Matter "falling into" a black hole despite infinite time dilation

[Yora]

A very popular question that you can find asked all throughout the internet, but as far as I can tell never gets answered.

The question is "If time dilation at the event horizon becomes infinite (for an outside observer), how can anything ever pass the event horizon (as seen from the outside), and wouldn't that mean that we (as outside observers) never see a black hole form in the first place?"

Usually the parts in the bracets are not spelled out and so all the answers almost always come down that to anyone falling in time keeps on going normally. But we on Earth see "the effects of invisible sources of massive gravity that can only be explained as being black holes", and given that the universe is not infinitely old, these must have come into existance in a finite amount of time.

Now here's an idea how to possibly make sense of this: When mass aproaches infinitely close to the event horizon, the time dilation also aproaches infinity. Since any particles that are still an infinitesimal distance away from the event horizon would appear to move at infinitisimal speed as seen by an outside observer, couldn't we say that all particles basically pile up infinitely close to the event horizon in an infinitely thin layer?

Any mass that "falls into" a black hole causes the event horizon to grow in diameter. (And if I am not mistaken, it's the diameter that grows proportionally to the mass of the black hole, not the volume, so it grows rapidly in diameter any time mass is added.) And larger black holes also have less density. So couldn't we explain the apparent paradox by saying that what we call "falling into" is really "pilling up at the surface"?
That would seem like a workable explanation for me, who never did any actual calculations about any of it.

However, this would have the implication that no black hole has formed a singularity (region of infinite density) at its center yet, and never will.

(Also: If there is a singularity in the center of a black, would the space inside the event horizon be infinitely large? The funel shaped singularity in 2D space looks like it goes down infinitely far, thus having infinite volume.)

[factotum]

The question is "If time dilation at the event horizon becomes infinite (for an outside observer), how can anything ever pass the event horizon (as seen from the outside), and wouldn't that mean that we (as outside observers) never see a black hole form in the first place?"


In fact, infalling matter to a black hole *will* appear to stop at the event horizon. However, that doesn't mean it'll just stay there visible forever, because it would output an infinite amount of energy doing so--it will appear to stop at the event horizon and then gradually fade from view.

As for not seeing the black hole form, as far as I know we've never "seen" a black hole in any case. We see the consequences of a black hole being there--stars in ridiculously fast orbits at the galactic core, gravitational lensing effects from light passing nearby, or electromagnetic radiation coming from the accretion disk around an actively "feeding" hole--but you can't see a black hole directly because light doesn't escape from it.

[Yora]

But we assume that stelar black holes once were stars and now are black holes, so they have to have formed in a finite amount of time. They must be forming despite infinite time dilation.

[NichG]

It's a mistake to think of the entire black hole as an object with a single, global 'state'. In relativity, you lose the ability to have a unique 'now' that is universally simultaneous. So it's not like you'd get matter 'piling up' at the event horizon like a high density shell - even if from the outside you'll see that matter there for an infinite time hence, from its point of view it only sees a finite span of time pass as it's falling past the event horizon. So that in-falling matter is not interacting with stuff that falls in a million years later or whatever - it's already long gone by the time the new stuff catches up with its image.

[Binks]

But we assume that stelar black holes once were stars and now are black holes, so they have to have formed in a finite amount of time. They must be forming despite infinite time dilation.
There are lots of possible answers to your question, as this is an evolving field of physics. Two that I am familiar with are that there are no true black holes, and that there is no such thing as a mass at rest within the event horizon.

The idea of no true black holes sounds like what you're going for with your questions. It's basically the theory that, since time dilation heads to infinity as any matter approaches the event horizon, the object forming the black hole cannot have its matter pass that point as it would take infinite time. Hence, the matter "stops" (relative to the outside universe) just outside the event horizon.

There's no issue with this idea in our current understanding of black holes, as there is effectively no difference between a time dilation of 'this is going to take millions of years to move a planck length from your point of view' and 'this will never move from your point of view' as we've only been observing for a very short period of time and we really can't observe any of the matter near an event horizon. Basically, if the matter "stops" just short of becoming a true event horizon, the universe shrugs, mutters 'close enough', and moves on, acting like it's an event horizon (as, in any reasonable length of time, it may as well be).

The second means of handing the idea is that there is no 'at rest' near an event horizon. This comes from the Schwarzchild geometry stuff which I highly recommend looking into, I found a physics class page on it here (http://casa.colorado.edu/~ajsh/schwp.html) that did a pretty good job explaining it.

Effectively it boils down to this idea. Light cannot escape an event horizon as the escape velocity is c. The gravitational time dilation of a point at rest with escape velocity c should be infinite. We can't have the time dilation be infinite, so the point's not at rest. The idea is that gravity is so powerful at (and within) the event horizon that spacetime is moving, dragging everything inside towards the point singularity within the black hole. Thus, time dilation never actually goes to infinity, as the mass's rest escape velocity stays below c, but it's effective escape velocity goes above c.

From a beam of light's perspective, moving out from the singularity, the distance between it and the event horizon is infinite, as whatever distance it crosses in time T an equal or greater amount of spacetime is created between it and the horizon (and the same amount destroyed behind it by the singularity). So it's moving at c towards the event horizon, but will never get there, and thus will never escape the black hole.

Of course, that's all assuming there is an event horizon and that time dilation continues on to infinity within a black hole, which some physicists (including Stephen Hawking) disagree with. And there's the new quantum gravity stuff, and et cetera, et cetera.

EDIT: Ninja'd on the first theory by NichG :P. That's exactly what the idea of 'no true black holes' would look like, matter constantly building up at the horizon.

[Yora]

Yes, I did not mean that matter gets actually compacted in any way. With supermassive black holes you can have very large structures reaching the event horizon fully intact.

What I am thinking about is that what looks like the "flat surface" of the event horizon from the outside is actually just highly warped space. The event horizon is not a "thing", it's more like an optical illusion (though of couse it doesn't have anything to do with optics). When you're at the event horizon, you don't notice any distortions on yourself or nearby objects, because you're entire reference frame has been squeezed along with everything inside it.

Though I am not fully certain that I am not treating the event horizon as a 2D singularity right now.

But let me give an example of where my mind roaming: Assume you got a supermassive black hole with a huge event horizon that experiences negible tidal forces. (Objects at the EH, that is.) And you got a star or planet falling into the EH. Once the leading surface of the star touches the EH, the mass of the star in that region seems to have frozen in time to an outside observer. (And I read somewhere that for a stelar black hole, time dilation starts to become significant only milimeters away from the EH.) But the rest of the star is still continuing it's movement into the EH. So in my reasoning, the star has to become flat. Is there an important error here?

