It points out many flaws in Norton's reasoning, some fatal to his argument, some not. Putting it as simply as I can, Norton seems to claim that "Newton's Laws" are non-deterministic. That's not quite right. Rather, they are non-complete. I.e. they are incomplete. They're incomplete insofar as Newton's First Law ("An object at rest remains at rest, and an object in motion remains in motion at constant speed and in a straight line unless acted on by an unbalanced force") establishes first-order and second-order derivatives (momentum and acceleration) as state variables but places no constraints on higher-order derivatives. However, higher-order derivatives are (as many as are needed) among a system's state variables. In many real systems (but far from all), higher-order derivatives are zero and human experience with them is rare, so they're easy to overlook. Norton's (unphysical) Dome is a specific example of a general class of systems where higher-order derivatives are not zero. Given that, the two branches of Norton's equation of motion (for the stable and unstable trajectories) cannot both describe the same system (or the same particle) with the same set of state variables. That's the sleight-of-hand.
Again, all credit to Gareth Davies for working this out. I am absolutely not trying to pass off his work for my own. Just reporting it and trying to summarize it.
Reading the original article I immediately thought about higher order derivative, which made me wonder what laws apply to them, that I’ve never studied that, that’s odd
I think one can simply use the Euler-Lagrange method which is able to account for the constraint forces acting on the ball. Haven't worked that out for this particular problem but it should be relatively easy. Davies argument is a bit overcomplicated I think, the main challenge here is correctly accounting for the geometric constraints in the movement of the particle. I find the argument about the higher-order derivatives a bit weird as well, the system can be fully described using its potential and kinetic energy which are scalar (possibly time-dependent) fields and implicitly contain all forces, given some initial conditions (position and momentum) we can solve the equation of motion of the system with that.
I think it is worth comparing this problem with the question of the behaviour of a particle placed at the apex of a cone. I claim it is clear that in this case, the problem is clearly not well-posed because the apex is a singular point: the slope at the apex is undefined. The singular nature of the slope (first derivative) is the issue.
This "dome" is essentially the same issue just with the singularity buried one level deeper: you need to take second derivatives to see it. Indeed a planar cross section containing the vertical axis through its center is a graph of the equation $y^2 = |x|^3$ (up to constants) and this is not twice differentiable at $x = y = 0$. Newtonian mechanics is governed by a second order differential equation, so we need a C^2 regularity assumption to get uniqueness.
So for me there is not really any more philosophically interesting than the question about a particle balancing at the apex of a cone.
“Instead of imagining the mass starting at rest at the apex of the dome, we will imagine it starting at the rim and that we give it some initial velocity directed exactly at the apex. If we give it too much initial velocity, it will pass right over the apex to the other side of the dome. So let us give it a smaller initial velocity. We produce the trajectory T1:”
As acceleration, jerk, snap, crackle, pop, … all must approach 0, does it ever actually reach the apex with zero velocity in finite time? That seems like the most obvious solution here where if you start with non zero velocity it never quite reaches the apex, and if you start with zero velocity at the apex it never leaves it.
This is a really neat example. When I was in academia, I wondered if I could teach a advanced undergrad physics course that focused exclusively on breaking all the standard physics theories. There are three ways that theories typically break.
- Falsified by experiment. This is for example discussed a lot in the Modern Physics course with regards to classical theories.
- Broken math for the theory. This is an example of this, and most of the theories have some weird stuff like this that has to be ignored.
- Philosophical unpleasantness. Sometimes theories have weird philosophical consequences that indicate that we should come up with something better.
Interesting that this is making the rounds...maybe OP also had this recent video show up in their Youtube feed? It includes interview clips with John Norton, the author of this paper. https://www.youtube.com/watch?v=EjZB81jCGj4
A mass sits on a dome in a gravitational field. After remaining unchanged for an arbitrary time, it spontaneously moves in an arbitrary direction, with these indeterministic motions compatible with Newtonian mechanics
Well, no. "It" does not spontaneously move in an arbitrary direction. It remains in place forever.
No, that is just a mistake that Norton makes. The only physical trajectory for a particle starting at the apex at rest is that it will remain at rest.
The other equation that Norton comes up with is not a physical description of a particle at rest, it is a description of a particle which has complex motion (the fourth derivative of its position(t), sometimes called crackle, is 1/6). Basically that means that in every second, its acceleration increases by (1/12s³). An equation of motion that has any non 0 derivative of time is not describing a particle at rest.
Not to mention, branching functions are also not valid equations of motion. You can't take two valid equations of motions and stitch them at some arbitrary time. The same problem will appear on a perfectly flat surface if you do that. I could say that the function f(t) = 0 for t < T, 7t for t >= T is a valid solution to the equations of motion, by the same logic (its second derivative is 0, equal to the net force acting on the particle). This doesn't prove that Netwonian mechanics is non-deterministic, it shows that you can't use functions that arbitrarily change as equations of motion (they just violate Newton's first law).
