Calculating bulk modulus from interatomic potential on cubic lattice How do I account for the lattice structure of a solid when computing the bulk modulus, given an interatomic potential?
The bulk modulus, $B$, can be expressed as the second derivative of the interatomic potential energy, $u$, with respect to the volume per atom, $V$:
$$ B = V_0 \, \frac{d^2u}{dV^2}\Bigr|_{V=V_0},$$
where $V_0$ is the equilibrium volume.  It seems like both $V_0$ and therefore $dV$ should depend on the lattice structure of the solid.
I hoped that given a good enough model for the interatomic potential one could calculate $B$ for an actual crystal to 10% or better, and then use the calculation as a group problem solving activity in a sophomore level course.
NaCl has a simple cubic lattice.
Using a spherical volume element should not work, because of the spherical packing problem.  The right thing to do seems to be to use a cubic volume element:
$$ V_0 = xyz = {r_0}^3 \quad\quad\mathrm{and}\quad\quad dV = yz\,dx + xz\,dy + xy\,dz = 3r^2\, dr.$$
Carrying through the calculation leads to:
$$B = V_0 \frac{1}{3r^2}\frac{d}{dr}\left[ \frac{1}{3r^2}\frac{du}{dr}\right]\Bigr|_{r=r_0} = \frac{{r_0}^3}{3{r_0}^2}\left[ -\frac{2}{3{r_0}^3}\frac{du}{dr}\Bigr|_{r=r_0} + \frac{1}{3{r_0}^2}\frac{d^2u}{dr^2}\Bigr|_{r=r_0}\right]$$
At the equilibrium lattice spacing $\frac{du}{dr}|_{r=r_0}=0$, so
$$ B = \frac{1}{9 r_0}\,\frac{d^2u}{dr^2}\Bigr|_{r=r_0}$$
Using an exponentially screened interatomic potential model and actual data for NaCl this gives about a factor of two too big for the bulk modulus.
Digging around in some old books I found a relevant exercise in Ashcroft and Mermin, Solid State Physics (1976).  In chapter 20 of that text problem 2 begins:

Show that the bulk modulus for an ionic crystal with the NaCl structure is given by $$ B = \frac{1}{18 r_0}\,\frac{d^2u}{dr^2}\Bigr|_{r=r_0}$$

So there's my factor of two!  The question remains, why?  Why is the prefactor $18$ and not $9$?
The text isn't particularly well cross referenced, so I'm having trouble digging out how to set things up from earlier chapters.
There's a similar example problem using a face-centered cubic lattice.  The authors, pull some nearest neighbor factors out of thin air and write down the volume element $V_{FCC}=\frac{r^3}{\sqrt{2}}$ with little explanation.  But for the simple cubic lattice, the lattice constant just is the nearest neighbor distance.  $V_{SC}=2r^3$ will give the right answer, but where does that $2$ come from.
How do I correctly set up the volume element for a simple cubic lattice?
 A: A little bit of pedantry first: NaCl has a face-centered cubic lattice structure with 2 atoms in the basis (1 Na and 1 Cl), which looks like a simple cubic structure only if we ignore that there are two different kinds of atoms. In the following, $r$ will be the minimum separation between atoms, i.e. the shortest distance between a sodium and a chlorine atom. In the case of NaCl, the equilibrium distance of $r$ is exactly half the lattice constant $r_0 = a/2$, and would be the lattice constant of the simple cubic lattice, if the two types of atoms were indeed identical. Going off your question, I assume this is what you used.
Regarding your calculation, Ashcroft and Mermin point out the convention that threw you off in a footnote, in my edition it's in chapter 20, ionic crystals, page 403, footnote 11, which reads

It is customary to calculate the cohesive energy per ion pair, rather than by ion. If there are $N$ ions, then there will be $N/2$ ion pairs.

Since your cube of side length $r$ only contains a total of half a cation and half an anion, that's only half an ion pair. The correctly sized volume to go with the given convention for $u$ is therefore one that contains a full ion pair, which must then consequently be twice as large, i.e. $V=2r^3$.
