Is Norton's dome valid (or does $\frac{d^2 \vec{p}}{dt^2} = \vec{0} \implies \frac{d^n \vec{p}}{dt^n} = \vec{0} \ \forall \ n > 2$)?

Question

I came across Norton's dome and I don't agree that it proves anything.

First, here's an obviously ridiculous and completely nonsensical example that I constructed from thinking about simple harmonic motion:

Consider a particle of unit mass at rest in free space, at $x = 0$. Newton's second law states that $\ddot{x} = 0$ at $t=0$. There's an obvious way to describe the particle's motion that satisfies Newton's second law of motion, and that is $x(t) = 0$. Let us denote this solution by $s_1(t)$.

However, here's another non-obvious way to describe the particle's motion that satisfies Newton's second law: $$ x(t) = \begin{cases} \sin(t - T) - t + T \quad &\text{if }\ t \ge T \\ 0 & \text{if }\ t < T \end{cases} $$ Let us denote this solution by $s_2(t)$.

Notice that $s_2(t)$, $\dot{s_2}(t)$ and $\ddot{s_2}(t)$ all obviously equal $0$ for all $t < T$.

Now consider time $t = T$. Notice that: $$ \begin{align*} s_2(T) & = \sin(T - T) - T + T = 0 \\ \dot{s_2}(T) & = \cos(T - T) - 1 = 0 \\ \ddot{s_2}(T) & = -\sin(T - T) = 0 \end{align*} $$

Thus, $s_2(t)$ is a completely valid solution which obeys the constraint that the particle is at rest at time $t=0$. However, notice that as $t$ grows beyond $T$, the particle starts to spontaneously shoot off to infinity in the negative $x$ direction. Also notice that $T$ is an arbitrarily chosen constant, it could assume any positive real number as its value.

Thus, the second solution states that the particle could spontaneously start shooting off to infinity in the negative $x$ direction after some arbitrary amount of time $T$.

So we have proven that Newtonian mechanics isn't deterministic, can harbor uncaused events, yada yada.

Why are Norton's dome and my example not valid?

They're incorrect because in those solutions, the particle isn't really at rest when the spontaneous motion starts occurring. Notice that the fourth derivative of $r(t) = \frac{d^4}{dt^4} \left( \frac{(t - T)^4}{144} \right) = \frac{1}{6}$, and that the third derivative of $s_2(t)$ at $t=T$ is equal to $-\cos(T-T) = -1$.

Clearly Norton believes that an object being "at rest" does not necessitate that $\frac{d^n \vec{p}}{dt^n} = \vec{0} \ \forall \ n \ge 2 $. He believes that Newton's first law does not imply $ \frac{d^2 \vec{p}}{dt^2} \implies \frac{d^n \vec{p}}{dt^n} = \vec{0} \ \forall \ n \ge 2 $.

However, believing this leads to all sorts of completely nonsensical predictions for even the simplest system one can think of, a particle at rest in free space, as I have shown above.

Therefore, it only makes sense to believe that $ \frac{d^2 \vec{p}}{dt^2} = \vec{0} \implies \frac{d^n \vec{p}}{dt^n} = \vec{0} \ \forall \ n > 2$.

So, coming to my questions: Are my arguments valid? Is Norton's second solution just plain incorrect? If you disagree with my arguments and instead agree with Norton, what do you think of my completely nonsensical second solution for a particle at rest in free space?

Answer 1

Your example is not analogous to Norton's dome.

Norton's point is that Newtonian mechanics says that the trajectory of a particle obeys $\ddot{x}(t) = F(x(t),\dot{x}(t), t)$ at every point in time $t$, and that physics hence is often interpreted to constitute a notion of "causation" between the force $F$ and the motion $x(t)$ as the solution $x(t)$ is usually claimed to be unique (as a consequence of various uniqueness theorems about solutions to differential equations, it often is) given an initial position $x(t_0)$ and velocity $\dot{x}(t_0)$.

What you're talking about has nothing to do with either Norton's dome - or, indeed, Newtonian mechanics - since your $s_2(t)$ only obeys the differential equation $\ddot{s}_2(t) = 0$ at $t \leq T$, and not for all times $t$. $s_2(t)$ is hence not a valid solution to the equations of motion.

In contrast to this, Norton's dome purposefully constructs a dome with a shape such that its associated $F$ acting on a particle sliding over it is not Lipschitz continuous, and hence the Picard-Lindelöf theorem can no longer guarantee uniqueness of the solution to the equatios of motion. Then, Norton shows explicitly that there are two different solutions $x(t)$ that both start with the particle at rest at the top of the dome (in the simple, colloquial sense $\dot{x} = 0$ - the particle isn't moving) and both solve the equations of motion for all times $t$.

Sure, you could force uniqueness in this case by says that you need to specify all derivatives of $x(t)$ at $t_0$, but that's not what physics does - a lot of classical physics is founded on the idea that position and velocity at a given time (or position and momentum) completely specify the state of a particle since they suffice to produce a unique trajectory via the equations of motion (there could be a digression about gauge theory here but that would be beside the point).

Also, note that e.g. for analytic functions, knowing all derivatives means knowing the value of the function at every point - this would mean the equations of motion lose their predictive power if we essentially need to know the solution already to fully specify it, and the idea of "causation" that Norton is concerned about would no longer enter.