[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).

[sudosysgen]

> There is no time when the particle is accelerating, or even moving, while not experiencing a net force in this setup.

I agree! But at the same time, there is no moment at which a net force compels the particle to leave a state of rest. The criteria you are putting forwards with that sentence isn't that of the first law, it's just a consequence of the second law. The first law makes a statement that force is required to cause a body to leave a state of rest, but that's not what we have here.

We just postulate that it must stop being at rest immediately after T, and then we find a force as a result of it not being at rest anymore. This inverts the causality that NFL requires, and because it doesn't have give a cause for the particle to leave the state of rest, which the 1st law requires, it's not a valid physical solution in Newtonian mechanics.

Here is Wikipedia's translation of the first law, which is as good as any:

> Every body continues in its state of rest, or of uniform motion in a straight line, unless it is compelled to change that state by forces impressed upon it.

In our trajectory, the body is not compelled to change its state by a force impressed on it. It's in a stable state, then we arbitrarily decide that it will change it's state immediately after T, and then we show that this leads to a trajectory which satisfies the second law. But since it changes state without being compelled by a force, it's no longer a trajectory which satisfies the first law.

> 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

I'm not changing Newtonian mechanics - Newton's first law is not a special case of the second law. It makes a statement about cause and effect. The second law doesn't - it's just a differential equation, which does not have such a content. The article redefines the first law to strip it of its causal content and instead make it a trivial statement, but that's not what the first law plainly states. In Newtonian physics, force is necessary to cause a change in velocity, it's not simply that force and acceleration co-occur.

> but actually says something nontrivial about all of the time-derivatives of position rather than just the first two

No, this isn't necessary. If you want, you can instead understand it as something about the nth derivative of position, and that way you can recover perfect reversibility, but you don't have to. You can just keep Newtonian mechanics as is, and recognize that they describe a causal system, and not just a system of differential equations (which make no statements about causation in and of themselves), but then you lose perfect reversibility for some pathological trajectories like this one.

> 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".

I'm not arguing for the infinite derivative modification either, but I don't see how that's the case - if you have a non-analytic function then some derivatives don't exist, and if they don't exist then they wouldn't be able to satisfy the infinite derivative formulation. Seems to me that such a law would directly eliminate all non-analytic trajectories.

[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