Choice of origin - Revision history

Kay: /* Examples */

2008-02-07T21:02:28Z

Examples

← Older revision		Revision as of 23:02, 7 February 2008
Line 8:		Line 8:

	* P1: There is only one operator (X, Y, Z) so there are no difference vectors between operators. In this case, any vector with components u,v,w may be added, meaning that any place in the unit cell may serve as an origin (or a molecule may be shifted anywhere and still give the same R-factor).		* P1: There is only one operator (X, Y, Z) so there are no difference vectors between operators. In this case, any vector with components u,v,w may be added, meaning that any place in the unit cell may serve as an origin (or a molecule may be shifted anywhere and still give the same R-factor).
	* P2<math>_1</math>: This has operators (X, Y, Z) and (-X, Y+1/2, -Z). Difference operator is (2X, 1/2, 2Z). Any change of (X,Y,Z) by adding (0, v, 0) or (0.5, 0, 0) or (0, 0, 0.5), or any combination of these, will leave the difference operator alone (because 2(X+0.5) = 2X +1 which is the same as 2X in Patterson space, it's just in the next unit cell of the Patterson). In other words: in P2<math>_1</math> you may shift a molecule along the b axis, or by half a unit cell in a or c, and still ~~get the same R-factor~~.		* P2<math>_1</math>: This has operators (X, Y, Z) and (-X, Y+1/2, -Z). Difference operator is (2X, 1/2, 2Z). Any change of (X,Y,Z) by adding (0, v, 0) or (0.5, 0, 0) or (0, 0, 0.5), or any combination of these, will leave the difference operator alone (because 2(X+0.5) = 2X +1 which is the same as 2X in Patterson space, it's just in the next unit cell of the Patterson). In other words: in P2<math>_1</math> you may shift a molecule along the b axis, or by half a unit cell in a or c, and still have a equally correct structure solution.


	== See also ==		== See also ==

	http://www.ccp4.ac.uk/dist/html/alternate_origins.html		http://www.ccp4.ac.uk/dist/html/alternate_origins.html

Kay: /* Examples */

2008-02-07T20:57:51Z

Examples

← Older revision		Revision as of 22:57, 7 February 2008
Line 9:		Line 9:
	* P1: There is only one operator (X, Y, Z) so there are no difference vectors between operators. In this case, any vector with components u,v,w may be added, meaning that any place in the unit cell may serve as an origin (or a molecule may be shifted anywhere and still give the same R-factor).		* P1: There is only one operator (X, Y, Z) so there are no difference vectors between operators. In this case, any vector with components u,v,w may be added, meaning that any place in the unit cell may serve as an origin (or a molecule may be shifted anywhere and still give the same R-factor).
	* P2<math>_1</math>: This has operators (X, Y, Z) and (-X, Y+1/2, -Z). Difference operator is (2X, 1/2, 2Z). Any change of (X,Y,Z) by adding (0, v, 0) or (0.5, 0, 0) or (0, 0, 0.5), or any combination of these, will leave the difference operator alone (because 2(X+0.5) = 2X +1 which is the same as 2X in Patterson space, it's just in the next unit cell of the Patterson). In other words: in P2<math>_1</math> you may shift a molecule along the b axis, or by half a unit cell in a or c, and still get the same R-factor.		* P2<math>_1</math>: This has operators (X, Y, Z) and (-X, Y+1/2, -Z). Difference operator is (2X, 1/2, 2Z). Any change of (X,Y,Z) by adding (0, v, 0) or (0.5, 0, 0) or (0, 0, 0.5), or any combination of these, will leave the difference operator alone (because 2(X+0.5) = 2X +1 which is the same as 2X in Patterson space, it's just in the next unit cell of the Patterson). In other words: in P2<math>_1</math> you may shift a molecule along the b axis, or by half a unit cell in a or c, and still get the same R-factor.


			== See also ==

			http://www.ccp4.ac.uk/dist/html/alternate_origins.html

Kay: /* Examples */

2008-02-07T20:51:42Z

Examples

← Older revision		Revision as of 22:51, 7 February 2008
Line 8:		Line 8:

	* P1: There is only one operator (X, Y, Z) so there are no difference vectors between operators. In this case, any vector with components u,v,w may be added, meaning that any place in the unit cell may serve as an origin (or a molecule may be shifted anywhere and still give the same R-factor).		* P1: There is only one operator (X, Y, Z) so there are no difference vectors between operators. In this case, any vector with components u,v,w may be added, meaning that any place in the unit cell may serve as an origin (or a molecule may be shifted anywhere and still give the same R-factor).
	* P2<math>_1</math>: This has operators (X, Y, Z) and (-X, Y+1/2, Z). Difference operator is (2X, 1/2, 2Z). Any change of (X,Y,Z) by adding (0, v, 0) or (0.5, 0, 0) or (0, 0, 0.5), or any combination of these, will leave the difference operator alone (because 2(X+0.5) = 2X +1 which is the same as 2X in Patterson space).		* P2<math>_1</math>: This has operators (X, Y, Z) and (-X, Y+1/2, -Z). Difference operator is (2X, 1/2, 2Z). Any change of (X,Y,Z) by adding (0, v, 0) or (0.5, 0, 0) or (0, 0, 0.5), or any combination of these, will leave the difference operator alone (because 2(X+0.5) = 2X +1 which is the same as 2X in Patterson space, it's just in the next unit cell of the Patterson). In other words: in P2<math>_1</math> you may shift a molecule along the b axis, or by half a unit cell in a or c, and still get the same R-factor.

Kay: New page: The "problem" of choosing between possible origins occurs in all spacegroups (I believe!). It normally manifests itself when independantly calculating phases (e.g. by [[Molecular replaceme...

2008-02-07T20:48:11Z

New page: The "problem" of choosing between possible origins occurs in all spacegroups (I believe!). It normally manifests itself when independantly calculating phases (e.g. by [[Molecular replaceme...

New page

The "problem" of choosing between possible origins occurs in all spacegroups (I believe!). It normally manifests itself when independantly calculating phases (e.g. by [[Molecular replacement]]) for a structure that was solved before, and finding, upon graphical comparison of the models, that the old and the new model are in different positions in space, even if crystallographic symmetry operations are considered.

This situation arises because the measured intensities are the same for two (or more) transformations of the model, and therefore all transformed models are equally valid. To understand the situation better, remember that the Patterson function (which may be calculated from the measured intensities) only depends on difference vectors. This means that all transformations of the model, which do not change the difference vectors, do not change the intensities. These transformations always correspond to addition of vectors.

To find all possible transformations which do not change the difference vectors, just calculate all differences between space group operators, and find out which transformations leave the difference vectors invariant.

== Examples ==

* P1: There is only one operator (X, Y, Z) so there are no difference vectors between operators. In this case, any vector with components u,v,w may be added, meaning that any place in the unit cell may serve as an origin (or a molecule may be shifted anywhere and still give the same R-factor).
* P2<math>_1</math>: This has operators (X, Y, Z) and (-X, Y+1/2, Z). Difference operator is (2X, 1/2, 2Z). Any change of (X,Y,Z) by adding (0, v, 0) or (0.5, 0, 0) or (0, 0, 0.5), or any combination of these, will leave the difference operator alone (because 2*(X+0.5) = 2*X +1 which is the same as 2X in Patterson space).