Comment by amluto

Comment by amluto a day ago

5 replies

> The two observers can't disagree on such matters!

Why not?

If a spaceship fell toward a black hole and, as it approached the event horizon, one observer saw it turn into a horse and the other saw it turn into a cat, that would be very strange indeed, and one would suspect at least one of the observers of being wrong.

But if one observer sees it fall through the event horizon and the other observer waits… and waits… and gets bored and starts doing some math and determines that they could spend literally forever and never actually observe the spacecraft falling through the event horizon, then what’s the inconsistency? You might say “well, the first observer could fire up their communication laser and tell the second observer that ‘yes, the spaceship fell in at such-and-such time’, and the second observer would now have an inconsistent view of the state of the universe”, but this isn’t actually correct: the first observer’s message will never reach the second observer!

jiggawatts a day ago

> Why not?

Because that's not how relativity works! Two observers can disagree only on the order and relative timing of events, not what the events are or the total number of events. There are far more restrictions than that, but those are sufficient for my point.

The whole quantum information loss problem is just this, but dressed up in fancy terminology. It's the problem with black holes that the "number of things" (particles, events, whatever) is "lost" when matter falls into them.

The modern -- accepted -- resolution to this problem is that this information is not lost, preserving quantum numbers, etc...

How exactly this occurs is still being debated, but my point is that if you believe any variant of QM information preservation, then the only logically consistent view is that nothing can fall past an event horizon from any perspective, including the perspective of the infalling observers.

If you disagree and believe the out-dated GR model that an astronaut can't even tell[1] that they've crossed the event horizon, ask yourself this simple question: When does the astronaut experience this "non-event"[1]? Don't start with the mathematics! Instead, start with this simple thought experiment: The non-victim partner far away from the black hole holds up a light that blinks on an off once a second. The victim is looking outward and is watching the blinking speed up. How many blinks do they count at the time they cross the horizon?

Now think through the scenario again, but this time assume the spaceship turns the light off when they observe that the black hole has finished evaporating. When does the in-falling astronaut observe the blinking stop? Keep in mind that every "toy model" makes the simplification here that the blinking rate goes to infinity as the astronaut falls in! (I.e.: "They see the entire history of the universe play out." is a common quote)

[1] Isn't that a strong enough hint for everybody that there is no horizon!?

  • amluto a day ago

    > Because that's not how relativity works! Two observers can disagree only on the order and relative timing of events, not what the events are or the total number of events.

    No, and this has nothing to do with quantum mechanics or the no-hair theorem or anything particularly fancy.

    As a toy example, suppose you have a frame with a (co-moving, but it doesn’t really matter) time coordinate t. A series of events happen at the origin (x=y=z=0 in this frame) at various times t.

    There’s another observer in a frame with a time coordinate t'. The frames are related by t' = t - 1/t for t<0. The t=-10 event happens at t'=-9.9. The t=-4 event happens at t’=-3.25. The t=-1 event happens at t’=0. t=-1/100 happens at t'=99.99. t=0 gets closer and closer to happening but never actually happens. t=1 doesn’t even come close.

    Critically, the t' observer does not observe t=0 or t=1 in some inconsistent manner. There is no disagreement between the observers as to what happens at t>=0. To the contrary, those events are simply not present in the t' observer’s coordinate system!

    Note that the transformation above isn’t about when light from an event gets to the t' observer — it’s the actual relativistic transformation between two frames.

    The Schwarzchild metric has a nastier transformation than this. If you toss a rock into an isolated black hole from far away, you will see the rock get progressively closer to the event horizon, and you will never see it fall in. But the rock is in trouble: its co-movimg coordinate system ends not long after it crosses the horizon. That latter phenomenon is called the “singularity”, it’s solidly inside the event horizon, and it’s not avoidable by coordinate system trickery. While general relativity does not explain what happens when one encounters the singularity, one might imagine that it’s fatal to the rock the reaches it.

    edit: FWIW, you also say:

    > The current "pop science" (nearly science fiction) statement is that it is possible to fall into a black hole and there is "nothing special" about the event horizon.

    I’m not sure what you’re talking about. In a pure GR model of an isolated black hole, as you fall in, you will observe tidal forces. In a smaller black hole, the tidal forces will squash you long before you reach the event horizon. In a large enough black hole, they will not! Your view of the sky would certainly look very, very distorted and delightfully and possibly dangerously blue-shifted, but we’re talking about an isolated black hole. Nothing to see in the sky, and you may well survive your visit to the event horizon. Then, dramatically less than one second later for any credibly sized black hole, you will meet the singularity, and IIRC you should probably expect to be squashed by tidal forces before that. Source: I took the class and did the math. I assume this is what the “pop science” you’re talking about is saying, and it’s not wrong.

    P.S. I’ve never tried to calculate how lethal the blue-shifted sky would be. Naively considering just the time transformation, it should be infinitely lethal at the event horizon. But trying to apply intuition based on only part of a relativistic transformation is a great way to reach incorrect conclusions.

    • Dylan16807 17 hours ago

      > Nothing to see in the sky, and you may well survive your visit to the event horizon. Then, dramatically less than one second later for any credibly sized black hole, you will meet the singularity, and IIRC you should probably expect to be squashed by tidal forces before that.

      Don't we know about a bunch of black holes best measured in AU? Won't you have a good chunk of time inside those? Does time dilation work severely against you?

      • amluto 10 hours ago

        I vaguely recall doing this calculation on a problem set. For a black hole roughly the size of the one at the center of our galaxy, IIRC you have well under a microsecond. I could be remembering wrong.

        Even worse: the way to maximize how long you have before you hit the singularity, you should do nothing. Firing your rocket in any direction gets you to the singularity in an even smaller amount of proper time: the singularity isn’t in front of you in space — it’s ahead of you in time.

        Keep in mind that all of this is for the Schwarzchild metric, which is a nice solution to Einstein’s equations in the sense that you can derive it on a blackboard. It can’t describe what we think of as a real black hole for plenty of reasons, including the major one that a Schwartzchild black hole has existed forever and therefore could not have formed in a supernova. You need a different solution for a black hole that has only existed for a finite time.

        • Dylan16807 6 hours ago

          > For a black hole roughly the size of the one at the center of our galaxy, IIRC you have well under a microsecond.

          You only get one microsecond as you cover over half a light minute? Huh.