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 p
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
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
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
> sh
It is mostly because in the higher angles intensity of the reflection is lower,
precision is lower and anomalous signal is washed out by counting statistics.
For very well diffracting test crystals anomalous signal is MEASURABLE to very
high resolution providing good enough I/sigma(I) is generate
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
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 L
On Thu, 2010-10-14 at 23:31 +0200, 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
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
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
> >> wit
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 pur
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 scatte
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
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 wavele
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 fr
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 exagger
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 o
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"
To:
Sent: Thursday, October 14, 2010 11:28 AM
Subject: Re: [ccp4bb] embarrassingly simple MAD phasin
To:
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
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
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 th
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 sing
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 am
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 d
> 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 sc
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 a
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 ph
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 reflect
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 wav
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
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