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|
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\end_header
\begin_body
\begin_layout Title
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vskip -3em
\end_layout
\end_inset
CrystFEL matrix notation conventions
\end_layout
\begin_layout Standard
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vskip -4em
\end_layout
\begin_layout Plain Layout
\backslash
thispagestyle{empty}
\end_layout
\end_inset
\end_layout
\begin_layout Standard
CrystFEL's UnitCell structure doesn't expose its contents as a matrix, only
as named components (
\begin_inset Formula $a_{x}^{*}$
\end_inset
,
\begin_inset Formula $c_{y}$
\end_inset
an so on).
However, the arrangement of the transformation matrices given to functions
such as cell_transform_rational() are important.
This document aims to make the conventions explicit.
The notation used in CrystFEL follows that used in the International Tables
for Crystallography, volume A, chapter 5.1 (
\begin_inset Quotes eld
\end_inset
Transformations of the coordinate system (unit-cell transformations)
\begin_inset Quotes erd
\end_inset
) and 11.1 (
\begin_inset Quotes eld
\end_inset
Point coordinates, symmetry operations and their symbols
\begin_inset Quotes erd
\end_inset
).
\end_layout
\begin_layout Itemize
\emph on
cell_transform_gsl_direct
\emph default
,
\emph on
cell_transform_rational
\emph default
and
\emph on
cell_transform_intmat
\emph default
all do what is written in equation
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:cell_transform"
plural "false"
caps "false"
noprefix "false"
\end_inset
.
These functions all take a
\begin_inset Formula $\boldsymbol{\hat{P}}$
\end_inset
matrix.
The
\begin_inset Quotes eld
\end_inset
intmat
\begin_inset Quotes erd
\end_inset
and
\begin_inset Quotes eld
\end_inset
rational
\begin_inset Quotes erd
\end_inset
versions will additionally determine the centering of the resulting unit
cell (and hopefully eventually the lattice type and unique axis).
\end_layout
\begin_layout Itemize
\emph on
transform_indices
\emph default
does what is written in equation
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:indices"
plural "false"
caps "false"
noprefix "false"
\end_inset
.
It takes a vector of (reciprocal space) Miller indices and a
\begin_inset Formula $\boldsymbol{\hat{P}}$
\end_inset
matrix.
\end_layout
\begin_layout Itemize
A SymOpList is essentially a list of
\begin_inset Formula $\boldsymbol{\hat{W}}^{-1}$
\end_inset
matrices, which, as shown below, behave like
\begin_inset Formula $\boldsymbol{\hat{P}}$
\end_inset
matrices.
\end_layout
\begin_layout Itemize
transform_fractional_coords_rtnl takes a
\begin_inset Formula $\boldsymbol{\hat{P}}$
\end_inset
matrix and a column vector of fractional coordinates, calculates the
\begin_inset Formula $\boldsymbol{\hat{Q}}$
\end_inset
matrix and evaluates equation
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:fractional-coord-transform-basis-change"
plural "false"
caps "false"
noprefix "false"
\end_inset
(NB
\series bold
not
\series default
equation
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:fractional-coord-transform-symmetry"
plural "false"
caps "false"
noprefix "false"
\end_inset
).
\emph on
Implementation detail:
\emph default
it actually solves a matrix-vector equation rather than going all the way
to the inverted matrix.
\end_layout
\begin_layout Itemize
transform_fractional_coords_rtnl_inverse takes a
\begin_inset Formula $\boldsymbol{\hat{P}}$
\end_inset
matrix and a column vector of fractional coordinates
\begin_inset Formula $\boldsymbol{v}$
\end_inset
, and does the reverse of transform_fractional_coords_rtnl.
That means it just has to evaluate
\begin_inset Formula $\boldsymbol{\hat{P}}\boldsymbol{v}$
\end_inset
.
This looks like equation
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:fractional-coord-transform-symmetry"
plural "false"
caps "false"
noprefix "false"
\end_inset
.
\end_layout
\begin_layout Subsection*
Matrices of unit cell components
\end_layout
\begin_layout Standard
When matrix calculations are performed in CrystFEL, which is usually only
inside certain libcrystfel functions, basis vectors
\begin_inset Formula $(\boldsymbol{a},\boldsymbol{b},\boldsymbol{c})$
\end_inset
are written as the columns of the matrix.
