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
On Wed, Nov 24, 2010 at 2:27 PM, Kropf, Arthur Jeremy
Anatoly,
I think that may be exactly the point. If you have half the beam on a 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 the same? 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 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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