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Suppose, I have a 2D simulation box whose corners are $A(-2,-1)$, $B(-2, 4)$, $C(5,4)$, $D(5,-1)$; I have a particle $P(6, 3)$ inside the simulation box.

The $X$ dimension of the box: $7$.

Therefore,

  • Original position of $P(6, 3)$
  • Image 1: $P'(6 - 7, 3) = P'(-1, 3)$ (wraparound in the x-axis direction)
  • Image 2: $P^"(-1 - 7, 3) = P^"(-8, 3)$ (wraparound in the opposite x-axis direction)
  • Image 3: $P^{'''}(6 + 7, 3) = P^{'''}(13, 3)$ (wraparound in the opposite x-axis direction)

According to the nearest image convention,

  • The nearest image of $P$ should be $P'(-1, 3)$

Duh! Isn't it common sense?

I haven't found anything special in this concept.

If so, why should the concept of "nearest/minimum/closest image" even come into the discussion of molecular simulation?

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    $\begingroup$ You need to add a second particle to your example, otherwise there's nothing for the minimum image convention to affect. It's not related to which "box" your atom is in, nothing cares about that (translational symmetry), it's related to which image of the other atoms you should consider. $\endgroup$ Jun 13, 2023 at 0:49

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I think you might be missing an important detail about the minimum image convention. Essentially, it tells you which images should be considered as the "real" particle for the purposes of computing interactions.

That is, if you take the main simulation cell and pick two points at opposite ends of the cell, those points would be far apart if there were no periodic boundaries. However, because there are periodic boundaries, those points are actually close to one another because the periodic image of one point is closer than the "actual" location of that point in the main simulation box.

The wikipedia page has a nice section discussing how the minimum image convention can be helpful. In short, using the minimum image convention is nice because it allows you to let particles drift past the boundary of the main simulation cell. Every time you compute a distance, you then have to find the shortest distance between a particle and all images of the other particle. Code to do this is given in the wikipedia page as well. Letting the particles continue past the cell boundary rather than explicitly wrapping the particles is convenient because it means the distance between all pairs of atoms is continuous. That is, particles don't ever appear to "jump" large distances in space.

So, you're right that there is nothing particularly profound about the minimum image convention, but many people find it counter-intuitive that the interaction between two atoms is not necessarily the interaction between two atoms in the main simulation cell.

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In your own example, the distance from $P'''$ to $P$ is the same as from $P'$ to $P$, namely seven units. So why should $P'$ be the minimum image and not $P'''$?

So no, as you yourself have just demonstrated, the minimum image concept is not that simple.


Minimum images arise in molecular dynamics because (1) we want to model bulk properties using periodic boundary conditions on, physically, absurdly small volumes (2) we often truncate forces and neglect them past some cutoff.

Under condition (1) alone, in theory, every pair of particles would have an infinite set of interactions (since particle 1 in the main box would interact with particle 2, and particle 2's adjacent image, and particle 2's next image, and so on ...). But under condition (2) pair interactions are now manageable: we have each particle keep a list of other particles whose main box images or adjacent images are within the cutoff, and simply evaluate those forces.

For example, let particle 1 be at the origin (0,0) and let particle 2's main image be at (8,0), in a box of width 10 with a force cutoff of 3. Then particle 1 experiences a force from particle 2's "minimum image" at (-2,0). Equivalently, particle 2 experiences a force from particle 1's "minimum image" at (10,0). (By the way, note how Newton's Third Law remains obeyed.)

Now, we are finally in a position to explain the minimum image convention: it is that only minimum image pairs are considered when calculating truncated image pairs. The most important effect of this convention is actually its corollary, that (roughly) the box size must be larger than twice the force cutoff in every dimension. This is necessary because the "neighbour list" is often evaluated by searching for all particles within the cutoff distance from a particle, and if the box size were less than twice the force cutoff, particle pairs would start to be multiply-counted in the neighbour list.

Note that this is neither trivial nor automatic. For example, the MD package LAMMPS does not follow the minimum image convention for pair forces. If the pair cutoff is large enough to warrant it, LAMMPS will simply "stack up" enough images to fill the neighbour list out to the cutoff. As such, LAMMPS can handle much smaller boxes relative to pair cutoffs (down to numerical error). Furthermore, long-ranged electrostatic corrections (such as Ewald, PME and PPPM) explicitly consider the entire set of infinite periodic images when calculating charges (using Fourier transforms into reciprocal space, and then truncating there at a frequency cutoff), so they don't follow the minimum image convention either. Variations of this concept are important for steered molecular dynamics simulations (when moving the centre of mass of one group away from another, their distance cannot be allowed to exceed half the box size, or there suddenly becomes no natural way to define a force between them).

So no, it is neither trivial nor simple. Either when it is applied, or when it is not, the minimum image convention is really important to understand.

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  • $\begingroup$ So why should P′ be the minimum image and not P′′′? --- Coz, one is inside the box, the other isn't. $\endgroup$
    – user366312
    Jun 12, 2023 at 11:16
  • $\begingroup$ So after all its just a computational trick? I mean lets say we want to calculate the interaction energy between atom 1 and the rest atoms. This would be $U = \sum_{\mathbf{r}_i \leq r_{cut}} U(1, i)$. With this technique we are just looking at most at one periodic image and as such we avoid finding the neighbors that satisfy the inequality (which can be computationally heavy). $\endgroup$
    – ado sar
    Aug 1, 2023 at 21:23

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