# Particle in a box

Say I have a point particle located at the center of a box and imagine that I give it a velocity v in some direction. It will bounce back and forth in different directions maintaining the same speed |v|. The question is: Given an infinite amount of time, will it ever pass through all of the points in the box.

The rule of bouncing is just that it reverses its motion along the normal component of the wall (ordinary reflection).

• What is the rule of bouncing? I guess mirror. Commented Apr 28, 2016 at 12:05
• – user57159
Commented Apr 28, 2016 at 12:13
• How many articles have you tried? ...No, not a spelling mistake.
– user328032
Commented Apr 28, 2016 at 12:24

Velocity in a 2-dimensional system is a 2-tuple $$v=\left( \dfrac{dx}{dt},\dfrac{dy}{dt} \right)$$, where $$(x,y)$$ is the position of the particle. There are certainly cases where the particles keeps bouncing back and forth between the same points (for example $$\dfrac{dx}{dt}=0$$ and $$\dfrac{dy}{dt}=0$$) without covering all the points in the box. From the screensaver on old televisions, we know that there are other paths for which the particle goes in 'endless circles' without visiting all the points in the box.
Many patters emerge from such type of motion. For example, the time spent at each point depending on different initial velocities $$t(x,y,v)$$. Particles often spend less time at the boundary because there are fewer ways of reaching there: a point at center can be reached from all directions while a point at the left boundary can only be reached from the right.
Between times $$0$$ and $$N$$, the particle traces out a path that is a union of line segments. Since the particle has constant speed, the total length of the path is finite. Hence, the number of those line segments with length greater than $$1/n$$ is finite. Taking a union over all $$n$$ and all $$N$$ shows that the total number of line segments in the path over the entire infinite span of time is countable. Each line segment has measure $$0$$ and a countable union of sets of measure $$0$$ has measure $$0$$. Therefore, the set of points that the particle will visit has measure $$0$$. Since the box has positive measure, it will not visit every point in the box.