Re: [ccp4bb] [QUAR] Re: [ccp4bb] embarrassingly simple MAD phasing question (another)
Am 20:59, schrieb James Holton: ... The loss of the 1/r^2 term arises because diffraction from a crystal is compressed into very sharp peaks. That is, as the crystal gets larger, the interference fringes (spots) get smaller, but the total number of scattered photons must remain constant. The photons/area at the tippy-top of this transform-limited peak is (theoretically) very large, but difficult to measure directly as it only exists over an exquisitely tiny solid angle at a very precise still crystal orientation. In real experiments, one does not see this transform-limited peak intensity because it is blurred by other effects, like the finite size of a pixel (usually very much larger than the peak), the detector point-spread function, the mosaicity of the crystal, unit cell inhomogeneity (Nave disorder) and the spread of angles in the incident beam (often called divergence or crossfire). It is this last effect that often tricks people into thinking that spot intensity falls off with 1/r. However, if you do the experiment of chopping down the beam to a very low divergence, choosing a wavelength where air absorption is negligible, and then measuring the same diffraction spot at several different detector distances you really do find that the pixel intensity is the same: independent of distance. ... Hi James, as always, I enjoy your explanations a lot. Just one minor point - I would not quite agree that the solid angle of a reflection is _that_ exquisitely small, even in the absence of finite pixel size, mosaicity, unit cell inhomogeneity and crossfire (wavelength dispersion could be added to the list of non-ideal conditions). In fact a crystal is composed of mosaic blocks (size roughly 1 um, maybe bigger for some space-grown crystals), and the coherence length is (taking numbers from Bernhard) several 0.1 um to several 10 um. Thus, if we assume a wavelength of 1A, then the angle arising from the finite size of the mosaic-block-and-coherence-length-combined is on the order of 1A/1um, which at 100mm distance means a width of about 0.1mm - the size of a typical detector pixel. (It follows that if we build detectors with much smaller pixels than 0.1mm this won't help much in increasing the signal/noise ratio; in particular, it won't help to measure data from tiny crystals unless these are single mosaic blocks.) best, Kay smime.p7s Description: S/MIME Cryptographic Signature
Re: [ccp4bb] [QUAR] Re: [ccp4bb] embarrassingly simple MAD phasing question (another)
Actually, people forget the 1/r term because it is gone by the end of Chapter 6 of Woolfson. Yes, it is true that, for the single reference electron the scattered intensity falls off with the inverse square law of distance (r) and, hence, the amplitude falls off with 1/r. However, the units of intensity here is photons/area. This is not the same as the units of intensity for an integrated diffraction spot (photons). That is, spots are an integral over solid angle, so the 1/r^2 term goes away. A shame, really, that we use the word intensity to describe so many different things. Leads to a lot of confusion like this! It is also a real shame that certain individuals are so draconianly eristic about units because this discourages the use of colloquial units (like photons/um^2 or electrons) as an educational tool. The loss of the 1/r^2 term arises because diffraction from a crystal is compressed into very sharp peaks. That is, as the crystal gets larger, the interference fringes (spots) get smaller, but the total number of scattered photons must remain constant. The photons/area at the tippy-top of this transform-limited peak is (theoretically) very large, but difficult to measure directly as it only exists over an exquisitely tiny solid angle at a very precise still crystal orientation. In real experiments, one does not see this transform-limited peak intensity because it is blurred by other effects, like the finite size of a pixel (usually very much larger than the peak), the detector point-spread function, the mosaicity of the crystal, unit cell inhomogeneity (Nave disorder) and the spread of angles in the incident beam (often called divergence or crossfire). It is this last effect that often tricks people into thinking that spot intensity falls off with 1/r. However, if you do the experiment of chopping down the beam to a very low divergence, choosing a wavelength where air absorption is negligible, and then measuring the same diffraction spot at several different detector distances you really do find that the pixel intensity is the same: independent of distance. Yes, I have actually done this experiment! -James Holton MAD Scientist On 10/14/2010 8:33 PM, William G. Scott wrote: 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
