Crystal coordinates

Crystal structures published by X-ray crystallographers are almost always expressed in terms of reduced or crystal coordinates. These are defined to be the atomic coordinates of atoms in the unit cell scaled such that they correspond to a unit cell that is a cube of edge length 1.

Let ${\bf R}= (x,y,z)$ be the position vector of an atom in the unit cell. How do we find the corresponding crystal or reduced coordinate, which we denote ${\bf S}= (s_x,s_y,s_z)$? The connection lies in the scaling relation satisfied by ${\bf R}$ and ${\bf S}$:

$$x = a_x s_x + b_x s_y + c_x s_z$$

$$y = a_y s_x + b_y s_y + c_y s_z$$

$$z = a_z s_x + b_z s_y + c_z s_z$$

Thus, we simply need to solve the above three equations for the coordinates $s_x$, $s_y$ and $s_z$, which is three equations in three unknowns.

Example: In a particular crystal, the lattice vectors are ${\bf a}= (a,0,0)$, ${\bf b}= (0,b,0)$, and ${\bf c}= (0,0,c)$, and the atoms are in a bcc-like arrangement with

$${\bf R}_1 = (0,0,0)\;\;\;\;\;\;\;\;\;\;\;\;{\bf R}_2 = (a/2,b/2,c/2)$$

What are the reduced coordinates of the atoms?

For ${\bf R}_1$, the problem is trivial

$$0 = as_x\;\;\;\;\;\;0 = bs_y\;\;\;\;\;\;0=cs_z$$

so that ${\bf S}_1 = (0,0,0)$, not surprisingly. For ${\bf R}_2$, we have

$${a \over 2} = as_x\;\;\;\;\;\;{b \over 2} = bs_y\;\;\;\;\;\;\;{c \over 2} = cs_z$$

so that the reduced coordinates are ${\bf S}_2 = (1/2,1/2,1/2)$.

For an fcc crystal, the reduced coordinates are ${\bf S}_1 = (0,0,0)$, ${\bf S}_2 = (1/2,1/2,0)$, ${\bf S}_3 = (1/2,0,1/2)$, and ${\bf S}_4 = (0,1/2,1/2)$.