[Max™]

With a large enough black hole the tidal forces at the event horizon can be well below the human ability to perceive them, so you can quite sensibly ponder what it would be like to fall into one.

You wouldn't really notice much except that the light from objects that should be on the far side of the hole is visible at what you would call the horizon, there would be bands of light circling around it which you would fall through, and as you got deeper and deeper the visible horizon would begin to fold all the light in the universe up into a neat little circle and then thread and then a single point behind you as you fell.

From your perspective you fell through the event horizon without really noticing, it wouldn't be until you started getting some interesting fraction of the radius inside the hole that you'd get more obvious effects and eventually shredded.

From the OUTSIDE observer perspective, someone hanging just outside the hole, you would slow down and freeze and your afterimage would appear to linger there and redshift out of visibility, frozen at the moment you crossed the hole.

Things don't race away to infinity at the horizon, that's just the point where worldlines all point inwards towards the center, which is where things start to race off to infinite values.

Things can indeed appear to be infinitely dilated though, but there is a difference between the frame of an observer outside of such a deep gravity well and that of one inside of it. This is one of the key parts of GR, after all.

I can't say I'm sitting still and you're moving if we're both moving at a steady velocity, but if one of us is undergoing acceleration we can determine which one it was. You don't seem to experience g-forces and rush away just because I am experiencing g-forces and moving away. You only seem to move slower through time relative to my clocks, and vice versa.


Editing to note: an event horizon should increase in area as matter is added, which is to say that the entropy of the mass within the area bounded by an event horizon can be directly related to the area of said event horizon (Bekenstein and uh... dammit I always forget the other one who discovered this for some reason!) and all sorts of fascinating stuff related to information worked out by Shannon as I recall.

If you don't see why that makes it a little odd to suggest that the event horizon is the physical surface of an object, imagine putting a ball into a soap bubble (soapy ball so it won't pop it) and the bubble gets larger according to how massive the ball was. So a 1 cm ball of aerogel wouldn't enlarge it as much as a 1 cm ball of lead, even though both displace the same volume of air inside said bubble, the soap bubble does not behave like a physical object, which it shouldn't as it is a surface in the relativistic sense, with all the weirdness involved therein.

[NichG]

Yes, I did not mean that matter gets actually compacted in any way. With supermassive black holes you can have very large structures reaching the event horizon fully intact.

What I am thinking about is that what looks like the "flat surface" of the event horizon from the outside is actually just highly warped space. The event horizon is not a "thing", it's more like an optical illusion (though of couse it doesn't have anything to do with optics). When you're at the event horizon, you don't notice any distortions on yourself or nearby objects, because you're entire reference frame has been squeezed along with everything inside it.

Though I am not fully certain that I am not treating the event horizon as a 2D singularity right now.

But let me give an example of where my mind roaming: Assume you got a supermassive black hole with a huge event horizon that experiences negible tidal forces. (Objects at the EH, that is.) And you got a star or planet falling into the EH. Once the leading surface of the star touches the EH, the mass of the star in that region seems to have frozen in time to an outside observer. (And I read somewhere that for a stelar black hole, time dilation starts to become significant only milimeters away from the EH.) But the rest of the star is still continuing it's movement into the EH. So in my reasoning, the star has to become flat. Is there an important error here?

The frozen in time thing is for the infinitely distant observer. The observer on an infalling geodesic just above the external observer's event horizon would 'see' the event horizon as being further in (or perhaps just absent entirely - I'd have to check what happens if you transform to infalling coordinates).

[Yora]

That's also something I've been wondering. As an infalling observer, do you ever see yourself passing through the event horizon? That it shrinks once you get closer would seem pretty logical.
Though in that case, how can we even say an event horizon has a diameter?


From the OUTSIDE observer perspective, someone hanging just outside the hole, you would slow down and freeze and your afterimage would appear to linger there and redshift out of visibility, frozen at the moment you crossed the hole.

Things don't race away to infinity at the horizon, that's just the point where worldlines all point inwards towards the center, which is where things start to race off to infinite values.

Things can indeed appear to be infinitely dilated though, but there is a difference between the frame of an observer outside of such a deep gravity well and that of one inside of it. This is one of the key parts of GR, after all.

So what you are saying is that what we "see" frozen at the surface of the event horizon is not actually what is happening there, but simply an afterimage of something that no longer exists?

That's an alternative solution I had been pondering.

The path of photons does get bend by gravity, as we see in gravitational lensing. So does Newton's Cannon work with protons as well? Isn't the event horizon the distance at which the speed c is a stable orbit?
Then let's say a photon leaves the object at the event horizon, right before the moment of "falling in". But because of the massive gravity the photon has to travel in an outward spiral around the event horizon several times before it actually reaches an observer. The closer the photon was to the horizon at the moment of emission, the more orbits it will have to do before it really "escapes" and moves in something similar to a straight line. Make that some billions of trillions of revolutions and the distance travelled from the emmision point to the observer becomes so huge that millions of years can be pass before the observer is reached. The light from the "falling in" event reaches us delayed, but that doesn't say anything about what's actually happening at or inside the horizon.
However, I am not sure if that lines up with the observed red shift.

And this seems too simple and obvious an answer to explain why this question comes up all the time. "Time dilation is not infinite at the event horizon" would be a very simple statement to end all debate.

[Max™]

I've been learning about relativity and getting used to these concepts and the math underlying them for 30 years, it's still pretty weird stuff.

People catch on to the "I see you going slower and you see me going slower" stuff a lot easier than the "my accelerating frame is equivalent to a curvature in space and time, and the results of this are... " stuff.

When you see "frozen infinitely at the EH" that is always from the perspective of an outside observer... maybe the infalling observer in the case of a white hole solution... but those hurt my head, and I am not going to delve any further into that sort of confusing territory.

[Sith_Happens]

redshift out of visibility

Singling out this part because it seems important for the "Why can't we see a big pileup of everything that's ever fallen into the black hole" issue.

[Yora]

While the redshift is interesting (because continous and potentially continuing forever), there is actually also the matter of the "last photon". Photons are particles and have a countable number. An object "falling through" the event horizon will emit only a finite number of photons that can still escape the black hole. Once all those photons have flown out into space, no more will come. And I've read in two places that the last photon will reach an observer within a fraction of a second after the "falling through" event.
(Though I have a hunch that it's actually similar to a half-life effect. Potentially that last photon could reach you after trillions of years, but there's a high chance that it will already have gone after less than a second.)

