# In the lattice model, what's a nearest neighbor?

I was reviewing my grad Thermodynamics textbook to go over a concept that I never learned well: the Lattice model. One set of equations1 has: \begin{align} zN_1 &= 2N_{11} + N_{12}\\ zN_2 &= 2N_{22} + N_{12}\\ \end{align} where $z$ is the coordination number, or the number of nearest neighbors (defined twice).

My trouble is a simple thought experiment:

• pure solution of $1$
• $z = 10$
• $N = 100$

So then the first equation becomes $10{\times}100=2{\times}N_{11}+0$, so $N_{11}=500$. Five-hundred nearest pairs, right? But if I actually had a solution of 100 species 1 each with 10 neighbors, then I'd expect (ignoring boundaries) to have about $100{\times}10=1000$ pairs of $1-1$.

I can't seem to figure out what is meant by $N_{11}$. Could someone explain it?

1 J. Prausnitz, R. Lichtenthaler, and E. Gomes de Azevdeo. Molecular Thermodynamics of Fluid-Phase Equilibria, 3rd Edition, p.331, Eq.7-68.

A nearest neighbour in general terms is literally that: Find the closest atom of any given element, that is your nearest neighbour distance for that element in the lattice. The number of nearest neighbours for that element is the number of atoms that are at this distance from your starting atom.

$N_{11}$ is the number of nearest neighbour pairs of atoms of element 1, and $N_{12}$ is the number of nearest neighbour pairs between any atom of 1 and any atom of 2 etc. For the $N_{11}$ or $N_{22}$ case, you have to consider that each atom of 1 or 2 in the equation has a contribution to $zN$, hence the term being multiplied by 2 when viewed from the perspective of element 1 or 2.

In the case of your thought experiment, your expected value neglects edges (you can't just neglect edges!) so will be too high by a good margin.

The dissertation of Alexander Zhou1 covers these equations (p.34) and explains them slightly differently (and more clearly I think) than de With's textbook.2

1. Alexander Zhou. Explicit Non-Random Contribution in Lattice Fluid Equation of State. Explizite nicht zufällige Wahrscheinlichkeitsverteilungen in gitterbasierten Zustandsgleichungen. Doktor Ingenieur, Technische Fakultät der Universität Erlangen-Nürnberg, Nürnberg, 2013. See UB Friedrich-Alexander-Universität Erlangen-Nürnberg (FAU)
2. Gijsbertus de With. Liquid-State Physical Chemistry: Fundamentals, Modeling, and Applications; John Wiley & Sons: 2013. See Google Books
• It appears that you are linking "This link may help" to content inside a book. You may be lucky enough to reside in a country where this resource is available (it is not where I am), but in general it would be preferable to include a citation in readable form, so that other means of obtaining that text could be performed. – Martin - マーチン Feb 14 '17 at 6:44
• Looking back at this question, it looks like I accepted your answer, but forgot to +1 it. Ohh... right, I probably didn't have the rep to +1 back then. Hah, silly me. – Nat Mar 23 '19 at 4:11

The discrepancy between 500 and 1000 is due to double counting: the method of multiplying 100 by 10 counts each pair twice: one for each atom it is a part of. So we get the total number of pairs by dividing that number by 2: 500 pairs, still neglecting edges.