[Ifeffit] Is good EXAFS data possible with 0.01 transmission coefficient

[Matthew Marcus]

Just to amplify a bit on Carlo's point:  The transmitted signal is the sum of the transmissions through different areas of the sample, thus

	T/I0 = < exp(-mu(E)*t) >

where <...> is the spatial average.  However, what you measure is

	tmu_eff(E) = ln(I0/T) = ln<exp(-mu(E)*t)>  .

If t is everywhere the same, then the averaging brackets go away and you recover something proportional to mu(E).  However, if t varies
a lot, then you're doing something non-linear.

Consider the following scenario:  The sample consists of 90% nice, uniform material and 10% pinhole.  Now imagine that the transmissions
as a function of energy for the two different regions look like this (read this in a constant-width font):

	below edge		Above_edge(chi=0)	Above_edge(chi=0.1)
hole		1		1			1				(I'm ignoring non-resonant, background abs)
Mn		1		exp(-4)			exp(-4.4)			edge jump is 4, modulation of mu = 0.1

Now, the total transmission is 0.90*(transmission of Mn)+0.10*(transmission of hole):

	below edge		Above_edge(chi=0)	Above_edge(chi=0.1)
		1		0.1+0.9*exp(-4)	0.1+0.9*exp(-4.4)
		1		0.116484		0.11105

Taking the log, we have the percieved absorbance:
	below edge		Above_edge(chi=0)	Above_edge(chi=0.1)
		0		2.150001		2.19778
If there were no hole effect, the numbers here would be 0, 4 and 4.4.  Not only is the edge jump reduced, so the sample
"looks" a lot thinner than it is, but the change in absorbance above the edge now reads only 1.2% of the edge jump,
when it was supposed to read 10%.  The XAS spectrum is highly flattened on top.  Thus, a spectrum which should look like this
           *
          * *
          * *
          *  *
          *    *
         *
********

may wind up looking like this:

           *
          *  *
         **    *
         *
********

If you have harmonics or any sort of stray light getting into the transmission chamber, that's  effectively like
a hole in your sample and causes the same problem.  If you have a monochromator glitch, the energy distribution
and harmonic content change during the glitch and the beam shape can change too, resulting in a non-normalization
which gets worse for thick samples.

This reminds me of a brain-teaser Anatoly Frenkel posted here a while back.  He showed two spectra for Rh foil (K-edge) taken on
the same beamline on different runs, and the second had attenuated EXAFS and distorted XANES relative to the first, and he asked why that was.
The answer was that the second time, the beamline was misadjusted in such a way as to admit some white light with the monochromatic.

I hope this clears it up.  What to do about it is a lot less clear.
	mam
		

On 7/25/2013 3:43 PM, Carlo Segre wrote:
>
> Hi Damon:
>
> On Thu, 25 Jul 2013, Damon Turney wrote:
>
>> Thank you Matthew Marcus.  To answer your questions: Most of the
>> transmission loss is due to the Mn particulates.  The 0.01
>> transmission coefficient (a.k.a. ~4 absorption lengths) is in regards
>> to xray energies above the absorption edge.   Below the absorption
>> edge the transmission coefficient should be 0.1 to 0.4.  I don't
>> understand the definition of "edge jump" or how to calculate it (but
>> trust me I will be reading up on all of this ASAP).  Anyway, the
>> transmission coefficient contains all the required information.
>>
>
> The edge jump is basically the change in the absorption from just below the edge to the back-extrapolated baseline after the edge.  in your case, if the absorption is 0.4 below the edge and 0.001 above the edge (worst case, then assumint the variation of the absorption is all due to the Mn, you would have an edge jump of approximately
>
>        delta mu ~ ln(1/0.001) - ln(1/0.4) = 6.91 - 0.91 = 5.0
>
> This is  very large and would not work well for transmission at all because of the huge variation in intensity through the edge.  If you have your best case it would be
>
>        delta mu ~ ln(1/0.01) - ln(1/0.1) = 4.6 - 2.3 = 2.3
>
> This is acceptable even if not ideal for transmission.
>
>> I doubt that I have "pinhole effect" because the MnO2 is small
>> particulates (10 micron diameter) evenly dispersed throughout the 300
>> micron thick sample, so I doubt there are any pinholes.
>>
>
> The absorption length of 6.6keV photons in MnO2 is 6.9 microns so there is a good chance that your particles, even though they are relatively small will distort the spectrum because they are significantly bigger than one absorption length.  This could also cause some problems in fluorescence. however, if this is your particle size and it is constrained due to your experiment, you will have to work with it.
>
>> Matthew are these "distortions" that you speak of the same thing as
>> "harmonics" that I've read about in "XAFS for Everyone" that can
>> complicate XAFS analysis?
>>
>
> The distortions come because you have to take a ratio between the incident beam and the transmitted beam or the fluorescence signal and the incident beam intensity.   There are specific assumptions made in taking this ratio that, if not true, will distort your spectrum from what it really is. For transmission, the assumption is that any photon coming through the Io chamber has an equa probability of being absorbed in the sample.  If oyu have harmonice in teh beam (that is in stead of just having E, you have a component of photons with 3E or 5E), then this assumption is not valid. normally, when you have a relativley optically thin sample (not too much absorption, this is a good assumption.  When you have a highly absprbing sample, the harmonics will be the perdominant signal in the transmission chanber and thus you will ahve a problem with the ratio.
>
> For fluorescence, there are other assumptions which when violated will distort the spectrum.
>
> Carlo
>