Quoting Jacob Keller <j-kell...@md.northwestern.edu>:
Aha, so I have re-invented the wheel! But I never made sense of why f' is negative--this is beautiful! Just to make sure: you are saying that the real part of the anomalous scattering goes negative because those photons are sneaking out of the diffraction pattern through absorption-->fluorescence?
I doubt that this is a correct interpretation. It is f" which is related to absorption (and therefore to fluorescence) not f' ! In fact f' can be positive, even if there is absorption (and fluorescence). Examples: the f' factors of C, O, S, Cl and most other "lighter" elements are positive at the Cu K-alpha wavelength, but they are still absorbing.
The optical theorem relates \mu, the macroscopic absorption coefficient, to f", NOT to f' ! The amount that any material absorbs is in no way related to the f' factors of the atoms of which it is build up. But it is directly related to their f" factors. When you collect a fluorescence scan, you get a quantity which is directly related to f" and NOT to f' (the raw scan resembles already very much the spectral curve of f"). To get f', you have to perform a Kramers-Kronig transform.
The macroscopic counterpart of f' is dispersion, i.e. a change of phase velocity.
Marc
