After conversing with Bernhard a bit offline I think the relevant 
question is:

How far apart can two electrons in the crystal be before their 
scattering becomes "incoherent" ... as in no longer interfering with 
each other in the way Bragg described.


The answer to this is about 10 um if the source is 1 m away (or the 
detector whichever is closer), and about 3 um if the detector/source is 
100 mm away.

I derive this answer from the breakdown of Bragg's assumption that the 
incoming and outgoing waves are a perfect plane wave.  Since the 
distance to the source is not infinity, there must be some curvature to 
the wavefront.  This has nothing to do with beam divergence or spectral 
dispersion.  We are talking about one photon (emitted from the 
acceleration of some electron in the storage ring or in the copper 
anode) propagating through space in the quantum-wave mechanical way, and 
then being detected on the detector with some probability distribution 
that is the result of all these interfering scattered waves.  The Braggs 
simplified all this to a single geometric diagram, and that is why we 
all like them so much.

Here is my version of the Bragg diagram:
http://bl831.als.lbl.gov/~jamesh/pickup/Bragg.png

The fundamental principle behind Bragg's Law is that the length of the 
path taken by the wavefront from the source to an atom in the crystal 
and then on to a point on the detector must be the same (plus or minus 
an integral number of wavelengths) as some other path from the exact 
same point in the source to the exact same point on the detector but 
through a different atom (maybe in the same unit cell, maybe not).  
Bragg assumed that the source and the detector were REALLY far away and 
that simplifies the math, but that must break down eventually as we talk 
about atoms getting further and further apart in the crystal.  For 
example, if the atoms are on opposite sides of the room (d-spacing large 
compared with the distance to the source) then Bragg's Law will break down.

I have drawn a generalized Bragg construction (not necessarily to scale) 
here:
http://bl831.als.lbl.gov/~jamesh/pickup/Bragg-ish.png

So, all we have to do is calculate the distance traveled by "the photon" 
taking a path through atom #1 vs atom #2 and then require that these two 
distances be an integral multiple of the wavelength.  This is exactly 
what Bragg did, except he assumed d was very much smaller than the 
source-to-xtal distance.  You need some very high-precision 
floating-point arithmetic to do this "right", since the distance between 
the source and the detector is typically ten orders of magnitude larger 
than the distance between the atoms.  It is a rare thing in science when 
a technique spans such a large range in length scale.  I imagine Bragg 
didn't feel like working with 10-digit numbers, so he just made the 
lines parallel.
    Since I, unlike Bragg, have access to a very high precision 
calculating machine (gnuplot) I can calculate Bernhard's "coherence 
length" using the Pythagorean theorem:
  Simplest thing is to assume forward-scattering and consider two 
electrons that are always exactly the same distance from the source 
(stx).  Let's assume this is one meter or 10^10 Angstrom.  Also, for 
simplicity, let's assume the detector is also one meter from the sample 
(xtf).  Also, for simplicity assume that the path from the source to 
detector via atom #1 is straight (as in my second diagram).  Now, in 
Bragg's diagram, it doesn't matter how big "d" is: the lines are 
parallel and the same length (outside the diagram) so the x-rays always 
constructively interfere at the detector as long as 2*d*sin(theta) = 
n*lambda.  So, given that the lines must actually intersect at the 
source and the detector, how large can "d" be before the path from 
source to detector via atom #2 is 0.5 Angstrom longer than the path 
through atom #1?

sqrt( (1e10)**2 + 1e5**2 ) =  1e10 + 0.5

So, 100,000 Angstroms (10 um). 

