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The wrinkle in your question which sibling comments miss is this: it's difficult to reason about how a photon 'experiences' things under any circumstances.

No, not because they don't have sensory organs or the like, we can wave that aside, but because time doesn't pass for objects moving at the speed of light. It's so compressed that everything is instantaneous: a photon from another galaxy arrives in our eyeballs at the same moment it is emitted.

So you're asking a question about the duration experience around a black hole, of a particle which can't have duration experience.

Yes I was a bit imprecise and wandered about with my various examples... a photon doesn't experience time but an observers in a reference frame inertial-ish to the microwave background and generally far away from a gravity well will have a consistent agreement of when a photon is emitted, when it crosses an event horizon, and perhaps what happens to that photon as the black hole evolves. From the photon's experience it is emitted and absorbed at the same instance, my questions about duration are about how an external observer would assign times and locations to events for that photon.

As a particle, you continuously accelerate even after the event horizon (which you don't realize). Immediately after you passed the horizon, any photon you can send won't ever leave the black hole.

What outside observers will see "frozen" is the instant just before you cross the horizon. A bit like if your image 1ms before you cross the horizon will be seen by the observers years after, your image 1µs before will be seen millions of years after, etc...

But... let’s watch that image or at least keep a model of our friends falling into the black hole for 10^100 years. Forever is a long time.

If we keep watching that image the black hole eventually stops growing, the “image” never crosses the event horizon and when the universe cools down enough the event horizon starts getting smaller and hotter until we watch our friends getting roasted outside the evaporating black hole which eventually is gone and just normal matter.

The timeline of the “image” would seem to reconnect with the real article having never crossed the event horizon.

In other words it would seem if we waited long enough the image of our friends outside the event horizon would outlive the black hole and we could go say hello after it evaporated.

In other, other words, how do we see the universe outside aging as we fall into a black hole? Do we not get to watch the heat death of the universe as we approach and consequently the black hole very quickly evaporating in front of us as we fall towards it?

> it would seem if we waited long enough the image of our friends outside the event horizon would outlive the black hole and we could go say hello after it evaporated

You would see your friend's image, but your friend's image is not your friend. If you tried to go towards where the image appeared to be coming from, you would just find empty space, where the black hole that evaporated away (with your friend inside, having long before hit the singularity) used to be. Your friend would not be there any more.

Note, btw, that once you see the final image of your friend (the image of him just crossing the horizon, which you see in a big flash of light when you see the hole finally evaporate), you see nothing more; your friend, and the hole, and all of the other things that fell into the hole (and whose images as they crossed the horizon you also saw in the big flash of light) all vanish after that (no more images are coming).

This has been my line of thinking also. Many models of black holes in the past have been way oversimplified, to the point of absurdity. Their temporal existence is bounded in both directions, and neither is an instant cut-off!

1) When a star collapses, the black hole it becomes isn't formed instantly, that would violate the very theory of relativity that predicts their formation.

2) Black holes evaporate, as you've mentioned.

But this begs the question: Forget the hypothetical "probe" particle. What about the particles that made up the original star? They're like many probe particles!

Imagine the black hole starting at the centre of the collapsing star, expanding outwards, slowing down time for the infalling matter. Does any of the rest of the star fall in to the black hole? Or, like your test particle, does the rest of the star just hang there, frozen in time, a thin layer above the event horizon?

Regressing that back to the formation: Does anything ever truly fall into the black hole?

I suspect not. My thinking is that a black hole truly is a hole in spacetime, and contains nothing, not even spacetime. It's like poking your finger through a loose-knit wool jumper and making a hole. From the perspective of the threads, you've made a distortion that can be approached, but not entered.

Then the evaporation of the black hole is the hole in spacetime closing up, releasing the compressed matter near the horizon as Hawking radiation...

You’re correct: for an outside observer watching the formation of a black hole, the matter going into it will never really fall in. It’ll just continuously get dimmer and dimmer.

