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

Whoops.  On closer look sigma2 isn't very accurate either.

Jeremy


> -----Original Message-----
> From: ifeffit-bounces@millenia.cars.aps.anl.gov
> [mailto:ifeffit-bounces@millenia.cars.aps.anl.gov] On Behalf
> Of Kropf, Arthur Jeremy
> Sent: Wednesday, November 24, 2010 12:09 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 the
> question 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 the same
> > 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 the
> number of
> > 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
> 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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