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] protein ligand energy
Dear Colleagues, I see I should quote the last sentence of our abstract of Bradbrook et al 1998:- This work demonstrates the difficulty in relating structure to thermodynamics, but suggests that dynamic models are needed to provide a more complete picture of ligand - receptor interactions. Best wishes, John Prof John R Helliwell DSc On Wed, Oct 13, 2010 at 3:35 PM, Martyn Winn martyn.w...@stfc.ac.uk wrote: This is all true. And I think the bottom line is that it is extremely non-trivial to get a meaningful number. The Amber MM-PBSA script is the best established one. We have an equivalent CHARMM-based script at: http://www.cse.scitech.ac.uk/cbg/software/charmm/ But I guess this is beyond the original question. A simpler option (but more approximate) would be to run the PDB of the modelled complex through PISA (online or CCP4 version) and look at the results for the protein ligand interface. Cheers Martyn On Wed, 2010-10-13 at 15:15 +0100, Robert Esnouf wrote: Dear Rex, It certainly matters what you mean by the energy of a protein ligand complex. And whether you are comparing a series of related similar structures or looking for an absolute energy. The problem is that there is no such thing as an absolute energy, it is always relative to something else. Typically, you might calculate the the binding free energy (delta G) for the components in aqueous solution. If you were looking at the (small) differences between related structures then you'd look at the change (delta delta G) and hope the other errors largely cancel out. One method for which there is substantial literature is based on Amber simulations. There are even sample scripts to do the correct job. You simulate the complex in a water box and sample the conformation every so many steps. You then discard the waters and use something like the Poisson-Boltmann method to estimate solvation free energies for the complex and the isolated components. The difference is then your estimation of the binding free energy. In all such simulations it is the effect of the solvent (partial charges, dielectric properties and entropic effects) that are likely to dominate the calculation. You have to do your best to include them as realistically as possible. Amber is not free, but not expensive and your institution probably already has a site licence. Other simulation programs would also do the job (probably just as well!) but I am not aware they have available scripts. Best wishes, Robert -- Dr. Robert Esnouf, University Research Lecturer and Head of Research Computing, Wellcome Trust Centre for Human Genetics, Roosevelt Drive, Oxford OX3 7BN, UK Emails: rob...@strubi.ox.ac.uk Tel: (+44) - 1865 - 287783 and rob...@esnouf.com Fax: (+44) - 1865 - 287547 -- *** * * * Dr. Martyn Winn * * * * STFC Daresbury Laboratory, Daresbury, Warrington, WA4 4AD, U.K. * * Tel: +44 1925 603455 E-mail: martyn.w...@stfc.ac.uk * * Fax: +44 1925 603634 Skype name: martyn.winn * * URL: http://www.ccp4.ac.uk/martyn/ * *** -- Professor John R Helliwell DSc
Re: [ccp4bb] cad, freerflag, uniqueify : free set when making anomalous from merged data
To try to answer the Q I think you are asking.. If you keep anomalous seperate you will get a file from ctruncate with h k l F+ SIGF+ F- SIGF- I+ SIGI+ I- SIGI- The observations flagged as F- or I- etc are actually measured for the reflection -h-k-l So uniqueify generates markers for each possible h k l (but not for -h-k -l) to a given resolution limit and outputs a file h k l Marker Then cad combines that file with the ctruncate output and if requested also picks up a FreeRflag generated for another data set. for that entry so both hkl, and -h -k -l will belong either to the Free or the working set. freerflag is a tool which generates Free flags for a requested % of this data set. If some Free flags are already assigned it a) guesses the % chosen for that set b) generates FreeR flags for the same % but only assigns ones to those h k l which do not aready carry a FreeR flag. Is that clear!! It is a bit complex.. Gets even more so if you want to consider twinning.. In that case the FreeR flags must be generated in the highest possible pointgroup and extended in cad to the actual one, This means that twinned pairs also belong either to the Free or the working set. Eleanor # On 10/13/2010 08:07 PM, Lepore, Bryan wrote: [ ccp4 6.1.13 ] [ gui 2.0.6 ] if a data set with merged Bijvoet pairs and a free set is then made anomalous - i.e. re-scaled with separated Bijvoet pairs - i do not understand how acentric reflections in the free set are extended by cad or freerflag or uniqueify. the easy question : is [uniqueify with extension] -[cad with combination but not extension] adequate? i.e. i gather that cad preserves FreeR records between two resolution limits (e.g. 500-3.5) and only extends them in the higher resolution segment (e.g. 3.5 - 2.1). so do i have this right : for acentrics, -h-k-l will be preserved while hkl go to the work set. That will decrease the fraction of reflections in the Free set by half the number of acentrics in the merged free set, because they would be in the work set. centrics do not meet this problem. -bryan
[ccp4bb] Problem installing CCP4 on MacOSX behind firewall?
