Chemistry:Fractional coordinates

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In crystallography, a fractional coordinate system is a coordinate system in which the edges of the unit cell are used as the basic vectors to describe the positions of atomic nuclei. The unit cell is a parallelepiped defined by the lengths of its edges $a,b,c$ and angles between them $\alpha ,\beta ,\gamma$.

General case

Consider a system of periodic structure in space and use $\mathbf{𝐚}$ , $\mathbf{𝐛}$, and $\mathbf{𝐜}$ as the three independent period vectors, forming a right-handed triad, which are also the edge vectors of a cell of the system. Then any vector $\mathbf{𝐫}$ in Cartesian coordinates can be written as a linear combination of the period vectors

$\mathbf{𝐫}=u\mathbf{𝐚}+v\mathbf{𝐛}+w\mathbf{𝐜}.$

Our task is to calculate the scalar coefficients known as fractional coordinates $u$, $v$, and $w$, assuming $\mathbf{𝐫}$, $\mathbf{𝐚}$, $\mathbf{𝐛}$, and $\mathbf{𝐜}$ are known.

For this purpose, let us calculate the following cell surface area vector

${\mathit{\sigma }}_{\mathbf{𝐚}}=\mathbf{𝐛}\times \mathbf{𝐜},$

then

$\mathbf{𝐛}\cdot {\mathit{\sigma }}_{\mathbf{𝐚}}=0,\mathbf{𝐜}\cdot {\mathit{\sigma }}_{\mathbf{𝐚}}=0,$

and the volume of the cell is

$\mathrm{\Omega }=\mathbf{𝐚}\cdot {\mathit{\sigma }}_{\mathbf{𝐚}}.$

If we do a vector inner (dot) product as follows

$\begin{array}{rl}\mathbf{𝐫}\cdot {\mathit{\sigma }}_{\mathbf{𝐚}}& =u\mathbf{𝐚}\cdot {\mathit{\sigma }}_{\mathbf{𝐚}}+v\mathbf{𝐛}\cdot {\mathit{\sigma }}_{\mathbf{𝐚}}+w\mathbf{𝐜}\cdot {\mathit{\sigma }}_{\mathbf{𝐚}}\\ & =u\mathbf{𝐚}\cdot {\mathit{\sigma }}_{\mathbf{𝐚}}\\ & =u\mathrm{\Omega },\end{array}$

then we get

$u=\frac{1}{\mathrm{\Omega }}\mathbf{𝐫}\cdot {\mathit{\sigma }}_{\mathbf{𝐚}}.$

Similarly,

${\mathit{\sigma }}_{\mathbf{𝐛}}=\mathbf{𝐜}\times \mathbf{𝐚},\mathbf{𝐜}\cdot {\mathit{\sigma }}_{\mathbf{𝐛}}=0,\mathbf{𝐚}\cdot {\mathit{\sigma }}_{\mathbf{𝐛}}=0,\mathbf{𝐛}\cdot {\mathit{\sigma }}_{\mathbf{𝐛}}=\mathrm{\Omega },$

$\mathbf{𝐫}\cdot {\mathit{\sigma }}_{\mathbf{𝐛}}=u\mathbf{𝐚}\cdot {\mathit{\sigma }}_{\mathbf{𝐛}}+v\mathbf{𝐛}\cdot {\mathit{\sigma }}_{\mathbf{𝐛}}+w\mathbf{𝐜}\cdot {\mathit{\sigma }}_{\mathbf{𝐛}}=v\mathbf{𝐛}\cdot {\mathit{\sigma }}_{\mathbf{𝐛}}=v\mathrm{\Omega },$

we arrive at

$v=\frac{1}{\mathrm{\Omega }}\mathbf{𝐫}\cdot {\mathit{\sigma }}_{\mathbf{𝐛}},$

and

${\mathit{\sigma }}_{\mathbf{𝐜}}=\mathbf{𝐚}\times \mathbf{𝐛},\mathbf{𝐚}\cdot {\mathit{\sigma }}_{\mathbf{𝐜}}=0,\mathbf{𝐛}\cdot {\mathit{\sigma }}_{\mathbf{𝐜}}=0,\mathbf{𝐜}\cdot {\mathit{\sigma }}_{\mathbf{𝐜}}=\mathrm{\Omega },$

$\mathbf{𝐫}\cdot {\mathit{\sigma }}_{\mathbf{𝐜}}=u\mathbf{𝐚}\cdot {\mathit{\sigma }}_{\mathbf{𝐜}}+v\mathbf{𝐛}\cdot {\mathit{\sigma }}_{\mathbf{𝐜}}+w\mathbf{𝐜}\cdot {\mathit{\sigma }}_{\mathbf{𝐜}}=w\mathbf{𝐜}\cdot {\mathit{\sigma }}_{\mathbf{𝐜}}=w\mathrm{\Omega },$

