Note: This post is an English adaptation of my original Chinese article (URL). Some parts have been modified for clarity, cultural relevance, or to better fit the English-speaking audience.
Let’s explore this from a mathematical perspective:
First, when we say that classical mechanics follows determinism, mathematically, it means the following:
Within the frame of classical mechanics, the mechanical differential equations of any system must always have a solution, and that solution is unique.
However, it is important to note that this does not imply that these mechanical differential equations must satisfy the sufficient conditions of the Existence and Uniqueness Theorem (for ODEs, this would be the Picard-Lindelöf Theorem or the Cauchy Existence and Uniqueness Theorem, among others; for PDEs, it could be the Cauchy-Kowalevski Theorem or the Lax-Milgram Theorem, among others). Since the sufficient conditions of the Existence and Uniqueness Theorem are not necessary conditions, we cannot prove that a differential equation lacks a unique solution. For instance, some ODEs may not satisfy Lipschitz continuity, but their solutions may still exist and be unique.
Now, for the Norton’s Dome problem, its mechanical differential equation is as follows (note, we are only considering the tangential equation, as we are more interested in how the ball moves along the surface of the dome. Here, $\vec{r}(t)$ represents the displacement vector from the apex of the dome to the position of the ball along its surface. Due to the radial symmetry of the geometric model, we can use the scalar $r(t)$ to simplify the calculations):
$$
\frac{d^2 r(t)}{dt^2} = \sqrt{r(t)}, \quad \left. \frac{dr(t)}{dt} \right|_{t=0} = 0 , \quad r(0) = 0\\
$$
Mathematically, we can verify (details omitted here for brevity) that the function on the right-hand side of the differential equation satisfies the Continuity for $r\ge0$, which guarantees the existence of at least one solution. However, the function does not satisfy the Lipschitz Continuity at $r=0$, which means we cannot guarantee the uniqueness of the solution.
Once again, to emphasize: since these conditions are sufficient but not necessary, we cannot prove that the differential equation lacks a unique solution.
So, let’s try solving it anyway (perhaps there is a unique solution, who knows? Haha). But in the end, we find, unfortunately, that there are infinitely many solutions.
$$
r(t) = \begin{cases} 0, & \forall \ t<t_0 \\ \dfrac{1}{144}(t – t_0)^4, & \forall \ t \geq t_0 \end{cases}, \quad \text{where } t_0 \in \mathbb{R}\\
$$
From a physical standpoint, constructing such a perfect system is likely impossible. It’s akin to assuming a perfect spherical object placed on a perfectly flat surface and calculating the pressure distribution. However, we know that in reality, the surface of the sphere must curve, otherwise, we would obtain an infinite pressure value.
Updated: 2024-09-11
I’ve seen some responses and comments suggesting that Norton’s Dome can be resolved by applying Newton’s 1st Law, leading to the conclusion that classical mechanics does adhere to determinism. I’d like to offer my perspective on this from a mathematical standpoint.
First, mathematically speaking, Newton’s 2nd Law is essentially a definition of force. I recall reading in a book that if we were to remove the concept of force entirely and rely solely on $\dfrac{\mathrm{d} \vec{p}}{\mathrm{d} t}$, all the results of physics would remain unchanged.
Similarly, Newton’s 1st Law is simply a special case of the 2nd Law where $\vec{F} = 0$. In fact, Newton’s 1st Law cannot hold prior to the 2nd Law, because before we’ve defined force, we cannot even interpret $\vec{F} = 0$. Thus, mathematically, Newton’s 2nd Law is more fundamental. Moreover, Newton’s 1st Law is not a definition but rather a proposition, and it is a true proposition, meaning it’s a theorem (based on its importance level).
As such, Newton’s 1st Law cannot solve this issue. The core problem is that, based on the displacement solution $r(t)$, we can derive the acceleration $a(t)$:
$$
a(t) = \frac{d^2 r(t)}{dt^2} = \begin{cases} 0, & \forall \ t<t_0 \\ \dfrac{1}{12}(t – t_0)^2, & \forall \ t \geq t_0 \end{cases}, \quad \text{where } t_0 \in \mathbb{R}\\
$$
We can further derive the force $F(t)$:
$$
F(t)=ma(t) = m\frac{d^2 r(t)}{dt^2} = \begin{cases} 0, & \forall \ t<t_0 \\ \dfrac{m}{12}(t – t_0)^2, & \forall \ t \geq t_0 \end{cases}, \quad \text{where } t_0 \in \mathbb{R}, \quad m \in \mathbb{R}^+ \\
$$
Clearly, for $t \geq t_0$, the force $F(t)$ is not a constant but a variable. Thus, if you wish to apply Newton’s 1st Law, it only holds for $t \le t_0$, since during this time $F(t)=0$, satisfying the conditions of Newton’s 1st Law. However, as soon as $\ t > t_0$, the force becomes $F(t)\ne0$, and Newton’s 1st Law no longer applies.
This shows that Newton’s 1st Law alone is insufficient to resolve the Norton’s Dome problem.
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