Hi Anatoly, Your example is a slightly different model than the one I just suggested. I'm taking the limit in which the lattice atoms are fixed in place. In that case, symmetry demands the third cumulant to be zero. In a case such as you describe, the lattice atoms themselves can move "outward" (e.g. the material has a positive thermal expansion coefficient), which then yields a positive third cumulant. Naturally, no material has lattice atoms that are entirely fixed. But I can fairly easily conceive of a circumstance in which the lattice atoms have much less disorder than the interstitial atoms, and the distribution of the interstitial atoms is symmetric between two lattice atoms in crystallographically identical sites. In that case, the third cumulant may be considered "negligible," while the fourth cumulant might not be. --Scott Calvin Sarah Lawrence Collehe On Jan 21, 2009, at 10:12 AM, Frenkel, Anatoly wrote:
Hi Scott,
Third cumulant in your example will not be zero because this arrangement is symmetric only on the average. Locally, the interatomic pair potential (and the cumulants are the measures of the effective pair potential) which is the sum of the two potentials - between the interestitial and its neighbors on the opposite sides)- is still asymmetric, since the repulsive bruch of the potential is steeper than the attractive brunch. You can model your situation using two anharmonic pair potentials, e.g., Morse potential (see, for example, Rehr-Hung's paper in Phys Rev B in the 1990's, and I've done such calculations too, just in the case you described) and you will obtain that the effective pair potential is still analytically anharmonic and it has a non-zero third cumulant.
Anatoly