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I am thinking of this Norton's dome.

The author guesses a solution,

$$ r(t)=\left\{ \begin{array}{c l} \frac{1}{144}(t-T)^4 & ,t\geq T\\ 0 & ,t\leq T \end{array}\right. $$

for this second order of differential equation

$$\ddot{r}=r^{\frac{1}{2}}$$

with $r(0)=0, \dot{r}(0)=0$

and then argue that classical mechanics does not have to be deterministic, given $T$ can be any number we want (as long as greater than zero).

I have a background of physics major, and understand (somewhat by this argument) that determinism as an assumption in classical physics - that we have to assume the conditions of uniqueness theorem always exist in nature for classical physics - but I would like a complete understanding of this topic, and hence I would like to ask,

Does the solution exist? Does the author's argument stands in an rigorous mathematical standing point?

Shing
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    http://physics.stackexchange.com/a/39673/25575 – user5402 Sep 08 '15 at 18:05
  • Yes these are solutions of the differential equation, is this your question? – Did Sep 08 '15 at 18:25
  • @Did Yes, but a professor of mine told me that T has to be any value rather than greater than 0 in order to be a valid solution. but I am not sure what he meant. – Shing Sep 09 '15 at 10:06
  • What's amazing is that even if the mass is an object composed of many particles, provided that the object has perfect xyz symmetry with an imaginary axis going through the dome vertically, the mass will not roll down the slope. Net torque will be 0. – Armend Veseli Aug 05 '18 at 06:17
  • Take a look to this Wiki Singular solutions – Joako Jun 02 '22 at 22:49

2 Answers2

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The function is a valid solution of the differential equation. There is nothing wrong mathematically.

Whether it follows that classical mechanics is not deterministic depends on what you consider to be classical mechanics. Note that the equation of motion was derived using a constraint (that the trajectory lies on the surface of the dome), which is an idealization of the forces actually acting between the particle and the dome. If the dome itself were modelled as a classical body of finite mass, the particle could not suddenly gather momentum out of nowhere in violation of the law of conservation of momentum. Thus, for one, you have to consider idealized constraints as part of classical mechanics to reach your conclusion based on this example.

For other reflections on and possible objections to the argument from Norton's dome, see e.g. Malament, D.B., Norton's Slippery Slope, Philosophy of Science, $75$ (December $2008$), pp. $799$-$816$, and for other arguments in favour of indeterminism in classical mechanics, see e.g. the article on causal determinism in the Stanford Encyclopedia of Philosophy.

joriki
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  • I am kind of confused, as a professor of mine (physics department though) told me that T has to be any value rather than greater than 0 in order to be a valid solution (in other words, he claims that that solution does not exist). but I do not know how this is so, and, no offense meant, it seems to conflict with you and Did, would you mind elaborating it a bit? – Shing Sep 09 '15 at 10:14
  • @Shing: No offense taken. I don't know what this "greater than zero" thing is about and why you added that to the question. $T$ can be any value; the entire setup is invariant under time translations; there's no special significance attached to time $0$, it's just an arbitrary reference point with respect to which time is measured. Also I don't understand what you mean by "that solution does not exist". It exists; you've written it down. To check that it's a solution of the differential equation, you just have to differentiate it twice, check that the second derivative exists and plug it in. – joriki Sep 09 '15 at 10:35
  • Thanks for your reply, I mean T has to be greater than 0 in order to satisfy the initial condition $r(0)=0, \dot{r}(0)=0$. otherwises, we will have a moving point right at the beginning. – Shing Sep 09 '15 at 10:39
  • @Shing: Ah, I see, sorry, I overlooked the initial conditions. Yes, in that case we need $T\ge0$ (not $T\gt0$). But that doesn't affect the validity of the overall argument. – joriki Sep 09 '15 at 10:41
  • So basely, for any differential equation, that's always true, "I guess, and if I feed it into the equation, and it satisfies the equation , then I win (a valid one).", do I get it right? (although it may not be the general solution.) – Shing Sep 09 '15 at 10:54
  • @Shing: It depends on what you're trying to do. If you're looking for all solutions, that's obviously not the right way to make sure you find them all. But here the author is trying to make a point based on a certain set of solutions -- if there were further solutions, that could at most mean more indeterminism, not less. So it doesn't matter whether these are all solutions -- all that matters (for his argument) is that they're solutions. – joriki Sep 09 '15 at 10:56
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    I see, that makes more sense to me. thank you for the elaborating! – Shing Sep 09 '15 at 11:01
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Norton's dome is not evidence for non-determinism. The dome curve equation admits solutions which are non-Newtonian in one singular point in the infinity of positions in state space.

