Let me try again.
First of all, perhaps you missed that after my first post, I
acknowledged that my statement in the first post was mistaken? In
fact, I later said:
> 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.
The Lu and Stern paper, which I now agree with, comes up with an
equation I have attached as a png file.
That equation is derived for spheres of diameter D in a monolayer. It
does, incidentally, include a typo; the part inside the ln should
start with mu_2^2/mu_1^2. But that's just typographical, and doesn't
affect their subsequent argument.
Lu and Stern then make the following claim:
> Finally, the attenuation in N layers is given by (I/I0)^N, where I
> is the transmitted intensity through one layer. Chi_eff for N layers
> is then the same as for a single layer since N will cancel in the
> final result.
That's the statement I initially thought was incorrect, and thought
was equivalent to a non-random stacking of layers. Since I'm now
convinced that the statement is correct, I don't want to get bogged
down in a discussion of why I still find it a bit puzzling. Other than
my first post, no one else in this thread has yet challenged the Lu
and Stern statement, and some, such as Matthew Marcus, have supported
it.
The statement can be rephrased to say that, for a given particle
diameter D, the ratio of chi_eff to chi_real is independent of the
number of layers; that is, the distortion due to nonuniformity is not
reduced by adding additional identical layers.
I found this conclusion so startling that at first I rejected it.
Several others in this thread indicated that they too tended to think
that additional identical layers would reduce distortions in the
spectrum due to nonuniformity.
I then tested this principle that emerges from the Lu and Stern paper
in two other situations, and found that it holds in those as well.
Here are the details:
The first case I'll examine is a Gaussian distribution of standard
deviation sigma and average width x_o, with a distribution narrow
enough so that we can ignore the negative-thickness tail.
Grant Bunker has investigated this case, for example on page 88 of his
new textbook. The derivation is straightforward, and he finds that
(mu x)_eff = mu x_o - (mu sigma)^2/2
Your simulation beautifully demonstrates that when I use N layers of
this type, we can expect that the new mean thickness x_new = N x_o,
and sigma_new = sqrt(N) sigma. This gives:
(mu x)_new_eff = mu N x_o - [mu sqrt(N) sigma]^2/2 = N [mu x_o - (mu
sigma)^2/2] = N (mu x)_eff
Thus the new spectrum is exactly the same shape as the old spectrum--
just scaled up by a factor of N.
Therefore, after normalization, the spectrum of the multi-layer sample
will be exactly the same as for the single layer. (Note that this
analysis does not include the effect of signal to noise, harmonics,
etc.). Thus the distortion due to nonuniformity is independent of the
number of identical layers used.
My next thought was to see if this was a peculiarity of normally
distributed samples and monolayers of spheres. After all, extremely
nonuniform samples will tend to approach a Gaussian thickness
distribution as more layers are randomly stacked up--perhaps they
asymptotically approach the distortion present in the Gaussian case?
So I investigated pinholes. This time I did the calculation all the
way from scratch, just to make sure I understood all the assumptions.
Take a sample that has a fraction p of pinholes, with the remaining 1-
p of the sample being of uniform thickness x. In that case,
I = p I_o + (1-p) I_o exp(-mu x)
mu_eff = ln (I/I_o) = ln [p + (1-p) exp(-mu x)]
Now take two layers of that type, with the pinholes randomly
distributed. A fraction p^2 will still be pinhole, a fraction 2p(1-p)
will have thickness x, and a fraction (1-p)^2 will have thickness x^2.
So now
I/I_o = p^2 + 2p(1-p) exp(-mu x) + (1-p)^2 exp(-2 mu x)
mu_eff = ln[p^2 + 2p(1-p) exp(-mu x) + (1-p)^2 exp(-2 mu x)] = ln[p+(1-
p)exp(-mu x)]^2 = 2ln[p + (1-p) exp(-mu x)]
And that is exactly twice the mu_eff of the single layer--i.e., after
normalization it will be identical. Any distortions caused by pinholes
will be identical, despite the fraction of sample having pinholes all
the way through dropping from p to p^2.
