[Ifeffit] Third cumulant in DWF

[Matthew Marcus]

The lambda is in the amplitude.
I assume FEFF does not integrate over the distribution of distances, which is where the term in question comes from.  That is, FEFF
calculates A (complex) in

    chi(k) = Im(A(k)*exp(-lambda*R+2*i*k*R)/R^2)

for a single R, but what you want with a non-0 ss (Artemis' notation for sigma2) is

    chi(k,ss) = <Im(A(k)*exp(-lambda*R+2*i*k*R))/R^2>

where <...> is the average over a Gaussian distribution of R with second moment equal to ss.  The derivation, for
any who are interested, starts with setting R = R0+dr and approximating 1/R^2 as (1/R0^2)*exp(-2*dr/R0) to get

    chi(k,ss) = Im[(A(k)/R^2)*Integral{Exp(-[(lambda+2/R0)+2*i*k]*dr-dr^2/(2*ss)),dr}/Integral{Exp(-dr^2/(2*ss),dr}] (Mathematica notation).

Completing the square and doing the integral yields the cited term (except with a 2/R0 not 1/R0) as well as the familiar
exp(-2*ss*k^2) amplitude.  Typically, the term is question is pretty small.
It results in a distance shift of ss*(1/lambda+2/R0).  If ss=0.01A^2 (fairly hefty), R0=2A and lambda = 0.1A^-1, then we get
a shift of 0.01A.  With an ss=0.01A^2, one might expect anharmonic effects of that order.  I also did not include in the above
any R dependence of A(k), either in magnitude or phase due to curved-wave effects.  Bruce, any idea if those are ever significant
in this context?  I suspect not.

Therefore, the question is whether, if you tell FEFF that there's a ss, it adds in the above factors, and whether IFEFFIT/Artemis
do that, assuming you care about a <~0.01A shift which would probably exist in any reference spectrum as well.
    mam

----- Original Message ----- 
From: "Bruce Ravel" <bravel at bnl.gov>
To: "XAFS Analysis using Ifeffit" <ifeffit at millenia.cars.aps.anl.gov>
Sent: Monday, April 04, 2011 2:33 PM
Subject: Re: [Ifeffit] Third cumulant in DWF


> On Monday, April 04, 2011 05:25:39 pm Ping, Yuan wrote:
>> Does the math expression in IFEFFIT include the term -4k*sigma2*(1/labmda
>> +1/R) in the phase? If yes, the 1st cumulant is sigma1= R+dR. If no,
>> sigma1= R+dR+2*sigma2*(1/labmda +1/R). It this correct?
> 
> When Ifeffit evaluates the exafs equation, it includes lamdba from the
> Feff calculation.
> 
> B
> 
> -- 
> 
> Bruce Ravel  ------------------------------------ bravel at bnl.gov
> 
> National Institute of Standards and Technology
> Synchrotron Methods Group at NSLS --- Beamlines U7A, X24A, X23A2
> Building 535A
> Upton NY, 11973
> 
> My homepage:    http://xafs.org/BruceRavel
> EXAFS software: http://cars9.uchicago.edu/~ravel/software/exafs/
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