I have a problem installing CCP4 on a MacOSX 10.6.4. When I follow the procedure on Bill Scott's web-pages to install the precompiled version, I get an error message saying Err http://sage.ucsc.edu stable/main Packages 403 Forbidden When trying to install it via the normal Fink way, I get curl (56) FTP response reading failed ###execution of curl failed, exit code 56 Downloading the file ccp4-6.1.13-core-src.tar.gz failed. I think there is an issue with our firewall, but apart from setting the http_proxy to what has been suggested by our IT services I don't really know how to proceed. Does anyone have any suggestions? I know I can download the disk images from the ccp4 pages, but I would like to understand what is going wrong. Cheers, Mads
Re: [ccp4bb] embarrassingly simple MAD phasing question
Formally, a complex number (e.g. a structure factor) is not a vector. Just because the addition subtraction rules (i.e. 'a+b' 'a-b') are defined for real numbers, complex numbers and vectors doesn't make a complex number a vector, any more than it makes a real number a vector (or vice versa). Entities are defined according to the rules of algebra that they obey, thus real and complex numbers obey the same rules, i.e. the familiar addition, subtraction, multiplication, division raising to a power. Hence real and complex numbers are both scalars: a real number is a special case of a complex scalar with zero imaginary part (one could program an algorithm for reals using only complex variables functions and still get the right answer). This also means that the transcendental functions (sin, cos, tan, exp, log etc) are all defined equally well for both real and complex scalars, but not for vectors, a property that programmers in Fortran, C C++ (and probably others) will be familiar with. Of the addition, subtraction, multiplication, division power rules, vectors only obey the first two, but unlike real complex scalars they also obey the scalar product and exterior product rules. The general rule is that if and only if it looks like a duck, waddles like a duck and quacks like a duck, then it is a duck - complex numbers might look like vectors but they neither waddle nor quack like them! Cheers -- Ian On Wed, Oct 13, 2010 at 9:57 PM, Yong Y Wang wang_yon...@lilly.com wrote: It is already vertical, relative to the real part of Fa (in red), i.e. the blue vector is always vertical to the red vector in this picture (and counter-clockwise). Yong William Scott wgsc...@chemistry.ucsc.edu Sent by: CCP4 bulletin board CCP4BB@JISCMAIL.AC.UK 10/13/2010 01:48 PM Please respond to William Scott wgsc...@chemistry.ucsc.edu To CCP4BB@JISCMAIL.AC.UK cc Subject [ccp4bb] embarrassingly simple MAD phasing question Hi Citizens: Try not to laugh. I have an embarrassingly simple MAD phasing question: Why is it that F in this picture isn't required to be vertical (purely imaginary)? http://www.doe-mbi.ucla.edu/~sawaya/tutorials/Phasing/phase.gif (Similarly in the Harker diagram of the intersection of phase circles, one sees this.) I had a student ask me and I realized that there is this fundamental gap in my understanding. Many thanks in advance. -- 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)
Re: [ccp4bb] embarrassingly simple MAD phasing question
On Thu, Oct 14, 2010 at 12:34:30PM +0100, Ian Tickle wrote: Formally, a complex number (e.g. a structure factor) is not a vector. Formally, C is isomorphous to R^2 (at least that's what math departments in Germany teach, and it's not difficult to prove), therefore complex numbers are vectors. That's is unaffected by whether there is a ring-isomorphism between C and R^2, and it's correct that the elements of a field are usually not called 'vectors', but that does not mean that it is wrong to consider a complex number a vector. Tim Just because the addition subtraction rules (i.e. 'a+b' 'a-b') are defined for real numbers, complex numbers and vectors doesn't make a complex number a vector, any more than it makes a real number a vector (or vice versa). Entities are defined according to the rules of algebra that they obey, thus real and complex numbers obey the same rules, i.e. the familiar addition, subtraction, multiplication, division raising to a power. Hence real and complex numbers are both scalars: a real number is a special case of a complex scalar with zero imaginary part (one could program an algorithm for reals using only complex variables functions and still get the right answer). This also means that the transcendental functions (sin, cos, tan, exp, log etc) are all defined equally well for both real and complex scalars, but not for vectors, a property that programmers in Fortran, C C++ (and probably others) will be familiar with. Of the addition, subtraction, multiplication, division power rules, vectors only obey the first two, but unlike real complex scalars they also obey the scalar product and exterior product rules. The general rule is that if and only if it looks like a duck, waddles like a duck and quacks like a duck, then it is a duck - complex numbers might look like vectors but they neither waddle nor quack like them! Cheers -- Ian On Wed, Oct 13, 2010 at 9:57 PM, Yong Y Wang wang_yon...@lilly.com wrote: It is already vertical, relative to the real part of Fa (in red), i.e. the blue vector is always vertical to the red vector in this picture (and counter-clockwise). Yong William Scott wgsc...@chemistry.ucsc.edu Sent by: CCP4 bulletin board CCP4BB@JISCMAIL.AC.UK 10/13/2010 01:48 PM Please respond to William Scott wgsc...@chemistry.ucsc.edu To CCP4BB@JISCMAIL.AC.UK cc Subject [ccp4bb] embarrassingly simple MAD phasing question Hi Citizens: Try not to laugh. I have an embarrassingly simple MAD phasing question: Why is it that F in this picture isn't required to be vertical (purely imaginary)? http://www.doe-mbi.ucla.edu/~sawaya/tutorials/Phasing/phase.gif (Similarly in the Harker diagram of the intersection of phase circles, one sees this.) I had a student ask me and I realized that there is this fundamental gap in my understanding. Many thanks in advance. -- 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) -- -- 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] Problem installing CCP4 on MacOSX behind firewall?
Dear Mads, the first error message _might_ be due to a dead link, but without further information that's difficult to tell. The second error message says that curl tried to get the archive using ftp. When you download something through the ccp4 web site your computer is probably using http, and maybe our IT services allow http but not ftp. You should address your IT services with your questions first, because they set their firewall rules, and it is probably more likely that it is indeed these rules that cause the error. Tim On Thu, Oct 14, 2010 at 10:20:17AM +0100, Mads Gabrielsen wrote: I have a problem installing CCP4 on a MacOSX 10.6.4. When I follow the procedure on Bill Scott's web-pages to install the precompiled version, I get an error message saying Err http://sage.ucsc.edu stable/main Packages 403 Forbidden When trying to install it via the normal Fink way, I get curl (56) FTP response reading failed ###execution of curl failed, exit code 56 Downloading the file ccp4-6.1.13-core-src.tar.gz failed. I think there is an issue with our firewall, but apart from setting the http_proxy to what has been suggested by our IT services I don't really know how to proceed. Does anyone have any suggestions? I know I can download the disk images from the ccp4 pages, but I would like to understand what is going wrong. Cheers, Mads -- -- 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
