Continuity & Differentiability of Two Particles in Collision

Question

Suppose two particles A and B collide. Consider the position vs time plot of the trajectories of A and B, shown in green and red in the following diagram. The two trajectories "meet" at coordinates (x', t') indicating collision, although according to the Pauli exclusion principle, two particles cannot be at the same position at the same time.

Now, it is clear that the trajectories A and B are continuous and differentiable, indicating that both A and B have well-defined velocities throughout their trajectories. But now consider trajectory C formed by the part of the trajectory A before collision, and the part of the trajectory B after collision (at the point of collision, assume trajectory C has the value of trajectory A). Trajectory C is shown in blue:

My question is this: is trajectory C continuous and differentiable at the point of collision (x', t')? That is, does it have a well-defined velocity at point (x', t')?

My own thoughts on this problem so far is that trajectory C is not continuous at (x', t'), because the left-hand limit of trajectory C is the point on A's green trajectory near (x', t'), while the right-hand limit of trajectory C is the point on B's red trajectory near (x', t'). But I want to understand/formalize/prove this more rigorously.

Answer 1

I think I have found a solution to my own problem. The trick is, instead of considering infinitesimal particles directly, we first consider the case of circular particles with a non-zero radius r colliding at time t'. The following image shows two circular particles, A and B, colliding at time t':

Now, the composite trajectory C is formed by the part of the trajectory A before collision, and the part of the trajectory B after collision (at the point of collision, trajectory C has the value of trajectory A). Trajectory C is shown in blue:

Clearly for a non-zero radius, trajectory C is neither continuous nor differentiable at time t'. There is a jump in the position of the trajectory at t', and the left-hand derivative and right-hand derivative of C at t' are not equal:

$$ \lim_{h\to 0^-} \frac{C(t' + h) - C(t')}{h} \neq \lim_{h\to 0^+} \frac{C(t' + h) - C(t')}{h}$$

As long as the circular particles have a non-zero radius- even if the radius is an infinitesimal radius dr- or the particles have a non-zero separation in position, there will be a jump discontinuity in trajectory C at the time of their collision.

Of course, if the particles have exactly zero radius r = 0, which means that at the time of collision t' they are exactly at the same position with zero separation, than indeed trajectory C will be perfectly continuous and differentiable. But for all real-world situations, there will always be some separation between the colliding particles, with a minimum distance of a Planck length.

Answer 2

Correct me if I am wrong, but my thought process (a little physics inclined) to prove that the trajectory C is also continuous and differentiable was as follows :

Let A and B be two identical point particles with same mass who are initially moving in the trajectory given (before collision point). Now we know that point particles can only have a head on collision and assume the their collision is perfectly elastic and we also know that in a perfectly elastic head on collision of two particles, the particles exchange their velocities.

So here, after collision, we find that A and B have exchanged velocities and although B moves on it's own path and A on it's own path, since the particles are exactly identical and have exchanged velocities, it is as if B is now moving in the path of A (after the collision point) as if A did not collide with B at all (and vice versa).

Now, we know that the path of A (if it did not collide with B) and velocity would be continuous at all points (thus the path is differentiable at all points). Coincidentally, the path C that you have drawn combining A trajectory before collision and B trajectory after collision (by using previous reasoning) is just A trajectory when it would not have collided with B which we just proved is continuous and differentiable at all points.

Thus path C is differentiable at collision point as well.