[wumpus]

Can't this concept be boiled down into two questions?

Is Zeno's paradox about Achilles and the Tortoise valid? No. Both by experiment and by calculus proves this wrong.

Does 100% time dilation in one field of reference change anything for external fields of reference? No. All photons "experience" zero time change, and yet they still appear to move from all other fields of reference.

[Max™]

You can be an outside observer and see the infalling observer slow to a halt and redshift into oblivion while the infalling observer isn't aware of anything unusual taking place. Relativity doesn't require these frames to be symmetrical, one is clearly undergoing acceleration which the other is not experiencing.

[Yora]

Let's say a star collides with a black hole. As outside observers we would see the star disappear (possibly with a good fraction of its matter being flung out into space because the black hole is a messy eater). And at the same time we would also see the diameter of the event horizon increase.
Because we see the diameter of the event horizon increase, we can also say that the star has fallen trough it. Not just in the frame of reference of the infalling star, but also in our frame of reference as outside observers. Therefore we must conclude that the matter of the star is not perpetually frozen just outside the event horizon because of time dilation. And therefore time dilation at the event horizon can not be infinitely high.

Bam!
Okay, I admit that to my knowledge the event of a star falling into a black hole and the diameter of its event horizon increasing has never been directly observed. But still I don't see how this supposed infinitely high time dilation can be a thing.

[Max™]

Wait, I just realized that I've been mentally correcting something that is key to your question.

Time dilation does not go to infinity at the event horizon, it increases, and is observable, but I've been mentally subbing in the fact that it uses the uh... lkjasdlfj what the hell was it, square root of 1 - 2GM/rc^2 equation... hang on, lemme find you a link with it in case you wanna pore over the math.

Well, I wasn't too far off, been a while since I went over them.
http://i.imgur.com/lFdoE6H.jpg

That doesn't tend to infinity at the event horizon unless you're using Schwarzschild coordinates in and out.

They behave fine on either side of the horizon, but get wonky right at the horizon itself, so you need to choose a more sensible coordinate system when doing silly things like falling into a black hole.

Ahhh, this is a wonderful site!

http://casa.colorado.edu/~ajsh/schwp.html

So wonderful! Look at it! Go! Do it!

[weaseldust]

Isn't the question confusing time dilation with observing a decrease in speed? A body that falls into a black hole should appear to age increasingly slowly to outsider observers, and they will be able to see light from it after it falls in, but it still has a positive and increasing speed in their frame of reference.

It's just like how, if a clock is moving at 0.6c relative to me and ticking at 0.8 times the rate of my clock, I don't conclude that it's actually only moving at 0.8 times that speed. There's a difference between observing a slowed ticking of the clock and observing a reduction in its speed of travel. That's why we can see photons, which we observe to experience no internal change, travelling from place to place.

So it's not contradictory for us to detect an object has fallen in even though it can undergo no internal change in our frame and though we can see light from its past (which are different but related effects). Not that detecting that it has fallen in is easy, but I suppose the recent gravitational wave detection did exactly that, albeit the thing falling in was another black hole.

[Yora]

Alright, so the commonly made statement that "matter takes forever to fall into a black hole" is simply wrong?


This seems to support it (http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/fall_in.html), and also appears to explain where the mistaken assumption comes from.


On my worldline as I fall into the black hole, it turns out that the Schwarzschild coordinate called t goes to infinity when I go through the event horizon.* That doesn't correspond to anyone's proper time, though; it's just a coordinate called t.* In fact, inside the event horizon, t is actually a spatial direction, and the future corresponds instead to decreasing r.* It's only outside the black hole that t even points in a direction of increasing time.* In any case, this doesn't indicate that I take forever to fall in, since the proper time involved is actually finite.

[...]

Light beams aimed directly outward from just outside the horizon don't escape to large distances until late values of t.* For someone at a large distance from the black hole and approximately at rest with respect to it, the coordinate t does correspond well to proper time.




I wonder why nobody ever says that any time this question shows up?

[NichG]

Alright, so the commonly made statement that "matter takes forever to fall into a black hole" is simply wrong?

This seems to support it (http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/fall_in.html), and also appears to explain where the mistaken assumption comes from.

I wonder why nobody ever says that any time this question shows up?

I think explanations tend to arise from what people personally find makes sense to them as the way to understand it. That's often very individualized, so it doesn't always transfer well and you also get all sorts of different takes on the same idea with varying levels of precision and accuracy. There can be a lot of explanations that aren't wrong but don't tell the full story or don't capture the mathematical detail or whatever.

[Max™]

Also as I just illustrated, it is easy to overlook that something which I might consider basic or common or implied regarding a subject like relativistic weirdness has no reason to actually be basic/common/implied for all readers.

I learned Penrose coordinates in 1986, I'm accustomed to thinking of what goes on in a black hole in terms of them, it seems obvious to me to do so, but I'm pretty sure most of you reading this have no clue what they are. I assume NichG knows about them, as they seem to be pretty well informed about the subject, but as I have to remind myself, it is by no means a regular thing for people to encounter and deal with them.

That site I linked seems like the best reference to me because it helps illustrate the difference between Schwarzschild, Penrose, and a few other coordinate systems in the vicinity of a black hole.

Then I think "wait, what does it mean when I say 'coordinate system' to everyone else?" and feel derpy.

[BannedInSchool]

In relativity it seems there's often the question of what one sees via light and what one can then deduce is "actually" happening in various frames of reference. It's all gloopy.

[Yora]

In this case "photons reaching you" is not the same as "light cone".

[MLai]

Please help me, this question buuurns-!!

Assume floating astronaut falling through event horizon of supermassive black hole. Tidal forces this far out of a supermassive black hole is negligible, so human being is not spaghettified just for crossing event horizon. Meaning, gravity this far out is neglible, right?

(1) If so, then why can't astronaut just rocket his way back out, if it's such a "weak" event horizon?
(2) Suppose there's a chain attached to him, being pulled by someone else outside the event horizon. The other person isn't within the event horizon, so why can't he pull the victim back out if the event horizon is so "weak"?

[Lord Torath]

Spaghetti-fication is not due to strong gravity. It's due to a strong gravity differential; gravity pulling much harder on your feet than it is pulling on your head. Once you cross the event horizon, there's no going back. But for a super massive black hole, just past the event horizon, gravity doesn't vary much over a 6-foot distance. It's still pulling hard enough that not even light can escape it, though.

