I Black hole event horizon confusion

[Yukterez]

Depending on the type of simultaneity convention that you choose, you will end up with different conclusions

Sure, but in this case we were talking about the external observer, so it should be clear that it is his plane of simultaneity in which nothing falls through. Whenever you ask the external observer "where is the test-particle now?" (and this question has nothing to do with "where do you see the test-particle right now") he will always answear "just above the horizon".

For example see the trajectory of a test particle, which in it's own frame of reference falls through the horizons in a finite proper time (on the display τ propr):

while in the frame of an external observer, it freezes at the horizon and corotates there forever with the black hole, the clocks frozen short before the proper time when it penetrates the horizon:

 

[PAllen]

Sure, but in this case we were talking about the external observer, so it should be clear that it is his plane of simultaneity in which nothing falls through. Whenever you ask the external observer "where is the test-particle now?" (and this question has nothing to do with "where do you see the test-particle right now") he will always answear "just above the horizon".

Now is a convention for any observer, even in SR. The only psysical distinction is between causal future, causal past, on possible present i.e. spacelike separation. For any distant observer, there arises a time when part of the history of the horizon is possibly now.

Further, even insisting on the convention of 4 orthogonal geodesic definition of now, consider a distant observer oscillating radially. Then, when heading toward the BH, the horizon exists now.

 

[PAllen]

And why shouldn’t a distant observer simply use Gaussian normal coordinates, which for Schwarzschild metric are Lemaitre coordinates? Then, any external observer considers various horizon crossing events to be now.

 

[Yukterez]

And why shouldn’t a distant observer simply use Gaussian normal coordinates, which for Schwarzschild metric are Lemaitre coordinates? Then, any external observer considers various horizon crossing events to be now.

Because Lemaitre coordinates don't use the time he can read off his clock (t), but the proper time of the infalling observer (τ), see Wikipedia. We are still talking about the external observer, not the infalling one.

 

[PAllen]

Because Lemaitre coordinates don't use the time he can read off his clock (t), but the proper time of the infalling observer (τ), see Wikipedia. We are still talking about the external observer, not the infalling one.

We can be talking about free fall observer a billion light years from the BH, colocated with a stationary one, at relative speed 1 millimeter per century relative to the stationary observer. Then, you claim it’s objectively true that the horizon exists for one and not the other. Sorry, this is completely absurd.

 

[Yukterez]

We can be talking about free fall observer a billion light years from the BH, colocated with a stationary one, at relative speed 1 millimeter per century relative to the stationary observer. Then, you claim it’s objectively true that the horizon exists for one and not the other. Sorry, this is completely absurd.

I am not the one who claims that the horizon exists for any of those two observers 1 Gyr away from the BH, even for the 2nd observer who is moving slowly towards the black hole the formation of the horizon is in the future, but for sure not in his present. There is no way that any moving observer could ever tell the stationary observer that the horizon has formed right now, and especially not in the scenario that you have constructed.

 

[PAllen]

I am not the one who claims that the horizon exists for any of those two observers 1 Gyr away from the BH, even for the 2nd observer who is moving slowly towards the black hole the formation of the horizon is in it's own future, but for sure not in his present.

Wrong. For both, the horizon formation is not in their causal future. However, extending a orthogonal geodesic from one, you don’t reach the horizon while for the other you do. I claim both should say the horizon formation is possibly now. Either can choose a simultaneity convention to make this so. Using one based on extending a local Minkowski frame, it happens that one would say the horizon has already formed, while the other wouldn’t, even though they are colocated with relative speed 1 millimeter per century.

 

[PeterDonis]

Lemaitre coordinates don't use the time he can read off his clock (t), but the proper time of the infalling observer (τ)

You can have an infalling observer--one free-falling radially towards the hole--1 GLy away from the hole. Lemaitre coordinates use his proper time. So if such an observer happens to be passing a static observer (for whom Schwarzschild coordinate time is his proper time, at least to a good approximation--note that for this statement to be exact, the static observer would have to be at infinity, so it's never exactly true), as @PAllen pointed out, your position implies that the horizon exists "now" for the infalling observer but not for the static observer. That is absurd.