The issue is that those equations just fall out of the three laws of motion, gravity, and a surface with a particular shape. The equations of motion here aren’t things Norton came up with, they’re what you get when you analyze Newtonian motion on this surface.
Likewise, the branching function isn’t a result of stitching together two equations of motion. It’s a result of analyzing the motion of a particle with a certain initial velocity. Work the math in an entirely conventional way and the result is that the particle travels up the dome, stops at the top, and stays there forever. Since Newtonian physics is symmetrical with respect to time, that means you should be able to reverse the particle’s motion at any point and have it traverse the same points at the same times, but backwards. This means that you can have a particle that stays still for an arbitrary time and then spontaneously starts to move.
You can’t object to the equations of motion unless you object to Newtonian physics. The particular equations of motions here are derived from Newton’s laws. The only thing that’s chosen is the shape of the surface, and it seems sensible to say that a surface can have any continuous shape in Newtonian physics.
There are a few ways out of this.
1. Declare the problem to be unphysical and therefore ignorable. This has the notable benefit of being objectively true (totally smooth surfaces don’t exist, solid matter is made of discrete particles, blah blah blah).
2. Decide (or accept?) that time reversibility in laws of motion does not necessarily imply time reversibility in a specific evolution under those laws. I’d bet this one is attractive to programmers: of course the integrator isn’t required to return to the initial conditions when you run it backwards. Time reversibility? Take a backup before you run, and restore if you want to go back.
3. Declare that Newtonian physics forbids certain shapes. This works but seems totally arbitrary.
The surface is only relevant here in that it gives us a natural reason to have an equation of motion with a 4th power of time. But the equations used are not the correct description of Newton's laws (even though they are very commonly used).
When we say that a particle is at rest or moving with a constant velocity, we commonly translate that to dx/dt = v, some constant scalar (or equivalently, dv/dt = 0). But this is almost a shorthand: in reality, we also have to ask that d²x/d²t = 0, and in general that the Nth derivative is 0 for all N. And the function 1/144(t-T)^4 actually has a non-0 4th derivative, so it isn't describing a particle that no forces are acting on. It is instead describing a particle that reaches the summit and then immediately falls away, as you would expect for any point of unstable equilibrium.
What the construction of the surface does is to make it impossible for a particle that didn't start at rest on the apex to ever reach the apex and remain at rest there. Any particle that can reach the apex will do so with an "accelerated" motion (one where some derivative of it's momentum is non-0), and so will immediately fall away.
You can also look at this just by analyzing the requirement for time reversibility: if a particle is nudged so that it arrives at the apex at a time t = T, and was not at the apex at any time t < T, then it can't be at the apex at any time t > T either.
Or, you can see that the stitch is not natural, or even valid, by looking at those derivatives. The branch with t < T has all derivatives of t equal to 0. The other branch has the 4th derivative equal to a constant 1/6. How would the particle acquire this non-0 [derivative of] acceleration without the action of any force?
We can also look at the dome problem by analyzing the force acting on the particles. For a moving particle, whose trajectory is r(t) = (1/144) (t-T)^4 (that is, one that reaches the summit at t=T), the net force acting on it from gravity + surface normal is F(t) = r(t) ^ (1/2) = (1/12) (t-T)². The force at time t=T is 0, but we can see that the force at time t = T+dt is not 0, so the particle can never rest on the apex. Conversely, the net force acting on a particle whose trajectory is r(t) = 0 is F(t) = r^(1/2) = 0, for any t > 0. So a particle starting at rest can never leave.
By the way, we can reproduce a similar system with a much more mundane surface. Say we have a car rolling on a perfectly flat plane with some friction coefficient k, moving along with speed V. The brakes are connected to an oscillator that gradually activates and releases them. The engine is run at a constant rate, constantly producing a forward force Fe = kV. When the oscillator starts activating the brakes, the car will start experiencing an extra force, Fb, that gradually increases from 0 to kV.
The car's motion will continue in fits for as long as the engine and oscillator have fuel. But, the car will frequently reach 0 velocity and 0 acceleration (when Ff + Fb = Fe, the car has 0 acceleration), and then it will start again. This is because the car is experiencing a non-constant force, which means that its acceleration is also a function of time, and temporarily reaching 0 acceleration and 0 velocity doesn't imply that it will stay in this state forever.
In your car example, there is something producing a force that makes it start moving again.
You say that a particle that reaches the apex must immediately fall away. When it reaches the apex with zero velocity, what causes it to fall away? The only way a particle’s velocity can change is if a force acts on it. What force acts on this particle to make it fall away? And a more practical question, what direction does it fall?
The net force acting on it the instant it reaches the very apex is 0, but only for an instant. The force will necessarily increase afterwards, because, through its trajectory, the (2nd) derivative of the force is not 0. To put it another way, nothing changes for the particle when it reaches the apex, the net force acting on it is still F(t) = (1/144)(t-T)^2. Just because F(T) happens to be 0 doesn't mean that F(T+dt) has to be 0, it will be (1/144)(dt)², in the same direction as it was before reaching the summit.