Re: [ccp4bb] embarrassingly simple MAD phasing question (another)
Hi Bernhard, On Wed, Oct 13, 2010 at 08:07:04PM -0700, Bernhard Rupp wrote: [...] BR PS: Just in case it might come up - there is NO destructive interference between F000 and direct beam - the required coherence that leads to extinction/summation of 'partial waves' is limited to a single photon. Standard (non-FEL) X-ray sources are (with minor exceptions in special situations) not coherent. Has been discussed many times on bb. This sounds as though you are saying that a single photon interacts with several electrons to give rise to a reflection. If I understand Feynman diagrams correctly, this does not conform with current notion of photons. If your statement refers to a chapter in your book, please point that out before we discuss this on the bb so that I can set myself right. Cheers, Tim -- -- Tim Gruene Institut fuer anorganische Chemie Tammannstr. 4 D-37077 Goettingen phone: +49 (0)551 39 22149 GPG Key ID = A46BEE1A signature.asc Description: Digital signature
Re: [ccp4bb] embarrassingly simple MAD phasing question (another)
On Thu, 2010-10-14 at 08:41 +0200, Tim Gruene wrote: This sounds as though you are saying that a single photon interacts with several electrons to give rise to a reflection. Not only with several - it shouldn't be much of an exaggeration to say that the photon senses all the electrons in the Universe as it travels between the source and detector. Once it hits detector, it's trajectory magically collapses into a specific one. Quantum physics is undeniably crazy stuff :) Cheers, Ed. -- I'd jump in myself, if I weren't so good at whistling. Julian, King of Lemurs
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 Liulijun@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 Scottwgsc...@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 + iFA (vector addition) The vector FA 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
Re: [ccp4bb] embarrassingly simple MAD phasing question (another)
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 Liulijun@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 Scottwgsc...@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
Re: [ccp4bb] embarrassingly simple MAD phasing question (another)
yes there is a F000, it is always in diffraction condition independent of crystal orientation (hx+ky+lz) is always zero for any xyz when hkl = 000 There are no Miller planes but I guess you can think of a Miller volume F000 is normally not use in map calculations and that is why the average value of any such map is zero (all other Fourier terms are cosines that have just as much signal above as below zero). If you want to create a density map that actually represent electrons per cubic Angstrom, you need to add F000 which you can approximate as the number of electrons in the unit cell (assuming all other reflections are on absolute scale already) There is never really interference with scattered photons. Like James last message, a single photon can be scattered by two slits simultaneously. Likewise a photon is scattered by all scatterers in a certain volume of the crystal which includes many unit cell. This is the same for F000 reflections with the difference being that all atoms scatter in phase What took me a bit by surprise is that F000 would have a non-zero phase but I guess it is correct. If there were no imaginary component to F000 in the presence of anomalously diffracting atoms than the imaginary part of the density map would have an average of zero whereas it needs to be positive. Bart On 10-10-14 10:59 AM, Jacob Keller wrote: 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
Re: [ccp4bb] embarrassingly simple MAD phasing question (another)
Power on scattering by atoms is angle dependent, which is true for both the real and imaginary parts. (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, 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 Liulijun@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 Scottwgsc...@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
Re: [ccp4bb] embarrassingly simple MAD phasing question (another)
