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It’s frustrating as the text does not go anywhere near demonstrating any “simple violation of determinism in Newtonian mechanics”. The core issue of the discuss

It’s frustrating as the text does not go anywhere near demonstrating any “simple violation of determinism in Newtonian mechanics”. The core issue of the discussion linked in the story is that there are mathematical solutions to Newton’s second law that are inconsistent with other bits of classical physics (in this case that a particle at rest in a given Galilean frame of reference cannot just start moving without something else applying some force to it). That is entirely true, interesting, and there is nothing wrong with this. What this does not tell us is how it somehow violates any kind of determinism. > Then there are many solutions of Newton’s law F = ma. In one the ball remains at rest on top of the dome. But in others, it starts to roll down the dome in some arbitrary direction! Moreover it can start rolling at any time. That is just not going to happen in (classical) reality, though. Because once you properly set the initial state of the ball (force=velocity=0, or any other values), then the solution becomes unique and that’s it, there is one possible trajectory. The ball starting to move without anything acting on it would violate other principles of classical mechanics. It’s not going to happen regardless of whether that trajectory is consistent with Newton’s second law.

SlySherZ 3y ago
The article is hard to follow for me, but if I understood it correctly, this is not true: > That is just not going to happen in (classical) reality, though. Because once you properly set the initial state of the ball (force=velocity=0, or any other values), then the solution becomes unique If you set velocity = velocity = 0, then the ball staying at the top is a valid solution, AND the ball rolling down the hill (in any direction) is also a valid solution. If this sounds confusing (it did for me), look at the example at the end, it's possible to do the reverse - send the ball rolling up the hill with perfect velocity, such that it stops at the very top after time T. And if that is possible, the opposite is also possible because NM is time reversible.
- lisper 3y ago
  > If you set velocity = velocity = 0, then the ball staying at the top is a valid solution, AND the ball rolling down the hill (in any direction) is also a valid solution. Yes, that is exactly right. Not only in any direction, but beginning at any time. The easiest way to see this is described at the end: imagine the ball is initially in motion and the initial conditions are precisely those that bring it precisely to rest at the apex of the dome at some time T. (Making this possible is the reason the dome has to be a specific shape. Not all shapes allow this.) The time-reversal of this motion is the ball beginning to move in some arbitrary direction at some arbitrary time.
  - eesmith 3y ago
    > The time-reversal of this motion ... at some arbitrary time. The "ball rolling to the top of the sphere" requires infinite time. "Some arbitrary time" is an expression of a finite time. You cannot simple mix ideas of finite and infinite and have the result make sense, as anyone who has stayed at the Hilbert Hotel knows. https://en.wikipedia.org/wiki/Hilbert's_paradox_of_the_Grand_Hotel https://en.wikipedia.org/wiki/Hilbert's_paradox_of_the_Grand...
    - xyzzyz 3y ago
      The article explicitly discusses the fact that this is possible specifically because of the shape of the dome, and does not work on a hemisphere, precisely for reason you bring up.
      - kergonath 3y ago
        The article explicitly discusses this without demonstrating anything. On it’s face this argument has the weight of these demonstrations.
      - eesmith 3y ago
        You are right - I misread it. The next step would be to verify that the paths always stay on the surface. The mathematics shown says the point always follows the surface, but I don't see a demonstration that that's true. I no longer have the skills to easily do this calculation. EDIT: Oh man, I used to be a lot better at this. I remember the mgh = 1/2 m v^2 and the slope calculation, but can't figure out how tell when the falling point mass detaches from the slope. If it detaches at h=0 then there's no physically viable reversed path on the surface.
    - crazygringo 3y ago
      > The "ball rolling to the top of the sphere" requires infinite time. No it doesn't, because it's not a sphere. The dome is specifically designed so that it takes finite time. There's zero involvement of infinity, or mixing infinity, here.
