[Ifeffit] Cumulant expansion fittings

Scott Calvin SCalvin at slc.edu
Wed Jan 21 09:55:33 CST 2009

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

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