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
\boldsymbol{\hat{M}}=\left(\begin{array}{ccc}
a_{x} & b_{x} & c_{x}\\
a_{y} & b_{y} & c_{y}\\
a_{z} & b_{z} & c_{z}
\end{array}\right)
\]
\end_inset
\end_layout
\begin_layout Standard
Reciprocal basis vectors are written as the rows of the matrix.
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
\boldsymbol{\hat{R}}=\left(\begin{array}{ccc}
a_{x}^{*} & a_{y}^{*} & a_{z}^{*}\\
b_{x}^{*} & b_{y}^{*} & b_{z}^{*}\\
c_{x}^{*} & c_{y}^{*} & c_{z}^{*}
\end{array}\right)
\]
\end_inset
\end_layout
\begin_layout Standard
Real and reciprocal spaces are sometimes said to be
\begin_inset Quotes eld
\end_inset
inverse transposes of each other
\begin_inset Quotes erd
\end_inset
.
The components of the two matrices are already transposed with respect
to one another (one has vectors in columns, the other in rows).
Therefore
\begin_inset Formula $\boldsymbol{\hat{M}}=\boldsymbol{\hat{R}}^{-1}$
\end_inset
and
\begin_inset Formula $\boldsymbol{\hat{R}}=\boldsymbol{\hat{M}}^{-1}$
\end_inset
.
\end_layout
\begin_layout Subsection*
Transformation of unit cell basis
\end_layout
\begin_layout Standard
Given a transformation matrix
\begin_inset Formula $\boldsymbol{\hat{P}}$
\end_inset
which means
\begin_inset Formula $\boldsymbol{a}^{\prime}=\boldsymbol{a}-\boldsymbol{b}$
\end_inset
,
\begin_inset Formula $\boldsymbol{b}^{\prime}=\boldsymbol{b}$
\end_inset
,
\begin_inset Formula $\boldsymbol{c}^{\prime}=\boldsymbol{c}$
\end_inset
:
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{equation}
\boldsymbol{\hat{M}}^{\prime}=\boldsymbol{\hat{M}}\boldsymbol{\hat{P}}=\left(\begin{array}{ccc}
a_{x} & b_{x} & c_{x}\\
a_{y} & b_{y} & c_{y}\\
a_{z} & b_{z} & c_{z}
\end{array}\right)\left(\begin{array}{ccc}
1 & 0 & 0\\
\bar{1} & 1 & 0\\
0 & 0 & 1
\end{array}\right)\label{eq:cell_transform}
\end{equation}
\end_inset
\end_layout
\begin_layout Standard
Note that the
\begin_inset Quotes eld
\end_inset
amount of old a,b,c to make new a
\begin_inset Quotes erd
\end_inset
is found in the first column of the
\begin_inset Formula $\boldsymbol{\hat{P}}$
\end_inset
matrix, and so on.
\end_layout
\begin_layout Subsection*
Transformation of reciprocal basis
\end_layout
\begin_layout Standard
With the same transformation as before (
\begin_inset Formula $\boldsymbol{a}^{\prime}=\boldsymbol{a}-\boldsymbol{b}$
\end_inset
,
\begin_inset Formula $\boldsymbol{b}^{\prime}=\boldsymbol{b}$
\end_inset
,
\begin_inset Formula $\boldsymbol{c}^{\prime}=\boldsymbol{c}$
\end_inset
), the reciprocal basis transforms as follows:
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
\boldsymbol{\hat{R}}^{\prime}=\left(\begin{array}{ccc}
a_{x}^{*} & a_{y}^{*} & a_{z}^{*}\\
b_{x}^{*} & b_{y}^{*} & b_{z}^{*}\\
c_{x}^{*} & c_{y}^{*} & c_{z}^{*}
\end{array}\right)^{\prime}=\boldsymbol{\hat{P}}^{-1}\boldsymbol{\hat{R}}=\boldsymbol{\hat{Q}}\boldsymbol{\hat{R}}=\left(\begin{array}{ccc}
1 & 0 & 0\\
\bar{1} & 1 & 0\\
0 & 0 & 1
\end{array}\right)^{-1}\left(\begin{array}{ccc}
a_{x}^{*} & a_{y}^{*} & a_{z}^{*}\\
b_{x}^{*} & b_{y}^{*} & b_{z}^{*}\\
c_{x}^{*} & c_{y}^{*} & c_{z}^{*}
\end{array}\right)=\left(\begin{array}{ccc}
1 & 0 & 0\\
1 & 1 & 0\\
0 & 0 & 1
\end{array}\right)\left(\begin{array}{ccc}
a_{x}^{*} & a_{y}^{*} & a_{z}^{*}\\
b_{x}^{*} & b_{y}^{*} & b_{z}^{*}\\
c_{x}^{*} & c_{y}^{*} & c_{z}^{*}
\end{array}\right)
\]
\end_inset
\end_layout
\begin_layout Subsection*
Miller indices (reciprocal space) under transformation of unit cell basis
(real-space)
\end_layout
\begin_layout Standard
From ITA chapter 5.1, this is how the reflections would be re-labelled if
the axes were changed.