On Oct 14, 2010, at 7:40 AM, Ed Pozharski wrote: On Thu, 2010-10-14 at 08:41 +0200, Tim Gruene wrote: This sounds as though you are saying that a single photon interacts with several electrons to give rise to a reflection. Not only with several - it shouldn't be much of an exaggeration to say that the photon senses all the electrons in the Universe as it travels between the source and detector. Once it hits detector, it's trajectory magically collapses into a specific one. Quantum physics is undeniably crazy stuff :) Cheers, Ed. Less ephemerally, the photon scatters from every scattering center in the crystal lattice. Under these (incoherent scattering) experimental conditions, it is my understanding that the individual photon only interferes with itself. The quantum weirdness creeps in from the fact that the wave describing the scattering is spherically symmetric, sampled by the reciprocal lattice. But if a photon is a particle, and you were to do a single photon experiment, the particle of light can only wind up in one of the diffraction spot locations, but the diffracted wave determines the propensity of the photon to wind up in that location. It is basically the generalization of the single photon double-split paradox. I've found the headaches start to go away if you don't take the duality part of wave-particle duality too seriously. -- Bill
Re: [ccp4bb] embarrassingly simple MAD phasing question (another)
On Thursday, October 14, 2010 10:41:17 am Lijun Liu wrote: Power on scattering by atoms is angle dependent, which is true for both the real and imaginary parts. Actually, no. The f' and f terms are independent of scattering angle, at least to first approximation. This is why the signal from anomalous scattering increases with resolution. cheers, Ethan (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, 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 Liulijun@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
Re: [ccp4bb] [QUAR] Re: [ccp4bb] embarrassingly simple MAD phasing question (another)
On Thursday, October 14, 2010 12:12:18 pm Lijun Liu wrote: I think I need make it clear. Not their changes (f' and f) but their contribution to reflection intensities changes. f' and f are not changes. They are the real and imaginary components of anomalous scattering. They are wavelength dependent but not angle dependent. It is right at higher resolution, it turned to be increased. Changes against resolution is itself an evidence to that the contribution is angle dependent. The lower the resolution, the lower the contribution from those guys. The contribution from normal scattering, f0, is strong at low resolution but becomes weaker as the scattering angle increases. The contribution from anomalous scattering, f' + f, is constant at all scattering angles. Let us define the contribution from FAS = (f' + f). At low resolution: FAS / f0(angle) is a small number At high resolution: FAS / f0(angle) is a bigger number. To the lowest one (000), the contribution is 0. The contribution to all reflections including F[0,0,0] is the non-zero constant FAS. To see the effect this has on phasing power, etc, you might have a look at http://skuld.bmsc.washington.edu/scatter/AS_signal.html Lijun On Oct 14, 2010, at 11:13 AM, Ethan Merritt wrote: On Thursday, October 14, 2010 10:41:17 am Lijun Liu wrote: Power on scattering by atoms is angle dependent, which is true for both the real and imaginary parts. Actually, no. The f' and f terms are independent of scattering angle, at least to first approximation. This is why the signal from anomalous scattering increases with resolution. cheers, Ethan (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, 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
Re: [ccp4bb] [QUAR] Re: [ccp4bb] embarrassingly simple MAD phasing question (another)
On 10-10-14 01:34 PM, Ethan Merritt wrote: ... The contribution from normal scattering, f0, is strong at low resolution but becomes weaker as the scattering angle increases. The contribution from anomalous scattering, f' + f", is constant at all scattering angles. ... My simple/simplistic mental picture for this is that electrons form a cloud surrounding the atom's nucleus. The larger the diameter of the cloud the more strongly the atomic scattering factor decreases with resolution (just like increased B-factors spread out the electrons and reduce scattering). Anomalous scattering is based on the inner electron orbitals that are much closer to the nucleus and thus their scattering declines more slowly with resolution. By this reasoning f' and f" would still decline with resolution but perhaps the difference is so substantial that within the resolution ranges we work with they can be considered constant. By the same reasoning you'd expect neutron diffraction to have scattering factors that are for all practical purposes independent of resolution, assuming b-factors of zero. In addition, the different fall-off in the scattering factors for f0 and f' or f" will be much less noticeable for anomalous scatters with high B-values where the latter dominates the 3D distribution of the electrons. Bart Bart Hazes (Associate Professor) Dept. of Medical Microbiology Immunology University of Alberta 1-15 Medical Sciences Building Edmonton, Alberta Canada, T6G 2H7 phone: 1-780-492-0042 fax:1-780-492-7521
