Re: [Ifeffit] Distortion of transmission spectra due to particlesize

24 Nov 2010


      That looks right except for a minor quibble.  Your calc is for 1/3 foil and 2/3 "pinhole".  I think the equation should have an extra set of parenthesis:

set pinhole.xmu    = -ln((1 + exp(-good.xmu))/2)

Jeremy

PS.  That is a whole lot easier than what I did for the plot - export data to another program, do some calculations, and then reimport the results to Athena.  I really need to learn to use ifeffit better.  Thanks for the demonstration. :)
...
-----Original Message-----
From: ifeffit-bounces@millenia.cars.aps.anl.gov
[mailto:ifeffit-bounces@millenia.cars.aps.anl.gov] On Behalf
Of Matt Newville
Sent: Wednesday, November 24, 2010 3:01 PM
To: XAFS Analysis using Ifeffit
Subject: Re: [Ifeffit] Distortion of transmission spectra due
to particlesize
Hi Jeremy,
For a sample of thickness t in the beam, and a good measurement being
   I_t = I_0 * exp(-t*mu)
I think having 1/2 the sample missing (and assuming uniform
I_0) would be:
I_t_measured = (I_0 / 2)*exp(-0*mu)  + (I_0/2) * exp(-t*mu)
                        =  I_0 * (1 + exp(-t*mu) / 2
So that
   t*mu_measured =  -ln(I_t_measured/I_0) =  -ln (1 + exp(-t*mu)/2)
An ifeffit script showing such an effect would be:
   read_data(cu.xmu, group =good)
   plot good.energy, good.xmu
   set pinhole.energy = good.energy
   set pinhole.xmu    = -ln(1 + exp(-good.xmu)/2)
spline(pinhole.energy, pinhole.xmu, kmin=0, kweight=2, rbkg=1)
   spline(good.energy, good.xmu, kmin=0,  kweight=2, rbkg=1)
newplot(good.k, good.chi*good.k^2, xmax=18, xlabel='k (\A)',
          ylabel='k\u2\d\gx(k)', key='no pinholes')
   plot pinhole.k, pinhole.chi*pinhole.k^2, key='half pinholes'
With the corresponding plot of k^2 * chi(k)  attached.
Corrections welcome,
--Matt
...
Anatoly,
I think that may be exactly the point.  If you have half
...
foil and half off, even with a uniform beam, you cant get the same
spectrum as with the whole beam on the foil.
I tried to come up with a quick proof by demonstration, but I got
bogged down on normalization.  That will have to wait.
Jeremy
________________________________
From: ifeffit-bounces@millenia.cars.aps.anl.gov
[mailto:ifeffit-bounces@millenia.cars.aps.anl.gov] On Behalf Of
Frenkel, Anatoly
Sent: Wednesday, November 24, 2010 1:33 PM
To: XAFS Analysis using Ifeffit
Subject: RE: [Ifeffit] Distortion of transmission spectra due to
particlesize
Jeremy:
In your simulation, "(c) 1/2 original, 1/2 nothing (a large
"pinhole")" it appears that chi(k) is half intensity of the
original spectrum.
Does it mean that when the pinhole is present, EXAFS
wiggles are half
of the original ones in amplitude but the edge step remains
...
Or, equivalently, that the wiggles are the same but the
edge step doubled?
Either way, I don't think it is the situation you are describing (a
large pinhole). If there is a large pinhole made in a perfect foil
(say, you removed half of the area of the foil from the
footprint of
the beam and it just goes through from I0 to I detector,
unaffected).
Then, if I0 is a well behaving function of energy, i.e.,  the flux
density is constant over the entire sample for all
energies, EXAFS in
the both cases should be the same.
Or I misunderstood your example, or, maybe, the colors?
Anatoly
________________________________
From: ifeffit-bounces@millenia.cars.aps.anl.gov on behalf of Kropf,
Arthur Jeremy
Sent: Wed 11/24/2010 1:08 PM
To: XAFS Analysis using Ifeffit
Subject: Re: [Ifeffit] Distortion of transmission spectra due to
particlesize
It's not that I don't believe in mathematics, but in this
case rather
than checking the math, I did a simulation.
I took a spectrum of a copper foil and then calculated the
following:
(a) copper foil (original edge step 1.86)
(b) 1/3 original, 1/3 with half absorption, and 1/3 with 1/4
absorption
(c) 1/2 original, 1/2 nothing (a large "pinhole")
(d) 1/4 nothing, 1/2 original, 1/4 double (simulating two randomly
    stacked layers of (c))
Observation 1: Stacking random layers does nothing to