- kergonath 3y ago
  > The article is hard to follow for me, but if I understood it correctly, this is not true You are right, I was missing some conditions. The higher order derivatives need to be zero as well. > If you set velocity = velocity = 0, then the ball staying at the top is a valid solution, AND the ball rolling down the hill (in any direction) is also a valid solution. It is a valid solution to the f=ma equation. It is not a valid trajectory in Newtonian physics because it violates other principles. It is a “gotcha” only if you think that Newton’s second law is the entirety of classical mechanics. > If this sounds confusing (it did for me), look at the example at the end, it's possible to do the reverse - send the ball rolling up the hill with perfect velocity, such that it stops at the very top after time T. This paragraph is confusing. And does not demonstrate much of anything, instead asserting facts that we are supposed to believe. In the time-reversal “experiment”, where the particle comes from the rim towards the apex, it ends up at the apex with a non-zero fourth derivative, because of the pathological shape of the dome. It cannot stay on the apex for any length of time, even with a velocity of 0. It is completely different from a particle starting at rest on the apex. > And if that is possible, the opposite is also possible because NM is time reversible. It is not.
  - sgregnt 3y ago
    > It is a valid solution to the f=ma equation. It is not a valid trajectory in Newtonian physics because it violates other principles. It is a “gotcha” only if you think that Newton’s second law is the entirety of classical mechanics Could you please elaborate which Newtonian principles it does violate?
    - kergonath 3y ago
      The simplest one is that a particle on its own keeps a linear trajectory with a constant speed. A change in that (like going from rest to any motion) requires interacting with another particle: things do not start moving for no reason. This is a generalisation of one of the formulations of Newton’s first law, which states that things that don’t move don’t start moving without being pushed (rough translation). This is related to another formulation of Newton’s first law: if there is a force that pushes the ball at some time T, it implies that there is another body that felt the opposite force. Another one is a bit more involved, but basically a mechanical system cannot change its symmetry by itself. In this case, the initial state with a ball at rest has a radial symmetry with a centre on the apex of the dome. This is not true anymore if the ball moves in one direction. This is related to the conservation of momentum. There are a couple of points that can be solved easily, but are clearly defects in the original formulation of the problem. for example, the height according to the equations is not a length, which is not a problem itself (we can just multiply by an arbitrary factor with the right dimensions) but an indication of sloppy thinking and hand waving. Similarly, the force is not bounded in the original formulation. Again, this can be fixed by restricting the valid range for r, but is rather messy.
lisper 3y ago
> once you properly set the initial state of the ball (force=velocity=0, or any other values), then the solution becomes unique and that’s it No, you've missed the point entirely. There are circumstances under which the solution is not unique, and the article describes such a circumstance.
- deleted 3y ago
  [deleted]
azeemba 3y ago
> Because once you properly set the initial state of the ball (force=velocity=0, or any other values), then the solution becomes unique and that’s it, The original tumblr post linked to this wikipedia article: https://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem#Example_of_non-uniqueness https://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_t... The initial conditions here are _not_ sufficient to pin to a unique solution. > anything acting on it would violate other principles of classical mechanics What other principles of classical mechanics does it violate though? I think that's the point of the exercise. To reflect on part of our mathematical modeling would prevent this.
tsimionescu 3y ago