Re: [ccp4bb] [QUAR] Re: [ccp4bb] embarrassingly simple MAD phasing question (another)
On Thursday, October 14, 2010 01:18:04 pm Bart Hazes wrote: On 10-10-14 01:34 PM, Ethan Merritt wrote: ... The contribution from normal scattering, f0, is strong at low resolution but becomes weaker as the scattering angle increases. The contribution from anomalous scattering, f' + f, is constant at all scattering angles. ... My simple/simplistic mental picture for this is that electrons form a cloud surrounding the atom's nucleus. The larger the diameter of the cloud the more strongly the atomic scattering factor decreases with resolution (just like increased B-factors spread out the electrons and reduce scattering). Anomalous scattering is based on the inner electron orbitals that are much closer to the nucleus and thus their scattering declines more slowly with resolution. By this reasoning f' and f would still decline with resolution but perhaps the difference is so substantial that within the resolution ranges we work with they can be considered constant. I don't trust my intuition as things start getting quantum mechanical. A detailed theoretical treatment and summary of earlier results is given in Acta Cryst. (1997). A53, 7-14[ doi:10.1107/S0108767396009609 ] Investigation of the Angle Dependence of the Photon-Atom Anomalous Scattering Factors P. M. Bergstrom Jnr, L. Kissel, R. H. Pratt and A. Costescu To the extent that I dare attempt a summary, my understanding of this analysis is: At energies near the absortion edge in question the anomalous scattering is essentially independent of angle. At much higher energies this breaks down, but even at these higher energies the deviation is not substantial for scattering angles less than ~60 degrees. Ethan By the same reasoning you'd expect neutron diffraction to have scattering factors that are for all practical purposes independent of resolution, assuming b-factors of zero. In addition, the different fall-off in the scattering factors for f0 and f' or f will be much less noticeable for anomalous scatters with high B-values where the latter dominates the 3D distribution of the electrons. Bart -- Ethan A Merritt Biomolecular Structure Center, K-428 Health Sciences Bldg University of Washington, Seattle 98195-7742
Re: [ccp4bb] [QUAR] Re: [ccp4bb] embarrassingly simple MAD phasing question (another)
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
Re: [ccp4bb] embarrassingly simple MAD phasing question (another)
Good evening citizens and non-citizens, On Thu, Oct 14, 2010 at 08:21:19AM -0700, William G. Scott wrote: On Oct 14, 2010, at 7:40 AM, Ed Pozharski wrote: On Thu, 2010-10-14 at 08:41 +0200, Tim Gruene wrote: This sounds as though you are saying that a single photon interacts with several electrons to give rise to a reflection. Not only with several - it shouldn't be much of an exaggeration to say that the photon senses all the electrons in the Universe as it travels between the source and detector. Once it hits detector, it's trajectory magically collapses into a specific one. Quantum physics is undeniably crazy stuff :) Cheers, Ed. Less ephemerally, the photon scatters from every scattering center in the crystal lattice. Under these (incoherent scattering) experimental conditions, it is my understanding that the individual photon only interferes with itself. I would like to understand how the notion of a photon being scattered from all electrons in the crystal lattice explains the observation that radiation damage is localised to the size of the beam so that we can move the crystal along and shoot a different location. The quantum weirdness creeps in from the fact that the wave describing the scattering is spherically symmetric, sampled by the reciprocal lattice. But if a photon is a particle, and you were to do a single photon experiment, the particle of light can only wind up in one of the diffraction spot locations, but the diffracted wave determines the propensity of the photon to wind up in that location. It is basically the generalization of the single photon double-split paradox. The