Cumulant expansion fittings
Dear all, Cumulant expansion fittings are generally used, if there are more than one close shells of similar nearest neighbors (if I have understood it correctly). Third and fourth cumulants are used in the fittings to account for non-Gaussian distribution of the neighbors. I would like to known, can one use only fourth cumulant (i.e. without using the third cumulant) in the fittings? I will be highly grateful to get the answer to my query. Umesh _________________________________________________________________ What can you do with the new Windows Live? Find out http://www.microsoft.com/windows/windowslive/default.aspx
Hi Umesh, At first, that seems to me like a very odd thing to do. What motivates it in your case? I say that because the cumulant expansion is a series expansion like, for example, a Taylor expansion: each successive term is typically supposed to be a smaller correction than the previous one. Of course, the third cumulant implies a radially skewed distribution, which the fourth cumulant does not (i.e. the third cumulant implies that the distribution is not symmetric about some distance). If I think about it, I can perhaps construct a situation where I expect that distribution to be symmetric, but not harmonic. For example, suppose there is a rigid lattice of atoms, with other atoms arranged interstitially half-way between lattice points. If the lattice atoms are treated as nearly fixed and are all of one type, then the interstitial atoms are equally likely to be less than half-way as more than half-way from one to the next. Thus, the third cumulant is zero by symmetry. The distribution may be very far from harmonic, though; in fact, it may have two minima that are not resolved (one slightly closer to one lattice atom, one slightly closer to the other). In that case, I'd think fitting a fourth cumulant but not a third would be justified. The bottom line is that you should have a good physical reason, having to do with symmetries, for using the fourth without the third. --Scott Calvin Sarah Lawrence College On Jan 21, 2009, at 9:35 AM, umesh palikundwar wrote:
Dear all,
Cumulant expansion fittings are generally used, if there are more than one close shells of similar nearest neighbors (if I have understood it correctly). Third and fourth cumulants are used in the fittings to account for non-Gaussian distribution of the neighbors. I would like to known, can one use only fourth cumulant (i.e. without using the third cumulant) in the fittings?
I will be highly grateful to get the answer to my query.
Umesh
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 ________________________________ From: ifeffit-bounces@millenia.cars.aps.anl.gov on behalf of Scott Calvin Sent: Wed 1/21/2009 10:00 AM To: XAFS Analysis using Ifeffit Cc: Scott Calvin Subject: Re: [Ifeffit] Cumulant expansion fittings Hi Umesh, At first, that seems to me like a very odd thing to do. What motivates it in your case? I say that because the cumulant expansion is a series expansion like, for example, a Taylor expansion: each successive term is typically supposed to be a smaller correction than the previous one. Of course, the third cumulant implies a radially skewed distribution, which the fourth cumulant does not (i.e. the third cumulant implies that the distribution is not symmetric about some distance). If I think about it, I can perhaps construct a situation where I expect that distribution to be symmetric, but not harmonic. For example, suppose there is a rigid lattice of atoms, with other atoms arranged interstitially half-way between lattice points. If the lattice atoms are treated as nearly fixed and are all of one type, then the interstitial atoms are equally likely to be less than half-way as more than half-way from one to the next. Thus, the third cumulant is zero by symmetry. The distribution may be very far from harmonic, though; in fact, it may have two minima that are not resolved (one slightly closer to one lattice atom, one slightly closer to the other). In that case, I'd think fitting a fourth cumulant but not a third would be justified. The bottom line is that you should have a good physical reason, having to do with symmetries, for using the fourth without the third. --Scott Calvin Sarah Lawrence College On Jan 21, 2009, at 9:35 AM, umesh palikundwar wrote: Dear all, Cumulant expansion fittings are generally used, if there are more than one close shells of similar nearest neighbors (if I have understood it correctly). Third and fourth cumulants are used in the fittings to account for non-Gaussian distribution of the neighbors. I would like to known, can one use only fourth cumulant (i.e. without using the third cumulant) in the fittings? I will be highly grateful to get the answer to my query. Umesh
