4 ms·

> Once they cross the event horizon, they are moving at the speed of light toward certain doom. Can that be correct? I thought the falling man experienced no c

> Once they cross the event horizon, they are moving at the speed of light toward certain doom. Can that be correct? I thought the falling man experienced no change on crossing the event horizon (other than the effects of spaghettification, which begins long before arriving at the horizon). I thought that you'd continue accellerating toward the singularity, asymptotically approaching both the speed of light and infinite inertial mass. Am I wrong?

pdonis 3y ago
> Can that be correct? The "towards certain doom" part is correct. The "at the speed of light" is not. Unfortunately this kind of thing is typical for pop science articles. > I thought the falling man experienced no change on crossing the event horizon 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. > (other than the effects of spaghettification, which begins long before arriving at the horizon).* Not for the typical black holes we observe in the universe. Tidal gravity at their horizons is too small for spaghettification; that doesn't happen until the infaller is well inside and approaching the singularity. > I thought that you'd continue accellerating toward the singularity, asymptotically approaching both the speed of light and infinite inertial mass. Am I wrong?* Yes. None of these things are true. Your acceleration is zero--you're in free fall. There is no well-defined notion of "speed" at all, but your worldline remains timelike. Your inertial mass is unchanged. And, finally, the singularity is not a place. It's a moment of time. You can't "accelerate towards" it any more than you can accelerate towards next week.
- denton-scratch 3y ago
  > 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.
  - ninkendo 3y ago
    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.
    - smallnamespace 3y ago
      > 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.
      - pdonis 3y ago
        > 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.
- smallnamespace 3y ago
  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'.
  - pdonis 3y ago
    > 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.
  - ithkuil 3y ago
    From the point of view of somebody outside the black hole the infaller never reaches the event horizon.
    - pdonis 3y ago
      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.
- ryanSrich 3y ago
  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?
  - jleahy 3y ago
    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.
    - ben_w 3y ago
      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.
  - russdill 3y ago
    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.
  - didgeoridoo 3y ago
    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.
- FrustratedMonky 3y ago
  Typically I thought Quanta magazine did better than typical pop-science. Guess this was an exception?
  - pdonis 3y ago
    Unfortunately not; my experience is that Quanta is just as prone to these kinds of errors as other pop science sites.
- beebeepka 3y ago
  > 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.
- denton-scratch 3y ago
  > Not for the typical black holes we observe in the universe. Ah. Are the typical black holes too small, or too big?
  - wiml 3y ago
    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.
  - pdonis 3y ago
    Too big.
- cvoss 3y ago
  > 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?
  - DavidSJ 3y ago
    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.
    - pdonis 3y ago
      > I think Bob would only see Alice's (N+1)st photon after he himself has crossed the event horizon. Yes, this is correct.
  - pdonis 3y ago
    > 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.
- gigatexal 3y ago
  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!
deleted 3y ago
[deleted]
downWidOutaFite 3y ago
No reason you would continue accelerating. Seems like you would would reach a stable speed while orbiting inside the event horizon.
- vikingerik 3y ago
  No - this doesn't happen, because the speed required for a stable orbit would be greater than the speed of light.
- renewiltord 3y ago
  The definition of the EH is that no such orbit exists within.
samstave 3y ago
When I was in 7th grade we were learning about Black Holes - and the movie "Explorers" had come out where the kids are able to make a forcefield sphere surrounding an abandoned car from a carnival ride/roller coaster --- and they fly into space and meet aliens and stuff... Welp - 7th grade me postulated that if you had one of those force spheres, and you could accelerate at the same rate or slightly faster than the speed at which matter is pulled into a black hole, could you stay ahead of the gravity wave and "surf" it safely into the black hole? https://www.youtube.com/watch?v=al75vXD6koc https://www.youtube.com/watch?v=al75vXD6koc -- Trailer https://www.youtube.com/watch?v=RqQKGxrfwsk https://www.youtube.com/watch?v=RqQKGxrfwsk -- Trivia
hilbert42 3y ago
"> Once they cross the event horizon, they are moving at the speed of light toward certain doom." I claim no expertise about the internals of black holes but that comment also drew my attention as questionable. First point, are we talking about a physical man comprising of atoms or the aftermath of spaghettification when much of him would have turned into EM radiation? Physical matter would be subject to Relativity therefore his v < c whereas the parts of him that had turned into radiation would actually travel at c. That said, the point about astronauts moving at c speed to their doom and the article's thrust about a second law of quantum complexity leads me to ask what is c inside a black hole and to conjecture it's likely very different to that in vacuo. c in vacuum is one of the well defined physical constants but it's also directly related to other physical constants such as vacuum permittivity, ε0, and vacuum permeability, μ0, by the relationship c=1/(ε0μ0)^½. The inside of a black hole is vastly different to the vacuum of free space, not only is gravity vastly contorting spacetime but the permittivity and permeability would no longer have the values they do in free space. In fact, we'd have to treat them as relative permittivity and permeability and given that black holes contain 'stuff'—which is more than does a vacuum. Light therefore, would travel more slowly than in a vacuum. The value of the relative permittivity and permeability would likely be determined by the size of the black hole. This leaves the possibility the remains of the astronauts could actually travel faster than the speed light as in the case of Cherenkov radiation. In effect, their velocity in the 'dialectic' of 'black hole volume' could exceed the phase velocity of light through that medium. (Of course, I'm just extrapolating physics as we understand it from outside a black hole, those more familiar with the subject may wish to comment.) This article also allows one to conjecture the possibility that the values of physical constants such as c may actually arise either directly from quantum information/"complexity equilibrium" sans heat or are emergent phenomena therefrom. With ε0, μ0 and c the 'quantum vacuum' providing a minimum 'circuit complexity' thus c obtains its maximum value in this universe.

