If time doesn't pass at the event horizon, how is it crossed?

[Martin_K]

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?

 

[Orodruin]

Time in the Schwarzschild metric is just a coordinate, not something physical. There are other coordinates that are not singular at the event horizon and where it is clear that there is absolutely no problem passing it.

This falls back to the inherent ambiguity in defining what ”simultaneous” means in relativity.

 

[Ibix]

There is no unique definition of "at the same time" for things not in the same place in relativity. This turns out to be a generalisation of the familiar notion that something that is directly infront of me isn't necessarily directly infront of you - directions in spacetime are relative, as are directions in space.

As Orodruin says, this bites you hard if you try to describe something crossing the event horizon using Schwarzschild's procedure for defining "at the same time". It simply doesn't work at the event horizon, so you can't describe "at the same Schwarzschild time as something crosses the event horizon". That doesn't mean you can't cross - simply that you can't describe it in these terms.

You are free to switch to (for example), Kruskal-Szekeres coordinates, which do describe horizon crossing in finite time. However, they are somewhat mathematically complex and less convenient for describing paths that don't cross the horizon. Swings and roundabouts.

 

[haushofer]

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?

"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.

 

[Dale]

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?

 

[Martin_K]

"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.

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''}).

 

[Martin_K]

Why should the time relative to an outside observer matter for someone going inside?

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?

 

[Demystifier]

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?

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.

To demystify such things and gain intuitive understanding, I recommend to think in terms of analog acoustic black holes, where acoustic horizon is the place at which the waterfall falls with the speed of sound and all communication by distant observers is performed with sound waves.

 

[Ibix]

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''}).

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.

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?

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.

 

[Martin_K]

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.

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.

 

[Ibix]

A former mentor here, @bcrowell, wrote a non-mathematical introduction to special and general relativity. It's free for download from http://lightandmatter.com/poets/. He's also got more mathematical treatments on the same site. The GR one is good - haven't read the SR one. Taylor and Wheeler's Spacetime Physics is a pretty good SR textbook and is probably accessible if you have maths and physics courses up to A level (or whatever you call 18-year-old school leaver exams in your country). Some chapters are free online if you want to take a look. Sean Carroll's lecture notes on general relativity are very good and free for download from https://www.preposterousuniverse.com/grnotes/, but the maths is heavy (it's a course for physics graduate students) and you will need to be confident in special relativity first.

 

[Grinkle]

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?

 

[Ibix]

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?

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.

 

[PeterDonis]

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).

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).

 

[Ibix]

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).

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).

 

[Dale]

Infinity is a scary (to me)

$$B\Large{\infty}!$$
Happy Halloween

Unless it truly doesn't matter even IF the time is infinite for the obeserver?

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.

 

[PeterDonis]

Presumably the form of the horizon is explainable that way too (otherwise it would be leaking information about the hole interior).

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.

 

[1977ub]

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 ?

 

[PeterDonis]

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 ?

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.

 

[Ibix]

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.

Black holes continue to fascinate me. There are so many unexpected (to me, anyway!) implications of such strongly curved spacetime.

You are right, of course, because you can only know what's in your past light cone, and the event horizon is an outward-going null surface. So if it's in your past light cone you are inside the hole.

 

[1977ub]

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.

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.

 

[Ibix]

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 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.

A heuristic way to see this is to drop a chain of observers into the hole. Each one sees spacetime as locally flat, so a light pulse from outside overtakes them. And if the previous observer is close enough, he'll be able to see the pulse catch up to that observer. Rinse and repeat.

 

[PeterDonis]

I'm trying to imagine the outer observer and the traveler exchanging light pings as the traveler approaches the black horizon.

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.

It is true that the pings coming back from the traveler will take longer and longer to reach the outer observer. But that does not mean the outer observer can continue to send pings forever and have the traveler be able to ping them back. The situation is not symmetric between outgoing pings and ingoing pings.

As long as traveler doesn't cross the horizon, the pings can be exchanged, right?

Yes.

But if he cuts his engines, plunging inside, the remote observers pings no longer reach him.

No. The remote observer's pings still reach him; but he can no longer send pings back to the remote observer.

 

[Grinkle]

@1977ub If some phenomena, like outside light suddenly no longer being visible, could be used as a horizon detector, as far as I understand things, that phenomena probably doesn't happen per GR.

So 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.

 

[PeterDonis]

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

Andrew Hamilton has created some nice visualizations of what an infalling observer would actually see here:

http://jila.colorado.edu/~ajsh/insidebh/intro.html

Note that the observer does, in fact, see a horizon in front of them, even though they have already passed the horizon. This is because ingoing light from the horizon bends around and comes at the observer from the front. (In fact, even light from the outside universe bends around and comes at the observer from the front.)

Note also that these simulations are not of an observer falling radially; this observer spirals in with a nonzero angular momentum.