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It's pretty clear GP was using the reference frame of a far-away observer.

It's fine to switch to the infalling observer's frame, as you're doing here, but that choice isn't objectively any more correct than GP's. For example:

> Your acceleration is zero--you're in free fall.

That's in the falling observer's frame. It's totally fine to say 'a astronaut in freefall is accelerating towards the earth', explicitly invoking Earth's reference frame.

The freedom to choose a frame is literally why it's called 'Relativity'.

> It's pretty clear GP was using the reference frame of a far-away observer.

Even in such a frame (for example, Painleve coordinates), not all of the statements are correct. In those coordinates, the infaller's coordinate acceleration is nonzero, as you say, but their coordinate speed inside the horizon is greater than the speed of light. Their inertial mass is still unchanged. And the coordinate acceleration in this frame is still not "towards" the singularity, since the singularity is still a moment of time, not a place in space.

> It's totally fine to say 'a astronaut in freefall is accelerating towards the earth', explicitly invoking Earth's reference frame.

In terms of coordinate acceleration, yes. But coordinate acceleration, precisely because it is frame-dependent, is not considered a physically meaningful quantity in relativity. Only invariants can be physically meaningful quantities. Proper acceleration--what an accelerometer attached to the infaller reads--is an invariant, and that is zero for a free-faller.

From the point of view of somebody outside the black hole the infaller never reaches the event horizon.

This is not correct. Someone outside will never see the infaller reach the horizon, but that doesn't mean the infaller never reaches the horizon. It just means the outside observer never sees it happen.

I think the inability of an outsider to detect an event is precisely what it means to say "from the point of view of sn outsider, this event never happens".

What happens but can never be observed, may aswell never happen from the p.o.v. of the observer. It has no consequences to say it does, or does not, happen (for if it did, those consequences would make it observable).

> I think the inability of an outsider to detect an event is precisely what it means to say "from the point of view of sn outsider, this event never happens".

No, it isn't. The fact that a particular observer cannot detect an event does not in any way justify the claim that that event never happens.

> may aswell never happen from the p.o.v. of the observer

Whether or not an event "happens" is not a matter of any particular observer's point of view. At least, not in General Relativity. In GR, an event happening means there is a point in spacetime at which it happens. That point is either there in spacetime or it isn't, independent of what any observer can or cannot detect.

You can say an event has no causal effect on a particular observer, but that's not the same as saying the event never happens.

> The infalling person has no way of knowing, locally, that they have crossed the horizon

Is there such a thing as "the" event horizon? Suppose Alice and Bob are both free falling into a block hole, but Alice is a little bit further in than Bob. Alice sends a message photon straight back out of the gravity well once per second by her pocket watch. There is a distant observer, Dana. Dana believes the event horizon to lie at a certain radius, and as Alice approaches that radius, she appears to Dana to slow down and to redshift so that Dana does not see Alice cross the horizon. However, there is a particular second, N, on Alice's pocket watch which marks the final messenger photon that can get back out to Dana; all messages > N are trapped.

Now, if Bob is right behind Alice, falling into the black hole, isn't it possible that he can observe Alice's (N+1)st photon? That photon need not escape the black hole all the way out to Dana in order for Bob to see it, since he's falling toward it. Therefore, doesn't Bob disagree with Dana about where the event horizon is? Alice hasn't crossed it yet as of N+1 as far as Bob can see.

I guess my bigger point/question is, if you decide to fall into a black hole, won't its horizon recede before your eyes as you approach?

I haven't ever decided to fall into a black hole so I'm not speaking from experience, but I think Bob would only see Alice's (N+1)st photon after he himself has crossed the event horizon.

> I think Bob would only see Alice's (N+1)st photon after he himself has crossed the event horizon.

Yes, this is correct.

> Is there such a thing as "the" event horizon?

Yes. It's a global feature of the spacetime geometry. It's not observer dependent.

> if Bob is right behind Alice, falling into the black hole, isn't it possible that he can observe Alice's (N+1)st photon?

If he himself crosses the horizon, yes, he will see Alice's N+1st photon after he has crossed the horizon. But not until then.

> doesn't Bob disagree with Dana about where the event horizon is?

No. See above.

> if you decide to fall into a black hole, won't its horizon recede before your eyes as you approach?

There are optical effects due to light rays traveling tangentially, not radially, that (if we imagine the horizon as a light-emitting surface, which it actually isn't) could make this seem to happen, but this would be an optical illusion; you would still fall through the horizon regardless of what appearances your eyes were seeing.