Hi Anatoly, Your example is a slightly different model than the one I just suggested. I'm taking the limit in which the lattice atoms are fixed in place. In that case, symmetry demands the third cumulant to be zero. In a case such as you describe, the lattice atoms themselves can move "outward" (e.g. the material has a positive thermal expansion coefficient), which then yields a positive third cumulant. Naturally, no material has lattice atoms that are entirely fixed. But I can fairly easily conceive of a circumstance in which the lattice atoms have much less disorder than the interstitial atoms, and the distribution of the interstitial atoms is symmetric between two lattice atoms in crystallographically identical sites. In that case, the third cumulant may be considered "negligible," while the fourth cumulant might not be. --Scott Calvin Sarah Lawrence Collehe 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
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:
/
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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Thanks, Matt--you said that much more clearly than I did. I'd add that, personally, I avoid using cumulants to account for unresolved features when possible. For example, suppose that a central metal atom is coordinated to six oxygen atoms in an octahedral arrangement. But suppose also that I have a "hunch" that the octahedron may be stretched along one axis, so that two of the oxygen atoms are a bit further away than the other four. This hunch may come from other reports in the literature, other structural probes, chemical modeling, etc. Depending on how much usable data I have in k-space and in how small the splitting is, the metal-oxygen scattering may appear as only one peak in the magnitude of the Fourier transform. One approach would be to use cumulants to try to handle the "disorder"--the radial distribution function is not symmetric, after all, and while I might not have the resolution to show splitting, the shape and position of the peak and its sidebands will be affected. A different approach would be to use two different paths, one for the equatorial atoms and one for the axial atoms, with appropriate coordination numbers. Then create a guessed parameter for the separation in distance between the two paths, and constrain the paths' MSRD's ("sigma2's") to be the same. Mathematically, this is only introducing one new free parameter, just as using a third cumulant introduces one new free parameter. The cumulant method has the advantage that it makes no judgements about what is causing the anharmonicity. The multiple-path method has the advantage that if the model is correct, it may provide a somewhat better fit, and it also provides information that may be of use in understanding the system (such as a value for the difference in average distance to atoms at the two kinds of sites). The disadvantage is that it can be misleading: the fit is "working" because it has a new variable to play with that models the anharmonicity, and a statistically good fit is not evidence that the particular model is correct. For example, just as anharmonicity of a single path can be used to approximate multiple unresolved paths, multiple unresolved paths can also be used to approximate the anharmonicity of a single path! So my personal approach is this. If I have a hunch as to what kind of splitting might be going on, I model that splitting, keeping in mind that a good fit in such a circumstance is not enough, on its own, to say that my model is physically correct. If I don't have a good hunch, I use cumulants. And, of course, if I suspect there is no splitting, but rather a path that is intrinsically anharmonic (e.g. all coordinated atoms at the same average distance, but the thermal variation about that average distance is not symmetric), then I also use cumulants. I hope that was helpful. --Scott Calvin Sarah Lawrence College On Jan 21, 2009, at 1:27 PM, Matt Newville wrote:
Umesh, Scott, Anatoly,
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 non-zero.
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.
--Matt
Matt, Scott:
There is a situation in one dimensional world, not so practically useful though, where the third cumulant is zero. It describes the 1st nearest neighbor interaction between a central atom (x) in the group of three atoms: A-----x-----A. Indeed, here the third cumulant is zero, the higher order - may be not, depending on the shape of the pair potential. Assume the pair potential as:
V(x) = 1/2*kx^2 + k3*x^3, (1)
Then, the asymmetric term vanishes since:
V_eff(x) = V(x) + V(-x) = kx^2. (2)
It is not obvious that there exist other examples. As I am sure you know, Hung, Hung/Rehr, Yokoyama, Pirog and others have studied behavior of cumulants in the most common structures, including fcc, hcp and bcc, using Morse potential. In all of them, the third cumulants were nonzero (albeit Hung/Rehr article contained an error in the third cumulant calculation, and these results were further generalized by Vila/Rehr in their recent paper.
