Fractional coordinates

General case

Let us 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

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

then

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

and the volume of the cell is

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

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

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

then we get

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

Similarly,

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

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

we arrive at

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

and

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

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

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

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

${\begin{aligned}u&={{\mathbf {r} }\cdot \mathbf {\sigma } _{\mathbf {a} }^{\prime }},\\v&={{\mathbf {r} }\cdot \mathbf {\sigma } _{\mathbf {b} }^{\prime }},\\w&={{\mathbf {r} }\cdot \mathbf {\sigma } _{\mathbf {c} }^{\prime }},\end{aligned}}$

where

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

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 simplicty, 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

${\begin{aligned}{\mathbf {a} }&=(a,0,0),\\{\mathbf {b} }&=(b\cos(\gamma ),b\sin(\gamma ),0),\\{\mathbf {c} }&=(c_{x},c_{y},c_{z}),\end{aligned}}$

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

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

Then

${\begin{aligned}c_{x}&=c\cos(\beta ),\\c_{y}&=c{\frac {\cos(\alpha )-\cos(\gamma )\cos(\beta )}{\sin(\gamma )}},\\c_{z}^{2}&=c^{2}-c_{x}^{2}-c_{y}^{2}=c^{2}\left\{1-\cos ^{2}(\beta )-{\frac {[\cos(\alpha )-\cos(\gamma )\cos(\beta )]^{2}}{\sin ^{2}(\gamma )}}\right\}.\end{aligned}}$

The last one continues

${\begin{aligned}c_{z}^{2}&=c^{2}{\frac {\sin ^{2}(\gamma )-\sin ^{2}(\gamma )\cos ^{2}(\beta )-[\cos(\alpha )-\cos(\gamma )\cos(\beta )]^{2}}{\sin ^{2}(\gamma )}}\\&={\frac {c^{2}}{\sin ^{2}(\gamma )}}\left\{\sin ^{2}(\gamma )-\sin ^{2}(\gamma )\cos ^{2}(\beta )-[\cos(\alpha )-\cos(\gamma )\cos(\beta )]^{2}\right\}\end{aligned}}$

where

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

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

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

the volume of the parallelepiped cell can also be expressed as

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

Once the volume is calculated as above, one has

$c_{z}={\frac {\Omega }{ab\sin(\gamma )}}.$

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

${\begin{aligned}{\mathbf {a} }&=({a}_{x},{a}_{y},{a}_{z})=(a,0,0),\\{\mathbf {b} }&=({b}_{x},{b}_{y},{b}_{z})=(b\cos(\gamma ),b\sin(\gamma ),0),\\{\mathbf {c} }&=({c}_{x},{c}_{y},{c}_{z})=(c\cos(\beta ),c{\frac {\cos(\alpha )-\cos(\beta )\cos(\gamma )}{\sin(\gamma )}},{\frac {\Omega }{ab\sin(\gamma )}}).\end{aligned}}$

Conversion from cartesian coordinates

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

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

where

${\begin{aligned}\mathbf {\sigma } _{\mathbf {a} ,x}&={b}_{y}{c}_{z}-{b}_{z}{c}_{y}=b\sin(\gamma ){\frac {\Omega }{ab\sin(\gamma )}}={\frac {\Omega }{a}},\\\mathbf {\sigma } _{\mathbf {a} ,y}&={b}_{z}{c}_{x}-{b}_{x}{c}_{z}=-b\cos(\gamma ){\frac {\Omega }{ab\sin(\gamma )}}=-{\frac {\Omega \cos(\gamma )}{a\sin(\gamma )}},\\\mathbf {\sigma } _{\mathbf {a} ,z}&={b}_{x}{c}_{y}-{b}_{y}{c}_{x}=b\cos(\gamma )c{\frac {\cos(\alpha )-\cos(\beta )\cos(\gamma )}{\sin(\gamma )}}-b\sin(\gamma )c\cos(\beta )\\&=bc\left\{\cos(\gamma ){\frac {\cos(\alpha )-\cos(\beta )\cos(\gamma )}{\sin(\gamma )}}-\sin(\gamma )\cos(\beta )\right\}\\&={\frac {bc}{\sin(\gamma )}}\left\{\cos(\gamma )[\cos(\alpha )-\cos(\beta )\cos(\gamma )]-\sin ^{2}(\gamma )\cos(\beta )\right\}\\&={\frac {bc}{\sin(\gamma )}}\left\{\cos(\gamma )\cos(\alpha )-\cos(\beta )\cos ^{2}(\gamma )-\sin ^{2}(\gamma )\cos(\beta )\right\}\\&={\frac {bc}{\sin(\gamma )}}\left\{\cos(\alpha )\cos(\gamma )-\cos(\beta )\right\}.\\\end{aligned}}$

Another surface area vector of the cell

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

where

${\begin{aligned}\mathbf {\sigma } _{\mathbf {b} ,x}&={c}_{y}{a}_{z}-{c}_{z}{a}_{y}=0,\\\mathbf {\sigma } _{\mathbf {b} ,y}&={c}_{z}{a}_{x}-{c}_{x}{a}_{z}=a{\frac {\Omega }{ab\sin(\gamma )}}={\frac {\Omega }{b\sin(\gamma )}},\\\mathbf {\sigma } _{\mathbf {b} ,z}&={c}_{x}{a}_{y}-{c}_{y}{a}_{x}=-ac{\frac {\cos(\alpha )-\cos(\beta )\cos(\gamma )}{\sin(\gamma )}}\\&={\frac {ac}{\sin(\gamma )}}\left\{\cos(\beta )\cos(\gamma )-\cos(\alpha )\right\}.\end{aligned}}$

The last surface area vector of the cell

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

where

${\begin{aligned}\mathbf {\sigma } _{\mathbf {c} ,x}&={a}_{y}{b}_{z}-{a}_{z}{b}_{y}=0,\\\mathbf {\sigma } _{\mathbf {c} ,y}&={a}_{z}{b}_{x}-{a}_{x}{b}_{z}=0,\\\mathbf {\sigma } _{\mathbf {c} ,z}&={a}_{x}{b}_{y}-{a}_{y}{b}_{x}=ab\sin(\gamma ).\end{aligned}}$

Summarize

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

As a result

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

where $(a$, $b$, $c)$ 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

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

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

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

Supporting file formats

• CPMD input
• CIF