In a circunference of any radius...

In a circunference of any radius, two particles start from the same point at the same time and start moving at the same constant speed. One goes through the diameter of the circunference, whipping back once it reaches the end, and the other one follows the line of the circunference. Why will they never meet?

Pi is irrational, so the relation between the distances travelled by both particles can not be expressed as a fraction (ratio). Therefore there is no combination of integer numbers of revolutions and straight travels.

Let the diameter = 1, wlog. No multiple of pi can ever be an integer, (if it could, pi would be rational) and the distance travelled by the diameter-bound particle will always be integer valued at the edge of the circle

I believe that for each real epsilon > 0, there's a time t for which the distance between the particles is inferior to that epsilon.
I don't know how to prove it because my math is a bit rusty so if someone a little clever than me could post a proof...

bump

Intuitively that seems true, but that doesn't mean it will ever be exactly 0 of course.

You could probably show it using the continued fraction of pi, and looking at the convergents.

Naturally, the details are left as an exercise.

could it be physically demonstrated?

potential for particles to meet is in starting position or the opposite position (half way in the circumference) these are the only places they can meet. Particle A travels distance x*π/2 Particle B travels distance x*1. The distance of the particle A from either of the potential crossing points can be expressed as (x*π/2)-trunc(x*π/2) [fractional part of x*π/2].
It seems to get closer to zero at certain intervals. These intervals seem to come further apart. It appears to never reach zero.
Who can explain why it gets closer to zero at certain intervals of the iteration? 0, 1, 7, 266, 33215, 364913 Why is this not random, rather these repeat several times before the next one?

This is true, but that doesn't mean they meet. They can only get arbitrarily close.

Perfect circles do not exist in the real world.

If |x-a| < epsilon for any epsilon, then x=a

Make epsilon a function of the number of revolutions, friend.

This is wrong.

It has to be less than epsilon for every epsilon greater than 0.

Also, this assumes x is constant (which is ok)

Let x1 be the position of the particle transiting the circumference of the circle and let x2 be the position of the particle transiting the diameter of the circle. Furthermore, let (x,y)=(r,0) be the position at which the two particles are initially located and let z be the angle.

X1 is given by:

X1(t)= r + rcosz*t

X2 (t)= r + (r/pi)cosz*t

Let t' be the time at which the two particles meet. Then:

X1 (t')=X2 (t')

And:

rcosz*t=(r/pi)cosz*t which implies that r=(r/pi) which is impossible unless r=0 which, incidentally, wouldn't correspond to a circle.

shut the fuck up you imbecile your post makes me fucking mad holy shit

I meant rcosz*t'=(r/pi)cosz*t' in the last sentence.

Seems right

why are you using the cos?

Polar coordinates. I oriented the circle so as to align the initial point of both particles with
(x,y)=(r,0). The x coordinate is used and any set of cartesian coordinates can be transformed as
(x,y)=(rcosz,rsinz).

Incidentally, the final equation r=(r/pi) does not depend on z-the cost in x1 and x2 cancel one another.

of course not, would require infinite force for the oscillating particle with constant speed

and mathematically?

Apologies. You were right to be confused; I accidentally used cos z instead of -sin z dz/dt

The argument beyond that mistake survives by replacing the cosz in x2 and x1 with -sin z dz/dt

>circunference

See