5 ms·

> 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 scie

by 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.
      - 867-5309 3y ago
        all this talk of GR and invariants, not to confuse a mathematical singularity with a gravitational one, at which point GR itself breaks down per the need for a quantum mechanical description. the truth is we don't know what happens within a black hole, and probably never will
        pdonis 3y ago
        > GR itself breaks down per the need for a quantum mechanical description This is the current predominant belief among physicists, yes, but that doesn't mean it's established fact. > the truth is we don't know what happens within a black hole, and probably never will This is way too pessimistic. The predictions of GR for this regime are precise and unequivocal. Even if you believe GR breaks down when spacetime curvatures reach the Planck scale (which is the current predominant belief, as I said above), that still leaves most of the black hole's interior within GR's domain of validity.
- akomtu 3y ago
  Falling in a black hole is a different experience. It must be more like being squeezed in a narrow tube with a glowing light that was once the universe on one end of the tube and something indescribable on the other end. Perception of time will change too, as all the particles that let us perceive the time will be messed up in strange ways. Perhaps time will seem multidimensional or there will be many timelines. In other words, your perception of a fall from an aeroplane isn't true or absolute, it's mediated by particles that define, for you, space and time, and if those particles get messed up, your reality will change completely.
  - pdonis 3y ago
    None of what you say is correct as a description of what General Relativity says will happen to someone who falls into a black hole.
  - denton-scratch 3y ago
    I think the "tube" business concerns what you see a you approach the speed of light. All the light/future events are in front of you, and you see nothing if you stand on the stern of your ship. I'm not sure how that relates to falling into a BH, bit I don't know; there's a lot of this that I'm struggling to understand. I'm not a physicist, and not clever, just very interested.
    - pdonis 3y ago
      > I'm not sure how that relates to falling into a BH The special relativistic effect you describe has nothing whatever to do with falling into a BH.
- pdonis 3y ago
  > if I jump out of an aeroplane, I don't accelerate towards the ground? Not in your rest frame. In your rest frame, the ground accelerates towards you. However, the acceleration described in both frames (yours and the Earth's) is coordinate acceleration, which is frame-dependent, and in relativity frame-dependent quantities don't have any direct physical meaning. The quantity that has physical meaning in terms of acceleration is proper acceleration--what an accelerometer strapped to you would read--and in free fall that is zero.
- AdamH12113 3y ago
  > So if I jump out of an aeroplane, I don't accelerate towards the ground? In a very real sense (although not a useful one for a skydiver), the answer is yes. This is a key insight of the Equivalence Principle and the curved-spacetime view of gravity. You (in free-fall) are on an inertial "straight line" path through spacetime towards the center of the Earth. When you reach the ground, it pushes on you, messily accelerating you off of that path.
- rocqua 3y ago
  In some sense, the ground is accelerating towards you. By jumping out of the plane, you've stopped the plane floor from accelrating you away from the ground at the same rate the ground is accelerating towards you. This is barely useful for humans since we are very used to 'standing still' actually involving the force of your weight, so we consider that distinct from acceleration. But physics gets much easier if you don't make that distinction.
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!

by pdonis 3y ago

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.
    - 867-5309 3y ago
      all this talk of GR and invariants, not to confuse a mathematical singularity with a gravitational one, at which point GR itself breaks down per the need for a quantum mechanical description. the truth is we don't know what happens within a black hole, and probably never will
      - pdonis 3y ago
        > GR itself breaks down per the need for a quantum mechanical description This is the current predominant belief among physicists, yes, but that doesn't mean it's established fact. > the truth is we don't know what happens within a black hole, and probably never will This is way too pessimistic. The predictions of GR for this regime are precise and unequivocal. Even if you believe GR breaks down when spacetime curvatures reach the Planck scale (which is the current predominant belief, as I said above), that still leaves most of the black hole's interior within GR's domain of validity.
akomtu 3y ago
Falling in a black hole is a different experience. It must be more like being squeezed in a narrow tube with a glowing light that was once the universe on one end of the tube and something indescribable on the other end. Perception of time will change too, as all the particles that let us perceive the time will be messed up in strange ways. Perhaps time will seem multidimensional or there will be many timelines. In other words, your perception of a fall from an aeroplane isn't true or absolute, it's mediated by particles that define, for you, space and time, and if those particles get messed up, your reality will change completely.
- pdonis 3y ago
  None of what you say is correct as a description of what General Relativity says will happen to someone who falls into a black hole.
- denton-scratch 3y ago
  I think the "tube" business concerns what you see a you approach the speed of light. All the light/future events are in front of you, and you see nothing if you stand on the stern of your ship. I'm not sure how that relates to falling into a BH, bit I don't know; there's a lot of this that I'm struggling to understand. I'm not a physicist, and not clever, just very interested.
  - pdonis 3y ago
    > I'm not sure how that relates to falling into a BH The special relativistic effect you describe has nothing whatever to do with falling into a BH.
pdonis 3y ago
> if I jump out of an aeroplane, I don't accelerate towards the ground? Not in your rest frame. In your rest frame, the ground accelerates towards you. However, the acceleration described in both frames (yours and the Earth's) is coordinate acceleration, which is frame-dependent, and in relativity frame-dependent quantities don't have any direct physical meaning. The quantity that has physical meaning in terms of acceleration is proper acceleration--what an accelerometer strapped to you would read--and in free fall that is zero.
AdamH12113 3y ago
> So if I jump out of an aeroplane, I don't accelerate towards the ground? In a very real sense (although not a useful one for a skydiver), the answer is yes. This is a key insight of the Equivalence Principle and the curved-spacetime view of gravity. You (in free-fall) are on an inertial "straight line" path through spacetime towards the center of the Earth. When you reach the ground, it pushes on you, messily accelerating you off of that path.
rocqua 3y ago
In some sense, the ground is accelerating towards you. By jumping out of the plane, you've stopped the plane floor from accelrating you away from the ground at the same rate the ground is accelerating towards you. This is barely useful for humans since we are very used to 'standing still' actually involving the force of your weight, so we consider that distinct from acceleration. But physics gets much easier if you don't make that distinction.

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!