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  • #11
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It can't. Your question makes no sense.

It's perfectly possible for a distant observer to assign a finite time at which an infaller crosses the horizon. They just choose something other than Schwarzschild coordinates, which can't describe a horizon crossing. An infalling observer assigns a finite time to their horizon crossing naturally.

For an eternal black hole, a distant observer will have to wait an infinite time to be certain they've received the last photon from an infaller, true. But that doesn't mean they have to believe the infaller hasn't crossed the horizon. Just that they haven't seen it.

For an evaporating black hole, you can't possibly have to wait longer than the lifetime of the black hole. Once it's gone it's gone, and the infaller must have fallen in earlier. Again, Schwarzschild coordinates can't describe this.
aha ok. So it is a paradox sort of in the context of schwarshild coordinates?
 
  • #12
PeroK
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aha ok. So it is a paradox sort of in the context of schwarshild coordinates?
It's Schwarzschild.
 
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  • #13
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When we approach a black hole, the effect of gravity is such that relative to us, time far from the black hole would approach infinity as the distance to the Event Horizon approaches zero. But what happens when we cross the Horizon? How do we measure time outside? Will infinite time have have passed outside once we crossed the Horizon?
I think this is one of those questions that can bogged down in the actual physics, when the basic question has a simple answer - if you get "frozen" in time by dilation, everything external to you appears to speed up until you get to a point of no longer observing anything. So your question is a little like asking what your bedroom would look like to you after you fell asleep - which clearly doesn't make any sense.
 
  • #14
Nugatory
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So it is a paradox sort of in the context of schwarshild coordinates?
It's not so much a paradox as that Schwarzschild coordinates are defined in such a way that it doesn't work to assign a Schwarzschild coordinate time at the event horizon, nor to compare Schwarzschild coordinate times for event on opposite sides of the event horizon.

Also this paper is worth reading: https://arxiv.org/abs/0804.3619

If you think about the actual physics, it may be easier to see what's going on. Suppose I am orbiting far from the black hole, and I have a clock that is constantly broadcasting its current reading. I also have a probe that has its own clock and a radio transponder so that it when it receives a signal from my clock it broadcasts back a signal saying "I received your pulse that said it was sent at time ##t_{outside}## when my clock read time ##t_{probe}##".

I drop the probe into the black hole and start my clock broadcasting. As the probe gets nearer to the event horizon, it takes more and more time for its replies to get back to me; I might not see a reply from just outside the event horizon until many billions of years into the future, and no matter how long I wait, I will never see a reply sent from the event horizon or below. However, the timestamps on these replies show that only a small amount of time is passing on the probe as it approaches the event horizon; the thing that's approaching infinity is not how long it takes for the probe to reach the horizon but how I long I have to wait until I hear back from it.

Nothing special happens to the probe as it passes through the horizon (assuming that the probe is small enough, or the black hole is massive enough, that the tidal forces won't damage the probe). It coninues to receive broadcasts from me, and it continues to respond to them, until it reaches the singularity, which happens very quickly (about 15 usec per solar mass says @Ibix) after it passes the horizon. However, none of these responses will escape from the black hole. so I never see them (unless I wait until the black hole evaporates, perhaps, depending on what assumptions we're making about how that works).

So for me outside, there are three interesting times:
- Time zero, when I drop the probe and start my clock
- A moment, not very long after that, when I send the last signal that will ever receive a reply from the probe. Any signal send after this point won't reach the probe until after it has crossed the horizon; the probe will reply but I won't see the replies (unless they escape when and if the black hole evaporates, very far in the future).
- And not very long after that, the moment when I send the last signal that the probe will ever receive. Any signal I send after this point won't reach the probe before the probe has reached the singularity; this signal will follow the probe into the singularity and arrive there after the probe is destroyed.
 
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  • #15
Nugatory
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I think this is one of those questions that can bogged down in the actual physics
This is exactly backwards. The actual physics is simpler and easier to understand than the coordinate-dependent concepts like time dilation - you'll note the disdain that @Orodruin and other are expressing above for simultaneity conventions and coordinate effects.
when the basic question has a simple answer - if you get "frozen" in time by dilation, everything external to you appears to speed up until you get to a point of no longer observing anything. So your question is a little like asking what your bedroom would look like to you after you fell asleep - which clearly doesn't make any sense.
The problem with this answer is that it is based on a faulty premise: Nothing gets frozen by time dilation, and time does not slow down for the infaller. The infaller sees everything that happens in their past light cone until they die at the singularity.
 
  • #16
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This is exactly backwards. The actual physics is simpler and easier to understand than the coordinate-dependent concepts like time dilation - you'll note the disdain that @Orodruin and other are expressing above for simultaneity conventions and coordinate effects.