Let's say you are R feet away from the center of the black hole (but still on the inside of the event horizon). Your feet are at R-3, and your head is at R+3. If R is very big (say you're 300,000 feet from the center), then R-3 is about the same as R+3, so there's very little difference to the pull felt by your feet and your head. You're still being pulled so hard that the chain the other guy is holding will break (regardless of the material it's made of) if it's connected to you. As you get closer to the center, R starts shrinking, so that the distance between your feet and your head starts becoming more important. That's when the tidal forces start pulling you apart.

Gravity goes as 1 over the square of your distance, while the tidal forces, the differential of gravity (the change of gravity over distance) goes as 1 over the cube of distance. So it's really weak far out, but starts getting really strong closer in.

Does that clear things up for you? Or do I need to explain it a different way?

[halfeye]

Does that clear things up for you? Or do I need to explain it a different way?
For me you need to explain it a different way.

The thing I'm thinking of is, that with big SMBHs (probably more massive than Sagittarius A*, possibly more massive than any we yet know), the gravity field at the EH is not hugely strong, but it's very deep. There theoretically has to be a possiblity of a BH with an EH circumference of a light hour. So, it isn't possible to get around it in an hour, but going in and out vertically seems theoretically possible, if you wanted to get it all the way out again you'd need a very complicated bucket chain to get fuel down to the vessel that dipped in, but it seems that an automated unintelligent probe could be sacrificeded, if we could find a big enough black hole. Time dilation might make the process a bit slow for observers if that theory is true, but wouldn't something come out eventually?

[MLai]

Does that clear things up for you? Or do I need to explain it a different way?
Yeah ok I get it now. Gravity is still inescapably strong at the event horizon. You simply don't get spaghettified because it's pulling at you "smoothly", that far out.

[Sith_Happens]

For me you need to explain it a different way.

Spaghettification doesn't depend on how strong the gravity is, it depends on how sharp the gravity gradient is.


[Snip]

Um, no. The event horizon is defined as the point it's impossible to get out from, so by definition the field strength at it is going to be the same no matter how big the black hole is.

[Max™]

I was thinking to myself "a light hour is not that large for a SMBH" and checked.

https://en.wikipedia.org/wiki/H1821%2B643 is a 30 billion solar mass hole, Sag A* at the center of our galaxy is a 4.3 million solar mass hole.

The EH on the big one is many times (~28x) larger than the orbit of Pluto, which itself is around 5 light hours from the Sun, so you would be damn near a light week out from the center when you crossed it.

By comparison, Sag A* is only a few times as large as the Sun (1.4 Gigameters vs 9 Gigameters, with the big one being 172 Terameters) but that fact is a handy way to see why the "it's just a compact object without a horizon" arguments are silly. Over 4 million bodies as massive as the Sun in a space that is swallowed by the orbit of Mercury.

[Lord Torath]

For me you need to explain it a different way.

The thing I'm thinking of is, that with big SMBHs (probably more massive than Sagittarius A*, possibly more massive than any we yet know), the gravity field at the EH is not hugely strong, but it's very deep. There theoretically has to be a possiblity of a BH with an EH circumference of a light hour. So, it isn't possible to get around it in an hour, but going in and out vertically seems theoretically possible, if you wanted to get it all the way out again you'd need a very complicated bucket chain to get fuel down to the vessel that dipped in, but it seems that an automated unintelligent probe could be sacrificeded, if we could find a big enough black hole. Time dilation might make the process a bit slow for observers if that theory is true, but wouldn't something come out eventually?The force of gravity on you depends on the mass of the massive object and your distance from it - the closer you are to the object, the greater the gravitational force it exerts on you. The event horizon is the distance from the center at which gravity is so strong that not even light can escape. There is no material you could make your cable out of that would be strong enough to pull anything, even itself, back out from the event horizon. There is no finite number of rockets you could put on a spaceship that could overcome that force.

Spaghetti-fication happens when gravity is accelerating your feet more quickly than it's accelerating your head (again, assuming a feet-first plunge). When part of you is closer to the massive object than the other, the closer part experiences a stronger tug than the farther part. At 30AU (http://www.onlineconversion.com/length_all.htm) (Neptune orbit, or about 4 light hours) the difference in force felt by your head is in the 13th decimal place from that felt by your feet (assuming you are 6' tall, and your feet are at 30 AU, with your head at 30 AU +6 feet. At 30,000 feet, your feet are feeling a force 1.004 times stronger than your head. At 300 feet, the difference is 1.04 times stronger. At 30 feet it's 1.4 times stronger, and at 3 feet, your feet are being accelerated 9 times faster than your head. But even at 30 AU, if you are on the wrong side of the Event Horizon, there's no going back. You've just got a while before you get stretched.

Does that clear it up?

[halfeye]

The force of gravity on you depends on the mass of the massive object and your distance from it - the closer you are to the object, the greater the gravitational force it exerts on you. The event horizon is the distance from the center at which gravity is so strong that not even light can escape. There is no material you could make your cable out of that would be strong enough to pull anything, even itself, back out from the event horizon.

...

Does that clear it up?
Not for me.

What I am talking about is that the EH (I could be confusing this with the ergosphere?) is the surface immediately above which light would orbit the black hole. If light orbits a black hole, then per degree of orbit, the force on the light must be the same, so when the orbit is longer, which is exactly what must happen with more massive black holes, the force per millimetre of orbit must be less.

I'm not suggesting something could completely escape, I'm saying it could dip in, punch upward again, transmit data, then fall back.

[NichG]

Stable orbits end at 1.5 times the EH radius by the way. The EH is the point where even if you're going directly outwards at c, you still fall in.

[Yora]

The distance at which lightspeed coresponds to a stable orbit is the photonsphere. Which is a considerable distance further from the center of the black hole than the event horizon.

https://sciencevspseudoscience.files.wordpress.com/2011/09/schwarzschild.png

[halfeye]

The distance at which lightspeed coresponds to a stable orbit is the photonsphere. Which is a considerable distance further from the center of the black hole than the event horizon.

https://sciencevspseudoscience.files.wordpress.com/2011/09/schwarzschild.png
Be that as it may, does anyone dispute that this:


If light orbits a black hole, then per degree of orbit, the force on the light must be the same, so when the orbit is longer, which is exactly what must happen with more massive black holes, the force per millimetre of orbit must be less.

is true? if so why and how?

