Thanks, Bruce. 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? Yuan On 4/1/11 2:18 PM, "Bruce Ravel" wrote:

On Friday, April 01, 2011 05:03:17 pm Ping, Yuan wrote:

...

Dear IFEFFIT experts:

Is it possible to add a k^3 term in the phase for fitting to take into account anharmonic effect? The fitted coefficient will be proportional to the 3rd cumulant of Debye-Waller factor. I work with high-temperature high-pressure systems where anharmonic effect is not negligible.

Thanks. Yuan Ping

Hi Yuan,

Here is the relevant page from the Ifeffit reference manual:

http://cars9.uchicago.edu/~ifeffit/refman/node51.html

You want to use the "3rd" path parameter, which adds a term proportional to C3*k^3 to the sine function in the exafs equation.

See the attached figure for the place in Artemis where this parameter is introduced.

HTH, B

So the term is included? Please see the attached paper. I meant the Coj(k) in Eq. (5). Thanks. Yuan On 4/4/11 2:33 PM, "Bruce Ravel" wrote:

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

OK, so if you don't add a sigma2 to your FEFF input file, then IFEFFIT will do the correction properly. Another question is whether, if you add a sigma2 to feff.inp, what does FEFF do? mam On 4/4/2011 3:37 PM, Bruce Ravel wrote:

On Monday, April 04, 2011 06:01:20 pm Ping, Yuan wrote:

...

So the term is included? Please see the attached paper. I meant the Coj(k) in Eq. (5).

The answer is "yes, Matt wrote Ifeffit correctly". But I am beginning to suspect that I am will never be able to answer your question to your satisfaction in words. Fortunately, you can go code diving. Once again, free software is the cat's pajamas.

Here is where the theoretical chi(k) is assembled:

http://cars9.uchicago.edu/svn/ifeffit/trunk/src/lib/chipth.f

Your question is answered between lines 132 and 173.

B

Hi Matthew, On Mon, Apr 4, 2011 at 5:51 PM, Matthew Marcus wrote:

OK, so if you don't add a sigma2 to your FEFF input file, then IFEFFIT will do the correction properly. Another question is whether, if you add a sigma2 to feff.inp, what does FEFF do? mam

FEFF's ff2chi module does apply sigma2 to the complex photoelectron wavenumber. It does not make the correction for the 1/R^2 term described by Tranquada and Ingalls. Like Bruce says, for general use, putting sigma2 and S02 in feff.inp can be somewhat confusing, as these are not applied to the feffNNNN.dat files that feffit/ifeffit/artemis uses, but only to the chi.dat file. In some sense, feffit/ifeffit/artemis were written as replacements for FEFF's ff2chi module, summing path contributions with sigma2, S02, and a few more bells and whistles. --Matt

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) = 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" To: "XAFS Analysis using Ifeffit" 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@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/ _______________________________________________ Ifeffit mailing list Ifeffit@millenia.cars.aps.anl.gov http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit