[Ifeffit] Cumulant expansion fittings

Scott Calvin SCalvin at slc.edu
Wed Jan 21 09:00:03 CST 2009


Hi Umesh,

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.

--Scott Calvin
Sarah Lawrence College

On Jan 21, 2009, at 9:35 AM, umesh palikundwar wrote:

>
> Dear all,
>
> 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.
>
> Umesh

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