improve chi(k)
amplitudes as has been discussed.  They are identical, but
I've offset
them by 0.01 units.
Observation 2: Pretty awful uniformity gives reasonable
EXAFS data. 
If you don't care too much about absolute N, XANES, or Eo
(very small
changes), the rest is quite accurate (R, sigma2, relative N).
Perhaps I'll simulate a spherical particle next with
absorption in the
center of 10 absorption lengths or so - probably not an uncommon
occurance.
Jeremy
Chemical Sciences and Engineering Division Argonne National
Laboratory
Argonne, IL 60439
Ph: 630.252.9398
Fx: 630.252.9917
Email: kropf@anl.gov
...
-----Original Message-----
From: ifeffit-bounces@millenia.cars.aps.anl.gov
[mailto:ifeffit-bounces@millenia.cars.aps.anl.gov] On
Behalf Of Scott
Calvin
Sent: Wednesday, November 24, 2010 10:41 AM
To: XAFS Analysis using Ifeffit
Subject: Re: [Ifeffit] Distortion of transmission spectra due to
particlesize
Matt,
Your second simulation confirms what I said:
...
The standard deviation in thickness from point to point in
a stack of
N tapes generally increases as the square root of N (typical
statistical behavior).
Now follow that through, using, for example, Grant
Bunker's formula
for the distortion caused by a Gaussian distribution:
(mu x)eff = mu x_o - (mu sigma)^2/2
where sigma is the standard deviation of the thickness.
So if sigma goes as square root of N, and x_o goes as N, the
fractional attenuation of the measured absorption stays
constant, and
the shape of the measured spectrum stays constant. There
is thus no
reduction in the distortion of the spectrum by measuring
additional
layers.
Your pinholes simulation, on the other hand, is not the scenario I
was describing. I agree it is better to have more thin
layers rather
than fewer thick layers. My question was whether it is
better to have
many thin layers compared to fewer thin layers. For the
"brush sample
on tape" method of sample preparation, this is more like
...
...
we face when we prepare a sample. Our choice is not to
spread a given
amount of sample over more tapes, because we're already
spreading as
thin as we can. Our choice is whether to use more tapes of
...
...
thickness.
We don't have to rerun your simulation to see the effect of using
tapes of the same thickness. All that happens is that the average
thickness and the standard deviation gets multiplied by
...
...
layers.
So now the results are:
For 10% pinholes, the results are:
# N_layers | % Pinholes | Ave Thickness | Thickness Std
Dev | #     1   
|  10.0      |    0.900      |    0.300          | #     5 
   |  10.0     
|    4.500      |    0.675          | #    25    |  10.0      |   
22.500      |    1.500          |
For 5% pinholes:
# N_layers | % Pinholes | Ave Thickness | Thickness Std
Dev | #     1   
|   5.0      |    0.950      |    0.218          | #     5 
   |   5.0     
|    4.750      |    0.485          | #    25    |   5.0      |   
23.750      |    1.100          |
For 1% pinholes:
# N_layers | % Pinholes | Ave Thickness | Thickness Std
Dev | #     1   
|   1.0      |    0.990      |    0.099          | #     5 
   |   1.0     
|    4.950      |    0.225          | #    25    |   1.0      |   
24.750      |    0.500        |
As before, the standard deviation increases as square root of N.
Using a cumulant expansion (admittedly slightly funky for such a
broad
distribution) necessarily yields the same result as the Gaussian
distribution: the shape of the measured spectrum is independent of
the number of layers used! And as it turns out, an exact
calculation
(i.e.
not using a cumulant expansion) also yields the same result of
independence.
So Lu and Stern got it right. But the idea that we can mitigate
pinholes by adding more layers is wrong.
--Scott Calvin
Faculty at Sarah Lawrence College
Currently on sabbatical at Stanford Synchrotron Radiation
Laboratory
On Nov 24, 2010, at 6:05 AM, Matt Newville wrote:
...
Scott,
...
OK, I've got it straight now. The answer is yes, the
distortion from
nonuniformity is as bad for four strips stacked as for
