On Oct 14, 2010, at 2:28 PM, Jacob Keller wrote:
> I have always found this angle independence difficult. Why, if the anomalous scattering is truly angle-independent, don't we just put the detector at 90 or 180deg and solve the HA substructure by Patterson or direct methods using the pure anomalous scattering intensities? Or why don't we see pure "anomalous spots" at really high resolution? I think Bart Hazes' B-factor idea is right, perhaps, but I think the lack of pure anomalous intensities needs to be explained before understanding the angle-independence argument.
>
> JPK
Yo Jacob:
I think one thing that got ignored as I followed the other irrelevant tangent is what f and F are.
f is the atomic scattering factor, and F is the corresponding Fourier sum of all of the scattering centers. This holds for f_0 vs. F_0, f' vs. F' and f" vs. F". The spots we are measure correspond to the capital Fs. Just like we add the f_o for each scatterer together and we get a sum (F) that has a non-zero phase angle, this also holds for F" (that is the part I missed when I posted the original question my student asked me).
The full scattered wave isn't given by f by the way. It is (1/r) * f(r) * exp(ikr) so the intensity of the scattered wave will still tail off due to the that denominator term (which is squared for the intensity). That holds for f_o, f' and f" unless I missed something fundamental.
People tend to forget that (1/r) term because we are always focusing on just the f(r) scattering factor.
-- Bill
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