double slit paradox is actually not a paradox, and a single photon is not scattered by both slits: if you reduce the light intensity so that you really detect single photons, you observe that each photon decides on exactly one slit that it goes through. It is only the sum of many photons that create the typical pattering of the double slit experiment. The photon knows it is both wave and particle, but depending on the experiment we carry out we observe only one of the two phenomena, but never both. That's also the idea behing Schroedinger's cat. Cheers, Tim I've found the headaches start to go away if you don't take the duality part of wave-particle duality too seriously. -- Bill -- -- Tim Gruene Institut fuer anorganische Chemie Tammannstr. 4 D-37077 Goettingen phone: +49 (0)551 39 22149 GPG Key ID = A46BEE1A signature.asc Description: Digital signature
Re: [ccp4bb] [QUAR] Re: [ccp4bb] embarrassingly simple MAD phasing question (another)
On Thu, Oct 14, 2010 at 04:28:26PM -0500, 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. We don't do this because your crystal is angle dependent - it usually does not have the required degree of order to scatter thus far, so the anomalous signal drowns in the noise. JPK -- -- Tim Gruene Institut fuer anorganische Chemie Tammannstr. 4 D-37077 Goettingen phone: +49 (0)551 39 22149 GPG Key ID = A46BEE1A signature.asc Description: Digital signature
Re: [ccp4bb] embarrassingly simple MAD phasing question (another)
On Thu, 2010-10-14 at 23:31 +0200, Tim Gruene wrote: you observe that each photon decides on exactly one slit that it goes through. That is if you observe which slit it goes through. -- I'd jump in myself, if I weren't so good at whistling. Julian, King of Lemurs
Re: [ccp4bb] [QUAR] Re: [ccp4bb] embarrassingly simple MAD phasing question (another)
On Thursday, October 14, 2010 02:28:26 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, Yes, certainly. But the fall-off of intensity at higher resolution due to imperfect ordering (modeled by a B factor) is a separate issue from have an angular dependence of the scattering factors. Both reduce the measured intensity, but for different reasons. Ethan JPK -- Ethan A Merritt Biomolecular Structure Center, K-428 Health Sciences Bldg University of Washington, Seattle 98195-7742
Re: [ccp4bb] embarrassingly simple MAD phasing question (another)
On Oct 14, 2010, at 2:31 PM, Tim Gruene wrote: I would like to understand how the notion of a photon being scattered from all electrons in the crystal lattice explains the observation that radiation damage is localised to the size of the beam so that we can move the crystal along and shoot a different location. Modify it to all points in the lattice bathed in the beam. The double slit paradox is actually not a paradox, I agree with that, but for different reasons than what follows ... and a single photon is not scattered by both slits: if you reduce the light intensity so that you really detect single photons, you observe that each photon decides on exactly one slit that it goes through. Really? Why do you get interference fringes then? You need two (or more) slits to create the interference pattern, and the location of the subsidiary maxima in the interference pattern do not change with the intensity of the light source. If you dim it to the point where one photon per second emerges, and you wait long enough, you still get the identical interference pattern. You do not observe single-slit diffraction. However, if you put your used chewing gum in one of the slits, the pattern changes to that of a single-slit experiment. It is only the sum of many photons that create the typical pattering of the double slit experiment. No. That is false. That would give you the scalar sum of two intensity peaks with no interference patterns. You have to add the amplitudes with phases, not the intensities, to get the interference pattern. The photon knows it is both wave and particle, but depending on the experiment we carry out we observe only one of the two phenomena, but never both. That's also the idea behing Schroedinger's cat. Schrödinger actually developed the cat gedanken-experiment to illustrate that the conventional (Copenhagen) interpretation leads to absurd conclusions. But it sounds like you are talking about the Heisenberg uncertainty principle or scatter relation. Sure, you can observe both, or we couldn't count photons in individual diffraction spots (which is what we do when we measure their intensities). The scatter relation simply means you can't measure both simultaneously to arbitrary precision.