-James Holton
MAD Scientist


James Holton wrote:
> Ethan Merritt wrote:
>> My impression is that the coherence length from synchrotron sources
>> is generally larger than the x-ray path through a protein crystal.
>> But I have not gone through the exercise of plugging in specific
>> storage ring energies and undulator parameters to confirm this
>> impression.  Perhaps James Holton will chime in again?
>>   
> Hmm.  I think I should point out that (contrary to popular belief) I 
> am not a physicist.  I am a biologist.  Yup. BS and PhD both in 
> biology.  However, since I work at a synchrotron I do have a lot of 
> physicists and engineers around to talk to.  Guess some of it has 
> rubbed off.
>
> I passed Bernhard's question along to Howard Padmore (who is 
> definitely a physicist) here at ALS and he gave me a very good 
> description of the longitudinal coherence length, similar to that 
> provided by Colin's posted reference:
>
> coherence_length = lambda^2/delta-lambda
>
> This made a lot of sense to me until I started to consider what 
> happens if lambda ranges from 1 to 3 A, like it does in Laue 
> diffraction.  One might expect from this formula that the coherence 
> length would be very small, smaller than a typical protein unit cell, 
> and then you would predict that none of the scattering from any of the 
> unit cells interferes with each other and that you should see the 
> molecular transform in the diffraction pattern.  But the oldest 
> observation in crystallography is that Laue patterns have sharp 
> spots.  You don't see the molecular transform, despite how nice that 
> would be (no more phase problem!).
>
> I think the coherence length is related to how TWO different photons 
> can interfere with each other, and this is a rare event indeed.  It 
> has nothing to do with x-ray diffraction as we know it.  No matter how 
> low your flux is, even one photon per second, you will eventually 
> build up the same diffraction pattern you get at 10^13 photons/s.  
> Colin is right that photons should be considered as waves and on the 
> length scale of unit cells, it is a very good approximation to 
> consider the electromagnetic wave front coming from the x-ray source 
> to be a flat plane, as Bragg did in his famous construction.
>
> So, I think perhaps Bernhard asked the wrong question?  I think the 
> question should have been "how far apart can two unit cells be before 
> they stop interfering with each other?"  The answer to this one is: 
> quite a bit.
>
> Consider a silicon crystal (like the ones in my monochromator).  These 
> things are about 10 cm across, but every atom is in perfect alignment 
> with every other.  It is one single mosaic domain that you can hold in 
> your hand.  And as soon as you shine an x-ray beam on a large perfect 
> crystal, lots of "weird" stuff happens.  Unlike protein crystals the 
> scattering of the x-rays is so strong that the scattered wave not only 
> depletes the incoming beam (it penetrates less than 1 mm and is nearly 
> 100% reflected), but this now very strong diffracted ray can reflect 
> again on its way out of the crystal (off of the same HKL index, but 
> different unit cells). Then some of that secondarily-diffracted ray 
> will be in the same direction as the main beam, and interfere with it 
> (extinction).  Accounting for all of this is what Ewald did in his 
> so-called "dynamical theory" of diffraction.  The important thing to 
> remember about perfect crystals is that a SINGLE PHOTON interacting 
> with my 10 cm wide silicon crystal will experience all these dynamical 
> effects.  It doesn't matter what the "coherence length" is.
>
> Now, if a perfect crystal is really really small (much smaller than 
> the interaction length of scattering), then there is no opportunity 
> for the re-scattering and extinction and all that "weird stuff" to 
> happen.  In this limiting case, the scattered intensity is simply 
> proportional to the number of unit cells in the beam and also to 
> |F|^2.  This is the basic intensity formula that Ewald showed how to 
> integrate over all the depleting beams and re-scattering stuff to 
> explain a large perfect crystal.  As I understand it, the fact that 
> there were large, macroscopic "single" crystals that were found to 
> still obey the formula for a microscopic crystal came as something of 
> a shock in the time of Darwin and Ewald.  They explained this 
> observation by supposing that these crystals were "ideally imperfect" 
> and actually made up of lots of little perfect crystals that were 
> mis-oriented with respect to one another enough so that the diffracted 
> ray from one would be very unlikely to re-reflect off of another 
> "mosaic domain" before it left the crystal.  Protein crystals are a 
> very good example of ideally imperfect crystals.
>
> I'm not sure where this rumor got started that the intensity reflected 
> from a mosaic block or otherwise perfect lattice is proportional to 
> the square of the number of unit cells.  This is never the case.  The 
> reason is explained in Chapter 6 of M. M. Woolfson's excellent 
> textbook, but the long and short of it is: yes the instantaneous 
> intensity (photons/steradian/s) at the near-infinitesimal moment when 
> a mosaic domain diffracts is proportional to the number of unit cells 
> squared, but this is not useful because x-ray beams are never 
> perfectly monochromatic nor perfectly parallel.  This means that for 
> all practical purposes the spot must always be "integrated" over some 
> angular width (such as beam divergence).  That is, you have to get rid 
> of the "steradians" in the units of intensity before you can get 
> simply photons/s.  The intrinsic "rocking curve" of this 
> near-infinitely-sharp peak from a single mosaic block is inversely 
> proportional to the number of unit cells in the mosaic block.  So, the 
> integrated intensity (photons/s * exposure time) is proportional to 
> the number of unit cells in the beam.  It doesn't matter how perfect 
> the crystal is.
>
> Okay, so I know there are a lot of people out there who don't agree 
> with me on this, but please have a look at Woolfson's Ch 6 before 
> flaming me.  I may be nothing more than a biologist, but I did take a 
> few math classes in college and think I do understand the math in that 
> book.
>
> So, I think the answer Bernhard was looking for is : the size of a 
> mosaic domain, which can be as much as 10 cm, or as little as a few 
> dozen unit cells.
>
> -James Holton
> MAD Scientist