Fractional coordinates

Consider a system of periodic structure in space and use ${\displaystyle {\mathbf {a} }}$ , ${\displaystyle \mathbf {b} }$, and ${\displaystyle \mathbf {c} }$ 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 ${\displaystyle \mathbf {r} }$ in Cartesian coordinates can be written as a linear combination of the period vectors

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

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

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

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

then

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

and the volume of the cell is

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

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

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

then we get

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

Similarly,

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

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

we arrive at

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

and

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

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

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

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

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

where

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

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

Unit cell definition using parallelepiped with lengths ${\displaystyle a}$, ${\displaystyle b}$, ${\displaystyle c}$ and angles between the sides given by ${\displaystyle \alpha }$, ${\displaystyle \beta }$, and ${\displaystyle \gamma }$ [1]

Then the edge vectors can be written as

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

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

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

Then

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

The last one continues

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

where

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

Remembering ${\displaystyle c_{z}}$, ${\displaystyle c}$, and ${\displaystyle \sin(\gamma )}$ being positive, one gets

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

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

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

the volume of the parallelepiped cell can also be expressed as

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

Once the volume is calculated as above, one has

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

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

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

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