# What's the average position of oscillating particles in a box with periodic boundary conditions?

Imagine an open box repeating itself in a way that a if a particle crossing one of the box boundary is "teleported" on the opposite boundary (typical periodic boundary position in 3D).

Now put a hundred of oscillating particles inside. Each particle position is identified in a uniq fashion through time.

I want to get the average position of each of these particles.

EDIT:

Thanks for your input. But I think I did not explain myself correctly. Please forgive me for that. That's why I removed most of my wordy description and created the following picture as a 1D reduction of my problem instead. Both pictures:

• This is a 1D projection of the 3D problem.
• The axis is infinite.
• The blue line represents the 1D unit cell repeated by translation along the axis without overlapping.
• The red dot represents the position of one unique and identified particle.
• The green dots represents the equivalent particles to the red one obtained by translation. These are really equivalent to the red one: same nature, velocity, force field, etc.
• As an input, I only have particle positions inside of the 1D unit cell (the blue line).

Top picture:

• The red particle is present in the unit cell for each time step.
• I take the particle position $x(t)$ and average it over $t$ to get $\bar{x} = \frac{1}{4}$ of the unit cell in this case (black arrow).

Bottom picture:

• The red particle is crossing the boundary of the unit cell.
• However, $x(t)$ is only defined in the unit cell, therefore in this case at:
• $t=0$, the position of the green particle (on the right) is taken into account ;
• $t=1$, the position of the red particle (on the left) is taken into account
• and so on;
• If the average $\bar{x}$ is calculated without much care, it is false since this average is done between two different particles (red arrow instead of black arrow).

I developed a small algorithm that tries to correct this assuming that $x(t+\Delta t) - x(t) < \frac{\Delta x}{2}$, with $\Delta x$, the size of the unit cell.

However this algorithm is quite slow. I know I can improve the algorithm itself and my implementation of it. But I also know that this is a trivial (and inefficient) way to do it. I also hate to reinvent the wheel again and again...

Do you know any smart implementation or algorithm able to calculate this kind of "periodic" average?

Just to be clear, it sounds like your system is a 3-torus, $$\mathbb{T}^3 = S^1\times S^1\times S^1$$
Position should then be modulo the length of the relevant dimension. The trick is an understanding that any particle can never be more than $\frac{L}{2}$ distance from the center point of the cube in any one direction (although total distance might be greater as calculated with Pythagorean Theorem, but this can never exceed $D = \sqrt{{(0.5L_x)}^2+{(0.5L_y)}^2+{(0.5L_z)}^2}$).
I would try something like this; define the position of the particles relative to the center of the cube and give the particle a sign flip each time it crosses a boundary. So if velocity is positive along a given axis, and the distance from the center reference goes above $\frac{L}{2}$ then multiply by $-1$. You should then be able to calculate and average position of particles relative to the center of the box.