For some time after Alice and Bob have both crossed, they can continue to communicate with one another, showing that they are not separated by a horizon. But since Alice is ahead, won't she and Bob get separated more and more over time? Isn't there a time after which Alice can't even get a message out to Bob? Can that not be characterized as an event horizon separating Alice and Bob, and one which is fundamentally distinct from the one separating Alice and Bob from Dana?

> For some time after Alice and Bob have both crossed, they can continue to communicate with one another

Yes.

> showing that they are not separated by a horizon.

No. They are separated by a horizon, but only for a short time--short enough that Bob crosses the horizon before seeing any light signals emitted upward by Alice after she has crossed the horizon.

> since Alice is ahead, won't she and Bob get separated more and more over time?

Yes.

> Isn't there a time after which Alice can't even get a message out to Bob?

Yes. If you draw a spacetime diagram of the hole with Alice's and Bob's worldlines on it (Kruskal coordinates work well for this), then at some point on Alice's worldline her future light cone will no longer include any portion of Bob's worldline; and once she passes that point, she can no longer send light signals to Bob (because he will hit the singularity before her signals can reach him). One can also make a similar argument for Bob.

> Can that not be characterized as an event horizon separating Alice and Bob, and one which is fundamentally distinct from the one separating Alice and Bob from Dana?

No.

> No.

This answer is pretty terse and doesn't provide a lot of understanding. Found these helpful articles instead, explaining the difference between an absolute horizon [0], which is what you're talking about, and an apparent horizon [1], which is the idea I've been asking about without knowing the magic word for.

[0] https://en.wikipedia.org/wiki/Absolute_horizon

[1] https://en.wikipedia.org/wiki/Apparent_horizon

> This answer is pretty terse and doesn't provide a lot of understanding.

Yes, sorry, I was pressed for time. I'll try to expand on it further below.

Unfortunately, "apparent horizon" is not the idea you were asking about. I don't know if there is a standard name for the idea you were asking about, but I do know that none of the "horizon" terms we have been using ("event horizon", "absolute horizon", or "apparent horizon") are it.

"Absolute horizon" and "event horizon" mean the same thing: they are the boundary of a black hole. This boundary is a global null surface that bounds a region of spacetime that cannot send light signals to infinity.

"Apparent horizon" means a surface at which radially outgoing light rays do not move outward. In other words, if you imagine a 2-sphere emitting light rays radially outward in all directions, normally those light rays, taken all together, will form their own 2-sphere that expands with time (its radius increases). If the 2-sphere emitting those light rays is part of an apparent horizon, however, the 2-sphere formed by the light rays does not expand--it keeps the same radius, which means it stays right at the apparent horizon instead of expanding outward away from it.

What you were describing, the boundary that describes when Alice can no longer send messages to Bob, is the future light cone of some event on Alice's worldline--the first event whose future light cone does not include any events on Bob's worldline. If you look at the light rays emitted by Alice that form her future light cone, they are not just radially outgoing light rays; the future light cone consists of all light rays emitted at a given event, in all directions, not just radially outward. AFAIK such a surface is never called a "horizon" of any kind; it's a future light cone, which is a different kind of surface.

> Your acceleration is zero--you're in free fall.

So if I jump out of an aeroplane, I don't accelerate towards the ground? Apparently my understanding of the natural world needs some enhancements.

Nope, you're in an inertial reference frame when in free-fall. Ignoring air resistance, if you were to take an apple out of your pocket and place it next to you in mid-air, it would be motionless relative to yourself. Put a box around you with no windows and you'll have no idea whether you're floating through space in the absence of gravity, or if you're free-falling inside a gravity well.

This is the main difference between general relativity and the newtonian notion of gravity: GR says gravity is just spacetime curvature, and by following an inertial (geodesic) path through spacetime, it appears to be acceleration to certain observers, but is in fact inertial. Newtonian mechanics says instead that gravity is a force and you're indeed accelerating, but it's a view that breaks down at high mass/speed, and GR makes correct predictions where Newtonian mechanics does not, so we generally prefer the GR interpretation.

> you're in an inertial reference frame when in free-fall.

But the falling person's frame isn't the only one available. Someone on the ground would rightfully describe the falling person as accelerating towards the ground (or them) (at least until they reach terminal velocity).

The most important lesson of relativity is that no reference frame is privileged. That means both descriptions are valid, provided you attach the reference frame to the description.

> The most important lesson of relativity is that no reference frame is privileged.

No, the most important lesson of relativity is that only invariants are physically meaningful. IIRC Einstein once commented that he wished his theory had been called the "theory of invariants" since that would have been a better description of its most important feature.

