# The Norton's dome suggesting non-determinism in classical physics

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?

• physics.stackexchange.com/a/39673/25575 – Paracosmiste Sep 8 '15 at 18:05
• Yes these are solutions of the differential equation, is this your question? – Did Sep 8 '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 9 '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 5 '18 at 6:17

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.

• 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 9 '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 9 '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 9 '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 9 '15 at 10:41
• I see, that makes more sense to me. thank you for the elaborating! – Shing Sep 9 '15 at 11:01

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!

• 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 7 '18 at 10:54
• It has nothing to do with the surface curve being smooth - why do you think it is? – Gruff Jun 7 '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 7 '18 at 21:21
• Well that's by definition non-Newtonian - it violates Newton's first law. – Gruff Jun 8 '18 at 9:25
• "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 5:15