Re: [ccp4bb] [QUAR] Re: [ccp4bb] embarrassingly simple MAD phasing question (another)
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
[ccp4bb] embarrassingly simple MAD phasing question (another)
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? Is it the centrosymmetric phases? Or a theoretical wave from the origin? 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 + iFA (vector addition) The vector FA 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 + iFA (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 ***
Re: [ccp4bb] embarrassingly simple MAD phasing question (another)
On Oct 13, 2010, at 4:21 PM, Jacob Keller wrote: 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? Is it the centrosymmetric phases? Or a theoretical wave from the origin? Conventionally, the phase difference is defined in terms of the path difference between the path traveled by a photon scattering from a point at the origin and a photon scattering from a point displaced by some finite amount (usually a multiple of a unit cell vector) from the origin. The key is the only physically meaningful quantity is the difference in path length, not the absolute values. Hence, if that difference is zero, then the phase angle will be zero. So it is compared to the wave from the origin (which is not theoretical, although by the theory we use -- the First-Order Born Approximation -- it appears to violate conservation of energy. Fortunately, the beamstop prevents us from having to face that horrific reality.) -- Bill
Re: [ccp4bb] embarrassingly simple MAD phasing question (another)
It is relative to a single point electron at the origin. -James Holton MAD Scientist On Wed, Oct 13, 2010 at 4:21 PM, Jacob Keller j-kell...@fsm.northwestern.edu wrote: 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? Is it the centrosymmetric phases? Or a theoretical wave from the origin? 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 + iFA (vector addition) The vector FA 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 + iFA (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 ***
Re: [ccp4bb] embarrassingly simple MAD phasing question (another)
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 + iFA (vector addition) The vector FA 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 + iFA (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
Re: [ccp4bb] embarrassingly simple MAD phasing question (another)
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 + iFA (vector addition) The vector FA 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 + iFA (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
Re: [ccp4bb] embarrassingly simple MAD phasing question (another)
Does f000 mean the direct beam? Having a hard time imagining such a miller index or the corresponding planes... No, F000 is NOT the direct beam. I may not have made that clear enough in some of my drawings and captions, and it will be emphasized in the second printing/ebook. There is in fact scattering/diffraction in forward (2theta zero) direction, which coincides with, but is entirely different from, the primary beam. Elastically scattered photons experience a 180 degree phase shift, unscattered direct beam photons just experience - nothing. F000 has a clear relation in amplitude and phase (0) to the remaining observable Fs. Not intuitive, agreed. BR PS: Just in case it might come up - there is NO destructive interference between F000 and direct beam - the required coherence that leads to extinction/summation of 'partial waves' is limited to a single photon. Standard (non-FEL) X-ray sources are (with minor exceptions in special situations) not coherent. Has been discussed many times on bb.
Re: [ccp4bb] embarrassingly simple MAD phasing question (another)
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? I do not think this could be done to get the origin correctly defined (in most cases), if good old ways were used as did in the good old days. There are many restrictions on the definition on origins (space group-, actually, symmetry-dependent), and a simple setting-it-to-0 on phase angle will cause, or at least complicates the problem (also see below)-this could be imagined from the reverse way. 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? The definition is not a simple setting it to 0. It comes from the origin of phases problem. (000) reflection is the sum of the total electrons (TE) in the unit cell, based on this physical/crystallographic meaning, the (000) phase angle gets to be 0 (otherwise the sum will be less than TE due to at least one cos(non-0)). There is no point a shift on this value is applied, but a counter balance for this shift will have to be applied during the calculation later if a shift had been applied. no any merits for this! Lijun 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 + iFA (vector addition) The vector FA 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 + iFA (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 Lijun Liu Cardiovascular
Re: [ccp4bb] embarrassingly simple MAD phasing question (another)
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 Liulijun@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 Scottwgsc...@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 + iFA (vector addition) The vector FA 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 + iFA (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
