"Frozen in time" is a observer-dependent (hence coordinate-dependent) statement.So when an object is falling towards a black hole, it's clock relative to us, outside observers, is slowing down. Until said object reaches the Schwarzschild radius and it stops completely. So how can that object ever cross the event horizon, if it is frozen in time?
Why should the time relative to an outside observer matter for someone going inside?So when an object is falling towards a black hole, it's clock relative to us, outside observers, is slowing down. Until said object reaches the Schwarzschild radius and it stops completely. So how can that object ever cross the event horizon, if it is frozen in time?
Is there an objective coordinate system that the black hole, the object falling into it and the observer can use? And if not - how would the same situation play out from the black hole's perspective (and its coordinate system, say {x'', y''})."Frozen in time" is a observer-dependent (hence coordinate-dependent) statement.
To give a purely spatial analogy: Say, I have a {xy}-frame and I move solely along my x-axis. I'd say I'm not moving in the y-direction, so "I'm frozen in the y-direction" if you want to rephrase this poetically. But another observer using a {x'y'}-frame which is e.g. just a rotated version of my {xy}-frame would probably say I'm moving in both the x' and y' direction.
Similary, if I'm standing outside a black hole and use my t-coordinate to describe the motion of an infalling person, I'll say that as the person reaches the horizon, the time t elapses more and more slowly. However, this person himself uses a different time coordinate t' (often denotes as the eigentime tau), and t' doesn't stop flowing at the horizon.
It really shouldn't, unless to the outside viewer's perspective, said time is infinite. Infinity is a scary (to me) concept. That's why I had trouble understanding the situation. Unless it truly doesn't matter even IF the time is infinite for the obeserver?Why should the time relative to an outside observer matter for someone going inside?
We the outside observers will never see that the object crosses the horizon. But the object itself will cross it in a finite proper time, which will be seen by an observer comoving with the object.So when an object is falling towards a black hole, it's clock relative to us, outside observers, is slowing down. Until said object reaches the Schwarzschild radius and it stops completely. So how can that object ever cross the event horizon, if it is frozen in time?
Coordinates aren't objective, more or less by definition. You can certainly pick coordinates that are well-behaved at the event horizon - I already mentioned one such in my last post. In those coordinates there's nothing special about crossing the event horizon - you just cross. However, once you've done so there is no way back out.Is there an objective coordinate system that the black hole, the object falling into it and the observer can use? And if not - how would the same situation play out from the black hole's perspective (and its coordinate system, say {x'', y''}).
Schwarzschild coordinates break down at the event horizon. The way Schwarzschild constructed them effectively hides that breakdown by describing the point of breakdown as always in the future. That's all.It really shouldn't, unless to the outside viewer's perspective, said time is infinite. Infinity is a scary (to me) concept. That's why I had trouble understanding the situation. Unless it truly doesn't matter even IF the time is infinite for the obeserver?
Wow that kind of blows my mind. I'm not studying physics currently, but I would really like to in the future. I've been incredibly interested in the field for a long time. Thank you and everyone else who responded to my inquiry. May I please ask you to recommend some literature on this topic? I would greatly appreciate it.Coordinates aren't objective, more or less by definition. You can certainly pick coordinates that are well-behaved at the event horizon - I already mentioned one such in my last post. In those coordinates there's nothing special about crossing the event horizon - you just cross. However, once you've done so there is no way back out.
Schwarzschild coordinates break down at the event horizon. The way Schwarzschild constructed them effectively hides that breakdown by describing the point of breakdown as always in the future. That's all.
From what I understand, no. The waves we detect are associated with the final moments of the separate holes before they collide. In principle you could see waves due to the horizon changing shape as the combined hole settles down, but I don't believe we've detected those (could be wrong). But either way there is no source of waves inside the horizon.Doesn't the gravitational wave detection count as us seeing an object cross the event horizon of a black hole (albeit another black hole) in finite time?
AFAIK the "ringdown" signal was part of what LIGO detected. However, these waves don't come from at or inside the horizon either. Describing them as coming from "the horizon changing shape" is thus somewhat misleading (though it's fairly common in descriptions for a lay audience).In principle you could see waves due to the horizon changing shape as the combined hole settles down, but I don't believe we've detected those (could be wrong).
Fair enough. The key point is that you can completely explain the presence of gravitational waves by looking at the metric and stress-energy (if any) outside the event horizon. Presumably the form of the horizon is explainable that way too (otherwise it would be leaking information about the hole interior).AFAIK the "ringdown" signal was part of what LIGO detected. However, these waves don't come from at or inside the horizon either. Describing them as coming from "the horizon changing shape" is thus somewhat misleading (though it's fairly common in descriptions for a lay audience).
$$B\Large{\infty}!$$Infinity is a scary (to me)
That is correct, the coordinates for some distant observer don't matter for the local observer. They would use their own coordinates, or better yet focus on invariants.Unless it truly doesn't matter even IF the time is infinite for the obeserver?
Only if you can directly observe "the form of the horizon" from outside. But you can't. You can observe objects and light going into a compact region and not coming out again within the time of your observation; but to know the true form of the horizon, you would have to know the exact boundary of the region from which nothing could possibly come out for the entire future history of the universe. And there's no way to know that.Presumably the form of the horizon is explainable that way too (otherwise it would be leaking information about the hole interior).
No. The traveler only receives light signals from his past light cone. The past light cone of the event where he crosses the horizon does not contain the entire future history of the universe.Every ping ever sent by us a distance from the black hole is received by the traveler before he crosses the horizon - including the one heralding a "big crunch" end of the universe ?
Black holes continue to fascinate me. There are so many unexpected (to me, anyway!) implications of such strongly curved spacetime.Only if you can directly observe "the form of the horizon" from outside. But you can't. You can observe objects and light going into a compact region and not coming out again within the time of your observation; but to know the true form of the horizon, you would have to know the exact boundary of the region from which nothing could possibly come out for the entire future history of the universe. And there's no way to know that.
I'm trying to imagine the outer observer and the traveler exchanging light pings as the traveler approaches the black horizon.No. The traveler only receives light signals from his past light cone. The past light cone of the event where he crosses the horizon does not contain the entire future history of the universe.
Let's say the traveler is moving very slowly, approaching the horizon, and he and the remote observer are sending pings back and forth. As long as traveler doesn't cross the horizon, the pings can be exchanged, right? But if he cuts his engines, plunging inside, the remote observers pings no longer reach him.No. The traveler only receives light signals from his past light cone. The past light cone of the event where he crosses the horizon does not contain the entire future history of the universe.
No. The traveller's pings cannot reach the outside observer, but the reverse is not true. At least some pings that cross the horizon after the infaller will catch him before he reaches the singularity.As long as traveler doesn't cross the horizon, the pings can be exchanged, right? But if he cuts his engines, plunging inside, the remote observers pings no longer reach him.
Yes, and there will be a particular time on the outer observer's clock, not very long after the traveler drops away, at which any light signal he sends to the traveler won't reach the traveler until the traveler has crossed the horizon, which means the traveler won't be able to send any ping back.I'm trying to imagine the outer observer and the traveler exchanging light pings as the traveler approaches the black horizon.
Yes.As long as traveler doesn't cross the horizon, the pings can be exchanged, right?
No. The remote observer's pings still reach him; but he can no longer send pings back to the remote observer.But if he cuts his engines, plunging inside, the remote observers pings no longer reach him.
Andrew Hamilton has created some nice visualizations of what an infalling observer would actually see here:the observer will see a black EH in front of them and an increasingly distorted outside universe behind them all the way to the crunch, I think