The second most important lesson of relativity is that yes, you can use whatever frame you wish, because invariants are the same no matter what frame you use to calculate them. So you will come up with the same physically meaningful quantities no matter what frame you choose.

all this talk of GR and invariants, not to confuse a mathematical singularity with a gravitational one, at which point GR itself breaks down per the need for a quantum mechanical description. the truth is we don't know what happens within a black hole, and probably never will

> GR itself breaks down per the need for a quantum mechanical description

This is the current predominant belief among physicists, yes, but that doesn't mean it's established fact.

> the truth is we don't know what happens within a black hole, and probably never will

This is way too pessimistic. The predictions of GR for this regime are precise and unequivocal. Even if you believe GR breaks down when spacetime curvatures reach the Planck scale (which is the current predominant belief, as I said above), that still leaves most of the black hole's interior within GR's domain of validity.

> that still leaves most of the black hole's interior within GR's domain of validity.

Thanks, that's what I (as a layman) supposed: that inside the event horizon isn't some space that you can't think about; it's qualitatively not distinguishable from outide.

Upthread, someone mentioned that the singularity wasn't a place in space, it was a location in time. That made me think; the mass of a BH isn't the mass of a volume of space, it's the mass of the "singularity", which (presumably?) is located at the center of the space described by the event horizon. So there must be a place where the singularity is located, and there must be a direction towards it.

Someone else suggested there are two things: a mathematical singularity and a gravitational singularity, and that they aren't the same. That hadn't crossed my mind.

> someone mentioned that the singularity wasn't a place in space, it was a location in time.

That was me.

> That made me think; the mass of a BH isn't the mass of a volume of space, it's the mass of the "singularity"

This is not correct. The mass of the hole is not somehow "concentrated" at the singularity. There is no matter anywhere inside a black hole. The "mass" of the hole is a global property of its spacetime geometry.

> "singularity", which (presumably?) is located at the center of the space described by the event horizon.

Your presumption is wrong. There is no "center" of the black hole in the ordinary sense. Its spacetime geometry simply doesn't work that way; it is not like the spacetime geometry inside an ordinary spherical object.

> So there must be a place where the singularity is located

Wrong. See above.

> and there must be a direction towards it.

There is such a spacetime direction, but it's a timelike direction--towards the future--not a spacelike direction.

> Someone else suggested there are two things: a mathematical singularity and a gravitational singularity, and that they aren't the same.

That is not correct.

but if most of the interior is concentrated into an infinite mass, i.e. impossible, that kinda invalidates its validity

So is it correct to say that Newtonian mechanics is "just" an approximation (based on our observations) for all practical purposes? (EDIT: Spelling)

Falling in a black hole is a different experience. It must be more like being squeezed in a narrow tube with a glowing light that was once the universe on one end of the tube and something indescribable on the other end. Perception of time will change too, as all the particles that let us perceive the time will be messed up in strange ways. Perhaps time will seem multidimensional or there will be many timelines. In other words, your perception of a fall from an aeroplane isn't true or absolute, it's mediated by particles that define, for you, space and time, and if those particles get messed up, your reality will change completely.

None of what you say is correct as a description of what General Relativity says will happen to someone who falls into a black hole.

I think the "tube" business concerns what you see a you approach the speed of light. All the light/future events are in front of you, and you see nothing if you stand on the stern of your ship.

I'm not sure how that relates to falling into a BH, bit I don't know; there's a lot of this that I'm struggling to understand. I'm not a physicist, and not clever, just very interested.

> I'm not sure how that relates to falling into a BH

The special relativistic effect you describe has nothing whatever to do with falling into a BH.

> if I jump out of an aeroplane, I don't accelerate towards the ground?

Not in your rest frame. In your rest frame, the ground accelerates towards you.

However, the acceleration described in both frames (yours and the Earth's) is coordinate acceleration, which is frame-dependent, and in relativity frame-dependent quantities don't have any direct physical meaning. The quantity that has physical meaning in terms of acceleration is proper acceleration--what an accelerometer strapped to you would read--and in free fall that is zero.

> So if I jump out of an aeroplane, I don't accelerate towards the ground?

In a very real sense (although not a useful one for a skydiver), the answer is yes. This is a key insight of the Equivalence Principle and the curved-spacetime view of gravity. You (in free-fall) are on an inertial "straight line" path through spacetime towards the center of the Earth. When you reach the ground, it pushes on you, messily accelerating you off of that path.

In some sense, the ground is accelerating towards you. By jumping out of the plane, you've stopped the plane floor from accelrating you away from the ground at the same rate the ground is accelerating towards you.

This is barely useful for humans since we are very used to 'standing still' actually involving the force of your weight, so we consider that distinct from acceleration. But physics gets much easier if you don't make that distinction.