I don't see any conflict here: all you're saying is that there's a 1-to-1 mapping between the complex scalar a+i*b in C and the 2-D vector (a,b) in R^2. However the vector does not have all the properties of the original complex scalar: for example I can happily compute a value for log(a+i*b) but if I try to compute log((a,b)) the compiler will throw a fit! Of course if you can get along without taking the log, or any other transcendental function, and without doing any algebra using the multiplication, division or power operators, then it won't matter in practice whether you regard it as a+i*b or (a,b). Personally I find it much cleaner to treat a complex number as the scalar a+i*b because then I have the full math library of the Fortran compiler (or whatever is your favourite) at my disposal. Cheers -- Ian On Thu, Oct 14, 2010 at 1:24 PM, Tim Gruene t...@shelx.uni-ac.gwdg.de wrote: On Thu, Oct 14, 2010 at 12:34:30PM +0100, Ian Tickle wrote: Formally, a complex number (e.g. a structure factor) is not a vector. Formally, C is isomorphous to R^2 (at least that's what math departments in Germany teach, and it's not difficult to prove), therefore complex numbers are vectors. That's is unaffected by whether there is a ring-isomorphism between C and R^2, and it's correct that the elements of a field are usually not called 'vectors', but that does not mean that it is wrong to consider a complex number a vector. Tim Just because the addition subtraction rules (i.e. 'a+b' 'a-b') are defined for real numbers, complex numbers and vectors doesn't make a complex number a vector, any more than it makes a real number a vector (or vice versa). Entities are defined according to the rules of algebra that they obey, thus real and complex numbers obey the same rules, i.e. the familiar addition, subtraction, multiplication, division raising to a power. Hence real and complex numbers are both scalars: a real number is a special case of a complex scalar with zero imaginary part (one could program an algorithm for reals using only complex variables functions and still get the right answer). This also means that the transcendental functions (sin, cos, tan, exp, log etc) are all defined equally well for both real and complex scalars, but not for vectors, a property that programmers in Fortran, C C++ (and probably others) will be familiar with. Of the addition, subtraction, multiplication, division power rules, vectors only obey the first two, but unlike real complex scalars they also obey the scalar product and exterior product rules. The general rule is that if and only if it looks like a duck, waddles like a duck and quacks like a duck, then it is a duck - complex numbers might look like vectors but they neither waddle nor quack like them! Cheers -- Ian On Wed, Oct 13, 2010 at 9:57 PM, Yong Y Wang wang_yon...@lilly.com wrote: It is already vertical, relative to the real part of Fa (in red), i.e. the blue vector is always vertical to the red vector in this picture (and counter-clockwise). Yong William Scott wgsc...@chemistry.ucsc.edu Sent by: CCP4 bulletin board CCP4BB@JISCMAIL.AC.UK 10/13/2010 01:48 PM Please respond to William Scott wgsc...@chemistry.ucsc.edu To CCP4BB@JISCMAIL.AC.UK cc Subject [ccp4bb] embarrassingly simple MAD phasing question Hi Citizens: Try not to laugh. I have an embarrassingly simple MAD phasing question: Why is it that F in this picture isn't required to be vertical (purely imaginary)? http://www.doe-mbi.ucla.edu/~sawaya/tutorials/Phasing/phase.gif (Similarly in the Harker diagram of the intersection of phase circles, one sees this.) I had a student ask me and I realized that there is this fundamental gap in my understanding. Many thanks in advance. -- 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) -- -- Tim Gruene Institut fuer anorganische Chemie Tammannstr. 4 D-37077 Goettingen phone: +49 (0)551 39 22149 GPG Key ID = A46BEE1A -BEGIN PGP SIGNATURE- Version: GnuPG v1.4.9 (GNU/Linux) iD4DBQFMtvZ0UxlJ7aRr7hoRAginAJ4ty3BHIL/h9BXe6U77+/64+cU5UgCY7NEn F64lkymlpzNz2b2BU3aOFQ== =dtZf -END PGP SIGNATURE-
Re: [ccp4bb] vector and scalars
The definition game is on! :) Vectors are supposed to have direction and amplitude, unlike scalars. Curiously, one can take a position that real numbers are vectors too, if you consider negative and positive numbers having opposite directions (and thus subtraction is simply a case of addition of a negative number). And of course, both scalars and vectors are simply tensors, of zeroth and first order :) Guess my point is that definitions are a matter of choice in math and if vector is defined as an array which must obey addition and scaling rules (but there is no fixed multiplication rule - regular 3D vectors have more than one possible product), then complex numbers are vectors. In a narrow sense of a real space vectors (the arrow thingy) they are not. Thus, complex number is a Vector, but not the vector (futile attempt at using articles by someone organically suffering from article dyslexia). Cheers, Ed. On Thu, 2010-10-14 at 14:24 +0200, Tim Gruene wrote: On Thu, Oct 14, 2010 at 12:34:30PM +0100, Ian Tickle wrote: Formally, a complex number (e.g. a structure factor) is not a vector. Formally, C is isomorphous to R^2 (at least that's what math departments in Germany teach, and it's not difficult to prove), therefore complex numbers are vectors. That's is unaffected by whether there is a ring-isomorphism between C and R^2, and it's correct that the elements of a field are usually not called 'vectors', but that does not mean that it is wrong to consider a complex number a vector. Tim Just because the addition subtraction rules (i.e. 'a+b' 'a-b') are defined for real numbers, complex numbers and vectors doesn't make a complex number a vector, any more than it makes a real number a vector (or vice versa). Entities are defined according to the rules of algebra that they obey, thus real and complex numbers obey the same rules, i.e. the familiar addition, subtraction, multiplication, division raising to a power. Hence real and complex numbers are both scalars: a real number is a special case of a complex scalar with zero imaginary part (one could program an algorithm for reals using only complex variables functions and still get the right answer). This also means that the transcendental functions (sin, cos, tan, exp, log etc) are all defined equally well for both real and complex scalars, but not for vectors, a property that programmers in Fortran, C C++ (and probably others) will be familiar with. Of the addition, subtraction, multiplication, division power rules, vectors only obey the first two, but unlike real complex scalars they also obey the scalar product and exterior product rules. The general rule is that if and only if it looks like a duck, waddles like a duck and quacks like a duck, then it is a duck - complex numbers might look like vectors but they neither waddle nor quack like them! Cheers -- Ian On Wed, Oct 13, 2010 at 9:57 PM, Yong Y Wang wang_yon...@lilly.com wrote: It is already vertical, relative to the real part of Fa (in red), i.e. the blue vector is always vertical to the red vector in this picture (and counter-clockwise). Yong William Scott wgsc...@chemistry.ucsc.edu Sent by: CCP4 bulletin board CCP4BB@JISCMAIL.AC.UK 10/13/2010 01:48 PM Please respond to William Scott wgsc...@chemistry.ucsc.edu To CCP4BB@JISCMAIL.AC.UK cc Subject [ccp4bb] embarrassingly simple MAD phasing question Hi Citizens: Try not to laugh. I have an embarrassingly simple MAD phasing question: Why is it that F in this picture isn't required to be vertical (purely imaginary)? http://www.doe-mbi.ucla.edu/~sawaya/tutorials/Phasing/phase.gif (Similarly in the Harker diagram of the intersection of phase circles, one sees this.) I had a student ask me and I realized that there is this fundamental gap in my understanding. Many thanks in advance. -- 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) -- Edwin Pozharski, PhD, Assistant Professor University of Maryland, Baltimore -- When the Way is forgotten duty and justice appear; Then knowledge and wisdom are born along with hypocrisy. When harmonious relationships dissolve then respect and devotion arise; When a nation falls to chaos then loyalty and patriotism are born. -- / Lao Tse /
