(Think about the plot of f vs sin(theta)/lamda).The f" contribution to anomalous scattering of F(000) is 0, just in contrast to that the real part in this (000) direction is the full number of electrons; i.e., electron does not anomalously scatter in this (000) direction. So, the phase of (000) stays safely at 0, or the symmetry-broken Friedel's law is broken (F000.ne.F-0-0-0).
(000) is not only centrosymmetric, but to itself, which is the only one in the diffraction space.
Lijun On Oct 14, 2010, at 9:28 AM, Dale Tronrud wrote:
Just to throw a monkey wrench in here (and not really relevant to the original question)... I've understood that, just as the real part of F(000) is the sum of all the "normal" scattering in the unit cell, the imaginary part is the sum of all the anomalous scattering. This means that in the presence of anomalous scattering the phase of F(000) is not zero. It is also the only reflection who's phase is not affected by the choice of origin. Dale Tronrud On 10/13/10 22:38, James Holton wrote:An interesting guide to doing phasing "by hand" is to look at direct methods (I recommend Stout & Jensen's chapter on this). In general there are several choices for the origin in any given space group, so for the "first" reflection you set about trying to phase you get to resolve the phase ambiguity arbitrarily. In some cases, like P1, youcan assign the origin to be anywhere in the unit cell. So, in general, you do get to phase one or two reflections essentially "for free", butafter that, things get a lot more complicated.Although for x-ray diffraction F000 may appear to be mythical (like the sound a tree makes when it falls in the woods), it actually plays a very important role in other kinds of "optics": the kind where the wavelengthgets very much longer than the size of the atoms, and the scatteringcross section gets to be very very high. A familiar example of this iswater or glass, which do not absorb visible light very much, but doscatter it very strongly. So strongly, in fact, that the incident beamis rapidly replaced by the F000 reflection, which "looks" the same as the incident beam, except it lags by 180 degrees in phase, giving theimpression that the incident beam has "slowed down". This is the originof the index of refraction.It is also easy to see why the phase of F000 is zero if you just look at a diagram for Bragg's law. For theta=0, there is no change in direction from the incident to the scattered beam, so the path from source to atomto direct-beam-spot is the same for every atom in the unit cell,including our "reference electron" at the origin. Since the structurefactor is defined as the ratio of the total wave scattered by astructure to that of a single electron at the origin, the phase of thestructure factor in the case of F000 is always "no change" or zero.Now, of course, in reality the distance from source to pixel via an atomthat is not on the origin will be _slightly_ longer than if you justwent straight through the origin, but Bragg assumed that the source anddetector were VERY far away from the crystal (relative to thewavelength). This is called the "far field", and it is very convenientto assume this for diffraction. However, looking at the near field can give you a feeling for exactly what a Fourier transform "looks like". That is, not just the before- and after- photos, but the "during". It is also a very pretty movie, which I have placed here: http://bl831.als.lbl.gov/~jamesh/nearBragg/near2far.html -James Holton MAD Scientist On 10/13/2010 7:42 PM, Jacob Keller wrote:So let's say I am back in the good old days before computers, hand-calculating the MIR phase of my first reflection--would I just set that phase to zero, and go from there, i.e. that wave willdefine/emanate from the origin? And why should I choose f000 over f010 or whatever else? Since I have no access to f000 experimentally, isn't it strange to define its phase as 0 rather than some other reflection?JPKOn Wed, Oct 13, 2010 at 7:27 PM, Lijun Liu<lijun....@ucsf.edu> wrote:When talking about the reflection phase:While we are on embarrassingly simple questions, I have wondered fora longtime what is the reference phase for reflections? I.e. a given phaseof say 45deg is 45deg relative to what? ========= Relative to a defined 0. Is it the centrosymmetric phases? ===== Yes. It is that of F(000). Or a theoretical wave from the origin? ===== No, it is a real one, detectable but not measurable. Lijun Jacob Keller ----- Original Message ----- From: "William Scott"<wgsc...@chemistry.ucsc.edu> To:<CCP4BB@JISCMAIL.AC.UK> Sent: Wednesday, October 13, 2010 3:58 PMSubject: [ccp4bb] Summary : [ccp4bb] embarrassingly simple MAD phasingquestion Thanks for the overwhelming response. I think I probably didn't phrase thequestion quite right, but I pieced together an answer to the question Iwanted to ask, which hopefully is right. On Oct 13, 2010, at 1:14 PM, SHEPARD William wrote:It is very simple, the structure factor for the anomalous scatterer isFA = FN + F'A + iF"A (vector addition)The vector F"A is by definition always +i (90 degrees anti- clockwise)with respect to the vector FN (normal scattering), and it represents the phase lag in the scattered wave.So I guess I should have started by saying I knew f'' was imaginary, the absorption term, and always needs to be 90 degrees in phase ahead ofthe f' (dispersive component).So here is what I think the answer to my question is, if I understoodeveryone correctly: Starting with what everyone I guess thought I was asking, FA = FN + F'A + iF"A (vector addition)for an absorbing atom at the origin, FN (the standard atomic scatteringfactor component) is purely real, and the f' dispersive term is purely real,and the f" absorption term is purely imaginary (and 90 degrees ahead).Displacement from the origin rotates the resultant vector FA in the complex plane. That implies each component in the vector summation is rotated by that same phase angle, since their magnitudes aren't changed fromdisplacement from the origin, and F" must still be perpendicular to F'. Hence the absorption term F" is no longer pointed in the imaginary axisdirection. Put slightly differently, the fundamental requirement is that the positive90 degree angle between f' and f" must always be maintained, but theirabsolute orientations are only enforced for atoms at the origin. Please correct me if this is wrong.Also, since F" then has a projection upon the real axis, it now has arealcomponent (and I guess this is also an explanation for why you don't getthis with centrosymmetric structures). Thanks again for everyone's help. -- Bill William G. Scott Professor Department of Chemistry and Biochemistry and The Center for the Molecular Biology of RNA 228 Sinsheimer Laboratories University of California at Santa Cruz Santa Cruz, California 95064 USA phone: +1-831-459-5367 (office) +1-831-459-5292 (lab) fax: +1-831-4593139 (fax) = ******************************************* Jacob Pearson Keller Northwestern University Medical Scientist Training Program Dallos Laboratory F. Searle 1-240 2240 Campus Drive Evanston IL 60208 lab: 847.491.2438 cel: 773.608.9185 email: j-kell...@northwestern.edu ******************************************* Lijun Liu Cardiovascular Research Institute University of California, San Francisco 1700 4th Street, Box 2532 San Francisco, CA 94158 Phone: (415)514-2836
Lijun Liu Cardiovascular Research Institute University of California, San Francisco 1700 4th Street, Box 2532 San Francisco, CA 94158 Phone: (415)514-2836
