# Dipole field issue in particle-mesh Ewald method with periodic boundary conditions

I am working on a thesis that makes a great use of molecular-dynamics simulations, and I am trying to understand how the particle-mesh Ewald method works. The problem is, I have difficulties understanding its very premise; now I will explain what I think I have learnt:

Long-range electrostatic forces do not converge if periodic boundary conditions are enforced; thus we cannot obtain them by summing pairwise interactions between every charge in the system if we take the periodic images of charges into account. The problem is, I see this as an issue with serious practical consequences, and I cannot imagine how a mathematical rearranging could solve it. I’ll make an example:

Let’s suppose we have an electrically neutral unit cell that gains an electric dipole moment during the simulation. This unit cell, and its dipole moment, would be instantly reproduced in all image cells. Let’s consider the electric potential in an arbitrary point: it would have an infinite number of shells of dipoles around, with area increasing as $r^2$ ($r$ is the shell radius) and the electric field dipole component would decrease as $r^{-2}$; the sum does not converge, so we would have an infinite electric potential in every point! Am I missing something?

I cannot see how PME or Ewald Summation, or any other algorithm, can solve a physical issue, unless that in some way those methods put additional boundary conditions. But I don’t see how. Can you help me understand? Thank you in advance.

EDIT: I was wrong about the infinite potential, because there is a cosine term in the dipole component that zeroes the potential in my proposed shell-by-shell calculation. Anyway, if we consider the electric field instead, we have it falling at $r^{-3}$; by adding the field produced by each shell we obtain a series of $1/r$ terms, that is still diverging at infinity, so my problem is still unsolved.

• Note that it’s totally fine to answer your own question. Also, your question does not have to reflect its own history – everybody can find it in the edit history if they need to. So please do not append edits at the and but correct what needs to be corrected (but take care that you do not invalidate existing answers). – Wrzlprmft Feb 6 '17 at 11:20

Well, the starting point is the fact that $1/r$ Coulomb interaction which diverges at infinity in k-space is only $1/k^2$ which converges at infinity, but diverges at 0. If one takes only parts of both which do converge, then it is possible to avoid divergences at all.

Do you agree with this and the problem appears at the next step or you do not agree with this starting point?

In fact, Ewald summation does not make magically divergent integrals converge. If you try to take the periodical system with "+" and "-" charges, the divergency at infinity stemming from Coulomb divergency is unphysical: at large distance, the average charge seen by distant probe charge is zero. However, if you break it in parts, the energy from interaction with "+" and "-" charges separately does diverge. Ewald summation is the rigorous mathematical trick to avoid calculation of interaction with "+"s and "-"s separately. One could as well compute the interaction with dipoles which would not diverge.

In your question, you are stressing the role of periodical boundary conditions. Are they really important? Have you tried to consider the infinite lattice instead?

• You said "then it is possible to avoid divergences at all" and "one could as well compute the interaction with dipoles which would not diverge" ; probably I am wrong, but I disagree in this points, because I think I've found a true physical divergence. I edited the question because I noticed that I was mistaking in calculating the potential in my example; but I still find the total field to be infinite in each point of space: am I wrong on this? Why? – data 1 Feb 4 '17 at 16:04
• @data1 Could you show, where the divergence appears? I mean, write the potential stemming from dipoles and obtain infinity somewhere. – Misha Feb 4 '17 at 17:00
• I put my example into a formula: $$\overrightarrow{E}\left(0\right)=\overset{\infty}{\underset{s=1}{\sum}}\overset{n_{d}}{\underset{i}{\sum}}\frac{3\left(\overrightarrow{p}\cdot\hat{r_{i}}\right)\hat{r}_{i}-\overrightarrow{p}}{4\pi\varepsilon_{0}r_{s}^{3}}$$ Where i is the dipole index in the shell $s$ we are considering, $r_s$ is the shell radius, $\hat{r}_i$ is the dipole position versor, $\overrightarrow{E}(0)$ is the electric field at origin, $\overrightarrow{p}$ the dipole moment (the same for all dipoles). $n_d$ increases as $r^2$, then no convergence, right? – data 1 Feb 4 '17 at 17:58
• Well, for the naive convergency of the field probably dipoles are not enough and one has to combine pairs of dipoles to quadrupoles. However, if we consider the potential for charges $$\phi = \sum_{s=1}^{\infty} \sum_{i=1}^{n_d} \frac{e_i}{r_s}$$ and compare with potential for dipoles $$\phi = \sum_{s=1}^{\infty} \sum_{i=1}^{n_d} \frac{\vec{d}\cdot\vec{r}_i}{r_s^3}$$ the second is definitely much better. – Misha Feb 4 '17 at 19:09
• Excuse me but I still don't get your point. If I had found a sum of two divergent series of opposite sign, we could argue if, once summed together properly, divergence may disappear; but I've found (or at least I think I have) a single diverging series: I see no mathematical escape from this. My idea is: or my calculation is wrong, or some additional conditions have to be enforced. – data 1 Feb 4 '17 at 19:19

I solved the issue. I was actually missing something. I didn't notice that the electric field at the center of a spherical shell of ideal dipoles is zero. I think it is not obvious if you look at the formula, but it can be demonstrated by using the continuous limit and integrating on the entire shell. Since we work with neutral simulation boxes (of course), and the other multipole components of the electric field are short-range, I think we can say that the total field converges. So the divergence is a mathematical issue indeed, not a physical one.