[Sith_Happens]

Be that as it may, does anyone dispute that this:

is true? if so why and how?

I'm going to a take a wild guess that you forgot to take relativistic adjustments to the relevant equations into account.

[Icewraith]

Be that as it may, does anyone dispute that this:



is true? if so why and how?

I think that talking about "the force on the light" is weird and not really the best way to think about it. We use "force" when talking about the gravitational attraction between two objects, usually in a classical sense. That's not valid for light, it would suggest that light has its own gravity- force is an equal and opposite reaction. The orbits of all massive objects are ever-so-slowly decaying. This is GR we're dealing with now, so it's probably better to talk about "the curvature of space".

There's some curvature such that if you shoot a beam of light on the right heading, it will be trapped in a stable orbit around a black hole. The distance from the black hole that reaches that curvature should be determined solely by the mass of the black hole, since light in a vacuum has the same speed for all observers.

[Yora]

Yes, gravity as an attracting force makes sense when you have objects with mass moving at relatively slow speed. You can think of it that way and you will get pretty accurate results when you try to make calculations with it.

But as it has been pointed out, light has no mass and so there isn't anything that gravity could pull on. And the important conclusion that Einstein made is that gravity is not really an attractive force. Instead gravity is a warping of space time. Everything that moves through this warped space in a "straight line" will actually follow the curvature and take a bend path.

(Bonus Quirk: If gravity does not pull but only makes things move on curved paths, why are things being pulled towards each other even when they are not moving relative to each other? While they may not be moving through space, they are still moving through time. And you can never not move through time [unless you move at the speed of light, but that's a whole nother topic]. The pull objects experience between each other is that movement through time along a curved path. Which is why we really have to think of space time as a single thing, not as two separate concepts of space and time.)

[halfeye]

I think that talking about "the force on the light" is weird and not really the best way to think about it. We use "force" when talking about the gravitational attraction between two objects, usually in a classical sense. That's not valid for light, it would suggest that light has its own gravity- force is an equal and opposite reaction. The orbits of all massive objects are ever-so-slowly decaying. This is GR we're dealing with now, so it's probably better to talk about "the curvature of space".

There's some curvature such that if you shoot a beam of light on the right heading, it will be trapped in a stable orbit around a black hole. The distance from the black hole that reaches that curvature should be determined solely by the mass of the black hole, since light in a vacuum has the same speed for all observers.



Yes, gravity as an attracting force makes sense when you have objects with mass moving at relatively slow speed. You can think of it that way and you will get pretty accurate results when you try to make calculations with it.

But as it has been pointed out, light has no mass and so there isn't anything that gravity could pull on. And the important conclusion that Einstein made is that gravity is not really an attractive force. Instead gravity is a warping of space time. Everything that moves through this warped space in a "straight line" will actually follow the curvature and take a bend path.

(Bonus Quirk: If gravity does not pull but only makes things move on curved paths, why are things being pulled towards each other even when they are not moving relative to each other? While they may not be moving through space, they are still moving through time. And you can never not move through time [unless you move at the speed of light, but that's a whole nother topic]. The pull objects experience between each other is that movement through time along a curved path. Which is why we really have to think of space time as a single thing, not as two separate concepts of space and time.)

Pardon me, but both of you are evading the question like politicians :smalltongue: :smallbiggrin:

Supposing that something non-light, say a particle with mass, fresh out of an accelerator, is going round a black hole, at a significant fraction of the speed of light, higher than light's lowest orbit, but in orbit. Would you dispute that:

"If a particle orbits a black hole at a known speed, then per degree of orbit, the [force on/curvature of space faced by] the particle must be the same, so when the orbit is longer, which is exactly what must happen with more massive black holes, the [force/curvature of space] per millimetre of orbit must be less".

is true? if so why and how?

[Yora]

You're giving politicians too much credit. We merely misunderstand the question. :smallbiggrin:


If a particle orbits a black hole at a known speed, then per degree of orbit, the [force on/curvature of space faced by] the particle must be the same

The same as what?

[halfeye]

The same as what?
Geometrically.

The object is in a circular orbit. So the angular change in direction is the same as the angle moved relative to the vertex/singularity.

Draw two points on a circle. The angle between the two lines from the points to the centre of the circle will be the same as the angle between the two tangents to the circle at those points.

[Yora]

On a larger circle the curvature for a given distance is lower than the curvature for the same distance on a smaller circle. Yes.

[halfeye]

On a larger circle the curvature for a given distance is lower than the curvature for the same distance on a smaller circle. Yes.

Good, I'm glad to hear that.

Thinking about it, the following was originally directed to someone else, so maybe you don't have the problem with it that they did.

Does it not therefore make sense, that the attraction/curved space is less at whichever horizon of a more massive black hole, than at a much smaller one?

[Max™]

Worth asking if you're thinking of these orbits as taking place in a flat space/volume/plane.

I've pointed this out before, but what the curvature means in the vicinity of a gravity well is a bit unusual.

Let's say you set up some way to measure a photon orbiting the photosphere of a black hole and determine how long it took, and then plug that in and calculate the radius of the hole.

The value you get would actually be shorter than the real depth (which tends towards infinity as you pass the EH in certain coordinate frames) and that is what type of curvature we're talking about. Excess radius, it being further in that direction *points towards the center of the hole* than those directions *points perpedicularly to the center* and the distortion in local spacetime as a result of that is why bodies follow the paths they do, rather than ones you would expect in a flat Euclidean volume.

http://www.spacetimetravel.org/expeditionsl/expeditionsl.html

Then we also need to ask if this black hole is rotating!

If it is then there is an ergosphere where you become constrained to rotate with the hole, and if you are near the equator there is a prograde and retrograde photon sphere.

[Sith_Happens]

Pardon me, but both of you are evading the question like politicians :smalltongue: :smallbiggrin:

Supposing that something non-light, say a particle with mass, fresh out of an accelerator, is going round a black hole, at a significant fraction of the speed of light, higher than light's lowest orbit, but in orbit. Would you dispute that:

"If a particle orbits a black hole at a known speed, then per degree of orbit, the [force on/curvature of space faced by] the particle must be the same, so when the orbit is longer, which is exactly what must happen with more massive black holes, the [force/curvature of space] per millimetre of orbit must be less".

is true? if so why and how?

