Dear James,
I have a slightly different way to think about transverse coherence. I
heard Mark Sutton give a talk about this at the APS a few years ago, and
here are graphics from a similar talk given by Alec Sandy at BNL:
http://www.bnl.gov/nsls2/workshops/docs/XPCS/XPCS_Sandy.ppt
The eqn. he gives on figure 8 is this:
L(coherence) =
Lambda*(source-to-observation-point-distance) / 2*pi*sigma(source)
This is the wavelength divided by the angle subtended by the source viewed
by the observer, with a 2.pi in there for some reason. The way I explain
it (though I cannot derive it this way) is that as you view the source,
consider that your eye sees a ray coming from the top of the source, and
one coming from the bottom. As you move your eye up and down, the two
rays will slide back and forth against one another. The coherence length
is how far you move your eye to have them slip about 1/4 of a wave. In
Mark's APS example you see the horizontal is 7 microns, the vertical is
200. This reflects the fact that the typical synchrotron source is much
wider than it is high.
Bob
> ________________________________________
> From: CCP4 bulletin board [[log in to unmask]] on behalf of James Holton [[log in to unmask]]
> Sent: Thursday, April 24, 2014 6:59 PM
> To: [log in to unmask]
> Subject: Re: [ccp4bb] AW: [ccp4bb] Twinning VS. Disorder
>
> There are two kinds of "coherence length": transverse and longitudinal. Longitudinal coherence is often quoted as delta-lambda/lambda, which is easy to calculate but unfortunately completely irrelevant for diffraction from crystals. If it weren't then Laue diffraction wouldn't produce spots.
>
> Transverse coherence tends to be around 3-10 microns with 1 A x-rays, depending on the detector distance. Yes, that's right, the detector distance. Longer detector distances give you a bigger coherence length, especially when the source is "very far away", like it is at a synchrotron.
>
> How this happens is easiest to picture if you consider the simplest possible diffraction situation: a "point" source of x-rays, two atoms, and a detector. As long as the atoms are very close together relative to the distances from the sample to the source and the detector, then you have the "far field" diffraction situation. This is where both atoms are within the "coherence length", Bragg's diagram for Bragg's Law holds: parallel incoming rays and parallel outgoing rays.
>
> But what if the atoms are very far apart? Obviously, the scattering from two atoms on different sides of the room will just add as intensities. And if they are very close together, then Bragg's Law holds and they scatter "coherently". What most people think of as the "coherence length" is the point of transition between these two kinds of scattering.
>
> This point is rather conveniently defined as the distance between two atoms when the path from the source to one atom to a given detector pixel becomes 0.5 wavelengths longer than the same path through the other atom. As long as both atoms lie in the "Bragg plane" (that's the plane perpendicular to the "s" vector, which is the vector difference between the incoming and outgoing beam directions), the far-field approximation tells us they should also be "in phase", but if they are far enough apart the 0.5 A change in total path length is enough to change the scattering completely from constructive to destructive interference. In ordinary optics, this is called the edge of the first "Fresnel zone".
>
> So, if your source is "very far away", emitting 1 A x-rays, and your detector is 1 meter away, then moving one atom 10 microns away from the centerline of the beam makes the path from that atom to the detector 1-sqrt(1^2+10e-6^2) = 0.5 A longer. So that implies the "coherence length" is 10 microns. But if the detector is only 100 mm away, that gives you 0.1-sqrt(0.1^2+3e-6^2) = 0.5 A, so 3 um is the "coherence length".
>
> Of course, this is for the ideal case of a point source very far away. In reality finite beam divergence will mess up the "coherence" inasmuch as a divergent source looks like an array of sources all viewing the sample through a pinhole. What you then get on the detector is the sum of the patterns from all those sources, so the "coherence" is not as clean. That is, you don't see the Fourier transform of the crystal shape in every spot. Mosaic spread also messes up "coherence" in this way. Some might even define the mosaic "domain size" as the inverse of the effective coherence length.
>
> But, the long and short of all this is that as long as your detector pixels are bigger than the "coherence length" the coherence doesn't really matter.
>
> Hope that makes sense,
>
> -James Holton
> MAD Scientist
>
>
>
> On Thu, Apr 24, 2014 at 2:32 PM, <[log in to unmask]<mailto:[log in to unmask]>> wrote:
> Dear Chen,
>
> Twinning can be thought of as of two or more macro-crystals glued or grown together. The reason that the reflections often overlap is that they share one common plane from which they grow in different directions. Many twinning tests are based on the fact that the two (or more) macro crystals do not interfere, which changes the intensity distributions. Since there is no interference, twinning cannot make spots disappear. Moreover, translational operations between twin domains would be equivalent to move the crystal a little in the beam, as with centering, which will not have any influence on the diffraction pattern (except for weak diffraction because of missing the beam).