Outside observers are more diverse than perhaps you suspect.

For instance, what does an observer hovering just above the horizon of a different black hole see?

There is more than one astrophysical black hole in our universe, and one should not forget that when reasoning based on the Schwarzschild exact solution. Worse, most of the astrophysical black hole candidates have non-negligible spin. Certainly it would be weird for a stellar black hole to have low spin, as they will have previously have been rotating stars, and so with any sort of trust in the exterior Kerr solution[1], it seems unwise to ignore the observables generated by the ergosphere, in particular the non-stationarity of objects within it.

It doesn't really change the thrust of my question two paragraphs up: can an accelerated observer see a blueshifted infall more similar (in terms of counting time by the ticks of the observer's wristwatch) to the left than to the right in https://en.wikipedia.org/wiki/File:Gravitational_time_dilati... ? Or, equivalently, are there families of observers who are not (in this case Boyer-Lindquist) coordinate observers ? (The second answer should be, (a) yes, we can find arbitrarily accelerated observers, and (b) we may not be able to find a mapping among the local inertial frames of the infaller, the coordinate observers, and arbitrarily accelerated observers, but we should not ascribe any physical meaning to our failure to do so. cf [2]).

Perhaps a more interesting way of thinking about it is that there is essentially no practical difference between a black hole with the infaller collided with the gravitational singularity and a black hole with the infaller at 2GM+epsilon (in units such that c=1 etc) above the gravitational singularity, and if there was a practical difference and it persisted longer than a light-crossing time, we would have disproven the no-hair conjecture. However, LIGO/VIRGO evidence shows a very clear ring-down for BH/BH and BH/NS mergers, which fails to support such a challenge to no-hair.

For us weakly accelerated Earthbound observers, neutron stars and black holes at a wide variety of distances from us completely fall into each other in finite (indeed, short in human terms) time.

Non-compact infallers don't raise much of a bump on the horizon of the black hole in comparison, but the result is the same: M changes (as does the spin parameter a), and we now "find" the horizon at a new set of points. (Here it's tempting to talk about slicing up spacetime into space and time so we can talk how it is easy to forget what we mean when we talk about e.g. the "before" horizon and the "after" horizon, or alternatively discuss whether, if we had a sensitive gravitational wave detector, we could use that to decide whether a large-mass infalling object was inside or outside the black hole).

- --

[1] There are objects like https://carinaemajoris.wordpress.com/2012/07/07/a-very-errat... whose observations are much closer matches to Kerr (exterior) solutions with spin parameter ~ 0.9 than to Kerr (exterior) solutions with spin parameter ~ 0 (in particular they're pretty clearly not generating the sorts of timelike geodesics that one would find in a Schwarzschild spacetime).

[2] https://physics.stackexchange.com/a/458855

> Black holes don't last until infinity though, they evaporate in a large but measurable time.

Yes, but things can still fall in and hit the singularity inside during that time. You can't see the inside from outside, but you can still calculate that it is there based on observations you can make outside.

> if you are a conscious very resilient little particle doomed to fall into a black hole... do you really appear from the outside to be falling in until the black hole evaporates?

Sort of. If you are watching from outside, when the hole finally evaporates, you will see a big flash of light that includes images of everything that fell into the hole before it evaporated away, at the instant those things crossed the hole's horizon. Up until that point, yes, you will see infalling objects appear to gradually slow down as they approach the horizon. But that is only a sort of optical illusion; the objects actually fell in and hit the singularity inside the hole long before you see the final image of them crossing the horizon.

> is a black hole sort of a time machine to the end of the universe where in the short process of falling in you get to watch the whole history of existence pass you by

No, it's not. If you fall into a black hole, you will only see a short portion of the future of the rest of the universe from the time you started falling. The rest of the future of the universe will never become visible to you.

> is there some third option

No.

> the objects actually fell in and hit the singularity inside the hole long before you see the final image of them crossing the horizon.

It's not an illusion at all, and neither observers reference frame is more privileged than the other.