> Because once you properly set the initial state of the ball (force=velocity=0, or any other values), then the solution becomes unique and that’s it, there is one possible trajectory. As the linked article points out, that is not true. Even for force = velocity = 0, there are solutions for which the ball starts rolling. Additionally, even adding the first law does not help, since the body has F = v = a = 0 at any time before T, and it has a > 0 only for times where it also has F > 0. So the body spontaneously starting to move is perfectly consistent with classical mechanics. A better attack on the example is that the shape of the dome may not be possible to construct from actual matter. The math only works with a shape that has a certain type of infinite smoothness with a single bump, which is probably not possible even in principle to construct from real matter.
sobellian 3y ago
> That is just not going to happen in (classical) reality, though. Because once you properly set the initial state of the ball (force=velocity=0, or any other values), then the solution becomes unique and that’s it, there is one possible trajectory. Not true, and the trick is very simple, you just contrive a force field that admits x''(0) = 0 but some non-zero higher derivative. Indeed, if you look at the proposed solution r = 1/144 (t - T)^4 for t > T, you'll see that: r(T) = 0 r'(T) = 0 r''(T) = 0 r'''(T) = 0 r''''(T) is ill-defined (but Newtonian mechanics AFAIK says nothing to forbid this)
- pistachiopro 3y ago
  Right, so for his nondeterministic path: r(t) = 0 ; t <= T r(t) = (1/144)(t-T)^4 ; t >= T We can see that r''''(T) is either 0 or 1/6, depending on if we go with the top or bottom equation. That does look like some sort of hidden state change, there. Interestingly, the spherical dome he mentioned (which doesn't yield to this non-determinism) forces all derivatives of r(t) to be continuous ...
- tkoolen 3y ago
  > but Newtonian mechanics AFAIK says nothing to forbid this In the Wikipedia article on Newton's laws of motion, the first law is stated as "A body remains at rest, or in motion at a constant speed in a straight line, unless acted upon by a force.". Here it would seem that we leave the state of rest not due to a force, but due to some other cause, which the first law would forbid. So I think that the particular interpretation of Newtonian mechanics used by the author is a bit of a strawman.
  - sobellian 3y ago
    We must ask ourselves what "at rest" means since there are multiple trajectories that give x(0) = 0, x'(0) = 0 and it seems that some posters believe this cannot fully constrain the particle to be either at rest or in motion at t=0. There are multiple ways to resolve this dissonance. We could demand that the higher derivatives are also zero. We could derive some elaborate rule excluding non-zero x''(t) in the neighborhood of t=0. Or something else altogether. The issue IMO with these resolutions is that they're quite complicated and Newton almost certainly did not have any of them in mind. It's much simpler to just content ourselves that NFL is a special case of F=ma (where F=0). I'm not sure why we should contort ourselves to preserve determinism since we know that gets thrown out the window with QM anyway.
H8crilA 3y ago
This is a (quite clever) comment on a physics theory that we know to be incomplete. Not a comment on reality. And the math does check out. There really are many solutions to this system, i.e. it is nondeterministic under the theory.
WindyLakeReturn 3y ago
>The ball starting to move without anything acting on it would violate other principles of classical mechanics. Isn't gravity acting on the ball? Ideally a ball balanced a the tip of a bowl has 0 net force, but only because it has the force of gravity pulling it into the bowl and an equal force from the bowl pushing back. This would require assuming a reality where there is no smallest particle, no atoms making up the bowl or ball as they are both perfect mathematical objects comprised of infinite points no matter what resolution you look at them at. The issue is that such a system, if it can be created by rolling the ball up the bowl with perfect precision from any direction, indicates that it is unstable and may reverse at any point, but without any obvious reason causing it to do so. So either the system has some non-deterministic factor which allows for the perfect stability to break at an arbitrary time in an arbitrary direction or it isn't time reversible as the ability to roll the ball up the bowl cannot be reversed. Looking at it another way, can you tell the difference between a ball I perfectly rolled to the top of the bowl 10 years ago and one I did 10 seconds ago? I do wonder if we've added so many assumptions, with mathematically perfect objects and infinite precision forces that we have created some sort of paradox. We have hit levels of perfectly spherical cows that cause even the perfectly spherical cows to complain about unrealistic standards.