Some in our community already realized that, but I didn't. In fact,
I'd go so far as to say I was shocked by the result. Judging by some
of the advice that is routinely given about sample preparation, I
suspect that this may come as a surprise to many others as well.
Hopefully that establishes what my understanding currently is.
Beyond that, we seem to be miscommunicating on some questions of
terminology:
On Nov 24, 2010, at 12:04 PM, Matt Newville wrote:
> Scott,
>
> You said:
>> the distortion from nonuniformity is as bad for four strips
>> stacked as for the single strip.
>
> As I showed earlier, a four layer sample is more uniform than a one
> layer sample, whether the total thickness is preserved or the
> thickness per layer is preserved.
>
>> 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 |
>
> Yes, the sample with 25 layers has a more uniform thickness.
>
>> 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.
>
> OK... The shape is the same, but the relative widths change.
>
> 24.75 +/- 0.50 is a more uniform distribution than 0.99 +/- .099.
> Perhaps this is what is confusing you?
The thicker sample is more uniform (in the sense of fractional
uniformity), but the distortion in the normalized mu(E) due to the
nonuniformity is identical. That is exactly what is surprising, and
what I initially did not believe. Now I am firmly convinced that it is
true.
>
>> So Lu and Stern got it right. But the idea that we can mitigate
>> pinholes by
>> adding more layers is wrong.
>
> Adding more layers does make a sample of more uniform thickness.
> Perhaps "mitigate pinholes" means something different to you?
By "mitigate pinholes" I mean that we can obtain a normalized energy
spectrum that is closer to the mu(E) we are trying to measure. But we
can't do that by adding more identically distributed layers if the
thick and thin spots are randomly stacked--we end up with exactly the
same normalized mu(E). (As usual, this analysis neglects signal to
noise, harmonics, and the like.)
I used a lot of words there to try to be precise. But basically,
stacking two layers of tape with the hope that pinholes will tend to
be cancelled out will not work.
>
> In your original message (in which you set out to "track down" a piece
> of "incorrect lore") you said that Lu and Stern assumed that layers
> were stacked "so that thick spots are always over thick and thin spots
> over thin". They did not assume that. Given that initial
> misunderstanding, and the fact that you haven't shown any calculations
> or simulations, it's a bit hard for me to fathom what you think Lu and
> Stern "got right" or wrong.
They got everything right. I was trying to save a different piece of
lore that I think is even more widely disseminated--that stacking thin
layers of tape reduces the amount of distortion due to nonuniformity
as compared to one thin layer of tape.
I thought I had shown calculations (the Gaussian case), and Jeremy has
shown simulations which confirm the result for the pinhole case.
> The main point of their work is that it is better to use more
> layers to get to a given thickness. You seem to have some objection
> to this, but I cannot figure out what you're trying to say.
I agree that your statement is a true one which is also consistent
with their paper, but would respectfully differ as to what the main
point of the paper is. The abstract reads:
> Powdered samples samples are commonly used to measure the extended X-
> ray absorption fine structure (EXAFS) and near-edge structure. It is
> shown that sizes of particles typically employed produce significant
> reduction in the EXAFS amplitude for concentrated samples. The
> distortion is calculated and found to be in good agreement with
> measurements on FeSi2 samples. To obtain accurate EXAFS amplitudes
> on powdered samples it is necessary to use particles significantly
> smaller than 400 mesh when the atom of interest is concentrated.
The use of increasing number of layers as the particles are sieved
more finely is done to provide a controlled experiment in which
differences in signal to noise and thickness effects are not a factor.
Suggesting that it is better to use more layers to get a given
thickness is an indirect way of getting at the real issue, which is
particle size. For a given particle size, multiple layers provide no
help whatsoever with nonuniformity-related distortions, but merely
allow for the desired signal to noise. Lu and Stern provide the
correct emphasis in their paper. It is only some of the subsequent
XAFS practitioners, including myself until about 24 hours ago, who
placed the stress on the multiple layers per se as addressing the
uniformity issue.
--Scott Calvin
Faculty at Sarah Lawrence College
Currently on sabbatical at Stanford Synchrotron Radiation Laboratory
