Dear All, I found in manual, that lamda ~ 1/sqrt(Ei) I want calculate lambda with fitting parameter Ei. How do it ? What is the exact dependence Ei and lambda ? During fitting I have example Ei=7,5, so I want calculate lambda. This dependence: lamda ~ 1/sqrt(Ei) is not a complitly. Yours faithfully Arkadisz Onyszko
On Saturday 13 May 2006 17:20, Arkadiusz Onyszko wrote:
Dear All, I found in manual, that lamda ~ 1/sqrt(Ei) I want calculate lambda with fitting parameter Ei. How do it ? What is the exact dependence Ei and lambda ? During fitting I have example Ei=7,5, so I want calculate lambda. This dependence: lamda ~ 1/sqrt(Ei) is not a complitly.
When fitting with Ifeffit, Ei is a correction to the broadening terms in Feff, one of which is the mean free path. The exact effect of Ei on the exafs equation as used in Feff and Ifeffit is explained in chapter 6 of this document: http://cars.uchicago.edu/~newville/feffit/feffit.ps Asyou can see from equation 6.2, Ei is actually a correction to the complex momentum. The complex momentum is related to the mean free path, but also to other things in Feff's energy model. This makes the precise interpretation of Ei rather subjective. Is Ei a correction to Feff's loss terms? Is it a correction to Feff's value of the mean free path? Both? Apparently, you are hoping that you can interpret Ei entirely as a correction to the mean free path. I'm not sure how you know that, but if so, Eq. 6.2 tells you how to do so. (And don't forget the constants -- 2m_e/hbar^2!) Hope that helps, B -- Bruce Ravel ---------------------------------------------- bravel@anl.gov Molecular Environmental Science Group, Building 203, Room E-165 MRCAT, Sector 10, Advanced Photon Source, Building 433, Room B007 Argonne National Laboratory phone and voice mail: (1) 630 252 5033 Argonne IL 60439, USA fax: (1) 630 252 9793 My homepage: http://cars9.uchicago.edu/~ravel EXAFS software: http://cars9.uchicago.edu/~ravel/software/
Dear Arkadisz, The ei parameter is a refinement to the imaginary part of Feff's complex photoelectron wavenumber, p, so that lambda(k) = 1 / Im[p(k)]. When Ei = 0, lamba(k) is easy to get as it is the lambda(k) straight from Feff. When Ei is not 0, ifeffit does not currently generate a 'refined lambda' -- it could. Also sadly, Ifeffit does not currently support complex math, so that calculating it yourself is somewhat harder than it should be. Here are some hints for how to do this. For a particular path, you could do this:
path(1, feff0001.dat) get_path(1, prefix=p1, do_arrays) plot p1.k, p1.lambda
With get_path(N, prefix=p1, do_arrays), arrays named p1.k, p1. amp, p1.phase, p1.caps, p1.rep, and p1.lambda are generated. 'rep' means 'The real part of p', and lamba is as above. All these are k-dependent functions. Also note that Re[p] is not exactly k, as the energy origins differ by a few eV. For non-zero Ei, the complex wavenumber p becomes p^2 = ( Re[p] - i / lambda ) ^2 - i * EI * 2m/(hbar*hbar) (where m = electron mass). Mapping that onto non-complex math is left as an exercise for the reader ;), but you'd still want to calculate lamba(k) as 1/ Im[p]. Hope that helps. If not, let me know, --Matt
participants (3)
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Arkadiusz Onyszko -
Bruce Ravel -
Matt Newville