Indeed, the potential in these cases is always a linear combination of crystallography-defined terms, e.g., for the fcc structure, assuming a displacement x along the nearest neighbor distance:
V_eff(x) = V(x)+2V(-x/2)+8V(x/4)+8V(-x/4)+4V(0) = 1/2*(5/2)*kx^2 + (3/4)*k3*x^3. (3)
If you take above equation for V(x) and plug it in, you will obtain that V_eff(x) is still anharmonic since the x^3 term is still present in the effective contribution. It does not matter in this case, whether the atoms surrounding the dopant are "fixed" or "not fixed", as long as their is an effective pair potential between the central atom and its neighbors described by equation (1).
Same results can be obtained for most other lattices. Note that even if for some displacement direction x the sum of the terms in Eq. (3) will have no x^3, there is no chance that it will be true for ALL directions, since there is nothing special in x pointing along the 1NN distance.
Thus, I am pretty much convinced, unless there is some mistake in my reasoning, that no case exists in 3D with a zero third cumulant.
Anatoly
________________________________
From: ifeffit-bounces@millenia.cars.aps.anl.gov on behalf of Matt Newville
Sent: Wed 1/21/2009 1:27 PM
To: XAFS Analysis using Ifeffit
Subject: Re: [Ifeffit] Cumulant expansion fittings
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:
/
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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Anatoly, You're right--3 dimensions ruins my symmetry argument. My mistake. On the other hand, I still suspect that there exists a realistic case where forcing the third cumulant to zero cause a much smaller increase in chi-square than forcing the fourth cumulant to zero; e.g., a broad, flat radial distribution function. For those of you out there who are relative novices, this is an entertaining and informative discussion, but I don't want to lose track of the practical point: It is very rare to find a system where the fourth cumulant is both necessary and sufficient. Either the potentials are close enough to harmonic that the fourth cumulant makes little difference, or they are so far from harmonic that the fourth cumulant alone is not enough. --Scott Calvin Sarah Lawrence College On Jan 21, 2009, at 10:11 PM, Frenkel, Anatoly wrote:
Thus, I am pretty much convinced, unless there is some mistake in my reasoning, that no case exists in 3D with a zero third cumulant.
Hi Scott, It could be an interesting direction, to use these type of lattice calculations to predict, as you suggested, what type of structures (or host compounds, for dopands), will, if not make it zero, which is probably impossible, but minimize third cumulant. Thus, it may be a rational way to design materials, at least hypothetically, with controlled thermal expansion, or even the lack of thereof. I do not care that someone may jump in and patent it, but I will appreciate a Porsche, if possible, when it is licensed. Anatoly ________________________________ From: ifeffit-bounces@millenia.cars.aps.anl.gov on behalf of Scott Calvin Sent: Wed 1/21/2009 10:30 PM To: XAFS Analysis using Ifeffit Subject: Re: [Ifeffit] Cumulant expansion fittings Anatoly, You're right--3 dimensions ruins my symmetry argument. My mistake. On the other hand, I still suspect that there exists a realistic case where forcing the third cumulant to zero cause a much smaller increase in chi-square than forcing the fourth cumulant to zero; e.g., a broad, flat radial distribution function. For those of you out there who are relative novices, this is an entertaining and informative discussion, but I don't want to lose track of the practical point: It is very rare to find a system where the fourth cumulant is both necessary and sufficient. Either the potentials are close enough to harmonic that the fourth cumulant makes little difference, or they are so far from harmonic that the fourth cumulant alone is not enough. --Scott Calvin Sarah Lawrence College On Jan 21, 2009, at 10:11 PM, Frenkel, Anatoly wrote:
Thus, I am pretty much convinced, unless there is some mistake in my reasoning, that no case exists in 3D with a zero third cumulant.
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OK, here's my alternative for modeling asymmetric distributions: I use for the effective RDF a Gaussian-broadened exponential tail
function. The tail function is defined as:
t(r) = (1/|w|)*exp(-r/w) r*sign(w) >0
0 r*sign(w) <= 0
This limits to a delta function for w->0. The complete function is the tail function convolved with a gaussian. The result has an
analytically-simple FT, so fitting in k-space is
easy. It adds in a string of cumulants with just one extra parameter. It has no physical significance, except that it looks sort
of like RDF's from asymmetric potentials.
mam
----- Original Message -----
From: "Frenkel, Anatoly"
Thus, I am pretty much convinced, unless there is some mistake in my reasoning, that no case exists in 3D with a zero third cumulant.
