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