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] vector and scalars
The definition of a vector as being something that has 'magnitude' and 'direction' is actually incorrect. If that were to be the case, a quantity like electric current would be a vector and not a scalar. Electric current is a scalar. A vector is something that transforms like the coordinate system, while a scalar does not. In other words, if you were to transform the coordinate system by a certain operator, a vector quantity in the old coordinate system can be transformed into the new one by using exactly the same operator. This is the correct definition of a vector. G. On Thu, 14 Oct 2010 10:22:59 -0400, Ed Pozharski epozh...@umaryland.edu wrote: The definition game is on! :) Vectors are supposed to have direction and amplitude, unlike scalars. Curiously, one can take a position that real numbers are vectors too, if you consider negative and positive numbers having opposite directions (and thus subtraction is simply a case of addition of a negative number). And of course, both scalars and vectors are simply tensors, of zeroth and first order :) Guess my point is that definitions are a matter of choice in math and if vector is defined as an array which must obey addition and scaling rules (but there is no fixed multiplication rule - regular 3D vectors have more than one possible product), then complex numbers are vectors. In a narrow sense of a real space vectors (the arrow thingy) they are not. Thus, complex number is a Vector, but not the vector (futile attempt at using articles by someone organically suffering from article dyslexia). Cheers, Ed. O -- ** Blow, blow, thou winter wind Thou art not so unkind As man's ingratitude; Thy tooth is not so keen, Because thou art not seen, Although thy breath be rude. -William Shakespeare **
Re: [ccp4bb] vector and scalars
The definition game is on! :) Vectors are supposed to have direction and amplitude, unlike scalars. I think that this is part of the problem here. Whilst vector quantities do possess both size and direction, not everything that possesses size and direction is necessarily a vector by definition. Thus, just because complex numbers possess an amplitude and a phase angle, that does not automatically make them vectors. The complex numbers are in fact a vector space over the real numbers, but that requires further justification. Curiously, one can take a position that real numbers are vectors too, if you consider negative and positive numbers having opposite directions (and thus subtraction is simply a case of addition of a negative number). And of course, both scalars and vectors are simply tensors, of zeroth and first order :) Guess my point is that definitions are a matter of choice in math and if vector is defined as an array which must obey addition and scaling rules (but there is no fixed multiplication rule - regular 3D vectors have more than one possible product), then complex numbers are vectors. In a narrow sense of a real space vectors (the arrow thingy) they are not. Thus, complex number is a Vector, but not the vector (futile attempt at using articles by someone organically suffering from article dyslexia). Cheers, Ed. On Thu, 2010-10-14 at 14:24 +0200, Tim Gruene wrote: On Thu, Oct 14, 2010 at 12:34:30PM +0100, Ian Tickle wrote: Formally, a complex number (e.g. a structure factor) is not a vector. Formally, C is isomorphous to R^2 (at least that's what math departments in Germany teach, and it's not difficult to prove), therefore complex numbers are vectors. That's is unaffected by whether there is a ring-isomorphism between C and R^2, and it's correct that the elements of a field are usually not called 'vectors', but that does not mean that it is wrong to consider a complex number a vector. Tim Just because the addition subtraction rules (i.e. 'a+b' 'a-b') are defined for real numbers, complex numbers and vectors doesn't make a complex number a vector, any more than it makes a real number a vector (or vice versa). Entities are defined according to the rules of algebra that they obey, thus real and complex numbers obey the same rules, i.e. the familiar addition, subtraction, multiplication, division raising to a power. Hence real and complex numbers are both scalars: a real number is a special case of a complex scalar with zero imaginary part (one could program an algorithm for reals using only complex variables functions and still get the right answer). This also means that the transcendental functions (sin, cos, tan, exp, log etc) are all defined equally well for both real and complex scalars, but not for vectors, a property that programmers in Fortran, C C++ (and probably others) will be familiar with. Of the addition, subtraction, multiplication, division power rules, vectors only obey the first two, but unlike real complex scalars they also obey the scalar product and exterior product rules. The general rule is that if and only if it looks like a duck, waddles like a duck and quacks like a duck, then it is a duck - complex numbers might look like vectors but they neither waddle nor quack like them! Cheers -- Ian On Wed, Oct 13, 2010 at 9:57 PM, Yong Y Wang wang_yon...@lilly.com wrote: It is already vertical, relative to the real part of Fa (in red), i.e. the blue vector is always vertical to the red vector in this picture (and counter-clockwise). Yong William Scott wgsc...@chemistry.ucsc.edu Sent by: CCP4 bulletin board CCP4BB@JISCMAIL.AC.UK 10/13/2010 01:48 PM Please respond to William Scott wgsc...@chemistry.ucsc.edu To CCP4BB@JISCMAIL.AC.UK cc Subject [ccp4bb] embarrassingly simple MAD phasing question Hi Citizens: Try not to laugh. I have an embarrassingly simple MAD phasing question: Why is it that F in this picture isn't required to be vertical (purely imaginary)? http://www.doe-mbi.ucla.edu/~sawaya/tutorials/Phasing/phase.gif (Similarly in the Harker diagram of the intersection of phase circles, one sees this.) I had a student ask me and I realized that there is this fundamental gap in my understanding. Many thanks in advance. -- 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) -- Edwin Pozharski, PhD, Assistant Professor University of Maryland, Baltimore
Re: [ccp4bb] vector and scalars
Electrical current is a 4-vector, is it not? Correct! - and an alternating electric current is represented as a complex number (then it's conventional to use the symbol 'j' for sqrt(-1) to avoid confusion with 'i', the symbol for electric current!). Since as you say electric current is a scalar not a vector, then a complex number has to be a scalar, not a vector! Cheers -- Ian On Thu, Oct 14, 2010 at 3:47 PM, Ganesh Natrajan natra...@embl.fr wrote: The definition of a vector as being something that has 'magnitude' and 'direction' is actually incorrect. If that were to be the case, a quantity like electric current would be a vector and not a scalar. Electric current is a scalar. A vector is something that transforms like the coordinate system, while a scalar does not. In other words, if you were to transform the coordinate system by a certain operator, a vector quantity in the old coordinate system can be transformed into the new one by using exactly the same operator. This is the correct definition of a vector. G. On Thu, 14 Oct 2010 10:22:59 -0400, Ed Pozharski epozh...@umaryland.edu wrote: The definition game is on! :) Vectors are supposed to have direction and amplitude, unlike scalars. Curiously, one can take a position that real numbers are vectors too, if you consider negative and positive numbers having opposite directions (and thus subtraction is simply a case of addition of a negative number). And of course, both scalars and vectors are simply tensors, of zeroth and first order :) Guess my point is that definitions are a matter of choice in math and if vector is defined as an array which must obey addition and scaling rules (but there is no fixed multiplication rule - regular 3D vectors have more than one possible product), then complex numbers are vectors. In a narrow sense of a real space vectors (the arrow thingy) they are not. Thus, complex number is a Vector, but not the vector (futile attempt at using articles by someone organically suffering from article dyslexia). Cheers, Ed. O -- ** Blow, blow, thou winter wind Thou art not so unkind As man's ingratitude; Thy tooth is not so keen, Because thou art not seen, Although thy breath be rude. -William Shakespeare **