SlySherZ 3y ago

The article is hard to follow for me, but if I understood it correctly, this is not true: > That is just not going to happen in (classical) reality, though. Because once you properly set the initial state of the ball (force=velocity=0, or any other values), then the solution becomes unique If you set velocity = velocity = 0, then the ball staying at the top is a valid solution, AND the ball rolling down the hill (in any direction) is also a valid solution. If this sounds confusing (it did for me), look at the example at the end, it's possible to do the reverse - send the ball rolling up the hill with perfect velocity, such that it stops at the very top after time T. And if that is possible, the opposite is also possible because NM is time reversible.

lisper 3y ago
> If you set velocity = velocity = 0, then the ball staying at the top is a valid solution, AND the ball rolling down the hill (in any direction) is also a valid solution. Yes, that is exactly right. Not only in any direction, but beginning at any time. The easiest way to see this is described at the end: imagine the ball is initially in motion and the initial conditions are precisely those that bring it precisely to rest at the apex of the dome at some time T. (Making this possible is the reason the dome has to be a specific shape. Not all shapes allow this.) The time-reversal of this motion is the ball beginning to move in some arbitrary direction at some arbitrary time.
- eesmith 3y ago
  > The time-reversal of this motion ... at some arbitrary time. The "ball rolling to the top of the sphere" requires infinite time. "Some arbitrary time" is an expression of a finite time. You cannot simple mix ideas of finite and infinite and have the result make sense, as anyone who has stayed at the Hilbert Hotel knows. https://en.wikipedia.org/wiki/Hilbert's_paradox_of_the_Grand_Hotel https://en.wikipedia.org/wiki/Hilbert's_paradox_of_the_Grand...
  - xyzzyz 3y ago
    The article explicitly discusses the fact that this is possible specifically because of the shape of the dome, and does not work on a hemisphere, precisely for reason you bring up.
    - kergonath 3y ago
      The article explicitly discusses this without demonstrating anything. On it’s face this argument has the weight of these demonstrations.
    - eesmith 3y ago
      You are right - I misread it. The next step would be to verify that the paths always stay on the surface. The mathematics shown says the point always follows the surface, but I don't see a demonstration that that's true. I no longer have the skills to easily do this calculation. EDIT: Oh man, I used to be a lot better at this. I remember the mgh = 1/2 m v^2 and the slope calculation, but can't figure out how tell when the falling point mass detaches from the slope. If it detaches at h=0 then there's no physically viable reversed path on the surface.
  - crazygringo 3y ago
    > The "ball rolling to the top of the sphere" requires infinite time. No it doesn't, because it's not a sphere. The dome is specifically designed so that it takes finite time. There's zero involvement of infinity, or mixing infinity, here.
kergonath 3y ago
> The article is hard to follow for me, but if I understood it correctly, this is not true You are right, I was missing some conditions. The higher order derivatives need to be zero as well. > If you set velocity = velocity = 0, then the ball staying at the top is a valid solution, AND the ball rolling down the hill (in any direction) is also a valid solution. It is a valid solution to the f=ma equation. It is not a valid trajectory in Newtonian physics because it violates other principles. It is a “gotcha” only if you think that Newton’s second law is the entirety of classical mechanics. > If this sounds confusing (it did for me), look at the example at the end, it's possible to do the reverse - send the ball rolling up the hill with perfect velocity, such that it stops at the very top after time T. This paragraph is confusing. And does not demonstrate much of anything, instead asserting facts that we are supposed to believe. In the time-reversal “experiment”, where the particle comes from the rim towards the apex, it ends up at the apex with a non-zero fourth derivative, because of the pathological shape of the dome. It cannot stay on the apex for any length of time, even with a velocity of 0. It is completely different from a particle starting at rest on the apex. > And if that is possible, the opposite is also possible because NM is time reversible. It is not.
- sgregnt 3y ago
  > It is a valid solution to the f=ma equation. It is not a valid trajectory in Newtonian physics because it violates other principles. It is a “gotcha” only if you think that Newton’s second law is the entirety of classical mechanics Could you please elaborate which Newtonian principles it does violate?
  - kergonath 3y ago
    The simplest one is that a particle on its own keeps a linear trajectory with a constant speed. A change in that (like going from rest to any motion) requires interacting with another particle: things do not start moving for no reason. This is a generalisation of one of the formulations of Newton’s first law, which states that things that don’t move don’t start moving without being pushed (rough translation). This is related to another formulation of Newton’s first law: if there is a force that pushes the ball at some time T, it implies that there is another body that felt the opposite force. Another one is a bit more involved, but basically a mechanical system cannot change its symmetry by itself. In this case, the initial state with a ball at rest has a radial symmetry with a centre on the apex of the dome. This is not true anymore if the ball moves in one direction. This is related to the conservation of momentum. There are a couple of points that can be solved easily, but are clearly defects in the original formulation of the problem. for example, the height according to the equations is not a length, which is not a problem itself (we can just multiply by an arbitrary factor with the right dimensions) but an indication of sloppy thinking and hand waving. Similarly, the force is not bounded in the original formulation. Again, this can be fixed by restricting the valid range for r, but is rather messy.

lisper 3y ago

> once you properly set the initial state of the ball (force=velocity=0, or any other values), then the solution becomes unique and that’s it No, you've missed the point entirely. There are circumstances under which the solution is not unique, and the article describes such a circumstance.

deleted 3y ago
[deleted]

azeemba 3y ago

> Because once you properly set the initial state of the ball (force=velocity=0, or any other values), then the solution becomes unique and that’s it, The original tumblr post linked to this wikipedia article: https://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem#Example_of_non-uniqueness https://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_t... The initial conditions here are _not_ sufficient to pin to a unique solution. > anything acting on it would violate other principles of classical mechanics What other principles of classical mechanics does it violate though? I think that's the point of the exercise. To reflect on part of our mathematical modeling would prevent this.