See later for how to apply symmetry operations.
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{equation}
\left(h^{\prime},k^{\prime},l^{\prime}\right)=\left(h^{\prime},k^{\prime},l^{\prime}\right)\boldsymbol{\hat{P}}=\left(h,k,l\right)\left(\begin{array}{ccc}
1 & 0 & 0\\
\bar{1} & 1 & 0\\
0 & 0 & 1
\end{array}\right)=(h-k,k,l)\label{eq:indices}
\end{equation}
\end_inset
\end_layout
\begin_layout Subsection*
Fractional coordinates under transformation of unit cell basis
\end_layout
\begin_layout Standard
Where u, v and w are fractional coordinates, i.e.
the cartesian coordinates are:
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
\boldsymbol{\hat{M}}\left(\begin{array}{c}
u\\
v\\
w
\end{array}\right)=\left(\begin{array}{ccc}
a_{x} & b_{x} & c_{x}\\
a_{y} & b_{y} & c_{y}\\
a_{z} & b_{z} & c_{z}
\end{array}\right)\left(\begin{array}{c}
u\\
v\\
w
\end{array}\right)
\]
\end_inset
From ITA chapter 5.1, the new coordinates are:
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{equation}
\left(\begin{array}{c}
u^{\prime}\\
v^{\prime}\\
w^{\prime}
\end{array}\right)=\boldsymbol{\hat{P}}^{-1}\left(\begin{array}{c}
u\\
v\\
w
\end{array}\right)=\boldsymbol{\hat{Q}}\left(\begin{array}{c}
u\\
v\\
w
\end{array}\right)\label{eq:fractional-coord-transform-basis-change}
\end{equation}
\end_inset
\end_layout
\begin_layout Subsection*
Symmetry operations acting on fractional coordinates
\end_layout
\begin_layout Standard
Following ITA chapter 11.1, the following matrix represents the symmetry
operation written as
\begin_inset Formula $-y,x,z$
\end_inset
, where for this time only, x,y and z are fractional coordinates:
\begin_inset Formula
\begin{equation}
\boldsymbol{\hat{W}}=\left(\begin{array}{ccc}
0 & \bar{1} & 0\\
1 & 0 & 0\\
0 & 0 & 1
\end{array}\right)\label{eq:wmatrix}
\end{equation}
\end_inset
\end_layout
\begin_layout Standard
It transforms the fractional coordinates as follows:
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{equation}
\left(\begin{array}{c}
u^{\prime}\\
v^{\prime}\\
w^{\prime}
\end{array}\right)=\boldsymbol{\hat{W}}\left(\begin{array}{c}
u\\
v\\
w
\end{array}\right)=\left(\begin{array}{ccc}
0 & \bar{1} & 0\\
1 & 0 & 0\\
0 & 0 & 1
\end{array}\right)\left(\begin{array}{c}
u\\
v\\
w
\end{array}\right)\label{eq:fractional-coord-transform-symmetry}
\end{equation}
\end_inset
\end_layout
\begin_layout Standard
Note also that we're ignoring the translational part of the symmetry operation
(there isn't one in this case).
\end_layout
\begin_layout Standard
Note that
\begin_inset Formula $\boldsymbol{\hat{W}}$
\end_inset
behaves like
\begin_inset Formula $\boldsymbol{\hat{P}}^{-1}$
\end_inset
above.
CrystFEL's
\begin_inset Quotes eld
\end_inset
SymOpList
\begin_inset Quotes erd
\end_inset
structure contains IntegerMatrix objects representing
\begin_inset Formula $\boldsymbol{\hat{W}}^{-1}$
\end_inset
matrices, also known as
\begin_inset Formula $\boldsymbol{\hat{P}}$
\end_inset
matrices.