On Wed, Nov 24, 2010 at 2:27 PM, Kropf, Arthur Jeremy
 wrote:
the beam on a
the same?
the question
the same
the number of
the single
...
...
...
...
strip.
I don't think that's correct.
...
This is surprising to me, but the mathematics is fairly clear.
Stacking
multiple layers of tape rather than using one thin layer
improves the
signal to noise ratio, but does nothing for uniformity.
So there's
nothing wrong with the arguments in Lu and Stern, Scarrow,
etc.--it's
the notion I had that we use multiple layers of tape to improve
uniformity that's mistaken.
Stacking multiple layers does improve sample uniformity.
Below is a simple simulation of a sample of unity thickness with
randomly placed pinholes.  First this makes a sample that
is 1 layer
of N cells, with each cell either having thickness of 1 or
0.  Then it
makes a sample of the same size and total thickness, but
made of 5
independent layers, with each layer having the same fraction of
randomly placed pinholes, so that total thickness for each
cell could
be 1, 0.8, 0.6, 0.4, 0.2, or 0.  Then it makes a sample with 25
layers.
The simulation below is in python. I do hope the code is
straightforward enough so that anyone interested can
follow. The way
in which pinholes are randomly selected by the code may not be
obvious, so I'll say hear that the "numpy.random.shuffle"
function is
like shuffling a deck of cards, and works on its array argument
in-place.
For 10% pinholes, the results are:
# N_layers | % Pinholes | Ave Thickness | Thickness Std
Dev | #    
1    |  10.0      |    0.900      |    0.300          |
#     5   
|  10.0      |    0.900      |    0.135          | #    25    | 
10.0      |    0.900      |    0.060          |
For 5% pinholes:
# N_layers | % Pinholes | Ave Thickness | Thickness Std
Dev | #    
1    |   5.0      |    0.950      |    0.218          |
#     5   
|   5.0      |    0.950      |    0.097          | #   
25    |  
5.0      |    0.950      |    0.044          |
For 1% pinholes:
# N_layers | % Pinholes | Ave Thickness | Thickness Std
Dev | #    
1    |   1.0      |    0.990      |    0.099          |
#     5   
|   1.0      |    0.990      |    0.045          | #   
25    |  
1.0      |    0.990      |    0.020          |
Multiple layers of smaller particles gives a more
uniform thickness
than fewer layers of larger particles. The standard deviation of
the thickness goes as 1/sqrt(N_layers).   In addition, one can
see that 5
layers of 5% pinholes is about as uniform 1 layer with
1% pinholes.
Does any of this seem surprising or incorrect to you?
Now let's try your case of 1 layer of thickness 0.4 with 4
layers of
thickness 0.4, with 1% pinholes.  In the code below, the
simulation
would look like
    # one layer of thickness=0.4
    sample = 0.4 * make_layer(ncells, ph_frac)
    print format % (1, 100*ph_frac, sample.mean(), sample.std())
    # four layers of thickness=0.4
    layer1 = 0.4 * make_layer(ncells, ph_frac)
    layer2 = 0.4 * make_layer(ncells, ph_frac)
    layer3 = 0.4 * make_layer(ncells, ph_frac)
    layer4 = 0.4 * make_layer(ncells, ph_frac)
    sample = layer1 + layer2 + layer3 + layer4
    print format % (4, 100*ph_frac, sample.mean(), sample.std())
and the results are:
# N_layers | % Pinholes | Ave Thickness | Thickness Std
Dev | #    
1    |   1.0      |    0.396      |    0.040          |
#     4   
|   1.0      |    1.584      |    0.080          |
The sample with 4 layers had its average thickness increase by a
factor of 4, while the standard deviation of that thickness only
doubled.  The sample is twice as uniform.
OK, that's a simple model and of thickness only.  Lu and
Stern did a
more complete analysis and made actual measurements of the
effect of
thickness on XAFS amplitudes.  They *showed* that many thin
layers is
better than fewer thick layers.
Perhaps I am not understanding the points you're trying to
make, but I
think I am not the only one confused by what you are saying.
--Matt
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