this post was submitted on 29 Dec 2023
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Natural Philosophy

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Crossing an event horizon (self.naturalphilosophy)
submitted 1 year ago* (last edited 1 year ago) by sudoreboot to c/naturalphilosophy
 

Tl;dr: Someone please explain to me why some physicists think something could ever cross the event horizon of a black hole.


There is a conflict between my understanding of what the event horizon of a black hole is vs the way that many theoretical physicists talk about them.

I understand that a result in general relativity is that time progresses more slowly in the presence of energy, and this is why light bends around massive objects.

The way I understand the dynamics around a black hole is that the surface of an event horizon is the region of space where the energy is so great that time literally grinds ~~to~~towards a halt (edit/clarification: from the perspective of an observer farther out). Light moves at the speed of causality, and when causality slows down, so does light. Light is bent and redshifted due to time dilation, and only when time stops does the wavelength of light go to zero. That's the event horizon as I understand it.

If an object falls towards a black hole, it shouldn't matter if we are that object or if we're just observing it from farther away, everyone should agree that it never crosses the boundary of the event horizon.

From a spectating observer's perspective, the object is redshifted until it fades entirely as it gradually stops moving through time (so light stops being emitted from it). But it will only ever approach the boundary asymptotically; it will never cross it.

~~From the perspective of the object itself, the universe around it will progressively speed up and the entirety of the history of the universe will play out behind it~~ (Edit: that only happens if the object accelerates to remain stationary). An infinite amount of time would pass everywhere else before it crosses the horizon. Now, that will never happen if black holes evaporate in finite time (and we have good reason to think they do). The black hole will evaporate long before any eternity passes anywhere. The more slowly you move through time, the faster this process will appear to you. When you are more or less frozen in time, the black hole will be evaporating at a rate that approaches 'instantaneously' - so the closer you get to it, the hotter it will appear and the faster it evaporates. You and everything else would literally radiate away from this noticeably shrinking event horizon before ever crossing it.

So, in this view, I feel utterly confused by physicists talking about "what it's like to cross the event horizon" or "what the interior of a black hole is like". Either my understanding is incorrect, or these physicists are just indulging themselves with hypotheticals rather than thinking about physics (or working on alternative models where black holes are fundamentally nothing like what I describe).

It's most likely me not understanding this properly, so.. what am I missing?


Update

As I mentioned in this comment, it has been shown that an event horizon may never form at all, and that all one ever sees is a shell of fading signatures followed by radiation from all the matter that falls into it.

I have more to learn about the particular dynamics around the area surrounding a black hole, but I believe I've managed to reduce my antecedents to the assumption that quantum information is conserved and the following counterfactuals, which appear promisingly independent of whichever dynamical model one might prefer:

  1. A sufficiently long-lived asymptotic (sufficiently distant) observer would be able to identify a particular point in time at which a black hole will have fully evaporated.
  2. A sufficiently long-lived asymptotic observer would be able to track the signature of something falling towards a black hole until it is radiated out.

Counterfactual (1) is supported by the prediction of Hawking radiation and means that the black hole has a finite life span. Counterfactual (2) is supported by the common claim that, to an observer far away, the wavelength of emitted light from an infalling object will go towards infinity as they get closer to the event horizon.

This means that the observer just has to wait long enough to detect each subsequent photon until the source of the emission has been radiated out, and so the observer is a witness of the fact that the infalling object was never inside the event horizon. For information to be conserved, there can never be disagreement between the objective experience of the witness and the information encoded in the radiation, and so if the infalling observer were to be reconstructed after being spat back out by the black hole, it would agree that it was never inside the horizon.

Feedback on my reasoning would be very welcome.

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[–] agamemnonymous@sh.itjust.works 6 points 1 year ago* (last edited 1 year ago) (2 children)

I think your understanding of the concept might be a bit off. The event horizon is just the distance around a black hole where nothing can escape the gravity well, even light. That's really it. Light doesn't "slow down", it just can't move fast enough to overcome the gravity.

Imagine it like a series of treadmills, each one going faster than the previous one. Far away, you can run fast enough to overcome the speed and you can get off. At a certain point, you reach a treadmill going at a speed that exceeds your top running speed, and after that you can no longer escape the treadmills. The event horizon of a black hole is simply the point in the gravity well where the fastest thing we can observe, light, can no longer escape the gravity.

It's not that time "grinds to a halt", but rather that we base time on causality, causality on observation, and we can no longer observe what happens past that point because light can't escape and we need light to observe. There is some funkiness right at the boundary, the photon sphere, where the gravity is exactly strong enough to hold photons in orbit around the black hole, which might be sort of analogous to the concept of time "stopping". Outside the photon sphere, gravity is strong enough to bend the path of photons without actually capturing them (this is the cause of gravitational lensing). Inside the photon sphere, gravity is strong enough to draw photons into an ever decaying orbit (the black hole itself).

This all refers to what we can see from outside the black hole. That's why physicists hypothesize about what happens inside one, past the event horizon: we can attempt to model the behavior, but since light can't escape we can't directly observe it.

[–] sudoreboot 1 points 1 year ago

I think you're probably correct that my understanding of the dynamics around the event horizon is not quite right. It's too late in the day for me to respond properly but I'll think and read on this some more.

[–] sudoreboot 1 points 1 year ago* (last edited 1 year ago) (1 children)

I'm a little confused here (clearly). Would an object spatially positioned at the centre of a perfectly isotropic (from its frame of reference) distribution of energy be subject to time dilation relative to an asymptotic observer positioned far away enough that the effects of said distribution of energy is negligible?

