OK. I don't find assigning causality a productive exercise, so I'll leave the field to those that do.The point is about how we like to assign causality to particular triggering events, but if a triggering event is almost sure to happen, then the particular loses its hallowed explanatory status. — apokrisis
OK. I don't find assigning causality a productive exercise, so I'll leave the field to those that do. — andrewk
The curve is constructed so that the displacement function is a constant multiple of (t-T)^4 for t>=T. The same would work for a displacement function proportional to (t-T)^n for any n>=4. In that case the non-differentiability won't appear until the (n-1)th derivative of the displacement function. So as long as n>=4 the nondifferentiability will be out of sight and out of mind.Any explanation for why this curve is somehow poised in a way that allows for the claimed indeterminism? — apokrisis
So I don't think this case does what it at first seems to do, which is to generate breaking symmetry out of nothing. The breaking symmetry is always there in the discontinuous Jounce, which we have simply assumed. The plausible physical solution is that which has smooth displacement and all derivatives are always zero - ie symmetry doesn't break. — andrewk
Comments suggest to me that the cause of the sudden spontaneous motion is a concealed fourth derivative jounce. So it is like the ball is set down on the apex in the middle of just being about to snap. — apokrisis
I suppose if we send it sliding up with exactly the correct initial velocity, and no touching it after we release it, all higher derivatives of displacement will be zero once it is on its way up. It follows that it will stop at the top rather than continuing down the other side, because it will have zero velocity and zero horizontal force on it at that time. — andrewk
The higher derivatives would have to be nonzero for the ball to pass the cime and go down the other side. If it stops there, there are no discontinuities because Jounce and Jerk were already zero on the way up.
True.The jounce wasn't zero on the way up. It was constant and equal to 1/6. — Pierre-Normand
Newton's first law says that an item will remain in its state of motion (which is interpreted to mean its velocity does not change) unless acted upon by a (net) external force. So the ball in a perfect, stationary position at the top will remain in its state of motion, which is stationary. It will not roll down. Hence the solution is non-Newtonian and must be rejected. It satisfies the second but not the first law. — andrewk
I got a bit lost here. Newton's third law is that for every action there is an equal and opposite reaction. I can't see how that law is relevant to the questions being examined in this scenario. Can you outline what you had in mind here?this construal of Newton's first law would also have the very unfortunate consequence that it makes it inconsistent with the third law in other cases. — Pierre-Normand
I got a bit lost here. Newton's third law is that for every action there is an equal and opposite reaction. I can't see how that law is relevant to the questions being examined in this scenario. Can you outline what you had in mind here? — andrewk
Could you imagine ceasing to care about the individual pushes and instead accepting that the generic impossibility of eliminating all disturbances is this deep truth? — apokrisis
So the act of placement is really a push, because placement cannot be precise. And, if the act of placement could be precise enough, or the surface flat enough, then a push would be needed. Therefore it's always a push. — Metaphysician Undercover
As apokrisis has said, the ball effectively vibrates, as its internal molecules move about (Unless the experiment takes place at absolute zero), so it 'pushes' itself, if nothing else does so first. No need even for QM, just Brownian motion is enough to explain it. — Pattern-chaser
An expansion that prevents bifurcation could be:
'Where there is more than one future movement pattern of an object that is compatible with the 2nd and 3rd laws and the conditions in place at time t, and one or more of those patterns involves the object's velocity remaining constant for the period [t,t+h) for some h>0, the pattern that occurs will be one of those latter patterns'.
Very wordy, I know, but it has to be in order to deal with nonphysical cases like this without just disappearing into Law 2. Note also that it leaves open the possibility that there may still be bifurcations possible with this law - not the one discussed in the paper, which would be ruled out, but other ones in even more pathological cases. I suspect it may be possible to prove there cannot be, but that's just a hunch. — andrewk
This states that the solution must have locally constant velocity both looking backwards and forwards if that is compatible with laws 2&3, else locally constant future velocity if compatible with 2&3, else locally constant past velocity if compatible with laws 2&3. Otherwise the law is silent. — andrewk
I'm not sure I understand the question. The above law would mandate that a ball sitting stationary exactly on top of the dome would not roll down. The second law does not mandate that. — andrewk
My expanded first law prohibits the ball rolling down (a solution not in U) because there exists a solution in U, ie in which it does not roll down, and the law requires that a solution in U be taken in preference to a solution outside it.both the second law, and your expanded law (if I understand it correctly) are silent regarding what happens next. — Pierre-Normand
My law prohibits the ball rolling down (a solution not in U) because there exists a solution in U, ie in which it does not roll down, and the law requires that a solution in U be taken in preference to a solution outside it. — andrewk
Yes. Say it reaches the apex at time t2. Then there is a path compatible with the 2nd law in which it remains there for the period [t2,t2+h) for any h>0. So that path must be what happens rather than a path in which it continues down the other side.Is your law mandating that the ball will stand still indefinitely after it has reached the apex — Pierre-Normand
Yes. Say it reaches the apex at time t2. Then there is a path compatible with the 2nd law in which it remains there for the period [t2,t2+h) for any h>0. So that path must be what happens rather than a path in which it continues down the other side. — andrewk
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