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From the second paragraph: A mass sits on a dome in a gravitational field. After remaining unchanged for an arbitrary time, it spontaneously moves in an ar

From the second paragraph: 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.

PeterWhittaker 2y ago
Not on Norton's Dome: it is the classic example of indeterminism in classical mechanics. While QM is sufficient for indeterminism, it is not necessary, as this example shows. Even in classical mechanics, physics is weirder than our intuition allows.
- tsimionescu 2y ago
  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).
  - wat10000 2y ago
    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.
    - tsimionescu 2y ago
      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.
- dventimi 2y ago
  > Not on Norton's Dome Really? Are you sure about that? Have you tried it? Where is this "Newton's Dome" so that the experiment can be replicated?
  - mrkeen 2y ago
    [flagged]
    - dventimi 2y ago
      I'll take that as a "no".
  - PeterWhittaker 2y ago
    Not Newton's Dome, Norton's: https://en.m.wikipedia.org/wiki/Norton%27s_dome https://en.m.wikipedia.org/wiki/Norton%27s_dome
    - dventimi 2y ago
      That was a typo from a phone keyboard.

PeterWhittaker 2y ago

Not on Norton's Dome: it is the classic example of indeterminism in classical mechanics. While QM is sufficient for indeterminism, it is not necessary, as this example shows. Even in classical mechanics, physics is weirder than our intuition allows.

tsimionescu 2y ago
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).
- wat10000 2y ago
  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.
  - tsimionescu 2y ago
    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.
dventimi 2y ago
> Not on Norton's Dome Really? Are you sure about that? Have you tried it? Where is this "Newton's Dome" so that the experiment can be replicated?
- mrkeen 2y ago
  [flagged]
  - dventimi 2y ago
    I'll take that as a "no".
- PeterWhittaker 2y ago
  Not Newton's Dome, Norton's: https://en.m.wikipedia.org/wiki/Norton%27s_dome https://en.m.wikipedia.org/wiki/Norton%27s_dome
  - dventimi 2y ago
    That was a typo from a phone keyboard.