There is no global reference frame that allows you to say that the objects actually fell in and hit the singularity inside the hole long before you see the final image

> neither observers reference frame is more privileged than the other.

I am not talking about any particular observers. I am talking about invariant properties of the spacetime geometry and invariant curves within that geometry. Nothing I have said depends on any particular choice of observers. I only phrased it in terms of what a particular observer sees because the post I was responding to was asking about what can be seen from outside the horizon. I could just as easily have phrased it in terms of what null curves intersect with a particular timelike curve; the underlying geometric facts, and the fact that they are invariant, are the same either way.

> There is no global reference frame that allows you to say that the objects actually fell in and hit the singularity inside the hole long before you see the final image

Yes, there is. There are a number of frames that cover both the exterior and interior regions; for example, Painleve coordinates, Eddington-Finkelstein coordinates, and Kruskal coordinates.

Thank you for your reply, going back to my original question that sparked all of this uninformed speculation... how much extra time and distance is covered/experienced traveling to an event horizon? (It's not exactly easy to construct the question correctly)

> how much extra time and distance is covered/experienced traveling to an event horizon?

I'm not sure what "extra" means, but a general rule of thumb is that the time to fall goes like the 3/2 power of the radial coordinate you are falling from. (Strictly speaking, this is the time to fall to the singularity, not the horizon, but for falling in from large radial coordinates, which will be the case for virtually all cases of interest, the difference is small.) The only technical complication is that this rule is true if the time and radial coordinate are in units of the Schwarzschild radius of the hole; you then have to convert to conventional units. A couple of specific examples:

Black hole of one solar mass, observer falling into the hole from a distance of 3 million kilometers (which is about 4.3 times the radius of the actual Sun--the distance is chosen for reasons that will appear in a moment). This makes the radial coordinate 1 million, 10^6, in units of the Schwarzschild radius of the hole (which is 3 km), so the time to fall will be 10^9, 1 billion, times the light-travel time across that distance, which is 10 microseconds (10^-5 seconds). So the time to fall will be 10^4 seconds, or about 3 hours.

Black hole of one million solar masses, observer falling into the hole from a distance of 30 billion kilometers (or about 5 times the semi-major axis of the orbit of Pluto around the Sun). Here the radial coordinate is only 10^4 in units of the Schwarzschild radius, so the time to fall will be 10^6 times the light travel time across that radius, which is 10 seconds (a million times 10 microseconds). So the time to fall will be about 10^7 seconds, or about 4 months.

The second option is accurate. As you fall from the event horizon to the singularity, you see the entire future of the universe unfold since all arrows of time now point toward the singularity so as you move along it, you are moving in time.

This is one of the things shown in interstellar before he hits the tesseract and they go into pure speculative scifi. He gets bombarded with extremely energetic blue shifted particles as he moves through time while falling.

Regarding what happens at the event horizon, your question is slightly inaccurate in that the photon is what you observe so it doesn't emit anything. If you assume, say, a torch falling in facing toward you, the reason that it feels delayed is ofc that it has to move out of the gravity well of the black hole. I haven't thought about the question myself, but basically think about the entire set of photons your torch emits and assume it is a magically resilient thing that never tears apart.

Then it emits a fair few trillion packets of energy before it crosses the event horizon. Most of this is caught in the photon sphere and never makes it out (depends on incident angle and some other calculations I forget).But we will assume photons are ejected in one line out on some radius vector of the black hole. Then all photos ejected before the event horizon will eventually escape and you will see them. The last one is supposed to escape asymptotically as you approach the horizon but it is only constrained by how much energy it loses traversing that piece of space time from the event horizon to infinity (i.e. where the observer is located).

So we can deduce that it gets emitted at some point in time. We assume infinity to make the math easier but real life is quantized so it'll likely be sooner than that. I can't do any better than that unfortunately.

>As you fall from the event horizon to the singularity, you see the entire future of the universe unfold since all arrows of time now point toward the singularity so as you move along it, you are moving in time.

