[Ifeffit] delR versus thermal expansion A and first cumulant C1*

[Anatoly Frenkel]

Matt is right - and the triangular inequality is a good way to think about
it. Fornasini and Dalba, as well as Ed Stern in J. Phys. paper, estimated
this term more accurately:  2*sigma2/R should be added to the Delta R to
account for the transverse vibration and obtain the difference between the
bond distance in the unknown sample and the standard (experimental or
theoretical). However, if sigma2 is of the order of 0.01 A2, and 1NN
distance is of the order of 2.8 A as in gold, platinum or palladium, this
correction is only 0.007 A, which is smaller than thermal expansion in many
materials (that can be on the order of 0.02-0.06 A, depending on the
temperature range, bond strength (or Debye/Einstein temperature). Of course,
since sigma2 varies with temperature linearly in classical limit (at
temperatures great than Einstein temperature), this correction is
temperature dependent. That means, at 300 K it may be 0.007 A2 (and the
correction is 0.005 A), and at 400 K it may be 0.01 A2 (and the correction
is 0.007 A). However, if the thermal expansion (R2-R1) is measured with and
without this correction, the difference is only 0.002 A, which is for sure
comparable or most likely less than the error bar in R.

Thus, transverse vibrations, ideally, should be included for more accurate
bond length determination. If they are small, the one-dimensional model of
Frenkel-Rehr is a reasonable approximation.

Anatoly

Anatoly Frenkel
Yeshiva University

-----Original Message-----
From: ifeffit-bounces at millenia.cars.aps.anl.gov
[mailto:ifeffit-bounces at millenia.cars.aps.anl.gov]On Behalf Of Matt
Newville
Sent: Wednesday, October 05, 2005 5:23 PM
To: XAFS Analysis using Ifeffit
Subject: Re: [Ifeffit] delR versus thermal expansion A and first
cumulant C1*


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


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