[Ifeffit] sigma^2 values for multiple scattering paths (Scott Calvin/Shelly Kelly/Abhijeet Gaur); Re: Ifeffit Digest, Vol 92, Issue 5
Scott Calvin
dr.scott.calvin at gmail.com
Thu Oct 7 13:38:01 CDT 2010
Not at all unusual, Han Sen. If you think about the EXAFS equation,
you'll see that sigma^2 and amplitude primarily affect the amplitude
of the signal, while distances affect the position of the peak in the
Fourier transform (or equivalently, the spacing of peaks in chi(k)).
So sigma^2 and amplitude can trade off without affecting distance-
based aspects of the fit much.
That's why I suggested you try forcing the sigma^2 to a "reasonable"
value to see what happened to your fit. Sometimes none of the aspects
of the fit you're interested in depend strongly on the sigma^2 of low-
amplitude paths--particularly if what you're interest in is distances
or information that is in part derived from distances, like phase
identification. In those cases, the anomalous sigma^2 can be a "yellow
flag" (think about what might be causing it and decide if it's a
problem to your scientific case) rather than a "red flag" (drop
everything and resolve the problem before proceeding).
Also, note from the EXAFS equation that sigma^2 is weighted by k^2,
and amplitude is not. If fits using different k-weights result in
significantly different values of sigma^2, that can be a clue that the
issue is actually one of amplitude, as in your case.
At any rate, I'm glad you solved your issue in such a satisfying way!
--Scott Calvin
Sarah Lawrence College
On Oct 7, 2010, at 11:22 AM, Han Sen Soo wrote:
> Hello Shelly and Scott,
> Thank you both again for your suggestions. It seems that after
> making the MS path more linear in my cif file, the FEFF calculation
> increased the amplitude value of the path and dramatically
> increased the sigma^2 value in the fit. Strangely, the fit values
> for the distances remain pretty much the same and the statistical
> figures of merit have improved, but the sigma^2 values are now much
> more reasonable (about twice as large, but I have a more triangular
> than linear model, so you're right Scott, your explanation does not
> work for my case). I guess the increased amplitude made a difference?
>
> Hello Abhijeet, I used a rudimentary geometrical way to get my bond
> angles. For a 3 atom triangle M-O-A, the effective MS path length
> (R_MOA) is twice the sum of the individual bond distances. So if you
> have the R_MOA, R_MO, and R_MA distances from your fits, you can use
> R_MOA - R_MO - R_MA to get the O-A bond length. And with the 3 sides
> of the triangle, you can use the geometrical Cosine Rule to get any
> of the 3 bond angles. This is just geometry so I don't know what the
> error propagation for this would be.
>
> Thanks again everyone!
> han sen
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