Derivation of ‘Unknown’ Symmetry Elements 3
When centring is combined with other symmetry elements, this can produce symmetry elements operating with the same unique axis. For example, consider the space group \(Ibca\). Combining the \(b\) glide with the body-centring takes the position \((x, y, z)\) to \((-x, y+\frac{1}{2}, z)\) to \((-x+\frac{1}{2}, y, z+\frac{1}{2})\). This position can also be generated by a \(c\) glide perpendicular to \(a\) at \(a = \frac{1}{4}\). Therefore, \(Ibca\) possesses both \(b\) and \(c\) glide perpendicular to \(a\). This is not the same as an \(n\) or an \(e\) glide, as they are positioned differently. We placed the \(b\) glide at \(a = 0\) and the \(c\) glide was at \(a= \frac{1}{4}\). The space group \(Icca\) is therefore identical to \(Ibca\), hence why this is not listed as a separate group.
Consider the space group \(C\frac{2}{m}\frac{2}{m}\frac{2}{m}\). What symmetry element is produced by combining the two-fold rotation axis parallel to \(a\) with the \(C\) centring? Scroll down for the answer.
Well the position arrived at after applying these two symmetry elements to \((x, y, z)\) is \((x + \frac{1}{2}, -y + \frac{1}{2}, -z)\). This is equivalent to applying a \(2_{1}\) screw axis located at \((a, \frac{1}{4}, 0)\) (remember the origin is moveable/where we chose it to be!).
A slightly harder one for you. Remembering that there are four lattice points in \(F\)-centring, what other mirror or glide planes must be present in the space group \(Fmmm\)? Determine their positions relative to the mirror plane. As the symmetry along each axis is identical, you can simply solve this for one axis and generalise. Scroll down for the answer.
\(F\)-centring means that the position \((x, y, z)\) is identical to the positions \((x+\frac{1}{2}, y+\frac{1}{2}, z)\), \((x+\frac{1}{2}, y, z+\frac{1}{2})\) and \((x, y+\frac{1}{2}, z+\frac{1}{2})\). Applying a mirror plane perpendicular to \(a\) to each position, the new positions are \((-x+\frac{1}{2}, y+\frac{1}{2}, z)\), \((-x+\frac{1}{2}, y, z+\frac{1}{2})\) and \((-x, y+\frac{1}{2}, z+\frac{1}{2})\).
The first two positions listed represent the two coordinates generated by an \(e\) glide plane perpendicular to \(a\) located at \(a = \frac{1}{4}\). Remember this is the ‘either’ glide plane. After reflection, translation can occur in either the \(b\) or \(c\) directions, generating two different positions. The final position generated represents an \(n\) glide parallel to \(a\) located at \(a = 0\). This is actually the same position as our original mirror plane! It does not get shown on the space group diagram due to this overlap, but it is present.