[Ifeffit] Cumulant expansion fittings
frenkel at bnl.gov
Wed Jan 21 09:12:01 CST 2009
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.
From: ifeffit-bounces at millenia.cars.aps.anl.gov on behalf of Scott Calvin
Sent: Wed 1/21/2009 10:00 AM
To: XAFS Analysis using Ifeffit
Cc: Scott Calvin
Subject: Re: [Ifeffit] Cumulant expansion fittings
At first, that seems to me like a very odd thing to do. What motivates it in your case?
I say that because the cumulant expansion is a series expansion like, for example, a Taylor expansion: each successive term is typically supposed to be a smaller correction than the previous one. Of course, the third cumulant implies a radially skewed distribution, which the fourth cumulant does not (i.e. the third cumulant implies that the distribution is not symmetric about some distance). If I think about it, I can perhaps construct a situation where I expect that distribution to be symmetric, but not harmonic. For example, suppose there is a rigid lattice of atoms, with other atoms arranged interstitially half-way between lattice points. If the lattice atoms are treated as nearly fixed and are all of one type, then the interstitial atoms are equally likely to be less than half-way as more than half-way from one to the next. Thus, the third cumulant is zero by symmetry. The distribution may be very far from harmonic, though; in fact, it may have two minima that are not resolved (one slightly closer to one lattice atom, one slightly closer to the other). In that case, I'd think fitting a fourth cumulant but not a third would be justified.
The bottom line is that you should have a good physical reason, having to do with symmetries, for using the fourth without the third.
Sarah Lawrence College
On Jan 21, 2009, at 9:35 AM, umesh palikundwar wrote:
Cumulant expansion fittings are generally used, if there are more than one close shells of similar nearest neighbors (if I have understood it correctly). Third and fourth cumulants are used in the fittings to account for non-Gaussian distribution of the neighbors. I would like to known, can one use only fourth cumulant (i.e. without using the third cumulant) in the fittings?
I will be highly grateful to get the answer to my query.
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