tsimionescu 3y ago

> Because once you properly set the initial state of the ball (force=velocity=0, or any other values), then the solution becomes unique and that’s it, there is one possible trajectory. As the linked article points out, that is not true. Even for force = velocity = 0, there are solutions for which the ball starts rolling. Additionally, even adding the first law does not help, since the body has F = v = a = 0 at any time before T, and it has a > 0 only for times where it also has F > 0. So the body spontaneously starting to move is perfectly consistent with classical mechanics. A better attack on the example is that the shape of the dome may not be possible to construct from actual matter. The math only works with a shape that has a certain type of infinite smoothness with a single bump, which is probably not possible even in principle to construct from real matter.

sobellian 3y ago

> That is just not going to happen in (classical) reality, though. Because once you properly set the initial state of the ball (force=velocity=0, or any other values), then the solution becomes unique and that’s it, there is one possible trajectory. Not true, and the trick is very simple, you just contrive a force field that admits x''(0) = 0 but some non-zero higher derivative. Indeed, if you look at the proposed solution r = 1/144 (t - T)^4 for t > T, you'll see that: r(T) = 0 r'(T) = 0 r''(T) = 0 r'''(T) = 0 r''''(T) is ill-defined (but Newtonian mechanics AFAIK says nothing to forbid this)

pistachiopro 3y ago
Right, so for his nondeterministic path: r(t) = 0 ; t <= T r(t) = (1/144)(t-T)^4 ; t >= T We can see that r''''(T) is either 0 or 1/6, depending on if we go with the top or bottom equation. That does look like some sort of hidden state change, there. Interestingly, the spherical dome he mentioned (which doesn't yield to this non-determinism) forces all derivatives of r(t) to be continuous ...
tkoolen 3y ago
> but Newtonian mechanics AFAIK says nothing to forbid this In the Wikipedia article on Newton's laws of motion, the first law is stated as "A body remains at rest, or in motion at a constant speed in a straight line, unless acted upon by a force.". Here it would seem that we leave the state of rest not due to a force, but due to some other cause, which the first law would forbid. So I think that the particular interpretation of Newtonian mechanics used by the author is a bit of a strawman.
- sobellian 3y ago
  We must ask ourselves what "at rest" means since there are multiple trajectories that give x(0) = 0, x'(0) = 0 and it seems that some posters believe this cannot fully constrain the particle to be either at rest or in motion at t=0. There are multiple ways to resolve this dissonance. We could demand that the higher derivatives are also zero. We could derive some elaborate rule excluding non-zero x''(t) in the neighborhood of t=0. Or something else altogether. The issue IMO with these resolutions is that they're quite complicated and Newton almost certainly did not have any of them in mind. It's much simpler to just content ourselves that NFL is a special case of F=ma (where F=0). I'm not sure why we should contort ourselves to preserve determinism since we know that gets thrown out the window with QM anyway.

H8crilA 3y ago

This is a (quite clever) comment on a physics theory that we know to be incomplete. Not a comment on reality. And the math does check out. There really are many solutions to this system, i.e. it is nondeterministic under the theory.

WindyLakeReturn 3y ago

>The ball starting to move without anything acting on it would violate other principles of classical mechanics. Isn't gravity acting on the ball? Ideally a ball balanced a the tip of a bowl has 0 net force, but only because it has the force of gravity pulling it into the bowl and an equal force from the bowl pushing back. This would require assuming a reality where there is no smallest particle, no atoms making up the bowl or ball as they are both perfect mathematical objects comprised of infinite points no matter what resolution you look at them at. The issue is that such a system, if it can be created by rolling the ball up the bowl with perfect precision from any direction, indicates that it is unstable and may reverse at any point, but without any obvious reason causing it to do so. So either the system has some non-deterministic factor which allows for the perfect stability to break at an arbitrary time in an arbitrary direction or it isn't time reversible as the ability to roll the ball up the bowl cannot be reversed. Looking at it another way, can you tell the difference between a ball I perfectly rolled to the top of the bowl 10 years ago and one I did 10 seconds ago? I do wonder if we've added so many assumptions, with mathematically perfect objects and infinite precision forces that we have created some sort of paradox. We have hit levels of perfectly spherical cows that cause even the perfectly spherical cows to complain about unrealistic standards.