Re: [ccp4bb] vector and scalars
Again, definitions are a matter of choice. Under your strict version I still may consider electric current as vector, if I introduce the coordinate system in the circuit. When I transform the coordinate system (from clockwise to counterclockwise), current changes direction with it. By the way, check the *current density* - it is a vector and it obeys, in generalized case of an inhomogeneous material, a tensor form of Ohm's law. There is no correct definition of anything. Ian is right in the narrow sense of the conventional vector in multiple dimensions and, especially, regarding the software implementation. But there is a legitimate (i.e. not self-contradictory) broader definition of a vector as an element of vector space, and complex numbers fall under it. Math is flexible, and there is definite benefit of consider complex numbers (and electric current under some circumstances) as vectors. Checking out of semantics hotel, Ed. On Thu, 2010-10-14 at 16:47 +0200, Ganesh Natrajan wrote: The definition of a vector as being something that has 'magnitude' and 'direction' is actually incorrect. If that were to be the case, a quantity like electric current would be a vector and not a scalar. Electric current is a scalar. A vector is something that transforms like the coordinate system, while a scalar does not. In other words, if you were to transform the coordinate system by a certain operator, a vector quantity in the old coordinate system can be transformed into the new one by using exactly the same operator. This is the correct definition of a vector. G. On Thu, 14 Oct 2010 10:22:59 -0400, Ed Pozharski epozh...@umaryland.edu wrote: The definition game is on! :) Vectors are supposed to have direction and amplitude, unlike scalars. Curiously, one can take a position that real numbers are vectors too, if you consider negative and positive numbers having opposite directions (and thus subtraction is simply a case of addition of a negative number). And of course, both scalars and vectors are simply tensors, of zeroth and first order :) Guess my point is that definitions are a matter of choice in math and if vector is defined as an array which must obey addition and scaling rules (but there is no fixed multiplication rule - regular 3D vectors have more than one possible product), then complex numbers are vectors. In a narrow sense of a real space vectors (the arrow thingy) they are not. Thus, complex number is a Vector, but not the vector (futile attempt at using articles by someone organically suffering from article dyslexia). Cheers, Ed. O -- Edwin Pozharski, PhD, Assistant Professor University of Maryland, Baltimore -- When the Way is forgotten duty and justice appear; Then knowledge and wisdom are born along with hypocrisy. When harmonious relationships dissolve then respect and devotion arise; When a nation falls to chaos then loyalty and patriotism are born. -- / Lao Tse /
Re: [ccp4bb] vector and scalars
Ed, The direction of current in an electrical circuit has nothing to do with any coordinate system. It is defined by convention in electricity as the direction opposite to that in which the electrons are moving. So the current is indicated as being from + to - in a circuit. Of course, you may change the convention but it still will not make the current (defined as the rate of change of charge i = dq/dt) a vector quantity. Current density is not the same as current. Current density is a directional quantity defined not with respect to any convention but with respect to a coordinate axis, and any transformation in that axis would result in a transformation in the current density by the same operator. Therefore current density is a vector. Mathematical tricks maybe, but all for a reason :) Ganesh On Thu, 14 Oct 2010 11:22:15 -0400, Ed Pozharski epozh...@umaryland.edu wrote: Again, definitions are a matter of choice. Under your strict version I still may consider electric current as vector, if I introduce the coordinate system in the circuit. When I transform the coordinate system (from clockwise to counterclockwise), current changes direction with it. By the way, check the *current density* - it is a vector and it obeys, in generalized case of an inhomogeneous material, a tensor form of Ohm's law. There is no correct definition of anything. Ian is right in the narrow sense of the conventional vector in multiple dimensions and, especially, regarding the software implementation. But there is a legitimate (i.e. not self-contradictory) broader definition of a vector as an element of vector space, and complex numbers fall under it. Math is flexible, and there is definite benefit of consider complex numbers (and electric current under some circumstances) as vectors. Checking out of semantics hotel, Ed. On Thu, 2010-10-14 at 16:47 +0200, Ganesh Natrajan wrote: The definition of a vector as being something that has 'magnitude' and 'direction' is actually incorrect. If that were to be the case, a quantity like electric current would be a vector and not a scalar. Electric current is a scalar. A vector is something that transforms like the coordinate system, while a scalar does not. In other words, if you were to transform the coordinate system by a certain operator, a vector quantity in the old coordinate system can be transformed into the new one by using exactly the same operator. This is the correct definition of a vector. G. On Thu, 14 Oct 2010 10:22:59 -0400, Ed Pozharski epozh...@umaryland.edu wrote: The definition game is on! :) Vectors are supposed to have direction and amplitude, unlike scalars. Curiously, one can take a position that real numbers are vectors too, if you consider negative and positive numbers having opposite directions (and thus subtraction is simply a case of addition of a negative number). And of course, both scalars and vectors are simply tensors, of zeroth and first order :) Guess my point is that definitions are a matter of choice in math and if vector is defined as an array which must obey addition and scaling rules (but there is no fixed multiplication rule - regular 3D vectors have more than one possible product), then complex numbers are vectors. In a narrow sense of a real space vectors (the arrow thingy) they are not. Thus, complex number is a Vector, but not the vector (futile attempt at using articles by someone organically suffering from article dyslexia). Cheers, Ed. O -- Edwin Pozharski, PhD, Assistant Professor University of Maryland, Baltimore -- When the Way is forgotten duty and justice appear; Then knowledge and wisdom are born along with hypocrisy. When harmonious relationships dissolve then respect and devotion arise; When a nation falls to chaos then loyalty and patriotism are born. -- / Lao Tse / -- *
Re: [ccp4bb] vector and scalars
