Leandro, This is a topic that seems to be slightly controversial. I don't want mis-represent the varying views/opinions, but today is a 'at the beamline' kind of day. So this reply is less well planned (coherent??) and a little more 'off the cuff' than I'd like it to be. Perhaps we start a wiki page about this discussion?? Also, everything below is 'in my opinion': The first thing is that EXAFS is sensitive to R. Period. It seems obvious, but it's sometimes easy to forget. As a result, I think notions like "perpendicular to the lattice planes" doesn't get you very far. Also, don't forget that the crystallography is an amazingly powerful technique for studying solids, and that the macroscopic thermal expansion parameter (dL/L)/dT is often determined from the crystallographically (atomic-scale) measurement (dA/A)/dT. (A= lattice parameter, L= bulk distance, T= temperature). To first approximation, dR/R from EXAFS is similar to dA/A from diffraction. Many people have equated these. In detail, and especially at high temperature, they are not the same. OK, so on to your questions:
Whenever we fit an EXAFS spectrum with Artemis/IFEFFIT, we get a /delR/ value. As far as my (very) limited comprehension goes, /delR/ is the difference between the theoretical (FEFF8) bond length, /R0/, and the result of the best fit to my data, /R/, which gives the experimental bond length. So, this should be the /first cumulant/ of my distance distribution, right?
Yes. Cumulants are simply one way to describe a distribution function of a variable (for us, R). They're especially convenient for functions that are exponential in that variable, and that's why they're often used in EXAFS. They're most useful when the distribution is near normal (or Gaussian). For complicated distributions, they don't work so well. delR is the first cumulant, C1, and is equal to the first moment of the distribution. It is the displacement from the starting center value, R0, Reff, etc. There is actually a subtlety in getting delR=C1 from EXAFS, as the EXAFS is not simply exponential in R, but also has a 1/R^2 term and R dependence in the mean-free-path term. These can be dealt with (and are dealt with in Ifeffit/Feffit), so that the delR, sigma2, third, and fourth *are* the cumulants of the atomic pair distribution. But that's not your question (yet??).
... when I came across some papers (P. Fornasini and G. Dalba in [PRL82, 4240, 1999; PRB70, 174301, 2004] and E. A. Stern in [J.Phys.IV 7, C2-137,1997]) where an issue is raised; it is said that the thermal evolution of the first cumulant /C1/* of the real distance distribution in a crystal (which was said to be equivalent to /delR/) is NOT equal to /A/, because in a crystal there are vibrations perpendicular to the bond direction which are not considered in the one-dimensional model. It is argued that the temperature dependence of these vibrations perpendicular to the bond direction give raise to a positive shift of the minimum of the effective pair potential, while the net thermal expansion /A/ accounts only for changes due to the asymmetry of the potential.
I haven't looked at their paper in a long time, but I think that Fornasini and Dalba have the math and basic explanation right. I don't so much like the 'perpendicular' v. 'parallel' distinction, but others do.
So, the main questions hammering my head are: - does /delR/ include any contribution from vibrations perpendicular to the bond direction? Can really /delR/ and /C1/* be the same quantity?
I think the answer to the first is 'No'. delR is the change in average bond length, and that's it. For the second, delR and C1 really are the same quantitiy.
- is the thermal evolution of /delR/ equal to the net thermal expansion /A/? Or should some correction for perpendicular vibrations be added to relate both quantities?
No. If you imagine two atoms vibrating independently about lattice points separated by distance A, you will be able to convince yourself that <R> >= A, just from the triangle inequality. In a simple approximation, the differnce is proportional to sigma2/R. The important result is that delR/R is not equal to delA/A. Hope that helps, or at least keeps the conversation going.... --Matt