[Ifeffit] Cumulant expansion fittings

Matt Newville newville at cars.uchicago.edu
Wed Jan 21 12:27:50 CST 2009

Umesh, Scott, Anatoly,

I'm not sure what Scott means by a "lattice atom" in "the limit in which
the lattice atoms are fixed in place".  Materials have atoms.  Crystals and
lattice points.  But I think I do agree with his approach to constructing a
non-harmonic distribution for which the third cumulant is zero.

Cumulants are simple combinations of the moments of a distribution that are
particularly useful when one has an exponential function of a variable and
wants to model a distribution function of that variable.  For XAFS, the
distribution we care about is r (interatomic distance), which has moments:

 <r^n>  = | (r-r_0)^n g(r) dr

where g(r) is the distribution function about a value r_0 (not
necessarily the mean value for g(r), but usually close to it).

To remind everyone, cumulants are combinations of the moments:
     C1 = <r>  ~= r_0
     C2 = <r^2> - <r>^2
     C3 = <r^3> - 3<r^2><r> + 2<r>^3
     C4 = <r^4> - 3<r^2>^2 - 4<r^3><r> + 12<r^2><r>^2 - 6 <r>^4

(NB: The Wikipedia page on cumulants has different coefficients for C4.  I
believe the ones above are as in Kendell's Advanced Theory of Statistics
and also what are used at http://mathworld.wolfram.com/Cumulant )

For a normal (aka Gaussian) distribution, the 1st and 2nd cumulants (mean
and variance) are non-zero and all higher cumulants are zero.  We often
call this the "harmonic" case.

It is definitely possible to have distributions that are "symmetric"
(having a g(r) function such that g(r_0 - x) = g(r_0 + x) ) but not
harmonic -- I think that is what Scott is saying.  It should also be
possible to construct a distribution that has C3 = 0 and a non-zero C4.

The real question was "can one use the fourth cumulant without the third
cumulant in fitting (XAFS)"?  The answer is: Yes.  As Scott and Anatoly
suggested, doing that may not make a great physical model for g(r), but
perhaps Umesh has a good reason to try this.

My advice is to try it and see if you get a fourth cumulant that is clearly

My experience (and all the experiences I've heard of) is that the fourth
cumulant rarely matters.  This is probably related to the idea that the
cumulant expansion diverges for very disordered systems and you would
either need many higher order cumulants to describe such a distribution or
are much better off using a finite set of atomic distances with some model
for the weighting of the different distances.  The first option (using many
cumulants) is impractical, and probably computationally dangerous (as you'd
quickly explore issues with computer precision of floating point numbers).
The second option (often called a "histogram" approach in the Ifeffit
world, and modeled somewhat after The GNXAS Approach) has been used
successfully a number of times.


> On Jan 21, 2009, at 10:12 AM, Frenkel, Anatoly wrote:
>> Hi Scott,
>> Third cumulant in your example will not be zero because this
>> arrangement is symmetric only on the average. Locally, the
>> interatomic pair potential (and the cumulants are the measures of
>> the effective pair potential) which is the sum of the two potentials
>> - between the interestitial and its neighbors on the opposite
>> sides)- is still asymmetric, since the repulsive bruch of the
>> potential is steeper than the attractive brunch. You can model your
>> situation using two anharmonic pair potentials, e.g., Morse
>> potential (see, for example, Rehr-Hung's paper in Phys Rev B in the
>> 1990's, and I've done such calculations too, just in the case you
>> described) and you will obtain that the effective pair potential is
>> still analytically anharmonic and it has a non-zero third cumulant.
>> Anatoly
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