Because of the above, the direction is also fully determined. It will fall back off on the same trajectory it came up on, if I have my signs straight (if not, then it will continue in the same trajectory on the other side of the dome).
Consider a modified version of the car experiment, where instead of an oscillator that is independent of the car, we set up some kind of pendulum pressing on the breaks such that they get activated when the car reaches a certain acceleration, and it gets depressed when the acceleration reaches some negative value. Also, instead of an engine and a flat plane, we can put the car with no engine on a sloped plane. So, the only energy we need to care about is still gravitational. Nevertheless, the car will start and stop continuously, temporarily reaching 0 acceleration and net force and then starting up again.
We don't have a term for this, but it's like an inertia for forces acting on the particle (gravity and the surface normal).
You can also view this as a missing piece in Newton's laws, as this is not explicitly covered as a possibility. But that is also true then for the entire motion, as all Newtonian equations of motion have the derivatives of acceleration equal to 0, while the motion along this surface necessarily doesn't.
No such term exists in Newton’s laws. If a particle accelerates, it must be because of a force acting on it. Not something “like an inertia for forces,” but an actual, physical force.
Two forces act on this particle, but they sum to zero. If you claim the particle begins to move from zero velocity, then either those two forces don’t sum to zero, or there’s a third force. But they do and there isn’t.
Yes, this "inertia-of-force" is not part of Newton's laws, but it's a straightforward extension of Newton's laws, not a contradiction. And it preserves all of the other interesting properties - energy conservation, time symmetry, determinism.
It's certainly less of a contradiction of Netwon's laws than the idea that the particle just sitting there can suddenly start moving, which is a straightforward contradiction of Netwon's first law, and implies that a particle can acquire higher order acceleration (jerk or snap) out of thin air.
Of course, this is all moot anyway, as this type of dome with a point-like apex is not a physical system to begin with: since matter is made of atoms, it can't have this type of singular point. Not even going into problems of the precision needed to give the particle just the right initial speed to actually have exactly 0 velocity at the top.
Inventing a different physics and declaring it to be more correct certainly works, but is kind of a boring answer to the problem.
It plainly is a contradiction given that there’s a system where it gives a different answer. If it wasn’t a contradiction then it would be equivalent (or somehow a superset?) and wouldn’t solve the problem.
Ignoring the problem as unphysical makes much more sense to me. The real world doesn’t follow Newtonian physics anyway, it’s effectively just a simplification useful for certain common situations. No need to care about edge cases that can’t happen.
With the time reversal trick, it almost seems like you could say the cause happens in the future and propagates backwards in time?
I'm curious that if it doesn't work for a hemispherical dome because of requiring infinite time, does that mean a ball on a hemispherical dome is stable and won't spontaneously roll off?
No. What all this shows is that both cases (hemispherical dome or the other one in OP) of a ball in stable equilibrium are outside the domain of applicability of the theory. You should simply not use the theory as is to make predictions about when and how the ball will roll off.
Actually, there is nothing mysterious here. The theory, applied correctly, predicts the expected thing: a particle that starts at rest at the apex will never leave it. The other "solution" is not actually a particle at rest (it has a non-0 6th derivative of position, that is the rate of change of the rate of change of its acceleration is 1/6). So it's actually describing a particle whose acceleration will increase at an increasing rate forever, but that just happens to be 0 at time T.
Just in case I'm completely wrong about reality, both cases are unstable equilibrium, aren't they? Equilibrium in that forces balance and zero acceleration but unstable in that any disturbance will lead to it rolling off rather than returning to where it was.
Time reversible systems make causality fuzzy. There’s nothing special about this example in that respect. You can’t object have ball A hit ball B. A causes B to move. You can then reverse it, end up with an equivalent but time reversed system where B hits A, and B causes A to move.
I am curious about the validity of (re)phrasing of Newton’s 2nd Law as “instantaneous.” The acceleration is non-differentiable at T=0. Can we really combine it with analysis of times t<T and t>T?
It points out many flaws in Norton's reasoning, some fatal to his argument, some not. Putting it as simply as I can, Norton seems to claim that "Newton's Laws" are non-deterministic. That's not quite right. Rather, they are non-complete. I.e. they are incomplete. They're incomplete insofar as Newton's First Law ("An object at rest remains at rest, and an object in motion remains in motion at constant speed and in a straight line unless acted on by an unbalanced force") establishes first-order and second-order derivatives (momentum and acceleration) as state variables but places no constraints on higher-order derivatives. However, higher-order derivatives are (as many as are needed) among a system's state variables. In many real systems (but far from all), higher-order derivatives are zero and human experience with them is rare, so they're easy to overlook. Norton's (unphysical) Dome is a specific example of a general class of systems where higher-order derivatives are not zero. Given that, the two branches of Norton's equation of motion (for the stable and unstable trajectories) cannot both describe the same system (or the same particle) with the same set of state variables. That's the sleight-of-hand.
Again, all credit to Gareth Davies for working this out. I am absolutely not trying to pass off his work for my own. Just reporting it and trying to summarize it.