 

[Nugatory]

Sure, but in this case we were talking about the external observer, so it should be clear that it is his plane of simultaneity...

What is this "plane of simultaneity" of which you speak? That concept only makes sense in the flat spacetime of special relativity. In a curved spacetime there will be more than one way of foliating spacetime into surfaces of simultaneity; none of these surfaces need be planes; and there's no reason to choose one over another.

 

[PAllen]

Note also, that if we speak of a distant observer in an unpowered space station far away from the BH, then Lemaitre coordinates are more in correspondence with the station clock than Schwarzschild.

 

[Yukterez]

Note also, that if we speak of a distant observer in an unpowered space station far away from the BH, then Lemaitre coordinates are more in correspondence with the station clock than Schwarzschild.

Far away from the black hole you will see almost no difference between the freefalling and the stationary clock at all, so I don't see what difference that is supposed to make.

In a curved spacetime there will be more than one way of foliating spacetime into surfaces of simultaneity; none of these surfaces need be planes; and there's no reason to choose one over another.

Sure, but the choice is not arbitrary (for example, just because we have a little gravity here on Earth you still can't reverse the order of events just as you please, and just because the gravity is a little bit stronger at a black hole still doesn't mean you can do it there).

 

[PAllen]

Far away from the black hole you will see almost no difference between the freefalling and the stationary clock at all, so I don't see what difference that is supposed to make.

I agree, but if you use coordinates adapted to the free faller, you find can find horizon formation in the past and crossing events in the present. I claim this all means the these coordinate differences irrelevant, that all that matters is the distinction between spacelike =possibly now, versus causal future. With this invariant classification, they both agree that both horizon formation and crossing events are possibly now. It is you that are trying to ascribe meaning to a coordinate artifact of Schwarzschild coordinates. I am pushing against this by showing that natural coordinates for a distant free faller lead to completely different conclusion. The resolution is reject both coordinate dependent claims and stick with invariant statements.

 

[Yukterez]

The resolution is reject both coordinate dependent claims and stick with invariant statements.

That would be a big loss, since they really help calculating stuff, for example when which observer reveives what time stamp from another. For the external observer there is always a signal sent right now, and a later time when he receives that signal. The light travel times from the freefaller just above the horizon to the observer outside are always finite, and the external observer keeps receiving signals from the freefaller all eternity long (given that the sent signal is continuous), so as far as the external observer is concerned, the freefaller is always still outside of the critical radius (if at any time he were inside, there would be a later time when the last signal is received). Since the light travel times are finite and known, one can always calculate what happened when in which frame of reference, given that it happened at all in that frame of reference.

 

[PAllen]

That would be a big loss, since they really help calculating stuff, for example when which observer reveives what time stamp from another. For the external observer there is always a signal sent right now, and a later time when he receives that signal. The light travel times from the freefaller just above the horizon to the observer outside are always finite, and the external observer keeps receiving signals from the freefaller all eternity long (given that the sent signal is continuous), so as far as the external observer is concerned, the freefaller is always still outside of the critical radius.

Lemaitre coordinates are equally easy to calculate in, for some problems easier. Causal relations are never reversed by coordinate changes, and those are all that matter. Note that your statements above are back to behavior of signal delays. You were the one claiming definitions of now have nothing to do with signaling delays. This is true, but definitions of now are entirely conventional beyond the classification of causal future, causal past, and possibly now. It is a coordinate independent fact that for any external observer there is point on their history such that horizon formation changes from causal future to possibly now. This is the critical point. All else convention.

 

[PeterDonis]

Far away from the black hole you will see almost no difference between the freefalling and the stationary clock at all, so I don't see what difference that is supposed to make.