pdonis 3y ago

> Can that be correct? The "towards certain doom" part is correct. The "at the speed of light" is not. Unfortunately this kind of thing is typical for pop science articles. > I thought the falling man experienced no change on crossing the event horizon 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. > (other than the effects of spaghettification, which begins long before arriving at the horizon).* Not for the typical black holes we observe in the universe. Tidal gravity at their horizons is too small for spaghettification; that doesn't happen until the infaller is well inside and approaching the singularity. > I thought that you'd continue accellerating toward the singularity, asymptotically approaching both the speed of light and infinite inertial mass. Am I wrong?* Yes. None of these things are true. Your acceleration is zero--you're in free fall. There is no well-defined notion of "speed" at all, but your worldline remains timelike. Your inertial mass is unchanged. And, finally, the singularity is not a place. It's a moment of time. You can't "accelerate towards" it any more than you can accelerate towards next week.

denton-scratch 3y ago
> 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.
- ninkendo 3y ago
  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.
  - smallnamespace 3y ago
    > 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.
    - pdonis 3y ago
      > 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.
smallnamespace 3y ago
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'.
- pdonis 3y ago
  > 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.
- ithkuil 3y ago
  From the point of view of somebody outside the black hole the infaller never reaches the event horizon.
  - pdonis 3y ago
    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.
ryanSrich 3y ago
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?
- jleahy 3y ago
  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.
  - ben_w 3y ago
    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.
- russdill 3y ago
  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.
- didgeoridoo 3y ago
  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.
FrustratedMonky 3y ago
Typically I thought Quanta magazine did better than typical pop-science. Guess this was an exception?
- pdonis 3y ago
  Unfortunately not; my experience is that Quanta is just as prone to these kinds of errors as other pop science sites.
beebeepka 3y ago
> 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.
denton-scratch 3y ago
> Not for the typical black holes we observe in the universe. Ah. Are the typical black holes too small, or too big?
- wiml 3y ago
  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.
- pdonis 3y ago
  Too big.
cvoss 3y ago
> 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?
- DavidSJ 3y ago
  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.
  - pdonis 3y ago
    > I think Bob would only see Alice's (N+1)st photon after he himself has crossed the event horizon. Yes, this is correct.
- pdonis 3y ago
  > 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.
gigatexal 3y ago
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!

deleted 3y ago

[deleted]

downWidOutaFite 3y ago

No reason you would continue accelerating. Seems like you would would reach a stable speed while orbiting inside the event horizon.

vikingerik 3y ago
No - this doesn't happen, because the speed required for a stable orbit would be greater than the speed of light.
renewiltord 3y ago
The definition of the EH is that no such orbit exists within.

samstave 3y ago

When I was in 7th grade we were learning about Black Holes - and the movie "Explorers" had come out where the kids are able to make a forcefield sphere surrounding an abandoned car from a carnival ride/roller coaster --- and they fly into space and meet aliens and stuff... Welp - 7th grade me postulated that if you had one of those force spheres, and you could accelerate at the same rate or slightly faster than the speed at which matter is pulled into a black hole, could you stay ahead of the gravity wave and "surf" it safely into the black hole? https://www.youtube.com/watch?v=al75vXD6koc https://www.youtube.com/watch?v=al75vXD6koc -- Trailer https://www.youtube.com/watch?v=RqQKGxrfwsk https://www.youtube.com/watch?v=RqQKGxrfwsk -- Trivia

hilbert42 3y ago

"> Once they cross the event horizon, they are moving at the speed of light toward certain doom." I claim no expertise about the internals of black holes but that comment also drew my attention as questionable. First point, are we talking about a physical man comprising of atoms or the aftermath of spaghettification when much of him would have turned into EM radiation? Physical matter would be subject to Relativity therefore his v < c whereas the parts of him that had turned into radiation would actually travel at c. That said, the point about astronauts moving at c speed to their doom and the article's thrust about a second law of quantum complexity leads me to ask what is c inside a black hole and to conjecture it's likely very different to that in vacuo. c in vacuum is one of the well defined physical constants but it's also directly related to other physical constants such as vacuum permittivity, ε0, and vacuum permeability, μ0, by the relationship c=1/(ε0μ0)^½. The inside of a black hole is vastly different to the vacuum of free space, not only is gravity vastly contorting spacetime but the permittivity and permeability would no longer have the values they do in free space. In fact, we'd have to treat them as relative permittivity and permeability and given that black holes contain 'stuff'—which is more than does a vacuum. Light therefore, would travel more slowly than in a vacuum. The value of the relative permittivity and permeability would likely be determined by the size of the black hole. This leaves the possibility the remains of the astronauts could actually travel faster than the speed light as in the case of Cherenkov radiation. In effect, their velocity in the 'dialectic' of 'black hole volume' could exceed the phase velocity of light through that medium. (Of course, I'm just extrapolating physics as we understand it from outside a black hole, those more familiar with the subject may wish to comment.) This article also allows one to conjecture the possibility that the values of physical constants such as c may actually arise either directly from quantum information/"complexity equilibrium" sans heat or are emergent phenomena therefrom. With ε0, μ0 and c the 'quantum vacuum' providing a minimum 'circuit complexity' thus c obtains its maximum value in this universe.