$w=\frac{1}{\mathrm{\Omega }}\mathbf{𝐫}\cdot {\mathit{\sigma }}_{\mathbf{𝐜}}.$

If there are many $\mathbf{𝐫}$s to be converted with respect to the same period vectors, to speed up, we can have

$\begin{array}{rl}u& =\mathbf{𝐫}\cdot {\mathit{\sigma }}_{\mathbf{𝐚}}^{\prime },\\ v& =\mathbf{𝐫}\cdot {\mathit{\sigma }}_{\mathbf{𝐛}}^{\prime },\\ \end{array}$

where

$\begin{array}{r}{\mathit{\sigma }}_{\mathbf{𝐚}}^{\prime }=\frac{1}{\mathrm{\Omega }}{\mathit{\sigma }}_{\mathbf{𝐚}},\\ {\mathit{\sigma }}_{\mathbf{𝐛}}^{\prime }=\frac{1}{\mathrm{\Omega }}{\mathit{\sigma }}_{\mathbf{𝐛}},\\ {\mathit{\sigma }}_{\mathbf{𝐜}}^{\prime }=\frac{1}{\mathrm{\Omega }}{\mathit{\sigma }}_{\mathbf{𝐜}}.\end{array}$

In crystallography

In crystallography, the lengths ($a$, $b$, $c$) of and angles ($\alpha$, $\beta$, $\gamma$) between the edge (period) vectors ($\mathbf{𝐚}$, $\mathbf{𝐛}$, $\mathbf{𝐜}$) of the parallelepiped unit cell are known. For simplicity, it is chosen so that edge vector $\mathbf{𝐚}$ in the positive $x$-axis direction, edge vector $\mathbf{𝐛}$ in the $x-y$ plane with positive $y$-axis component, edge vector $\mathbf{𝐜}$ with positive $z$-axis component in the Cartesian-system, as shown in the figure below.

Unit cell definition using parallelepiped with lengths $a$, $b$, $c$ and angles between the sides given by $\alpha$, $\beta$, and $\gamma$

Then the edge vectors can be written as

$\begin{array}{rl}\mathbf{𝐚}& =(a,0,0),\\ \mathbf{𝐛}& =(b\cos (\gamma ),b\sin (\gamma ),0),\\ \mathbf{𝐜}& =(c_x,c_y,c_z),\end{array}$

where all $a$, $b$, $c$, $\sin (\gamma )$, $c_z$ are positive. Next, let us express all $\mathbf{𝐜}$ components with known variables. This can be done with

$\begin{array}{rl}\mathbf{𝐜}\cdot \mathbf{𝐚}& =ac\cos (\beta )=c_{x}a,\\ \mathbf{𝐜}\cdot \mathbf{𝐛}& =bc\cos (\alpha )=c_{x}b\cos (\gamma )+c_{y}b\sin (\gamma ),\\ \mathbf{𝐜}\cdot \mathbf{𝐜}& =c^2=c_x^2+c_y^2+c_z^2.\end{array}$

Then

$\begin{array}{rl}{c}_x& =c\cos (\beta ),\\ c_y& =c\frac{\cos (\alpha )-\cos (\gamma )\cos (\beta )}{\sin (\gamma )},\\ c_z^2& =c^2-c_x^2-c_y^2=c^2\{1-{\cos }^2(\beta )-\frac{[\cos (\alpha )-\cos (\gamma )\cos (\beta ){]}^2}{{\sin }^2(\gamma )}\}.\end{array}$

The last one continues

$\begin{array}{rl}{c}_z^2& =c^2\frac{{\sin }^2(\gamma )-{\sin }^2(\gamma ){\cos }^2(\beta )-[\cos (\alpha )-\cos (\gamma )\cos (\beta ){]}^2}{{\sin }^2(\gamma )}\\ & =\frac{c^2}{{\sin }^2(\gamma )}\{{\sin }^2(\gamma )-{\sin }^2(\gamma ){\cos }^2(\beta )-[\cos (\alpha )-\cos (\gamma )\cos (\beta ){]}^2\}\end{array}$