Ed, I think you're confusing 'electric current' with 'electric current density'. The first is a scalar, the second a vector. The current I is defined as the surface integral of the density vector J with respect to the element of area dA: I = integral over S (J.dA) (how I wish we could use proper equations in e-mails!) in other words, the net flux of the current density vector field flowing through the surface S. J and dA are both vectors, J.dA is their scalar product, hence I must be a scalar. ( Source: http://en.wikipedia.org/wiki/Current_density ). You're right that in principle definitions are arbitrary. However there exist a whole set of conventional definitions which are completely self-consistent and which we would be wise to stick to. These have been honed over many years by people who know what they're doing, and which do not lead us into logical contradictions. If you're going to come up with an alternative set of definitions you are going to have to prove to everyone that they aren't going to lead us into contradictions now and in the future. I wish you luck with your task! For example how in your definition can both the current and the current density be vectors? What is your version of the above equation that relates them? Cheers -- Ian On Thu, Oct 14, 2010 at 4:22 PM, Ed Pozharski epozh...@umaryland.edu wrote: Again, definitions are a matter of choice. Under your strict version I still may consider electric current as vector, if I introduce the coordinate system in the circuit. When I transform the coordinate system (from clockwise to counterclockwise), current changes direction with it. By the way, check the *current density* - it is a vector and it obeys, in generalized case of an inhomogeneous material, a tensor form of Ohm's law. There is no correct definition of anything. Ian is right in the narrow sense of the conventional vector in multiple dimensions and, especially, regarding the software implementation. But there is a legitimate (i.e. not self-contradictory) broader definition of a vector as an element of vector space, and complex numbers fall under it. Math is flexible, and there is definite benefit of consider complex numbers (and electric current under some circumstances) as vectors. Checking out of semantics hotel, Ed. On Thu, 2010-10-14 at 16:47 +0200, Ganesh Natrajan wrote: The definition of a vector as being something that has 'magnitude' and 'direction' is actually incorrect. If that were to be the case, a quantity like electric current would be a vector and not a scalar. Electric current is a scalar. A vector is something that transforms like the coordinate system, while a scalar does not. In other words, if you were to transform the coordinate system by a certain operator, a vector quantity in the old coordinate system can be transformed into the new one by using exactly the same operator. This is the correct definition of a vector. G. On Thu, 14 Oct 2010 10:22:59 -0400, Ed Pozharski epozh...@umaryland.edu wrote: The definition game is on! :) Vectors are supposed to have direction and amplitude, unlike scalars. Curiously, one can take a position that real numbers are vectors too, if you consider negative and positive numbers having opposite directions (and thus subtraction is simply a case of addition of a negative number). And of course, both scalars and vectors are simply tensors, of zeroth and first order :) Guess my point is that definitions are a matter of choice in math and if vector is defined as an array which must obey addition and scaling rules (but there is no fixed multiplication rule - regular 3D vectors have more than one possible product), then complex numbers are vectors. In a narrow sense of a real space vectors (the arrow thingy) they are not. Thus, complex number is a Vector, but not the vector (futile attempt at using articles by someone organically suffering from article dyslexia). Cheers, Ed. O -- Edwin Pozharski, PhD, Assistant Professor University of Maryland, Baltimore -- When the Way is forgotten duty and justice appear; Then knowledge and wisdom are born along with hypocrisy. When harmonious relationships dissolve then respect and devotion arise; When a nation falls to chaos then loyalty and patriotism are born. -- / Lao Tse /
Re: [ccp4bb] vector and scalars
As I sit here listening to the giant whoosh sound of all the world's biologists unsubscribing from the CCP4BB, I wonder if anyone on this thread can explain to me the difference between a matrix and a tensor? I ask because I think stress and strain are mechanisms of radiation damage, but where I am stuck is that Young's modulus seems to always be represented by a tensor (as opposed to something that makes sense). This is not helped by the lack of a tensor class in stdlib! However, I do think it is interesting that this same Thomas Young performed a famous experiment in 1801 that (eventually) proved a single photon can scatter off of two slits at the same time. This is one of two experiments that can only be explained by quantum mechanics. -James Holton stressed and strained scientist On 10/14/2010 8:22 AM, Ed Pozharski wrote: Again, definitions are a matter of choice. Under your strict version I still may consider electric current as vector, if I introduce the coordinate system in the circuit. When I transform the coordinate system (from clockwise to counterclockwise), current changes direction with it. By the way, check the *current density* - it is a vector and it obeys, in generalized case of an inhomogeneous material, a tensor form of Ohm's law. There is no correct definition of anything. Ian is right in the narrow sense of the conventional vector in multiple dimensions and, especially, regarding the software implementation. But there is a legitimate (i.e. not self-contradictory) broader definition of a vector as an element of vector space, and complex numbers fall under it. Math is flexible, and there is definite benefit of consider complex numbers (and electric current under some circumstances) as vectors. Checking out of semantics hotel, Ed. On Thu, 2010-10-14 at 16:47 +0200, Ganesh Natrajan wrote: The definition of a vector as being something that has 'magnitude' and 'direction' is actually incorrect. If that were to be the case, a quantity like electric current would be a vector and not a scalar. Electric current is a scalar. A vector is something that transforms like the coordinate system, while a scalar does not. In other words, if you were to transform the coordinate system by a certain operator, a vector quantity in the old coordinate system can be transformed into the new one by using exactly the same operator. This is the correct definition of a vector. G. On Thu, 14 Oct 2010 10:22:59 -0400, Ed Pozharski epozh...@umaryland.edu wrote: The definition game is on! :) Vectors are supposed to have direction and amplitude, unlike scalars. Curiously, one can take a position that real numbers are vectors too, if you consider negative and positive numbers having opposite directions (and thus subtraction is simply a case of addition of a negative number). And of course, both scalars and vectors are simply tensors, of zeroth and first order :) Guess my point is that definitions are a matter of choice in math and if vector is defined as an array which must obey addition and scaling rules (but there is no fixed multiplication rule - regular 3D vectors have more than one possible product), then complex numbers are vectors. In a narrow sense of a real space vectors (the arrow thingy) they are not. Thus, complex number is a Vector, but not the vector (futile attempt at using articles by someone organically suffering from article dyslexia). Cheers, Ed. O
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] Problem installing CCP4 on MacOSX behind firewall?