That is the exact inverse of how orbit works. A mass at some specific distance r from the center of a black hole will experience a gravitational acceleration given by g=(GM)/r2.* It will be in a stable orbit if and only if it has a tangential velocity equal to that given by v=sqrt(gr).*

* Give or take relativistic adjustments that aren't important for the point I'm trying to make.

[Yora]

Good, I'm glad to hear that.

Does it not therefore make sense, that the attraction/curved space is less at whichever horizon of a more massive black hole, than at a much smaller one?

No. Because the event horizon and the photon sphere are both defined by having a certain strength of gravity. The event horizon is the distance from a singularity at which the gravity is strong enough that there are no more possible paths on which a photon (or anything else) could possibly avoid being pulled into the singularity.
The force of gravity at the event horizon is always the same. This amount of gravity is what makes the event horizon what it is.

[halfeye]

That is the exact inverse of how orbit works. A mass at some specific distance r from the center of a black hole will experience a gravitational acceleration given by g=(GM)/r2.* It will be in a stable orbit if and only if it has a tangential velocity equal to that given by v=sqrt(gr).*

* Give or take relativistic adjustments that aren't important for the point I'm trying to make.

Yeah I was assuming a circular orbit. Light was a nice example because you could set it in a place where the orbit had to be circular, but people had objections to it because it was too light.


No. Because the event horizon and the photon sphere are both defined by having a certain strength of gravity.

That can't be correct for the photon sphere at least. If the photon sphere is the minimum height at which a photon orbits the hole, then for millimetre wide black holes that must be more gravity than for light-year wide ones.

[Max™]

Uh, as I recall if you are below the photon sphere then the slower your orbit the less inertial pull you feel.

Black holes are weird, folks, from a distance they're just a strange dim gravity well (or OH GOD WHY IS EVERYTHING ON FIRE AND RADIOACTIVE WHY DID WE TRY TO SETTLE IN THE PATH OF A QUASAR ARGHHHH) but as you get in close the relativistic geometry stops behaving like you would expect... you know... geometry to be. Was just reading up on the work done recently (well, in the 90's, so more recent than some of the stuff I know) with rotating black holes and photon orbits. As they go through different trajectories you get crazy orbits that look like someone trying to draw the seams on a basketball while drunk, or spiral loops, and all sorts of stuff in between.

With a mm size hole you're going to be wanting to find the nearest Bomb Tech and run whichever way they are.

[Yora]

That can't be correct for the photon sphere at least. If the photon sphere is the minimum height at which a photon orbits the hole, then for millimetre wide black holes that must be more gravity than for light-year wide ones.

Don't quote anything of this in any kind of homework, but wouldn't the force of gravity be the same? It just would be a lot closer to the singularity.

[Max™]

Don't quote anything of this in any kind of homework, but wouldn't the force of gravity be the same? It just would be a lot closer to the singularity.

The g's in acceleration needed to hover at a given distance near the EH increases dramatically with a smaller hole.

With a hole the size of a galaxy cluster you could probably hover near the EH with a 1 g acceleration, with one like Sag A* you'd be getting into the million's I think, and you better be able to push out trillions of g's worth of acceleration to keep a stable distance from a stellar or smaller black hole.

[Yora]

Okay, I see why halfeye finds that confusing.

Me too... :smalleek:

I guess that's also science. Sometimes you talk with people and stop them from not understanding something. And sometimes you talk with people and start realizing you never understood it either.

[Max™]

Okay, I see why halfeye finds that confusing.

Me too... :smalleek:

Don't feel bad, I have fun reading random arxiv links and it's still pretty weird.

[halfeye]

With a mm size hole you're going to be wanting to find the nearest Bomb Tech and run whichever way they are.
Actually, not necessarily.

This book:

http://www.amazon.co.uk/gp/product/0708880142?keywords=jerry%20pournelle%20black%20ho les&qid=1457055753&ref_=sr_1_1&s=books&sr=1-1

It's mostly science fiction, with some science fact, and in one of the factual bits (called "fuzzy black holes have no hair") there's a lovely little table, with the sizes, masses and lifetimes of some theoretical black holes. One with the mass of the Sun would have a 3 kilometer radius and would las forever, one with a one kilogram mass would go blooewy in 10^-19 seconds, and one with the mass of Ceres would have a radius of 1.2 * 10^-4 cm (which even doubled for the diameter is still less than one mm) and would have a lifetime of <huge number> years.

I hope that paper is reprinted somewhere, that table is amazing. The article is probably out of date now, since it was published in 1978, we could do with a revised version from someone else who knows what they're talking about, they could put in theoretical thrust required to hover a mm above the EH and all sorts of other fascinating trivia.

[Erender]

Hey guys, love the discussion, it's very interesting and educating. I'm a layman in physics, but I'd like to suggest the resolution of the confusion as I see it - halfeye is correct in claiming that an orbiting object [/light] around a more massive black hole will experience less gravity [/space curvature] than one orbiting a smaller one. And Yora is correct in saying that the event horizon is defined by an invariant gravity force [/curvature].

However, the photon sphere radius is not defined by a gravity invariant, but rather by a stable orbit speed of c, and gravity force and stable orbit speed scale differently with respect to the distance from the singularity. the schwarzfield radius scales linearly with the BH's mass, and the photon sphere radius does as well (it's always 1.5 times larger, as NichG stated) so linearly with the distance. But gravity decays squarly (sp?), so the pull at the photon sphere is lesser, and shrinks considerably lesser the larger the BH is.

An object orbiting the event horizon will feel a lesser gravitational pull, too, as we scale the BH's size, which is perfectly consistent with the fact that such an object can not possibly exist :p.

[georgie_leech]

Halfeye, look at it this way. Forgetting about relativity for the moment, as that's not actually all that relevent to spaghettification, what would happen if your head was moving downwards at 1 m/s but your feet were moving at 10 m/s? Or at least, a force was trying to make them move at those speeds? You'd feel like you were being stretched, as that's literally what's happenning: a stronger force is acting on your feet then your head, and that's important. On Earth, you are being pulled downwards at more or less an equal strength everywhere, as the difference in height between your head and feet is negligible compared to the distance of the Earth. So you can feel it when someone, say, pulls up on your neck. It doesn't need to be much force before the difference in net forces becomes noticeable. How hard do you think someone would have to pull up on your head before you started being damaged by it? Enough force to lift a 10kg weight? 100kg? 1000kg? 1000000kg?

