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[ccp4bb]: Summary: anisotropic ellipsoids



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This is a summary to the following question I posted yesterday:

>According to many textbooks the first three of the thermal parameters
>U11 U22 U33 U12 U13 and U23 describe the displacements along the
>perpendicular principal axis of the ellipsoid and the latter three give
>the orientation of the principal axes with respect to the unit cell axes.
>However, I can't find anywhere how U12 U13 and U23 (apparently as
>direction cosines) exactly describe the orientation of the ellipsoid, say
>in a cartesian system.

First of all, my question was based on the false assumption that U11, U22 and
U33 are the components along the principal axes of the ellipsoid. The text on
page 533 of Glusker et al. "Crystal structure analysis for chemists and 
biologists" led me to that conclusion, although the example on page 536 
indicates that things are not as simple as that. U11, U22 and U33 are the
<u2> values along the reciprocal cell axes a*, b* and c*, respectively 
(e.g. Drenth, page 94).

The principal axes of the thermal ellipsoid can be obtained from the U values
via a principal axes transformation. This is described e.g. in Giacovazzo et
al., p. 75 ff. and 148 (don't rely on the index), in the ORTEP manual, and in 
some of the hints cited below.


Thanks to all who replied:

From: Manfred Buehner <buehner@biozentrum.uni-wuerzburg.de>
From: Eleanor J. Dodson <ccp4@ysbl.york.ac.uk>
From: Nicholas M Glykos <glykos@crystal2.imbb.forth.gr>
From: Witek Kwiatkowski <witek@octopus.salk.edu>
From: "Papiz, MZ (Miroslav)" <M.Z.Papiz@dl.ac.uk>
From: Phraenquex VD <loretta@scripps.edu>
From: M.D.Winn <M.D.Winn@dl.ac.uk>


Following are parts of the replies (in arbitrary order):

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Not really. U11 U22 U33 U12 U13 and U23 are the tensor elements
in the chosen coordinate system (usually orthogonal). Only if
the coordinate system is rotated to the principal axes system
are U11 U22 U33 the displacements along the principal axes, in
which case U12 U13 and U23 are zero. U12 U13 and U23 only define
the orientation of the principal axes in the very loose sense that
their non-zero-ness indicates the deviation between the orthogonal
and principal axes systems.

Hmmm, quick flip through ortep manual, if that's what you're using. 
There seem to be some definitions in terms of orientations (which
I've never used) but not couched in terms of U elements ....

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  Would the equation on page 327 of Vol.II of the international tables
(for the anisotropic thermal parameters) be of help to you ?

ps In case you can't find this volume, the equation is :

fr = exp{-(b11*h^2 + b12*hk + b13*hl + b22*k^2 + b23*kl + b33*l^2)} * fr0

where fr0 is the scattering factor for atom r at rest, fr is the
scattering factor of the anisotropically-treated atom, and
b11,b12,... are the atomic anisotropic thermal parameters of the atom.

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Apologies for repeating the obvious, but I thought that what you was
looking for was a qualitative answer to your question. If you are up to
calculating things, page 76 of "Fundamentals of Crystallography" by
Giacovazzo et al., has a worked (numerical) example for calculating the
principal axes of an atomic thermal ellipsoid starting from (b11,b12,...).

--------------------------------------------------------------------------

 Is this any help? Taken from $CLIBS/rwbrook.f

C  PDB files contain cartesian anisotropic temperature factors as
orthogonal Us.
C Stored as U11 U22 U33 U12 U13 U23
C The anisotropic temperature factors can be input/output to this
routine
C  as orthogonal or as crystallographic Us.
C 
C  Shelx defines Uf to calculate temperature factor as:
C T(aniso_Uf) = exp (-2PI**2 ( (h*ast)**2 Uf_11 + (k*bst)**2 Uf_22 + ...
C                            + 2hk*ast*bst*Uf_12 +..)
C
C   Note:   Uo_ji == Uo_ij and  Uf_ji == Uf_ij.
C
C  [Uo_ij] listed on ANISOU card satisfy  the relationship:
C  [Uo_ij] =   [RFu]-1 [Uf_ij] {[RFu]-1}T   C
C        where [Rfu] is the normalised [Rf] matrix read from the SCALEi
cards.
C        see code.   [ROu] ==  [RFu]-1
C
C T(aniso_Uo) = U(11)*H**2 + U(22)*K**2 + 2*U(12)*H*K + ...
C where H,K,L are orthogonal reciprocal lattice indecies. ( EJD: I
think????)
C
C Biso     = 8*PI**2 (Uo_11 + Uo_22 + Uo_33) / 3.0
C
C   [Uf(symm_j)] = [Symm_j] [Uf] [Symm_j]T
C

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  i think  to change into an orthonormal system you use
  U = [A .b. AT] where  B is the usual  anisotropic Dsip. tensor
  A is the cholensky factor and AT is its transpose
  A has the property     G = (AT.A)  where G is the real space
  metric tensor
         g       =     a**2,      a*b*cos(g), a*c*cos(b)
                       a*b*cos(G),  b**2, B*C*cos(a)
                       a*c*cos(b), b*c*cos(a), c**2

  the eignvalues and eigenvectors oF u are the principal
  axes (PA) of the elipsoides. I guess the angles they make
  with a, b, and c are the cosine of the vector.
  so with zero off diagonal terms the PA's are along the
  cell axes (in an orthoganal cell) and with non zero 
  values this tells you how to rotate these Pa's from the
  cell axes.
             
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exp(X^T U X) (where X is a vector from atom's equilibrium position
X^T is a transpose of this vector and U is an Uij matrix or
anisotropic displacement factor tensor if you prefer) is  a probability
function that describes the relative probability of finding nucleus
displaced by the vector X from equilibrium.
and equation X^T U X = C describes an ellipsoidal surface on which
such probability is constant.  What you are asking is finding principal
axis of this quadratic form - can be solved via the Lagrange
multipliers.

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-Norbert