This would suggest that as I see the entire future of the universe as I fall towards the event horizon I also see the entire future of the horizon itself... which means that I will never cross the horizon because it will at some point far in the future no longer exist.

i.e. there is no spacetime path which ever crosses the event horizon

I'm near the point where I'm going to have to learn some black hole math to convince myself otherwise. (I think I'm wrong but can't see why)

Relevant discussion: https://physics.stackexchange.com/a/79086

> i.e. there is no spacetime path which ever crosses the event horizon

As I understand it, there isn't, from the perspective of an observer external to the event horizon.

(warning: I am only an avid watcher of PBS Spacetime)

> As I understand it, there isn't, from the perspective of an observer external to the event horizon.

This is not correct. An observer outside the horizon cannot see anything at or inside the horizon; but that doesn't mean he can't calculate that there is a region of spacetime inside the horizon, from observations he can make from the outside. He can.

As I understand it at this point... there is no round trip spacetime path to the horizon and back and best-case light which hits (or is arbitrarily close to hitting the event horizon) can get back to you the moment the black hole finally evaporates. However the first half of that trip is possible and happens in finite time.

> As you fall from the event horizon to the singularity, you see the entire future of the universe unfold

No, you don't. Most of the universe will never be in your past light cone as you fall to the singularity. The fact that the singularity is unavoidably in your future when you are inside the hole, does not mean it is also to the future of everything outside the hole.

This is unnecessarily pedantic. What parent meant is more akin to "if Earth would exist forever, and you would live forever on it, you'd see the entire future of the universe unfold". Obviously you won't see what happens on the moon when it is behind the horizon, but that was not the point.

> What parent meant is more akin to "if Earth would exist forever, and you would live forever on it, you'd see the entire future of the universe unfold".

No, that's not what the post I responded to meant. That post was specifically talking about what an observer falling into a black hole and hitting the singularity would see. That is not the same as existing forever outside the hole's horizon.

Why do you think the post did not refer to its author's understanding, that your experience would be the same as hanging forever just outside the horizon except in finite time from your perspective? This would be my default interpretation of their comment, and since you are successfully attacking a different stronger interpretation you should reconsider that this strong interpretation is the right way to interpret the comment. That's scenario is actually covered in HN discussion rules.

To be fair until your comment I wrongly believed the moment you'd be crossing horizon you'd be hit by all light ever emitted toward that point of the horizon in the entirety of the future (assuming no hawking radiation of course).

It's basically all the future events that happen inside the black hole isn't it? At least that's what I meant.

> It's basically all the future events that happen inside the black hole isn't it?

No, you won't even see all of those if you fall in; the most you will see are the events that end up in your past light cone before you hit the singularity. That is only a portion of the entire spacetime region inside the horizon.

That's officially above my pay grade. :-)

Seriously my course in general relativity was about 30 years ago. And work on the information paradox work is about exactly what you're talking about.

I honestly don't fully know what happens when you're approaching a growing black hole. Do you cross the horizon then?

Well, the other wrinkle here is that the "singularity" is a mathematical approximation. There's no such thing as hitting it -- from the outside, you just move asymptotically slower as you approach it.

Not so fast. Black holes have a singularity in the middle, and an event horizon around them. There is no singularity at the event horizon, and that is all that we can see..

There might be naked singularities in case of a rapidly rotating black hole. The singularity would have a donut shape and might be outside of the event horizon. https://en.wikipedia.org/wiki/Naked_singularity

Of course, this is very theoretical and we don't actually expect that to exist.

Huh, I thought that naked singularities were thought to not exist. But apparently in 2018 someone found a case where they could.

Thanks!

You move asymptotically slower from the point of view of the particle falling in the black hole. From outside the black hole, the falling particle should fall into the black hole at a speed close to the speed of light. (but we can't observe that from the outside)

The image an outsider would see "printed" on the black hole has nothing to do with where the falling particle is.