> Disorder can have many causes, but is caused by different orientations of residues/molecules/whatever in different asymmetric units. It is close range, so there will be interference. However, since it is usually randomly distributed over the crystal, it will not cause disappearance of spots.
>
> The X-ray coherent length is depending on the crystal, not the synchrotron and my gut feeling is that it is at least several hundred unit cells, but here other experts may correct me.
>
> Disappearance of spots can occur due to a wrong space group assignment (e.g. screw axis have been overlooked) or translational non-crystallographic symmetry. In this case, I would first run a modern MR program to see if you get a solution and otherwise you will have to analyze very careful your space group, unit cell etc. to find out what is going on.
>
> My 2 cents,
> Herman
>
>
> Von: CCP4 bulletin board [mailto:[log in to unmask]<mailto:[log in to unmask]>] Im Auftrag von Chen Zhao
> Gesendet: Donnerstag, 24. April 2014 22:13
> An: [log in to unmask]<mailto:[log in to unmask]>
> Betreff: [ccp4bb] Twinning VS. Disorder
>
> Dear all,
> Hello! I am kinda confused and am thinking about the definition of twinning and disorder. I am just a starting student and might make some fundamental mistakes.
> 1) Twinning is a macroscopic phenomenon and the result is the addition of the intensity from different lattices; disorder is a microscopic phenomenon and the result is the addition of structure factors from different crystal "domains". Is this statement valid?
> 2) I am now very confused about how to define the macroscopic versus the microscopic level when I think of the systematic absences introduced by translational operation. Or in other words, can the translational operation between the twin domains create systematic absences? My answer is probably no because the distance between the two domains are too far away compared to the coherent length of the X-ray, i.e. the addition of the intensity alone cannot make some spots disappear. Is it true? If yes, what is the x-ray coherence length at the synchrotron in general?
>
> 3) If the statement in 2) is valid, then if a "twinning operation" can introduce systematic absences, this should be a disorder instead of a twin based on the definitions in 1). Is this right?
> Your answers will be greatly appreciated!
> Sincerely,
> Chen
On Thu, 24 Apr 2014, James Holton wrote:
> There are two kinds of "coherence length": transverse and longitudinal.
> Longitudinal coherence is often quoted as delta-lambda/lambda, which is
> easy to calculate but unfortunately completely irrelevant for diffraction
> from crystals. If it weren't then Laue diffraction wouldn't produce spots.
>
> Transverse coherence tends to be around 3-10 microns with 1 A x-rays,
> depending on the detector distance. Yes, that's right, the detector
> distance. Longer detector distances give you a bigger coherence length,
> especially when the source is "very far away", like it is at a synchrotron.
>
> How this happens is easiest to picture if you consider the simplest
> possible diffraction situation: a "point" source of x-rays, two atoms, and
> a detector. As long as the atoms are very close together relative to the
> distances from the sample to the source and the detector, then you have the
> "far field" diffraction situation. This is where both atoms are within the
> "coherence length", Bragg's diagram for Bragg's Law holds: parallel
> incoming rays and parallel outgoing rays.
>
> But what if the atoms are very far apart? Obviously, the scattering from
> two atoms on different sides of the room will just add as intensities. And
> if they are very close together, then Bragg's Law holds and they scatter
> "coherently". What most people think of as the "coherence length" is the
> point of transition between these two kinds of scattering.
>
> This point is rather conveniently defined as the distance between two atoms
> when the path from the source to one atom to a given detector pixel becomes
> 0.5 wavelengths longer than the same path through the other atom. As long
> as both atoms lie in the "Bragg plane" (that's the plane perpendicular to
> the "s" vector, which is the vector difference between the incoming and
> outgoing beam directions), the far-field approximation tells us they should
> also be "in phase", but if they are far enough apart the 0.5 A change in
> total path length is enough to change the scattering completely from
> constructive to destructive interference. In ordinary optics, this is
> called the edge of the first "Fresnel zone".
>
> So, if your source is "very far away", emitting 1 A x-rays, and your
> detector is 1 meter away, then moving one atom 10 microns away from the
> centerline of the beam makes the path from that atom to the detector
> 1-sqrt(1^2+10e-6^2) = 0.5 A longer. So that implies the "coherence length"
> is 10 microns. But if the detector is only 100 mm away, that gives you
> 0.1-sqrt(0.1^2+3e-6^2) = 0.5 A, so 3 um is the "coherence length".