where

$\begin{array}{rl}& {\sin }^2(\gamma )-{\sin }^2(\gamma ){\cos }^2(\beta )-[\cos (\alpha )-\cos (\gamma )\cos (\beta ){]}^2\\ & ={\sin }^2(\gamma )-{\sin }^2(\gamma ){\cos }^2(\beta )-{\cos }^2(\alpha )-{\cos }^2(\gamma ){\cos }^2(\beta )+2\cos (\alpha )\cos (\gamma )\cos (\beta )\\ & ={\sin }^2(\gamma )-{\cos }^2(\alpha )-{\sin }^2(\gamma ){\cos }^2(\beta )-{\cos }^2(\gamma ){\cos }^2(\beta )+2\cos (\alpha )\cos (\beta )\cos (\gamma )\\ & ={\sin }^2(\gamma )-{\cos }^2(\alpha )-[{\sin }^2(\gamma )+{\cos }^2(\gamma )]{\cos }^2(\beta )+2\cos (\alpha )\cos (\beta )\cos (\gamma )\\ & ={\sin }^2(\gamma )-{\cos }^2(\alpha )-{\cos }^2(\beta )+2\cos (\alpha )\cos (\beta )\cos (\gamma )\\ & =1-{\cos }^2(\alpha )-{\cos }^2(\beta )-{\cos }^2(\gamma )+2\cos (\alpha )\cos (\beta )\cos (\gamma ).\end{array}$

Remembering $c_z$, $c$, and $\sin (\gamma )$ being positive, one gets

$c_z=\frac{c}{\sin (\gamma )}\sqrt{1-{\cos }^2(\alpha )-{\cos }^2(\beta )-{\cos }^2(\gamma )+2\cos (\alpha )\cos (\beta )\cos (\gamma )}.$

Since the absolute value of the bottom surface area of the cell is

$\left|{\mathit{\sigma }}_{\mathbf{𝐜}}\right|=ab\sin (\gamma ),$

the volume of the parallelepiped cell can also be expressed as

$\mathrm{\Omega }=c_z\left|{\mathit{\sigma }}_{\mathbf{𝐜}}\right|=abc\sqrt{1-{\cos }^2(\alpha )-{\cos }^2(\beta )-{\cos }^2(\gamma )+2\cos (\alpha )\cos (\beta )\cos (\gamma )}$.

Once the volume is calculated as above, one has

$c_z=\frac{\mathrm{\Omega }}{ab\sin (\gamma )}.$

Now let us summarize the expression of the edge (period) vectors

$\begin{array}{rl}\mathbf{𝐚}& =(a_x,a_y,a_z)=(a,0,0),\\ \mathbf{𝐛}& =(b_x,b_y,b_z)=(b\cos (\gamma ),b\sin (\gamma ),0),\\ \mathbf{𝐜}& =(c_x,c_y,c_z)=(c\cos (\beta ),c\frac{\cos (\alpha )-\cos (\beta )\cos (\gamma )}{\sin (\gamma )},\frac{\mathrm{\Omega }}{ab\sin (\gamma )}).\end{array}$

Conversion from Cartesian coordinates

Let us calculate the following surface area vector of the cell first

${\mathit{\sigma }}_{\mathbf{𝐚}}=({\mathit{\sigma }}_{\mathbf{𝐚},x},{\mathit{\sigma }}_{\mathbf{𝐚},y},{\mathit{\sigma }}_{\mathbf{𝐚},z})=\mathbf{𝐛}\times \mathbf{𝐜},$

where

$\begin{array}{rl}{\mathit{\sigma }}_{\mathbf{𝐚},x}& =b_y{c}_z-b_z{c}_y=b\sin (\gamma )\frac{\mathrm{\Omega }}{ab\sin (\gamma )}=\frac{\mathrm{\Omega }}{a},\\ {\mathit{\sigma }}_{\mathbf{𝐚},y}& =b_z{c}_x-b_x{c}_z=-b\cos (\gamma )\frac{\mathrm{\Omega }}{ab\sin (\gamma )}=-\frac{\mathrm{\Omega }\cos (\gamma )}{a\sin (\gamma )},\\ {\mathit{\sigma }}_{\mathbf{𝐚},z}& =b_x{c}_y-b_y{c}_x=b\cos (\gamma )c\frac{\cos (\alpha )-\cos (\beta )\cos (\gamma )}{\sin (\gamma )}-b\sin (\gamma )c\cos (\beta )\\ & =bc\{\cos (\gamma )\frac{\cos (\alpha )-\cos (\beta )\cos (\gamma )}{\sin (\gamma )}-\sin (\gamma )\cos (\beta )\}\\ & =\frac{bc}{\sin (\gamma )}\{\cos (\gamma )[\cos (\alpha )-\cos (\beta )\cos (\gamma )]-{\sin }^2(\gamma )\cos (\beta )\}\\ & =\frac{bc}{\sin (\gamma )}\{\cos (\gamma )\cos (\alpha )-\cos (\beta ){\cos }^2(\gamma )-{\sin }^2(\gamma )\cos (\beta )\}\\ & =\frac{bc}{\sin (\gamma )}\{\cos (\alpha )\cos (\gamma )-\cos (\beta )\}.\\ \end{array}$