It makes a difference in their surfaces of simultaneity. The surfaces of simultaneity in Lemaitre coordinates go through the horizon, and in those coordinates, events where infalling observers cross the horizon have finite coordinate times, which can (and will, in the scenario we are discussing) be earlier than the coordinate time corresponding to the reading on the distant observers' clocks.

just because we have a little gravity here on Earth you still can't reverse the order of events just as you please

You can for events which are spacelike separated. In the scenario we are discussing, the event at which a previous infalling observer crossed the horizon is spacelike separated from the events where the distant observers are asking each other whether that previous infalling observer has crossed the horizon yet.

they really help calculating stuff, for example when which observer reveives what time stamp from another

Actually, many calculations are easier in Lemaitre (or Painleve) coordinates than they are in Schwarzschild coordinates.

he external observer keeps receiving signals from the freefaller all eternity long (given that the sent signal is continuous), so as far as the external observer is concerned, the freefaller is always still outside of the critical radius

No, because the fact that the external observer has not yet received a light signal from a given event does not mean that event has not happened.

 

[Yukterez]

Note that your statements above are back to behavior of signal delays. You were the one claiming definitions of now have nothing to do with signaling delays.

They are not the same, but you can calculate the one if you have the other. For example, you can always say "the signal I am receiving right now was emitted such and such years ago". In the case of a black hole, the stationary observer can do that all eternity long, and he will always receive signals that took a finite light travel time from the freefaller to him.

 

[PeterDonis]

For example, you can always say "the signal I am receiving right now was emitted such and such years ago".

No, you can't, at least, not if you want to calculate an invariant. "How long ago this light signal I am receiving was emitted" is a coordinate-dependent quantity.

 

[Yukterez]

No, you can't, at least, not if you want to calculate an invariant. "How long ago this light signal I am receiving was emitted" is a coordinate-dependent quantity.

Again, we are still speaking about the external observer, and the time we are talking about is the time he reads off his clock. The light ray itself does not even have a proper time at all. The fact that the external observer receives the timestamp τ=x from the freefaller when his own clock shows his wristwatch time t=y is invariant, all obervers agree with that.

 

[PeterDonis]

we are still speaking about the external observer, and the time we are talking about is the time he reads off his clock.

Which has nothing to do with the travel time of the light itself. It's just the observer's clock time when he receives the signal.

 

[PAllen]

They are not the same, but you can calculate the one if you have the other. For example, you can always say "the signal I am receiving right now was emitted such and such years ago". In the case of a black hole, the stationary observer can do that all eternity long, and he will always receive signals that took a finite light travel time from the freefaller to him.

Are you perhaps not aware that after the point where a horizon crossing changes from causal future to possibly now, for a given external observer, then any signal sent by the the external observer to the horizon crosser will only be received by the horizon crosser inside the horizon? This is an aspect of the change in causal relation that is wholly independent of coordinates.

 

[Yukterez]

Are you perhaps not aware that after the point where a horizon crossing changes from causal future to possibly now, for a given external observer, then any signal sent by the the external observer to the horizon crosser will only be received by the horizon crosser inside the horizon? This is an aspect of the change in causal relation that is wholly independent of coordinates.

I am well aware that any signal sent right at the horizon can only be received when the receiver crosses the horizon himself, that is because an outgoing lightray at the horizon has a constant radial coordinate. So the photon does not come to you, you have to come to the photon, so to speak:

In a spherical black hole, the river of space falls into the black hole at the Newtonian escape velocity, hitting the speed of light at the horizon. Inside the horizon, the river flows inward faster than light, carrying everything with it.

This quote is of course coordinate dependend.

 

[PAllen]

I am well aware that any signal sent right at the horizon can only be received when the receiver crosses the horizon himself, that is because an outgoing lightray at the horizon has a constant radial coordinate.