It's trivial to show Norton's extra solution, whilst being a correct solution to the differential equation, is not Newtonian at the 'singularity'. This is because it has a constant value for jounce or snap, with is a 4th order term that is the "acceleration of acceleration". This is what allows the particle to move off the apex despite zero velocity and acceleration there.

However, this in itself isn't anything to do with determinism, since his equation is still deterministic, it just isn't Newtonian. (It also violates conservation of energy and momentum, which nobody seems to notice but again is trivial to show).

The reason for the apparent non-deterministic result, is that he then creates a piecewise equation, stitching the two solutions together at arbitrary time t which is plain bizarre. It has no physical justification at all! The solutions are both mathematically correct, but even if both were Newtonian, you can't just stitch equations with different initial conditions together and claim that still represents physics. It's just nonsense.

I've given a more detailed breakdown of why he's wrong in this article if you're interested: https://blog.gruffdavies.com/2017/12/24/newtonian-physics-is-deterministic-sorry-norton/

Hope that helps!

Gruff
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    It's "non-Newtonian" only because you assumed that Newtonian physics only allows smooth curves. Norton clearly did not assume so, so his argument isn't wrong, but simply depend on an assumption that you disagree with. Also, the violation of conservation of energy is trivial to fix: instead of using a dome, just use a central force field with $F(r) = \sqrt{r}$. – MaudPieTheRocktorate Jun 07 '18 at 10:54
  • It has nothing to do with the surface curve being smooth - why do you think it is? – Gruff Jun 07 '18 at 20:51
  • I meant that the motion of the particle can be unsmooth, so that it can have a constant value for jounce or snap that suddenly becomes 0. – MaudPieTheRocktorate Jun 07 '18 at 21:21
  • Well that's by definition non-Newtonian - it violates Newton's first law. – Gruff Jun 08 '18 at 09:25
  • If you time-reverse it, then the ball rolls up the dome and... does it stay on the top, or not? According to Newton's law, at the top of the dome, the ball is at rest (has 0 velocity), and is not acted upon by a net force, so it would remains at rest. So this is Newtonian. Time-reverse it again, and it's still Newtonian. – MaudPieTheRocktorate Jun 08 '18 at 10:04
  • Time-reversibility isn't axiomatic in Newtonian mechanics (none of the three laws say anything about time-reversibility). So, if we we're talking about pure Newtonian mechanics, then the ball goes to the top and stays there. The fact that we have to lose time-reversibility at the apex (it's preserved everywhere else in this particular schema) tells us that Newtonian mechanics is incomplete. – Gruff Jun 08 '18 at 20:28
  • (BTW: the algebraic solution is continuous over the apex suggests that the ball should move away again, but this doesn't mean it's a correct interpretation of Newtonian law. It has no 'physical' justification. If we roll a ball with exactly the energy needed to reach the apex, it should remain there, not move away, and this is true for any shape with an apex, even a wedge. Being differentiable ["smooth"] or not, has no bearing on this. And none of this is new. Unstable equilibria in mechanics and electrodynamics were analysed in great detail by 18th century philosophers...) – Gruff Jun 09 '18 at 07:48
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    What do you mean by "Newton's First Law"? From my understanding, Newton's First Law is $F = 0$ if and only if $dv/dt = 0$, and says nothing about jounce or snap. Presumably, your Newton's First Law also requires $d^2v/dt^2=0$, and so on. Also, Newton's Laws of motion, as usually stated, are time-symmetric. It's not an axiom, but can be proven. – MaudPieTheRocktorate Jun 09 '18 at 08:19
