This F000 reflection is hard for me to understand:
-Is there a F-0-0-0 reflection as well, whose anomalous signal would have a
phase shift of opposite sign?
-Is F000 always in the diffraction condition?
-Is there interference between the scattered photons in F000?
-Does F000 change in amplitude as the crystal is rotated, assuming equal
crystal volume in xrays?
-Are there Miller planes for this reflection?
-Is it used in the Fourier synthesis of the electron density map, and if so,
do we just guess its amplitude?
JPK
----- Original Message -----
From: "Dale Tronrud" <det...@uoxray.uoregon.edu>
To: <CCP4BB@JISCMAIL.AC.UK>
Sent: Thursday, October 14, 2010 11:28 AM
Subject: Re: [ccp4bb] embarrassingly simple MAD phasing question (another)
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, you
can 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", but
after 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 wavelength
gets very much longer than the size of the atoms, and the scattering
cross section gets to be very very high. A familiar example of this is
water or glass, which do not absorb visible light very much, but do
scatter it very strongly. So strongly, in fact, that the incident beam
is rapidly replaced by the F000 reflection, which "looks" the same as
the incident beam, except it lags by 180 degrees in phase, giving the
impression that the incident beam has "slowed down". This is the origin
of 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 atom
to direct-beam-spot is the same for every atom in the unit cell,
including our "reference electron" at the origin. Since the structure
factor is defined as the ratio of the total wave scattered by a
structure to that of a single electron at the origin, the phase of the
structure 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 atom
that is not on the origin will be _slightly_ longer than if you just
went straight through the origin, but Bragg assumed that the source and
detector were VERY far away from the crystal (relative to the
wavelength). This is called the "far field", and it is very convenient
to 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 will
define/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?
JPK
On 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 for
a long
time what is the reference phase for reflections? I.e. a given phase
of 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 PM
Subject: [ccp4bb] Summary : [ccp4bb] embarrassingly simple MAD phasing
question
Thanks for the overwhelming response. I think I probably didn't
phrase the
question quite right, but I pieced together an answer to the question I
wanted 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 is
FA = 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 of
the f'
(dispersive component).
So here is what I think the answer to my question is, if I understood
everyone 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 scattering
factor 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 from
displacement from the origin, and F" must still be perpendicular to F'.
Hence the absorption term F" is no longer pointed in the imaginary axis
direction.
Put slightly differently, the fundamental requirement is that the
positive
90 degree angle between f' and f" must always be maintained, but their
absolute 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 a
real
component (and I guess this is also an explanation for why you don't
get
this 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
*******************************************
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
*******************************************
