4 ms·

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. Be

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...
  - 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
      > No it doesn't, because it's not a sphere. The dome is specifically designed so that it takes finite time. Can you explain why?
      - deleted 3y ago
        [deleted]
      - crazygringo 3y ago
        I don't know what kind of answer you're looking for. The equation was explicitly chosen/derived to have this property. I assume the mathematical proof of that isn't something that fits in a few sentences in an HN comment.
        eesmith 3y ago
        Back when I was a little smithling who knew more math than physics, I complained about an assignment whose solution didn't make mathematical sense. My teacher commented that I needed to think like a physicist, that is, understand that certain mathematical issues didn't exist in the real world, so could be ignored. The skit at https://www.youtube.com/watch?v=xPzR_D9qKeo https://www.youtube.com/watch?v=xPzR_D9qKeo gives some examples. The one at https://youtu.be/xPzR_D9qKeo?t=165 https://youtu.be/xPzR_D9qKeo?t=165 is pretty close to this example "if it's in physics, it's invertable." That doesn't mean that if it's invertable it's in physics. Are the inverse dynamics of this system still in Newtonian physics? For example, is is the inverse path actually on the described surface or does it detach? How does a moving mass have an instantaneous jerk with no change in velocity?
  - 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.
- kergonath 3y ago
  > 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. This is a red herring. It sounds plausible, but there is no trajectory that does this. This is the weakest paragraph in the original post, and I am not sure whether this is intentional (because the demonstration sounds truthy if you don’t go too deep in the details) or whether it was not entirely thought out. There is some discussion about the time-reversal thing here: https://blog.gruffdavies.com/2017/12/24/newtonian-physics-is-deterministic-sorry-norton/ https://blog.gruffdavies.com/2017/12/24/newtonian-physics-is... . There isn’t much to discuss however, because ultimately it is just a distraction.
  - hexane360 3y ago
    There's a lot of minor points in that post, but it seems like both authors largely agree on the meaning, but are using different language. From Dr. Davies' post: >To remain Newtonian and preserve determinism, we can exclude the singular point by constraining the higher orders to zero whenever the net force is zero. We lose time symmetry for this special case if we do this. If we wish to keep that, then we have to accept that Newtonian mechanics is incomplete and consider higher order differentials. And from Dr. Norton's article: >The solutions (3) are fully in accord with Newtonian mechanics in that they satisfy Newton's requirement that the net applied force equals mass x acceleration at all times. >An important feature of Newtonian mechanics is that it is time reversible, or at least that the dynamics of gravitational systems invoked here are time reversible. Dr. Davies is saying that there's three options: a) relaxing time-reversal symmetry (at singularities) from Newtonian mechanics, by interpreting Newton's First Law to apply to higher derivatives; b) considering Newtonian mechanics to be incomplete, and make (unspecified) choices about what trajectories of higher-order derivatives are acceptable; or c) accept non-determinism. Dr. Norton is defining "Newtonian mechanics" as necessarily having time-reversal symmetry, which prevents the first solution. He is also defining it as specifying acceleration only (which I think is quite reasonable), preventing the second solution. Therefore he's concluded the third solution: This mathematical stating of Newtonian mechanics is non-deterministic.
    - 3y ago
- wruza 3y ago
  If you throw a ball into a bowl, it will also find the (anti-)apex. And the time-reversal of that is the ball arbitrarily choosing a direction to jump off the center of the bowl. So what? Why is it important to mention in case of a non-stable equilibrium?
  - scatters 3y ago
    A ball rolls through the bottom of the bowl and out the other side. It doesn't come to rest.
  - cma 3y ago
    In a time reversal situation where the ball is at maximum magnitude momentum, it will just go the opposite way, right?
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

> 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...
- 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
    > No it doesn't, because it's not a sphere. The dome is specifically designed so that it takes finite time. Can you explain why?
    - deleted 3y ago
      [deleted]
    - crazygringo 3y ago
      I don't know what kind of answer you're looking for. The equation was explicitly chosen/derived to have this property. I assume the mathematical proof of that isn't something that fits in a few sentences in an HN comment.
      - eesmith 3y ago
        Back when I was a little smithling who knew more math than physics, I complained about an assignment whose solution didn't make mathematical sense. My teacher commented that I needed to think like a physicist, that is, understand that certain mathematical issues didn't exist in the real world, so could be ignored. The skit at https://www.youtube.com/watch?v=xPzR_D9qKeo https://www.youtube.com/watch?v=xPzR_D9qKeo gives some examples. The one at https://youtu.be/xPzR_D9qKeo?t=165 https://youtu.be/xPzR_D9qKeo?t=165 is pretty close to this example "if it's in physics, it's invertable." That doesn't mean that if it's invertable it's in physics. Are the inverse dynamics of this system still in Newtonian physics? For example, is is the inverse path actually on the described surface or does it detach? How does a moving mass have an instantaneous jerk with no change in velocity?
- 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.
kergonath 3y ago
> 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. This is a red herring. It sounds plausible, but there is no trajectory that does this. This is the weakest paragraph in the original post, and I am not sure whether this is intentional (because the demonstration sounds truthy if you don’t go too deep in the details) or whether it was not entirely thought out. There is some discussion about the time-reversal thing here: https://blog.gruffdavies.com/2017/12/24/newtonian-physics-is-deterministic-sorry-norton/ https://blog.gruffdavies.com/2017/12/24/newtonian-physics-is... . There isn’t much to discuss however, because ultimately it is just a distraction.
- hexane360 3y ago
  There's a lot of minor points in that post, but it seems like both authors largely agree on the meaning, but are using different language. From Dr. Davies' post: >To remain Newtonian and preserve determinism, we can exclude the singular point by constraining the higher orders to zero whenever the net force is zero. We lose time symmetry for this special case if we do this. If we wish to keep that, then we have to accept that Newtonian mechanics is incomplete and consider higher order differentials. And from Dr. Norton's article: >The solutions (3) are fully in accord with Newtonian mechanics in that they satisfy Newton's requirement that the net applied force equals mass x acceleration at all times. >An important feature of Newtonian mechanics is that it is time reversible, or at least that the dynamics of gravitational systems invoked here are time reversible. Dr. Davies is saying that there's three options: a) relaxing time-reversal symmetry (at singularities) from Newtonian mechanics, by interpreting Newton's First Law to apply to higher derivatives; b) considering Newtonian mechanics to be incomplete, and make (unspecified) choices about what trajectories of higher-order derivatives are acceptable; or c) accept non-determinism. Dr. Norton is defining "Newtonian mechanics" as necessarily having time-reversal symmetry, which prevents the first solution. He is also defining it as specifying acceleration only (which I think is quite reasonable), preventing the second solution. Therefore he's concluded the third solution: This mathematical stating of Newtonian mechanics is non-deterministic.
  - 3y ago
wruza 3y ago
If you throw a ball into a bowl, it will also find the (anti-)apex. And the time-reversal of that is the ball arbitrarily choosing a direction to jump off the center of the bowl. So what? Why is it important to mention in case of a non-stable equilibrium?
- scatters 3y ago
  A ball rolls through the bottom of the bowl and out the other side. It doesn't come to rest.
- cma 3y ago
  In a time reversal situation where the ball is at maximum magnitude momentum, it will just go the opposite way, right?

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.