>
> Of course, this is for the ideal case of a point source very far away. In
> reality finite beam divergence will mess up the "coherence" inasmuch as a
> divergent source looks like an array of sources all viewing the sample
> through a pinhole. What you then get on the detector is the sum of the
> patterns from all those sources, so the "coherence" is not as clean. That
> is, you don't see the Fourier transform of the crystal shape in every
> spot. Mosaic spread also messes up "coherence" in this way. Some might
> even define the mosaic "domain size" as the inverse of the effective
> coherence length.
>
> But, the long and short of all this is that as long as your detector pixels
> are bigger than the "coherence length" the coherence doesn't really
> matter.
>
> Hope that makes sense,
>
> -James Holton
> MAD Scientist
>
>
>
> On Thu, Apr 24, 2014 at 2:32 PM, <[log in to unmask]> wrote:
>
>> Dear Chen,
>>
>>
>>
>> Twinning can be thought of as of two or more macro-crystals glued or grown
>> together. The reason that the reflections often overlap is that they share
>> one common plane from which they grow in different directions. Many
>> twinning tests are based on the fact that the two (or more) macro crystals
>> do not interfere, which changes the intensity distributions. Since there is
>> no interference, twinning cannot make spots disappear. Moreover,
>> translational operations between twin domains would be equivalent to move
>> the crystal a little in the beam, as with centering, which will not have
>> any influence on the diffraction pattern (except for weak diffraction
>> because of missing the beam).
>>
>> Disorder can have many causes, but is caused by different orientations
>> of residues/molecules/whatever in different asymmetric units. It is close
>> range, so there will be interference. However, since it is usually randomly
>> distributed over the crystal, it will not cause disappearance of spots.
>>
>>
>>
>> The X-ray coherent length is depending on the crystal, not the synchrotron
>> and my gut feeling is that it is at least several hundred unit cells, but
>> here other experts may correct me.
>>
>>
>>
>> Disappearance of spots can occur due to a wrong space group assignment
>> (e.g. screw axis have been overlooked) or translational
>> non-crystallographic symmetry. In this case, I would first run a modern MR
>> program to see if you get a solution and otherwise you will have to analyze
>> very careful your space group, unit cell etc. to find out what is going on.
>>
>>
>>
>> My 2 cents,
>>
>> Herman
>>
>>
>>
>>
>>
>> *Von:* CCP4 bulletin board [mailto:[log in to unmask]] *Im Auftrag von
>> *Chen Zhao
>> *Gesendet:* Donnerstag, 24. April 2014 22:13
>> *An:* [log in to unmask]
>> *Betreff:* [ccp4bb] Twinning VS. Disorder
>>
>>
>>
>> Dear all,
>>
>> Hello! I am kinda confused and am thinking about the definition of
>> twinning and disorder. I am just a starting student and might make some
>> fundamental mistakes.
>>
>> 1) Twinning is a macroscopic phenomenon and the result is the addition of
>> the intensity from different lattices; disorder is a microscopic phenomenon
>> and the result is the addition of structure factors from different crystal
>> "domains". Is this statement valid?
>>
>> 2) I am now very confused about how to define the macroscopic versus the
>> microscopic level when I think of the systematic absences introduced by
>> translational operation. Or in other words, can the translational operation
>> between the twin domains create systematic absences? My answer is probably
>> no because the distance between the two domains are too far away compared
>> to the coherent length of the X-ray, i.e. the addition of the intensity
>> alone cannot make some spots disappear. Is it true? If yes, what is the
>> x-ray coherence length at the synchrotron in general?
>>
>>
>> 3) If the statement in 2) is valid, then if a "twinning operation" can
>> introduce systematic absences, this should be a disorder instead of a twin
>> based on the definitions in 1). Is this right?
>>
>> Your answers will be greatly appreciated!
>>
>> Sincerely,
>>
>> Chen
>>
>
--
========================================================================
Robert M. Sweet E-Dress: [log in to unmask]
Group Leader, PXRR: Macromolecular ^ (that's L
Crystallography Research Resource at NSLS not 1)
http://px.nsls.bnl.gov/
Photon Sciences and Biosciences Dept
Office and mail, Bldg 745, a.k.a. LOB-5
Brookhaven Nat'l Lab. Phones:
Upton, NY 11973 631 344 3401 (Office)
U.S.A. 631 344 2741 (Facsimile)
========================================================================
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