Given that four-velocity vectors are always of magnitude ±c, it would stand to reason that (in general relativity, at least) the answer is "no". If the distribution of the surrounding energy around some observer A is perfectly isotropic, the time component of A's four-velocity will be exactly ±c for any finite amount of time into the past or future. The asymptotic observer's four-velocity would be (approximately) identical to A's (again for any finite amount of time).

However, this answer is unsatisfactory and leaves me with further questions: the energy would induce a curvature which would affect the light emitted by A as it travels towards an asymptotic observer; is there a way to tell the difference between this description and one where time truly passes at different rates in the presence of energy?

Another thing I'm confused about: in the case where time truly doesn't pass slower in the presence of energy (i.e. light never slows down), wouldn't curvature affect light identically in both directions, inducing gravitational redshift both ways such that both observers would consider each other as moving more slowly through time?

Edit, adding the below and also restating a few things above.

Light doesn’t “slow down”, it just can’t move fast enough to overcome the gravity.

This statement I think presents a false dichotomy. I believe the story of GR can be told equivalently in more than one way:

  • "Light doesn't slow down, it just takes longer to traverse the expanding space in the presence of energy"
  • "Light moves more slowly due to time dilation in the presence of energy"

and which story you might choose to tell depends on your frame of reference. If you are a photon? You never slow down, space just sort of contorts in various ways along your path. If you are a human observing photons coming your way, you might say that time progressed at different rates in the region of space it travelled through.

Am I wrong?

[–] agamemnonymous@sh.itjust.works 1 points 1 year ago* (last edited 1 year ago) (1 children)

I'd have to dig up stuff I haven't touched in the better part of a decade to really get into this, but it seems at first glance that you're losing the forest for the trees, so to speak. Curvature = Energy is a simplification. At the level you're trying to analyze, you really need to get into the full picture.

Unfortunately, my plate's a bit full at the moment to be going into a differential geometry refresher. But that's where you should look for insight.

[–] sudoreboot 1 points 1 year ago

I appreciate the pointer. I hope to learn enough to get a more complete picture eventually.

[–] perestroika 5 points 1 year ago* (last edited 1 year ago) (1 children)

My take:

  • there is no universal time, there is your time and an external observer's time
  • if you fly straight at a planet, you crash into it in a finite your-time, for an observer behind you, the light of your crash will be delayed by a tiny amount due to the planet's gravity
  • if you fly straight at a highly massive neutron star, you crash into it in a finite your-time, just a bit quicker for youself due to its great mass, and the light of your crash will be delayed a bit harder, but will escape and show that you crashed
  • if you fly straight at a black hole, you crash into it in a finite your-time, just much quicker due to the high mass of the hole, and the light of your crash will be delayed indefinitely, and will not show that you disappeared
[–] sudoreboot 5 points 1 year ago

there is no universal time, there is your time and an external observer’s time

I assume you mean there's no universal "present" time, in which case you are correct. All observers agree on the order of causal interactions (which is what makes any 'clock' tick, and that is how we measure time), so time as a dimension is objective and universal.

if you fly straight at a black hole, you crash into it in a finite your-time

As a composite physical system of fermions and particle interactions, you would certainly die in finite time, but in my understanding, the individual particles that make up you would never cross the horizon. Depending on the ultimately fundamental quantum mechanical nature of spacetime, it could certainly be the case that a particle eventually reaches the horizon (however it may be defined in such a theory), but it would be causally isolated until it's finally turned into radiation.

[–] subverted_per@sh.itjust.works 2 points 1 year ago (1 children)

So this is my totally amateur thinking. From the observers perspective we never see the object fall in due to red shifting, the light of the falling object can't escape the back hole as it's falling in. However the object doesn't ever actually reach the speed of light upon falling into the black hole, depending on the trajectory it could be remarkably slow as it approaches the black hole. Time does slow down the deeper into a gravity well, but time never comes to a stop, not until crossing the event horizon which is where physics gets wonky anyway. Everything gets really slow right next to the black hole, but it would only become impossible to fall into it once an object is already at the event horizon. Once there it's hard to say what happens, but the object is already across the border.

[–] sudoreboot 0 points 1 year ago (1 children)

it would only become impossible to fall into it once an object is already at the event horizon. Once there it’s hard to say what happens, but the object is already across the border.

Well, I'd put it like this: in order to cross the event horizon, you must first arrive to it. Once arrived at the event horizon -- but before crossing -- you are frozen. While frozen, you can't cross it.

There's a relevant paradox called Zeno's dichotomy paradox:

That which is in locomotion must arrive at the half-way stage before it arrives at the goal.

[–] subverted_per@sh.itjust.works 1 points 1 year ago (1 children)

Like an asymptote which I will say rolled around in my head as i considered the question. The thing is that the point at which this all becomes relevant is too late, the object is at the event horizon.

[–] sudoreboot 1 points 1 year ago

Sorry but I just can't figure out what you're saying

[–] sudoreboot 1 points 1 year ago* (last edited 1 year ago)

There is a paper related to this that argues that an event horizon can never form in finite time, and that what one observes is only ever the fading signatures of infalling objects and what they call pre-Hawking radiation.

According to the paper, no matter whether you are an asymptotic or infalling observer (i.e. whether you are watching someone fall into a black hole from far away or falling into a black hole yourself), you never observe an event horizon, much less cross it. The infalling objects will be unambiguously radiated back out before an event horizon has time to form, so the black hole evaporates without anything ever falling into it.

Interestingly, while I would expect a very dramatic blast of radiation as I approach the Schwarzschild radius (of a Schwarzschild black hole), the one they predict is actually undetectable (but still there and still intense).

This isn't to say that my set of premises or understanding is correct, but the paper at least seems to reaffirm my intuition.

Edit: credit to @apsmith@fediscience.org for letting me know about the paper.