You don't need Relativity to tell you that the strength of gravity varies with distance. Newton tells us that. Technically, we experience the difference in gravitational force between our feet and head constantly, but the difference is so small that it's not worth even noticing. With most Black holes though, the forces are great enough that even a small proportional difference is a huge shift in the net forces on different parts of you. For the sake of example, imagine a black hole that at the event horizon, your feet were being pulled down with what feels like 1000000kg of force. Further, the difference in distance from your feet and your head means the gravity acting on your head is 1% weaker than that at your feet (I know that's not likely anywhere close to reasonable numbers but it's late and accurate mathematics aren't terribly important to the illustration :smalltongue: ). That difference in forces would be like if you were being pulled apart by a force that could lift 10000kg. How much damage do you think that would do to a person?

But as the mass of a black hole grows, the event horizon gets further and further away from the center. There's nothing in principal mysterious about this. It's the same way that at a certain distance from, say, Jupiter, you will experience similar gravitational attraction as you would on Earth. You'll just be much farther away from the planet. In the same way, the gravitational attraction at the event horizon is the same regardless of distance from the center of mass of the black hole. As the event horizon gets larger, the distance between your feet and head becomes less and less significant. That 1% difference ripped you apart with 10000kg of force, but if the event horizon was enough out that instead of a 1% difference it was 0.0001% weaker on your head? Suddenly the net difference in forces is only 1kg, about the weight of a dictionary. That isn't enough force to do any sort of damage to you, so you could pass the event horizon without adverse effects. With even larger event horizons, you could even pass through it and the difference in forces would be similar to those on earth; you wouldn't notice the difference in force at all.

[Yora]

That's true. But there's still the mystery why you can get very close to a hypermassive black hole and then fly away again with your spaceship, but that same spaceship would be unable to get away from a stelar blackhole when at a similar distance from the event horizon.

Assuming that's actually the case. What is this assumption based on?

[Max™]

That's true. But there's still the mystery why you can get very close to a hypermassive black hole and then fly away again with your spaceship, but that same spaceship would be unable to get away from a stelar blackhole when at a similar distance from the event horizon.

Assuming that's actually the case. What is this assumption based on?

The gravity well isn't as steep that far out, there is still a point where the escape velocity=c, but the gradient above it is such that a much smaller acceleration will lift you out of the well.

If you were the same distance from the geometric center of a stellar mass black hole as a SMBH you wouldn't need much acceleration either.

Case in point, Solar escape velocity is pretty fast, but not THAT fast, if the Sun were replaced with a black hole of the same mass the bodies around would still orbit it the same, it's when you get in closer that you start to fall into the much deeper well with no way out short of burning the mass of a galaxy cluster to accelerate your ship.

If there were an SMBH centered on the Sun, the edges might extend to Mercury, or for the real monsters, past Pluto, and you can have a much lower density within such a large Schwarzschild radius. If you filled the entire solar sytem with water it should collapse into a black hole.

[NichG]

That's true. But there's still the mystery why you can get very close to a hypermassive black hole and then fly away again with your spaceship, but that same spaceship would be unable to get away from a stelar blackhole when at a similar distance from the event horizon.

Assuming that's actually the case. What is this assumption based on?

Roughly speaking:

When you're hovering near the event horizon, most of your acceleration is going towards shifting your proper time rate relative to infinity rather than actually getting you out faster. At the event horizon, this becomes divergent - the apparent distance to the outside is dilating faster than you can travel even if you're a photon.

[Yora]

So, to get away from a larger black hole you need more energy (or force, or whatever; I am not really sure about the exact meaning of these terms), but you all the requied work is spread out over a much larger distance/longer time, so that in any "one moment" the amount of work is much less.
So you still would need more fuel to get away from a larger black hole, but you could do it with less efficient engines?

Similar to how it can be easier to move a larger load than a smaller load, when you have a leaver that is more than just proportionally larger?

[NichG]

So, to get away from a larger black hole you need more energy (or force, or whatever; I am not really sure about the exact meaning of these terms), but you all the requied work is spread out over a much larger distance/longer time, so that in any "one moment" the amount of work is much less.
So you still would need more fuel to get away from a larger black hole, but you could do it with less efficient engines?

Similar to how it can be easier to move a larger load than a smaller load, when you have a leaver that is more than just proportionally larger?

It's less about the fuel you need, and more about the tradeoff between changing the relative rate of passage of time/relative scale of space, versus actually changing your velocity to another point. So when you accelerate really hard towards Alpha Centauri, when you get near c relative to Alpha Centauri most of your acceleration is going towards slowing down your proper time relative to Alpha Centauri/contracting the distance between you and Alpha Centauri, rather than speeding up your motion. So from your point of view, you arrive there sooner because when you accelerated the distance instantaneously shrank, not because you moved faster. That's the weird relativity part of it.

So for the black hole, the issue is that, roughly speaking, space itself is falling into the black hole at some rate. If you just slow down your proper time, that means that you are (from your perspective) actually falling in faster. Your velocity relative to the singularity may be higher, but now space is effectively falling in faster since your rate of proper time passage relative to the fixed point of the singularity has changed. The event horizon is the point where those two effects would precisely balance at c - by accelerating, you appear to be going faster, but at the same time you fall in faster at exactly the rate at which you're speeding up.

That's an effect that cannot exist at all classically, so a picture designed to produce a classical intuition about it is going to be a little incorrect. I'm pretty sure that what I just said isn't mathematically exact, but its the rough picture of why you can't just accelerate a little more and get out. Below the event horizon, I think accelerating outwards will actually make you hit the singularity sooner (from your proper time perspective).

[halfeye]

Halfeye, look at it this way. Forgetting about relativity for the moment, as that's not actually all that relevent to spaghettification, what would happen if your head was moving downwards at 1 m/s but your feet were moving at 10 m/s? Or at least, a force was trying to make them move at those speeds? You'd feel like you were being stretched, as that's literally what's happenning: a stronger force is acting on your feet then your head, and that's important. On Earth, you are being pulled downwards at more or less an equal strength everywhere, as the difference in height between your head and feet is negligible compared to the distance of the Earth. So you can feel it when someone, say, pulls up on your neck. It doesn't need to be much force before the difference in net forces becomes noticeable. How hard do you think someone would have to pull up on your head before you started being damaged by it? Enough force to lift a 10kg weight? 100kg? 1000kg? 1000000kg?