I can't accelerate towards next week, but I can predict when next week will happen, with extreme accuracy. Is this the same for reaching the singularity? Does the person falling (for the sake of this example let's assume they maintain consciousness) experience the passing of time as they fall further "into" the black hole? If so, how long would one experiencing the falling until they reach the singularity?

Yes, and you can calculate exactly how much time it takes (it was an exam question in my final exams for undergrad iirc). The amount of time depends on the mass of the black hole but it's definitely less than a minute.

I'd always assumed that for TON 618 it would've been a few days… but I don't have a physics degree, and now I come to write this comment realise I'd been just presuming it was "Schwarzschild radius/c" and it's not necessarily going to be such an easy result.

You certainly can accelerate to next week as measured by Earth calendars. Climb into your local particle accelerator and ask them to get you accelerating around the loop as close to the speed of light as possible. By accelerating, you can get to next week in just a few seconds of your own subjective time.

Also, you can accelerate towards the black hole singularity and get there faster.

I’d love to understand the interaction effects between 1) acceleration toward singularity (prob approaching c?), 2) gravity (prob approaching infinity?), 3) distance compression experienced by infalling observer.

My instinct is that the observer would subjectively experience it to take exactly the time predicted by classical mechanics to “hit” the singularity — basically that all the considerations above would cancel out in the local reference frame — but that to an outside observer it would take an infinite amount of time.

That's my layman's model.

> I’d love to understand the interaction effects between 1) acceleration toward singularity (prob approaching c?), 2) gravity (prob approaching infinity?), 3) distance compression experienced by infalling observer.

None of these things are useful in the case under discussion. 1) is invalid--the singularity is a moment of time, not a place in space, and you aren't "accelerating towards" it. 2) is not well-defined inside the horizon (spacetime curvature is, and does increase without bound as you approach the singularity, but that's not quite the same as "gravity"). 3) is a special relativistic effect on others observing you; you yourself don't experience any "distance compression" regardless of your state of motion.

> I can't accelerate towards next week, but I can predict when next week will happen, with extreme accuracy. Is this the same for reaching the singularity?

If you know the mass of the hole, yes, you can predict how much time will elapse on your clock between now and when you hit the singularity.

> Does the person falling (for the sake of this example let's assume they maintain consciousness) experience the passing of time as they fall further "into" the black hole?

Yes.

> If so, how long would one experiencing the falling until they reach the singularity?

A reasonable estimate for free-falling into a black hole is that the time to reach the singularity is the 3/2 power of the radial coordinate you are falling from, in units of the hole's Schwarzschild radius. If you are just crossing the hole's horizon, you are at the hole's Schwarzschild radius, so your time to reach the singularity from there would be that radius divided by the speed of light (to convert to time units instead of distance units). For a hole of 10 solar masses, for example, this time is about 100 microseconds (Schwarzschild radius of about 30 kilometers divided by the speed of light).

What happens to my speed as I reach (hit?) the singularity 0.1s after crossing the event horizon? Do I jolt to a halt, do I continue to accelerate? Do I keep a fixed speed, but never change position?

> my speed

There is no such thing. Speed is not absolute in relativity. You can only meaningfully talk about your speed relative to some particular object or observer, and then, strictly speaking, only when you and that object or observer meet (i.e., your worldlines cross).

Similar remarks apply to all of your questions; they simply aren't meaningful.

I thought it’s after something passes the event horizon time and space flip. So one is not traveling through space at some speed say the speed of light but rather through time at the same speed (whatever that means).

Just as an aside these conversations are amazing and I’m learning a ton!

> That's correct. The infalling person has no way of knowing, locally, that they have crossed the horizon and are now doomed to hit the singularity. But that doesn't change their doom.

This is what happened to SG-10.

> Not for the typical black holes we observe in the universe.

Ah. Are the typical black holes too small, or too big?

Spaghettification is just the extreme effect of tidal forces. IIRC, objects made of normal materials falling into a stallar-mass BH will be ripped apart lengthwise (and compressed widthwise) before they cross the event horizon. The same object falling into a supermassive (galaxy-center) BH won't be ripped apart until some time after it passes the event horizon. Or an "object" which is a loose cloud of dust will be "ripped apart" much farther away (at the Roche limit). But the tidal force increases without limit as you approach the center of the BH, so no matter what the object is, it'll eventually be ripped apart.

Too big.

Typically I thought Quanta magazine did better than typical pop-science. Guess this was an exception?

Unfortunately not; my experience is that Quanta is just as prone to these kinds of errors as other pop science sites.