[SlySherZ]

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.

[kergonath]

> 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]

> 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]

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.

[eigenket]

Theres a couple of mistakes here. Firstly the particle is not on its own, it is being acted on by the dome and by gravity.

The thing about symmetry breaking also doesn't make much sense. I guess you're trying to appeal to Noether's theorem, but Noether's theorem in classical mechanics is a consequence of f = ma. You derive the Lagrangian formulation of mechanics from f=ma and Noether's theorem from that. However the weird solution when then ball suddenly randomly falls down the dome after staying put for an arbitrary time is completely consistent with f=ma, so that can't help you here.

In any case the radial symmetry you're looking for (the system is invariant under rotations around the peak of the dome) implies conservation of angular momentum about this point, and not about any other point (since the setup is manifestly not symmetric under rotations about any other point). However (one can easily check) that for both the static solution and the randomly starts moving solution, the angler momentum about the axis through the peak of the dome is always zero.

[sudosysgen]

The particle does not undergo any net force while at rest. If you understand "acted upon" to be a causal statement, then no, the particle should not be able to leave a resting position, because when it is at rest there is no net force acting upon it.

The argument is basically assuming that the particle moves, showing that it moves in a way that respects the second law, then restating the first law to be a special case of the second to avoid the causal language it contains and to make it completely redundant.

[eigenket]

There is no time when the particle is accelerating, or even moving, while not experiencing a net force in this setup. The argument you'd have to make would be to change Newtonian mechanics so the first law is no longer a special case of the second law, but actually says something nontrivial about all of the time-derivatives of position rather than just the first two. This (in my opinion) would no longer be Newtonian mechanics, but some extention.

Even then I'm not sure that would save you in general, since it should be possible to cook up examples where the motion is some non-analytic thing like a portion of a "bump function".

Edit: by the way, a more modern formulation of the first law is essentially that there are no privileged inertial frames. All inertial frames are equally valid. This is straight-up false in this setup since obviously the rest frame of the dome is privileged (since the dome is given infinite inertial mass).

[geysersam]

This is not correct. Momentum is conserved by the spurious solution and there's still an equal but opposite force on another body (the body producing the gravitational force).

I think this example just illustrates a case where the Newtonian model of reality simply does not describe reality itself

[tiberious726]

Even if it does, that would amount to a contradiction in Newtonian mechanics. You don't get to simply ignore that the ball starting to roll after arbitrary (non deterministic) time T is a solution to these equations.

(Note that the article goes on at length separating Newtonian mechanics from the "real world" or whatever)

[lisper]

> 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.

[kergonath]

> 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... . There isn’t much to discuss however, because ultimately it is just a distraction.

[hexane360]

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.

[kergonath]

You are entirely right, whether something that depends on higher-order derivatives can be called Newtonian is debatable. Personally I don’t really care either way, as this is just a label. Newton did not mention higher-order derivatives but on the other hand they are a trivial extension to the mathematical framework. It is difficult to call a body at rest if any of the derivatives of the position is not zero, because then it will start moving instantaneously so it is hard to read the first law otherwise. And the second law does not care about anything other than acceleration. And there certainly isn’t anything that prevents us from using clever shape to roll balls on, as long as the shape make physical sense.

What this does not change, however, is that the dome does not demonstrate non-determinism. The apparent demonstration hinges on logical errors that remain errors regardless of the framework used, be it classical or quantum mechanics, or relativity.

[archgoon]

It is not a trivial extension. If you need all infinite number of derivatives to predict motion than your theory is non predictive.

[wruza]

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]

A ball rolls through the bottom of the bowl and out the other side. It doesn't come to rest.

[cma]

In a time reversal situation where the ball is at maximum magnitude momentum, it will just go the opposite way, right?

[eesmith]

> 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...

[crazygringo]

> 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]

> 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?

[crazygringo]

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]

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 gives some examples. The one at 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]

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.

[eesmith]

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]

The article explicitly discusses this without demonstrating anything. On it’s face this argument has the weight of these demonstrations.