<Rant>
It shouldn't be called 'self-absorption'.  That's a misnomer, which seems to 
have come from a 1992
paper (Troger, et. al."Full correction of the self-absorption in soft-fluorescence 
extended x-ray-absorption fine structure", PRB 46,3283 (1992).
The effect was described and analyzed in a 1982 paper, which called it an "attenuation 
factor": Goulon, et. al. "On experimental attenuation factors of the amplitude
of the EXAFS oscillations in absorption, reflectivity and luminescence 
measurements", J. Physique 43, 539 (1982).
</Rant>
        mam

On 4/3/2016 4:56 PM, Matteo Busi wrote:
Hi Bruce,
Thanks for your help, it's really appreciated.
I was trying to keep it simple but it seems I'm just messing around.

What I am trying to do is to perform a new developed self-absorption correction 
using collected fluorescence absorption coefficient data on a CuSO4 
(pentahydrate) capillar (cylinder) and spherical sample with Cu as absorber, 
having different values of molarity and penetration depth.
The correction expression requires these measured quantity:
μX(E) : the absorption coefficient due to a given core excitation of the 
absorbing atom
               - I used the background function for this ( bkg(E))
μo = μ(E) : photoelectric total linear absorption coefficient of the sample at 
incident energy E
                    - I used the xmu(E) for this
μh = μ(E) :photoelectric total linear absorption coefficient of the sample at 
fluorescence emission Ef
                    - I used the xmu(Ef) with Ef the K absorption edge of Cu ( 
8.9789 eV)
χ: and here I used the chi(E) values of the exported ascii .xmu file

Hope it's clear now.
Matteo


2016-04-02 19:00 GMT+02:00 <ifeffit-requ...@millenia.cars.aps.anl.gov 
<mailto:ifeffit-requ...@millenia.cars.aps.anl.gov>>:

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        1. Re: Ifeffit Digest, Vol 158, Issue 1 (Matt Newville)


    ----------------------------------------------------------------------

    Message: 1
    Date: Fri, 1 Apr 2016 15:56:29 -0500
    From: Matt Newville <newvi...@cars.uchicago.edu 
<mailto:newvi...@cars.uchicago.edu>>
    To: XAFS Analysis using Ifeffit <ifeffit@millenia.cars.aps.anl.gov 
<mailto:ifeffit@millenia.cars.aps.anl.gov>>
    Subject: Re: [Ifeffit] Ifeffit Digest, Vol 158, Issue 1
    Message-ID:
             <CA+7ESbrsbsFQGMGLwGNPJmVO=fgzksbv1sezr3m5prkxann...@mail.gmail.com 
<mailto:fgzksbv1sezr3m5prkxann...@mail.gmail.com>>
    Content-Type: text/plain; charset="utf-8"

    Matteo,


    On Fri, Apr 1, 2016 at 8:35 AM, Matteo Busi <baseb...@gmail.com 
<mailto:baseb...@gmail.com>> wrote:

     > Hi Bruce,
     > Now this is clear.
     > In my case the correction I have to perform requires these
     > measured/evaluated parameters: chi, mu (not sure if it's better to work
     > with normmu here) and the background function.
     > Is it physically meaningless to have the mu(k) and bkg(k) data? If that 
is
     > not the case I would like to have these two columns exported in the ascii
     > chi(k) file.
     > So then I can perform the correction and re-import the new corrected
     > chi(k) and make comparison with the uncorrected.
     >
     > Kind regards,
     > Matteo
     >
     >
    It would certainly help (and by that I mean "help you get the answers
    you're looking for") if you gave us more details about what you're trying
    to do rather than ask for how to what you think you want to do.

    Just to be clear: Converting from E to k does not involve a Fourier
    transform, just
        k = sqrt((2m*e/hbar**2)*(E-E0))

    for k in Ang^-1 and E in eV, that's  k ~= sqrt(0.262468 
<tel:%280.262468>*(E-E0))

    I'm not sure what you're trying, but I would image you want to use
    normalized mu(E) or at least pre-edge subtracted mu(E).  That is, what we
    call mu(E) is typically really -ln(I/I0)
    where I and I0 are not the actual intensities before and after the sample,
    but counts or counts per time in some detector that samples the flux.
    These values include more or less arbitrary scale factors (amplifier gains,
    etc) included.  That means that (unless you very careful) mu(E) does not
    have meaningful units, certainly not cm^-1 or cm^2/gram.

    Normalized mu(E) doesn't have such units either, but it's an easier place
    to start.

    --Matt
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