[Ifeffit] Anharmonic correction (Aaron Slowey)

Scott Calvin SCalvin at slc.edu
Fri Mar 26 15:32:43 CDT 2010


I concur with Grant.

But I would like to also make a complementary observation. For  
moderately disordered systems such as nanoscale samples, there's also  
no a priori reason to assume that the distribution is not skewed. In  
other words, the question of convergence isn't avoided by just  
ignoring higher cumulants!

In general, a reasonably good process is:

1) Try the fit with the third cumulant forced to 0 (i.e. not fit).

2) Try the fit with the third cumulant guessed.

3a) If the third cumulant refines to 0 to within the reported  
uncertainty and the other parameters don't move outside their original  
error bars, then the third cumulant is not needed; go back to fit 1.

3b) If the third cumulant refines to a nonzero value to within the  
reported uncertainty, then look at the value. If it violates the limit  
Grant gives, then it's not appropriate, but you need to find another  
way of dealing with the fit. (Sometimes you may have a splitting  
between two different path lengths that you haven't modeled, for  
instance.) If it is within Grant's limit, then proceed with the usual  
caution you accord to EXAFS fits. For example, evaluate the physical  
sensibility of parameters, the stability of the fit to small changes  
in the data ranges, etc..

--Scott Calvin
Sarah Lawrence College


On Mar 26, 2010, at 4:09 PM, grant bunker wrote:

> Aaron -
>
> There are a couple of things you should watch out for when fitting  
> cumulants.
>
> First, you should make sure in the fitting process that the third  
> cumulant C3 doesn't get much more than twice C2^(3/2) (i.e. 2  
> sigma^3) - values much larger than that are probably unphysical,  
> even if they happen to give you a better fit.
>
> Second,  the cumulant expansion loses its utility if it doesn't  
> converge quickly enough.  It's essentially an expansion in terms of  
> order k*sigma, and if that approaches 1 the higher order cumulants  
> may be large enough that convergence is questionable.  If you are  
> lucky and the effective distribution is Gaussian, or most of the  
> variance is due to Gaussian broadening of a skewed distribution, it  
> may converge OK, but that shouldn't be assumed a priori.
>
> Grant Bunker
>
> http://gbxafs.iit.edu
> _______________________________________________
> Ifeffit mailing list
> Ifeffit at millenia.cars.aps.anl.gov
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