Two ideas: 1. Create a file in your home directory called ~/.curlrc and in it put the following line: -P - ftp 2. Use wget first, install wget with fink Then put the line DownloadMethod: wget into the file /sw/etc/fink.conf (or /sw64/etc/fink.conf ). I use wget. It seems to be more robust somehow. On Oct 14, 2010, at 2:20 AM, Mads Gabrielsen wrote: I have a problem installing CCP4 on a MacOSX 10.6.4. When I follow the procedure on Bill Scott's web-pages to install the precompiled version, I get an error message saying Err http://sage.ucsc.edu stable/main Packages 403 Forbidden When trying to install it via the normal Fink way, I get curl (56) FTP response reading failed ###execution of curl failed, exit code 56 Downloading the file ccp4-6.1.13-core-src.tar.gz failed. I think there is an issue with our firewall, but apart from setting the http_proxy to what has been suggested by our IT services I don't really know how to proceed. Does anyone have any suggestions? I know I can download the disk images from the ccp4 pages, but I would like to understand what is going wrong. Cheers, Mads
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 for the
[ccp4bb] Sweet and discreet tweets about helices and sheets - @PDBeurope
Following feedback from students, collaborators and other structural biologists, the Protein Data Bank in Europe (PDBe; pdbe.org) has become aware of a need for a social networking presence to strengthen the link with current and potential users of its resources. So if you would like to be kept up-to-date about new structure releases or PDBe news, features, jobs, tutorials, roadshows, resources, services, outreach activities and more, then follow us on twitter... http://twitter.com/PDBeurope We rely on you (the user) to tell us what we should be tweeting about (or if we should just twut up). Reply to our tweets or give us feedback using the Feedback button near the top of the PDBe web pages. --Gerard (Who has just kicked his Facebook habit and refuses to become a Tweethoven.) --- Gerard J. Kleywegt, PDBe, EMBL-EBI, Hinxton, UK ger...@ebi.ac.uk . pdbe.org Secretary: Pauline Haslam pdbe_ad...@ebi.ac.uk
Re: [ccp4bb] vector and scalars
On Thursday, October 14, 2010 09:11:50 am James Holton wrote: As I sit here listening to the giant whoosh sound of all the world's biologists unsubscribing from the CCP4BB, I wonder if anyone on this thread can explain to me the difference between a matrix and a tensor? In invoking the latter, one risks that Tension, apprehension, And dissension have begun. (and now that whoosh may encompass all our younger readers, biologists or not). As to electric current, note that its generalized description may include a vector corresponding to the spin state of the electrons that carry it. Relevant to electromagnets, if nothing else. Ethan I ask because I think stress and strain are mechanisms of radiation damage, but where I am stuck is that Young's modulus seems to always be represented by a tensor (as opposed to something that makes sense). This is not helped by the lack of a tensor class in stdlib! Isn't that just because real life materials and real life objects are easier to bend in in some directions than in other directions? However, I do think it is interesting that this same Thomas Young performed a famous experiment in 1801 that (eventually) proved a single photon can scatter off of two slits at the same time. This is one of two experiments that can only be explained by quantum mechanics. -James Holton stressed and strained scientist On 10/14/2010 8:22 AM, Ed Pozharski wrote: Again, definitions are a matter of choice. Under your strict version I still may consider electric current as vector, if I introduce the coordinate system in the circuit. When I transform the coordinate system (from clockwise to counterclockwise), current changes direction with it. By the way, check the *current density* - it is a vector and it obeys, in generalized case of an inhomogeneous material, a tensor form of Ohm's law. There is no correct definition of anything. Ian is right in the narrow sense of the conventional vector in multiple dimensions and, especially, regarding the software implementation. But there is a legitimate (i.e. not self-contradictory) broader definition of a vector as an element of vector space, and complex numbers fall under it. Math is flexible, and there is definite benefit of consider complex numbers (and electric current under some circumstances) as vectors. Checking out of semantics hotel, Ed. On Thu, 2010-10-14 at 16:47 +0200, Ganesh Natrajan wrote: The definition of a vector as being something that has 'magnitude' and 'direction' is actually incorrect. If that were to be the case, a quantity like electric current would be a vector and not a scalar. Electric current is a scalar. A vector is something that transforms like the coordinate system, while a scalar does not. In other words, if you were to transform the coordinate system by a certain operator, a vector quantity in the old coordinate system can be transformed into the new one by using exactly the same operator. This is the correct definition of a vector. G. On Thu, 14 Oct 2010 10:22:59 -0400, Ed Pozharski epozh...@umaryland.edu wrote: The definition game is on! :) Vectors are supposed to have direction and amplitude, unlike scalars. Curiously, one can take a position that real numbers are vectors too, if you consider negative and positive numbers having opposite directions (and thus subtraction is simply a case of addition of a negative number). And of course, both scalars and vectors are simply tensors, of zeroth and first order :) Guess my point is that definitions are a matter of choice in math and if vector is defined as an array which must obey addition and scaling rules (but there is no fixed multiplication rule - regular 3D vectors have more than one possible product), then complex numbers are vectors. In a narrow sense of a real space vectors (the arrow thingy) they are not. Thus, complex number is a Vector, but not the vector (futile attempt at using articles by someone organically suffering from article dyslexia). Cheers, Ed. O -- 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)
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-
[ccp4bb] Webinar -- Scientific inquiry, inference and critical reasoning in the macromolecular crystallography curriculum
Dear colleagues, I would like to draw your attention to an upcoming webinar to be presented by Bernhard Rupp titled Scientific inquiry, inference and critical reasoning in the macromolecular crystallography curriculum. In this webinar, Bernhard will expand on his recent Journal of Applied Crystallography article that discusses higher education curricula in the context of scientific analysis. Bernhard analyzes recent cases of high profile structure retractions and argues that With the great power of modern crystallography comes great responsibility for its appropriate use. This webinar is scheduled to occur on Thursday, October 21st at 10:00 AM PDT (13:00 EDT / 18:00 GMT+1). You can find more information, including a registration link at: http://www.rigaku.com/protein/webinars.html Best regards, Angela -- Angela R. Criswell, Ph.D. Rigaku Americas Corp angela.crisw...@rigaku.com
Re: [ccp4bb] vector and scalars
As I sit here listening to the giant whoosh sound of all the world's biologists unsubscribing from the CCP4BB, I wonder if anyone on this thread can explain to me the difference between a matrix and a tensor? Since when are there biologists on this bb? JPK p.s. Is whooshing biologist-specific?