That is not at all what I said. Consider an observer staying one light year away from a BH. They fire a test vehicle toward the BH. They send a light signal to it every second per their clock. There are then two signals they send such that the test vehicle receives one just outside the horizon and the next just inside the horizon. This is a coordinate independent fact. That transition for the external observer is when the crossing changes from being in the causal future to the possibly now = spacelike separation. Do you get this? Your answer above strongly suggests you do not.

 

[Yukterez]

Consider an observer staying one light year away from a BH. They fire a test vehicle toward the BH. They send a light signal to it every second per their clock. There are then two signals they send such that the test vehicle receives one just outside the horizon and the next just inside the horizon. This is a coordinate independent fact.

That is correct, I calculated that a lot. It is by the way also a coordinate independend fact what time the stationary observer reads off his clock when he receives a given timestamp from the freefaller, so that statement goes both ways, with the only difference that all the signals the external observer receives come from outside of the horizon.

That transition for the external observer is when the crossing changes from being in the causal future to the possibly now = spacelike separation. Do you get this? Your answer above strongly suggests you do not.

You are right, I really don't get how that helps your case, but I will sleep over it. Maybe I get it tomorrow (or the other way round).

 

[PAllen]

That is correct.You are right, I really don't get how that helps your case, but I will sleep over it. Maybe I get it tomorrow (or the other way round).

There is nothing for me to sleep on since I don’t see you as having any case at all, sorry. Do you understand the distinction between the future light cone, the past light cone, and what is in between? And that after the event described for the external observer, the horizon crossing event has moved out of the future light cone and can no longer be considered causal future irrespective of coordinates? And that for events between future and past light cones, what is now is entirely a coordinate convention? This is really basic to SR, let alone GR.

 

[Yukterez]

Do you understand...

I understand all of this, but what you don't seem to understand is that in the future light cone of the external observer, the freefaller is always outside of the horizon, so it would make no sense to say the crossing of the horizon was "possibly now". But whatever, better having you against me (of course not on a personal level, but regarding the subject) and Susskind on my side, than the other way round!

 

[PAllen]

I understand all of this, but what you don't seem to understand is that in the future light cone of the external observer, the freefaller is always outside of the horizon, so it would make no sense to say the crossing of the horizon was "possibly now".

This is just false. We have hit on your key misunderstanding. For any external observer, independent of coordinates, there is a precise moment when the infaller’s horizon crossing is no longer in the future light cone of the external observer. This should be easier to address now since you are making a false statement. The future light cone of an event is the set of all events that can receive a signal from that event. Once the horizon crossing event can no longer receive a light signal from the external observer, it is no longer in its future light cone.

 

[Yukterez]

Once the horizon crossing event can no longer receive a light signal from the external observer, it is no longer in its future light cone.

Correct, I didn't mean the freefaller is outside the external observer's future light cone, I meant that in his future you are always outside of the event horizon. The external observer is outside of the freefaller's future light cone once he passed the horizon, I made a mistake in my last posting but replace either the two observers or the future light cone with future and the statement is correct.

 

[PeterDonis]

in the future light cone of the external observer, the freefaller is always outside of the horizon

No, this is not correct, as @PAllen points out.

didn't mean the freefaller is outside the external observer's future light cone

You should have. He is.

I meant that in his future you are always outside of the event horizon

Sorry, this is still wrong. More precisely, there are two possible referents for "his future" in this statement, and the statement is wrong for both of them.

Possibility #1: You mean that in the future of the free-faller, the external observer is always outside of the horizon. This is false; the external observer's worldline exits the future light cone of the free-faller when the free-faller crosses the horizon. After that point all events on the external observer's worldline are spacelike separated from the free-faller, so anyone of them could possibly be "now".

Possibility #2: You mean that in the future of the external observer, the free-faller is always outside the horizon. This is also false, as @PAllen has already explained.

At this point I am closing the thread since this particular subthread is going nowhere, and you are not the OP in any case; the OP's questions have been addressed.