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Hi Anatoly, Scott, Of course, XAFS *is* a one-dimensional probe, not a three-dimensional one. At least ignoring for the moment the angular dependence of multiple scattering, XAFS is sensitive to g(r) only. Sadly, this is sometimes forgotten in the literature, and one sees attempts to distinguish "sigma^2_perpendicular" and "sigma^2_parallel", which is a good sign of a paper that is complete nonsense. Simply stated, there is a g(r), and this distributions has moments and cumulants. As Anatoly points out, one can (and many have) make models for pair potentials (giving the interaction between any two atoms) or effective pair potentials (more precisely called potentials of mean force and giving the net potential seen by a pair of atoms) and from these calculate g(r). This is not easy to do in general, and, as Anatoly points out, "success" in this business is rare for systems that are not mono-atomic close-packed metals. There are actually some good calculations of g(r) based on force constants for organic molecules as well, which tends to be a different extreme case. That's all to make two points. 1) a general approach to converting model potentials into g(r) is not a solved problem, and 2) the potentials used are still parametrized models. Given that, I don't buy any "proof" from discussion of pair potentials about what cumulants of g(r) can and cannot be. The more common approach is to parametrize g(r) instead of V_eff(r), as in Matthew Marcus' example, and also as the GNXAS approach does. This has the advantage of not needing to parametrize potentials. I agree that it seems unlikely to have C3 == 0 and a non-zero C4 for a real system. That's mostly because, in my experience, it is very rare to have C4 be clearly distinguishable from 0. On the other hand, I have seen C3 be both positive and negative, so I'd hesitate to say that it cannot be zero while C4 is not zero in a real system. If you're looking for a model g(r) that will give C3==0 and a non-zero C4, then g(r) = exp[-(r0+delta)^2/(2sigma2) ] + exp[-(r0-delta)^2/(2sigma2) ] will do the trick. The attached figure shows such a distribution, looking fairly realistic and like a flattened Gaussian. The script to generate this is and calculate its moments and cumulants is: #### #### script to generate symmetric, non-Gaussian distribution set my.r = range(1,4,0.01) set r0 = 2.50 set sig = 0.05 set delr = 0.05 set my.gr = gauss(my.r,r0+delr,sig) + gauss(my.r,r0-delr,sig) # calculate moments around the mean value (r0 here) set norm = vsum(my.gr) set r0 = vsum(my.gr * my.r) / norm set my.dr = my.r - r0 set mom_1 = vsum(my.gr * my.dr ) / norm set mom_2 = vsum(my.gr * my.dr^2 ) / norm set mom_3 = vsum(my.gr * my.dr^3 ) / norm set mom_4 = vsum(my.gr * my.dr^4 ) / norm # calculate cumulants: set cumul_2 = mom_2 - mom_1*mom_1 set cumul_3 = mom_3 - 3*mom_2*mom_1 + 2*mom_1^3 set cumul_4 = mom_4 - 3*mom_2*mom_2 - 4*mom_3*mom_1 + 12*mom_2*mom_1*mom_1 - 6*mom_1^4 show mom_1, mom_2, mom_3, cumul_2, cumul_3, cumul_4 newplot my.r, my.gr , xlabel = 'R (\A)', ylabel='g(R)' ###### The resulting moments and cumulants are: mom_1 = -0.139956462E-14 mom_2 = 0.005000000 mom_3 = -0.201455968E-16 cumul_2 = 0.005000000 cumul_3 = 0.847872485E-18 cumul_4 = -0.000012500 which, to within machine precision, means C3==0 and C4=/= 0. Now, whether such a C4 can be measured from real data, and whether it might be better to model this g(r) as two Gaussians would be good questions. --Matt
participants (5)
-
Frenkel, Anatoly -
Matt Newville -
Matthew Marcus -
Scott Calvin -
umesh palikundwar