\end_layout
\begin_layout Subsection*
Symmetry operations acting on Miller indices (reciprocal space)
\end_layout
\begin_layout Standard
Since
\begin_inset Formula $\boldsymbol{\hat{W}}^{-1}$
\end_inset
matrices are just
\begin_inset Formula $\boldsymbol{\hat{P}}$
\end_inset
matrices, simply change the
\begin_inset Formula $\boldsymbol{\hat{P}}$
\end_inset
into
\begin_inset Formula $\boldsymbol{\hat{W}}^{-1}$
\end_inset
in equation
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:indices"
plural "false"
caps "false"
noprefix "false"
\end_inset
:
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{equation}
\left(h^{\prime},k^{\prime},l^{\prime}\right)=\left(h,k,l\right)\boldsymbol{\hat{W}}^{-1}=\left(h^{\prime},k^{\prime},l^{\prime}\right)\left(\begin{array}{ccc}
0 & 1 & 0\\
\bar{1} & 0 & 0\\
0 & 0 & 1
\end{array}\right)=(\bar{k},h,l)\label{eq:symmetry}
\end{equation}
\end_inset
\end_layout
\begin_layout Standard
In this case, the symmetry operation is a rotation, so the matrix inverse
is just the same as a matrix transpose.
However, don't get confused: the matrix appearing here is the inverse of
the one appearing in equation
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:wmatrix"
plural "false"
caps "false"
noprefix "false"
\end_inset
.
\end_layout
\begin_layout Standard
Note that the
\begin_inset Quotes eld
\end_inset
amount of old h, k and l to make new h
\begin_inset Quotes erd
\end_inset
is found in the first column of the
\begin_inset Formula $\boldsymbol{\hat{W}}^{-1}$
\end_inset
matrix, and so on.
\end_layout
\begin_layout Subsection*
Transforming symmetry operations
\end_layout
\begin_layout Standard
Given a
\begin_inset Formula $\boldsymbol{\hat{P}}$
\end_inset
matrix which transforms the unit cell as written in equation
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:cell_transform"
plural "false"
caps "false"
noprefix "false"
\end_inset
, and a
\begin_inset Formula $\boldsymbol{\hat{W}}^{-1}$
\end_inset
matrix representing a symmetry operation, we seek to apply the symmetry
operation to the cell as it would appear after applying the transformation.
For example, let
\begin_inset Formula $\boldsymbol{\hat{P}}$
\end_inset
permute the cell axes, turning a
\begin_inset Quotes eld
\end_inset
unique axis b
\begin_inset Quotes erd
\end_inset
cell into a
\begin_inset Quotes eld
\end_inset
unique axis c
\begin_inset Quotes erd
\end_inset
one, and let
\begin_inset Formula $\boldsymbol{\hat{W}}^{-1}$
\end_inset
represent a twofold rotation around the c axis:
\end_layout
\begin_layout Standard
To transform a set of Miller indices
\begin_inset Formula $(h,k,l)$
\end_inset
, first use
\begin_inset Formula $\boldsymbol{\hat{P}}$
\end_inset
to get the indices referred to the
\begin_inset Quotes eld
\end_inset
unique axis c
\begin_inset Quotes erd
\end_inset
setting using equation
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:indices"
plural "false"
caps "false"
noprefix "false"
\end_inset
, then apply the symmetry operation using equation
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:symmetry"
plural "false"
caps "false"
noprefix "false"
\end_inset
, and finally transform back to the original cell using equation
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:indices"
plural "false"
caps "false"
noprefix "false"
\end_inset
and the inverse matrix:
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
(h^{\prime},k^{\prime},l^{\prime})=(h,k,l)\boldsymbol{\hat{P}}\boldsymbol{\hat{W}}^{-1}\boldsymbol{\hat{P}}^{-1}
\]
\end_inset
\end_layout
\begin_layout Standard
The combined matrix
\begin_inset Formula $\boldsymbol{\hat{P}}\boldsymbol{\hat{W}}^{-1}\boldsymbol{\hat{P}}^{-1}$
\end_inset
represents a twofold rotation symmetry operation along the b axis instead
of c.
\end_layout
\end_body
\end_document
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