Hi James,
Derek Logan wrote:
- When Rontgen discovered a new kind of light, he called it
x-rays. Now only the Germans call them Rontgen rays.
Thanks for a great essay! Since I have nothing of real value
contribute here, I won't pass over the opportunity to be a
besserwisser (as the Swedes
- When Rontgen discovered a new kind of light, he called it x-
rays. Now only the Germans call them Rontgen rays.
Thanks for a great essay! Since I have nothing of real value
contribute here, I won't pass over the opportunity to be a
besserwisser (as the Swedes say, using a borrowed
Thanks very much for this interesting discussion.
We should have that more often.
Marius
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Le 26 juin 08 à 18:49, Ethan Merritt a écrit :
On Thursday 26 June 2008 09:36:16 am Serge Cohen wrote:
Please some one tells me if I'm wrong ... but I
Ahh. The history of science. I've always wondered how these naming
conventions get decided. Who is the authority on what gets named after
who? Historically, it seems to vary a lot.
- When Patterson published his incredibly useful map he called it the
F-square synthesis. Does anyone NOT
Dear All,
I wonder about the conventions using Friedel vs Bijvoet pair.
a) there are no differences. As long as h = -h, it's a Friedel
or a Bijvoet pair. They are the same.
b) A Friedel pair is any reflection h = -h including hR = -h, i.e.
including centric reflections.
A Bijvoet pair
Friedel pair is strictly F(hkl) and F(-h,-k,-l).
Bijvoet pair is F(h) and any mate that is symmetry-related to F(-h), e.g.,
F(hkl) and F(-h,k,-l) in monoclinic.
There are always anomalous differences, though they can be unmeasurably
small.
Bernie Santarsiero
On Thu, June 26, 2008 10:55 am,
I've always thought that a Bijvoet pair is any pair for which an
anomalous difference could be observed. This includes Friedel pairs
(h h-bar), but it also includes pairs of the form h h', where h'
is symmetry-related to h-bar. Thus Friedel pairs are a subset of all
possible Bijvoet
On Thursday 26 June 2008 09:36:16 am Serge Cohen wrote:
Please some one tells me if I'm wrong ... but I though that indeed one
is NOT supposed to measure anomalous difference from reflections h and
h' if those are related by one of the symmetry operator of the point
group...
This statement is
Let's try this again, with definitions, and pls scream if I am wrong:
a) Any reflection pair hR = h forms a symmetry related pair.
R is any one of G point group operators of the SG.
This is a set of reflections (S). Their amplitudes
are invariably the same. They do not even show up
,
Hidong
Bernhard Rupp [EMAIL PROTECTED]
Sent by: CCP4 bulletin board CCP4BB@JISCMAIL.AC.UK
06/26/2008 11:35 AM
Please respond to
[EMAIL PROTECTED]
To
CCP4BB@JISCMAIL.AC.UK
cc
Subject
Re: [ccp4bb] Friedel vs Bijvoet
Let's try this again, with definitions, and pls scream if I am
I quote from these pages:
Bijvoet pairs are Bragg reflections which are true symmetry
equivalents to a Friedel pair. These true symmetry equivalents
have *equal amplitudes, even in the presence of anomalous scattering*.
Sounds more like centric or perhaps simply symmetry related to me.
A few
There was a mistake in the letter that listed the Bijvoet pairs
for a monoclinic space group and that is confusing you. Let me
try.
The equivalent positions for a B setting monoclinic are
h,k,l; -h,k,-l.
The Friedel mates for the general position (h,k,l) are (-h,-k,-l).
This means
Bernhard Rupp wrote:
I quote from these pages:
Bijvoet pairs are Bragg reflections which are true symmetry
equivalents to a Friedel pair. These true symmetry equivalents
have *equal amplitudes, even in the presence of anomalous scattering*.
This is poorly worded. I would change it to
A
On Thursday 26 June 2008 11:35:31 am Bernhard Rupp wrote:
Let's try this again, with definitions, and pls scream if I am wrong:
a) Any reflection pair hR = h forms a symmetry related pair.
???
Maybe you meanh' = hR
R is any one of G point group operators of the SG.
: CCP4BB@JISCMAIL.AC.UK
Subject: RE: [ccp4bb] Friedel vs Bijvoet
I quote from these pages:
Bijvoet pairs are Bragg reflections which are true symmetry
equivalents to a Friedel pair. These true symmetry equivalents have
*equal amplitudes, even in the presence of anomalous scattering
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Le 26 juin 08 à 18:49, Ethan Merritt a écrit :
On Thursday 26 June 2008 09:36:16 am Serge Cohen wrote:
Please some one tells me if I'm wrong ... but I though that indeed
one
is NOT supposed to measure anomalous difference from reflections h
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