You don't need Relativity to tell you that the strength of gravity varies with distance. Newton tells us that. Technically, we experience the difference in gravitational force between our feet and head constantly, but the difference is so small that it's not worth even noticing. With most Black holes though, the forces are great enough that even a small proportional difference is a huge shift in the net forces on different parts of you. For the sake of example, imagine a black hole that at the event horizon, your feet were being pulled down with what feels like 1000000kg of force. Further, the difference in distance from your feet and your head means the gravity acting on your head is 1% weaker than that at your feet (I know that's not likely anywhere close to reasonable numbers but it's late and accurate mathematics aren't terribly important to the illustration :smalltongue: ). That difference in forces would be like if you were being pulled apart by a force that could lift 10000kg. How much damage do you think that would do to a person?

But as the mass of a black hole grows, the event horizon gets further and further away from the center. There's nothing in principal mysterious about this. It's the same way that at a certain distance from, say, Jupiter, you will experience similar gravitational attraction as you would on Earth. You'll just be much farther away from the planet. In the same way, the gravitational attraction at the event horizon is the same regardless of distance from the center of mass of the black hole. As the event horizon gets larger, the distance between your feet and head becomes less and less significant. That 1% difference ripped you apart with 10000kg of force, but if the event horizon was enough out that instead of a 1% difference it was 0.0001% weaker on your head? Suddenly the net difference in forces is only 1kg, about the weight of a dictionary. That isn't enough force to do any sort of damage to you, so you could pass the event horizon without adverse effects. With even larger event horizons, you could even pass through it and the difference in forces would be similar to those on earth; you wouldn't notice the difference in force at all.

Pardon me, which post of mine, or portion of a post of mine, was that a response to? So called spaghettification is something that I generally accept, though I don't think it's likely that it makes you thinner, just pulls you to bits.

[Max™]

In the vicinity of a steep enough gravity well, particularly around a stellar mass black hole, fermions do actually get stretched out into tube shapes, fun fact.

Less fun if you're experiencing this though.

What NichG was saying is also stated another way by noting that the future inside an event horizon is towards the singularity. All worldlines which cross the EH end at the singularity, so the future is literally something you could point at, right down there, in the deepest part of the hole.

You could move laterally to some extent within the hole, but you will always be moving closer and closer to the center as you do so.

[georgie_leech]

Pardon me, which post of mine, or portion of a post of mine, was that a response to? So called spaghettification is something that I generally accept, though I don't think it's likely that it makes you thinner, just pulls you to bits.

The earlier bits about spaghettification were a convenient lead up to the point that, while unique in how they interact with light, event horizons (and the photon sphere) are just a defined distance away from the singularity that have defined gravitational strengths. Below the Photon sphere, there aren't stable orbits; light trying to orbit a black hole will steadily fall in, but light moving more directly away from the black hole will be red shifted but can escape. Behind the event horizon, there are no paths leading away from the black hole due to the degree of curvature. Even if on a sufficiently massive back hole the tidal forces wouldn't kill you, the strength of the gravitational field remains strong enough that nothing can escape.

I thought that the confusion about whether you could escape the event horizon was based on the idea that a large enough black hole wouldn't destroy you as you pass through it. I misunderstood your point in that regard. Apologies.

[Bulldog Psion]

So with spaghettification, would you end up being disintegrated from the feet up as it occurred, or end up temporarily in some bizarre Salvador Dali state, stretched out literally like a spaghetti strand but alive, intact, and functional to experience the horror? :smalleek:

[georgie_leech]

So with spaghettification, would you end up being disintegrated from the feet up as it occurred, or end up temporarily in some bizarre Salvador Dali state, stretched out literally like a spaghetti strand but alive, intact, and functional to experience the horror? :smalleek:

Unless you somehow teleported next to the black hole, almost certainly neither. You'd be ripped apart like you were attached to one of those torture racks and Hercules gave it a spin.

[Bulldog Psion]

Unless you somehow teleported next to the black hole, almost certainly neither. You'd be ripped apart like you were attached to one of those torture racks and Hercules gave it a spin.

Okay, so it's effectively a force being exerted on the body, rather than a spatial effect. Thanks, that answers my question.

So in other words, as you got closer, it would start pulling pieces of flesh off you, then reach some threshold where, as you say, the strain would overcome the ability of your body to hold together and rip you apart at the joints. Bad way to go. :smalleek:

In a way, it would be worse than the Herakles-and-rack thing, because the force is being applied at all points, not just at the wrists and ankles; in other words, the force would pull apart at every point in between, too, so it would be an endless series of successively more "detailed" "material failures" in fairly rapid succession, rather than just one big rip, until the person was reduced to mist, and then that vanished into a cloud of atoms.

Sorry for the pseudo-graphic description, but I was curious about whether our hypothetical observer/victim would experience bizarre spatial lengthening, or brutal physical destruction. Brutal physical destruction it is, evidently.

(EDIT: So "spaghettification" is a really inapplicable term. "Successive shredding" is more like it.)

[georgie_leech]

Okay, so it's effectively a force being exerted on the body, rather than a spatial effect. Thanks, that answers my question.

So in other words, as you got closer, it would start pulling pieces of flesh off you, then reach some threshold where, as you say, the strain would overcome the ability of your body to hold together and rip you apart at the joints. Bad way to go. :smalleek:

In a way, it would be worse than the Herakles-and-rack thing, because the force is being applied at all points, not just at the wrists and ankles; in other words, the force would pull apart at every point in between, too, so it would be an endless series of successively more "detailed" "material failures" in fairly rapid succession, rather than just one big rip, until the person was reduced to mist, and then that vanished into a cloud of atoms.

Sorry for the pseudo-graphic description, but I was curious about whether our hypothetical observer/victim would experience bizarre spatial lengthening, or brutal physical destruction. Brutal physical destruction it is, evidently.

(EDIT: So "spaghettification" is a really inapplicable term. "Successive shredding" is more like it.)

At least as I understand it yeah. I think the idea behind the term is that it's a graphic way of describing how strong the forces are without actually describing people being pulled apart.

Mind you, I don't know of anyone who's ever passed through an event horizon, so you never know :smallwink:

[Yora]

Though I think you probably pass out long before your body starts to disintegrate. The first effect from gravity being unequal on your forward and backward facing parts of your body should be that you become tidally locked. All the fluid in your body will be as well, preventing blood circulation and you pass out. Blood circulation would be severely impeded even before it fully stops and that alone would get you a good time earlier.

Though I am not sure of what scale of duration we would really be talking about here. I wouldn't be surprised if the whole stretching process would take only a fraction of a second.