  • No, the first law states that a body at rest will remain at rest unless acted upon by a force. – Gruff Jun 09 '18 at 13:45
  • (in an inertial frame) – Gruff Jun 09 '18 at 13:52
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    Consider a particle of mass $1$ at rest at $x=0$, and starting at time $t=0$, it's subject to a force $F(t)=t^2$. It's easy to integrate that its motion is $x(t)=t^4/12$ At time $t=0$, it is not subject to a force, so why doesn't it remain at rest after $t=0$? – MaudPieTheRocktorate Jun 09 '18 at 22:38
  • It moves because at positive $t$ the force is positive. There's no dilemma here because you've expressed your force in terms of time, not position (and as you haven't stated the physical justification for this I can only assume there is one - for example an external force field rising as $t^2$ - however, if there isn't one and you derived the time equation from a physical structure like a dome, then at $t=0$ your equation is invalid and we need to apply boundary conditions. – Gruff Jun 10 '18 at 09:55
  • Be careful not to confuse maths with physics. Maths is a tool we use to model physics. Just because you have an equation for something doesn't mean it represents something "physical". We have to apply constraints appropriate to our axioms and the schema that have physical meaning. – Gruff Jun 10 '18 at 09:55
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    In this situation, it still has a discontinuous jounce, so it surely violates Newton's First Law according to you? Also, Newton's three Laws are completely mathematical, and anything mathematical that satisfies the laws must be Newtonian. If you disagree, you are adding extra assumptions to physics that are beyond Newton's three Laws. – MaudPieTheRocktorate Jun 10 '18 at 21:33
  • If you say that the discontinuous jounce is due to the force having a discontinuous second-derivative at $t=0$, well how does the particle sense the second-derivative of force suddenly becoming nonzero, so that its jounce would suddenly become nonzero too? All it could sense is force. Assuming more requires you to assume non-Newtonian laws of motion. – MaudPieTheRocktorate Jun 10 '18 at 21:43
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    "it still has a discontinuous jounce, so it surely violates Newton's First Law according to you?" 1) You said that. I said nothing of the sort. I repeat: smoothness has nothing to do with a system being Newtonian or not. Only position and velocity are required to be continuous. Any other requirements are imposed by the physics we're trying to model. Newton's first law implies all time derivatives of velocity higher than two must be zero in the absence of a force. Otherwise objects start moving in the absence of a force. The first law forbids this. – Gruff Jun 11 '18 at 05:15
  • I can't answer the second question unless you frame it in terms of a physical system. What is the external/physical cause for the force to become non-zero? The answer to that is the answer to why the higher derivatives change... – Gruff Jun 11 '18 at 05:20
  • "all time derivatives of velocity higher than two must be zero in the absence of a force" but in my example, jounce becomes nonzero at $t=0$, in the absence of a force. And the physical cause of that force could be that, I could push on the particle myself with that force. – MaudPieTheRocktorate Jun 11 '18 at 05:36
  • I should qualify my statement as it applies to static forces. If you have forces with time derivatives then sure, you'll have non zero higher orders to match (otherwise you can't acquire acceleration). The point is that force causes motion in Newtonian physics. That's what we have to make the maths obey. – Gruff Jun 11 '18 at 09:48
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    I at first thought It violates the Newton's 1st law, but then I could not find the particular moment that it violates - therefore, it actually does NOT violates the 1st law.(if it does, I should have been able to find the violating moment) – Shing Aug 06 '18 at 04:54