Another surface area vector of the cell

${\mathit{\sigma }}_{\mathbf{𝐛}}=({\mathit{\sigma }}_{\mathbf{𝐛},x},{\mathit{\sigma }}_{\mathbf{𝐛},y},{\mathit{\sigma }}_{\mathbf{𝐛},z})=\mathbf{𝐜}\times \mathbf{𝐚},$

where

$\begin{array}{rl}{\mathit{\sigma }}_{\mathbf{𝐛},x}& =c_y{a}_z-c_z{a}_y=0,\\ {\mathit{\sigma }}_{\mathbf{𝐛},y}& =c_z{a}_x-c_x{a}_z=a\frac{\mathrm{\Omega }}{ab\sin (\gamma )}=\frac{\mathrm{\Omega }}{b\sin (\gamma )},\\ {\mathit{\sigma }}_{\mathbf{𝐛},z}& =c_x{a}_y-c_y{a}_x=-ac\frac{\cos (\alpha )-\cos (\beta )\cos (\gamma )}{\sin (\gamma )}\\ & =\frac{ac}{\sin (\gamma )}\{\cos (\beta )\cos (\gamma )-\cos (\alpha )\}.\end{array}$

The last surface area vector of the cell

${\mathit{\sigma }}_{\mathbf{𝐜}}=({\mathit{\sigma }}_{\mathbf{𝐜},x},{\mathit{\sigma }}_{\mathbf{𝐜},y},{\mathit{\sigma }}_{\mathbf{𝐜},z})=\mathbf{𝐚}\times \mathbf{𝐛},$

where

$\begin{array}{rl}{\mathit{\sigma }}_{\mathbf{𝐜},x}& =a_y{b}_z-a_z{b}_y=0,\\ {\mathit{\sigma }}_{\mathbf{𝐜},y}& =a_z{b}_x-a_x{b}_z=0,\\ {\mathit{\sigma }}_{\mathbf{𝐜},z}& =a_x{b}_y-a_y{b}_x=ab\sin (\gamma ).\end{array}$

Summarize

$\begin{array}{rl}{\mathit{\sigma }}_{\mathbf{𝐚}}^{\prime }& =\frac{1}{\mathrm{\Omega }}{\mathit{\sigma }}_{\mathbf{𝐚}}=(\frac{1}{a},-\frac{\cos (\gamma )}{a\sin (\gamma )},bc\frac{\cos (\alpha )\cos (\gamma )-\cos (\beta )}{\mathrm{\Omega }\sin (\gamma )}),\\ {\mathit{\sigma }}_{\mathbf{𝐛}}^{\prime }& =\frac{1}{\mathrm{\Omega }}{\mathit{\sigma }}_{\mathbf{𝐛}}=(0,\frac{1}{b\sin (\gamma )},ac\frac{\cos (\beta )\cos (\gamma )-\cos (\alpha )}{\mathrm{\Omega }\sin (\gamma )}),\\ {\mathit{\sigma }}_{\mathbf{𝐜}}^{\prime }& =\frac{1}{\mathrm{\Omega }}{\mathit{\sigma }}_{\mathbf{𝐜}}=(0,0,\frac{ab\sin (\gamma )}{\mathrm{\Omega }}).\end{array}$

As a result

$\left[\begin{array}{c}u\\ v\\ w\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{a}& -\frac{\cos (\gamma )}{a\sin (\gamma )}& bc\frac{\cos (\alpha )\cos (\gamma )-\cos (\beta )}{\mathrm{\Omega }\sin (\gamma )}\\ 0& \frac{1}{b\sin (\gamma )}& ac\frac{\cos (\beta )\cos (\gamma )-\cos (\alpha )}{\mathrm{\Omega }\sin (\gamma )}\\ 0& 0& \frac{ab\sin (\gamma )}{\mathrm{\Omega }}\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$

where $(x$, $y$, $z)$ are the components of the arbitrary vector $\mathbf{𝐫}$ in Cartesian coordinates.

Conversion to Cartesian coordinates

To return the orthogonal coordinates in ångströms from fractional coordinates, one can employ the first equation on top and the expression of the edge (period) vectors

$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{ccc}a& b\cos (\gamma )& c\cos (\beta )\\ 0& b\sin (\gamma )& c\frac{\cos (\alpha )-\cos (\beta )\cos (\gamma )}{\sin (\gamma )}\\ 0& 0& \frac{\mathrm{\Omega }}{ab\sin (\gamma )}\end{array}\right]\left[\begin{array}{c}u\\ v\\ w\end{array}\right].$

For the special case of a monoclinic cell (a common case) where $\alpha =\gamma =90^{\circ }$ and $\beta >90^{\circ }$, this gives:

$\begin{array}{rl}x& =au+cw\cos (\beta ),\\ y& =bv,\\ z& =\frac{\mathrm{\Omega }}{ab}w=cw\sin (\beta ).\end{array}$

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