The problem with this answer is that it is based on a faulty premise: Nothing gets frozen by time dilation, and time does not slow down for the infaller. The infaller sees everything that happens in their past light cone until they die at the singularity.
They see everything, and what they observe is the motion of the external universe is running faster. You can call that whatever you want, but if I observe that a nearby planet's orbital velocity has seemingly increased, I might rightly assume that my experience of time has slowed in comparison.

But the OP's question is basically - If my observation of the universe is slowed enough to approach an infinite amount of time passing outside in an instant for me, will I see the death of the universe? Which is a very similar question to the Arrow Dicotomy Paradox. And you can examine that question without physics by realizing that the dilated person's ability to experience can end at a discreet external time even if they are unable to witness it themselves.

Certainly, do the math. But even without the physics the events are rational. That's all I'm attempting to point out.


And maybe I am being foolish in thinking that there is no useful difference between physical destruction at the Event Horizon and infinite dilation, but I don't think so. You can't observe your own end.
 
  • #17
Ibix
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They see everything, and what they observe is the motion of the external universe is running faster.
That's not true in general for people crossing the event horizon. It's true for people hovering above the horizon if they choose to use Schwarzschild coordinates to interpret their experiences. But the question was about people crossing the horizon.

I once proposed the following syllogism as an analogy for the "time freezes at the event horizon" claim.

1. You can usually place yourself on the surface of the Earth so that "north" is forwards and "south" is backwards.
2. All directions are south at the north pole
3. You can only travel backwards at the north pole, and doing so causes you to explode in every direction at once.

Trying to use Schwarzschild coordinates at the event horizon (and standard "intuitions" about how time works near black holes are based on them) is like step 3.
 
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  • #18
Ibix
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until it reaches the singularity, which happens very quickly (about 15 usec per solar mass says @Ibix)
That's a maximum time, assuming you drop from rest just above the horizon and don't accelerate.
 
  • #19
PeterDonis
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For an observer outside, the other person will never enter the black hole, but "freeze" on the Horizon. But we also know that the hole will eventually disappear due to Hawking radiation.
You're confusing two different models here.

The model that says an observer outside will never see the infalling person cross the horizon is the classical "eternal" black hole model, where the hole stays the same mass forever. In this model, the hole never evaporates.

The model that says the hole will evaporate is the semi-classical model, in which we include the effects of quantum fields in a curved background spacetime. In this model, it is no longer true that the outside observer never sees the infaller cross the horizon. Instead, he sees the infaller cross the horizon at the same time that he sees the light emitted from the hole's final evaporation (i.e., when the hole emits its last Hawking radiation and disappears).

Note that the semi-classical model I have just described might still not be correct; there are a variety of proposals about that. But all of the proposed models have in common that the outside observer does see the infaller cross the horizon.
 
  • #20
PeterDonis
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the black hole will disappear before the person enter the black hole, viewed from outside. But from the perspective of the person entering the black hole, it will still exist.
In the semi-classical model I described in my last post just now, there is still a singularity inside the black hole, and anyone who falls into the hole will hit it and be destroyed before the hole evaporates. So by the time the outside observer sees the hole evaporate away, there will be no infalling observers left inside the horizon; they will all have hit the singularity and been destroyed.
 
  • #21
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In the semi-classical model I described in my last post just now, there is still a singularity inside the black hole, and anyone who falls into the hole will hit it and be destroyed before the hole evaporates. So by the time the outside observer sees the hole evaporate away, there will be no infalling observers left inside the horizon; they will all have hit the singularity and been destroyed.
Thanks for clarifying! I will have some study to do. :)
 
  • #22
vanhees71
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As usual, the confusion is due to using frame-dependent quantities to describe the situation (like a point particle falling into a black hole). The coordinate time of Schwarzschild coordinates is such a frame-dependent quantity. Physically it's the proper time for an observer very far from the singularity. The reason, why for this observer it takes an infinite amount of time for the particle to cross the event horizon at the Schwarzschild radius is that any signal sent out by the falling particle gets more and more red-shifted, and the "picture" of the particle when crossing the event horizon gets "frozen". It seems to this far-distant observer as if the particle would never cross the horizon, but that's an illusion. It's also an illusion that the event horizon is something special to begin with. There are no singularities, and you can just choose other coordinates (e.g., Kruskal coordinates) to see that there are indeed none. The apparent singularity at the event horizon is thus just a coordinate singularity of Schwarzschild coordinates.

Of course, it's also easy to understand, what happens to the particle by choosing the appropriate parameter for its world line. For a massive particle you can choose its proper time, which is an affine parameter of the worldline (for a massless particle you may choose any affine parameter to facilitate the calculation; in this case of course there's no proper time). Then you'll see that for a finite affine parameter the particle just moves through the event horizon and finally ends up in the true singularity of Schwarzschild spacetime.
 

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