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Here is a thought experiment. I have two steel balls on a steel surface. One is stationary at $(0, 0)$. The other is rolling towards it along the $x$-axis from the negative side. It will collide at time $0$. After that, the initially stationary ball will move along the $x$-axis. I am no longer interested in the other ball. I am interested in the position of the initially stationary ball: $x(t)$. This will be a $C^0$ (continuous) function since the ball cannot matter transport itself to a new location. It will also be $C^1$ (once differentiable) since the velocity won't change instantly either. However, we usually assume that the acceleration will suddenly change when the collision occurs. So, my position function is not $C^2$ (twice differentiable).

But if we look more carefully, the contact won't really be instantaneous. As the rolling ball approaches the stationary ball, the electrons of its outermost atoms will start to repel those of the stationary ball. So, the stationary ball will start to accelerate a little before contact. In fact, it might not be possible say exactly what is the moment of contact. So, in fact my position function will be at least $C^2$.

Now my question: how smooth will the function be if we look closely enough? Might it even be $C^{\infty}$?

badjohn
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1 Answers1

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Yes, it would be $C^\infty$, since at the lowest level the potential of repulsion will follow a $\propto \frac{1}{r^n}$ law and thus, it will be infinitely differenciable except for the value $r=0$, but, since Pauli's exlusion principle forbids this (and this is precisely the origin of the force), then we can say it is, indeed, $C^\infty$.

Jorge Casajus
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