Re: [ccp4bb] vector and scalars
On Thu, 2010-10-14 at 09:11 -0700, James Holton wrote: I wonder if anyone on this thread can explain to me the difference between a matrix and a tensor? Matrix is a 2nd order tensor. Tensors may have any number of dimensions, including zero. Tensor is just a fancy name for a multidimensional array + definition of the rule by which different tensors may be multiplied. As for Young modulus, it is a 2D tensor to account for material anisotropy. -- 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)
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
[ccp4bb] NUVISION 60GX glasses/emitters
Is there anybody out there who has a use for these glasses/emitters that would be willing to purchase some for a reduced fee? I have 16 pair (I think, I need to go check closely, but I have a lot of them), plus 6-7 emitters. I just need to go away from this type of system, probably to a mixed bag of zalman passive and strict 2D. Not all of the glasses work (3 of them do not), but I think the problem is just a small microswitch, and NuVision will repair them for next to nothing (I just haven't had time to deal with that issue). Sorry to make this board into a garage sale, but it seems like this group is the best place to put these. I would give them away if I could afford it, but I'm trying to use proceeds from this to get a couple of higher end monitors. Thanks. Feel free to contact me off list if interested (drobe...@depauw.edu) Dave
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
[ccp4bb] Faculty Position, Dept of Molecular Biology, Princeton University
Faculty Position Department of Molecular Biology Princeton University The Department of Molecular Biology at Princeton University invites applications for a tenure-track faculty position at the assistant professor level. We are seeking an outstanding investigator in the area of biochemistry and structural biology. We are particularly interested in candidates whose plans to address fundamental biological questions include the use of X-ray crystallography. Applicants must have an excellent record of research productivity and demonstrate the ability to develop a rigorous research program. All applicants must have a Ph.D. or M.D. with postdoctoral research experience and a commitment to teaching at the undergraduate and graduate levels. Applications must be submitted online at http://jobs.princeton.edu, requisition #1000770, and should include a cover letter, curriculum vitae and a short summary of research interests. We also require three letters of recommendation. All materials must be submitted as PDF files. For full consideration, applications should be received by December 1, 2010. Princeton University is an Equal Opportunity Employer and complies with applicable EEO and affirmative action regulations.
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
[ccp4bb] inflated BOND_RMSD with external restraints (refmac)
It appears that external restraints are included in bond_rmsd calculation. When they are used to restrain the hydrogen bonds to maintain the Watson-Crick pairing in a 3A resolution structure of a protein-DNA complex, the bond_rmsd is inflated about 5 times. To verify this, the refmac run was done with external restraints removed and zero refinement cycles. With external restraints I get 0.028A, without - 0.006A. This is with v.5.5.0109. I assume this may be classified as a bug - there is no limitation to using external restraints only for the covalent bonds, thus they should not be included in bond_rmsd calculation. On a practical side, I now have a misleading bond_rmsd value. The correct one can be calculated as described, but this may make geometry weight optimization cumbersome. Do I understand correctly that an alternative is to monitor Zbonds, with a rule that it should be around 1.0? And more generally, shouldn't we not look at rmsd_bonds at all and only use Zbonds instead (which is, I assume, an average bond length deviation to the target value ratio?) I suspect that acceptable bond_rmsd value is slightly affected by sequence. Ed. -- I'd jump in myself, if I weren't so good at whistling. Julian, King of Lemurs
Re: [ccp4bb] inflated BOND_RMSD with external restraints (refmac)
Hi Ed refmac 5.6 should not have this problem. Yes, you are right. It should be considered as a bug. I think I have fixed it. Could you please try 5.6version from: www.ysbl.york.ac.uk/refmac/latest_refmac.html You need to take experimental version (it should be stable enough, although I update it more often than older versions). In this version to make external restraints as covalent bonds you need to specify type 1 (type 0 means dictionary values will be overwritten and type 2 is external restraints for non-covalent bonds) I hope it helps regards Garib On 14 Oct 2010, at 21:51, Ed Pozharski wrote: It appears that external restraints are included in bond_rmsd calculation. When they are used to restrain the hydrogen bonds to maintain the Watson-Crick pairing in a 3A resolution structure of a protein-DNA complex, the bond_rmsd is inflated about 5 times. To verify this, the refmac run was done with external restraints removed and zero refinement cycles. With external restraints I get 0.028A, without - 0.006A. This is with v.5.5.0109. I assume this may be classified as a bug - there is no limitation to using external restraints only for the covalent bonds, thus they should not be included in bond_rmsd calculation. On a practical side, I now have a misleading bond_rmsd value. The correct one can be calculated as described, but this may make geometry weight optimization cumbersome. Do I understand correctly that an alternative is to monitor Zbonds, with a rule that it should be around 1.0? And more generally, shouldn't we not look at rmsd_bonds at all and only use Zbonds instead (which is, I assume, an average bond length deviation to the target value ratio?) I suspect that acceptable bond_rmsd value is slightly affected by sequence. Ed. -- 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)
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
