Crystallography and non-cubic crystals.

In this lecture we shall see how to plot the poles of planes (hkl) onto a stereographic projection when the crystal is not cubic. The Miller indices of planes in the crystal (hkl) are still defined in terms of the inverse of the intercept of the plane on the three crystal axes a,b, and c; while the crystal direction [uvw] is still expressed as integer multiples of the three crystal vectors a,b and c. However, to plot the normals to the plane (hkl) or the pole of the direction [uvw] onto a stereographic projection we need to recognise that the three crystal unit vectors are not necessarily equal nor need they be mutually perpendicular.

Since the analysis developed earlier for cubic crystals relied on the fact that the unit crystal vectors were each identical to the unit vectors of cartesian space, we now need to develop a transfoprmation of co-ordinates matrix that will transform a vector [uvw], expressed in terms of the three crystal unit vectors a,b, and c, into a vector [xyz] expressed in terms of the 3 unit vectors of regular cartesian space. We shall call this matrix M, the mapping matrix.

To simplify the analysis we shall position our triclinic crystal in a very special orientation defined as follows:-

Align the c-axis of the crystal (i.e. the crystal vector [001]) with the z-axis of cartesian space.
Align the normal to the crystal plane (010) with the y-axis of cartesian space (i.e. the plane defined by the crystal vectors a and c lies in the x-z plane of cartesian space).



We now draw a line from the end of the vector a perpendicular to the x-axis and parallel to the z-axis. The projected lengths of the vector a on the x-axis and z-axis are:-





We now drop a line from the end of the vector b to the x-y plane plane so that the line is perpendicular to the x-y plane and parallel to the z-axis as shown below. The vector b then has projected lengths in the x-y plane and in the z-axis of:-




Next we draw a line from the end of the vector a to the end of the vector b. This forms a triangle aob; two sides of which are of lengths a and b and the included angle is g. We can use the cosine rule to determine the length of the third side d.



The next task is to calculate the angle d in the figure below so that we can determine exactly where the perpendicular dropped from the end of vector b to the x-y plane intersects the x-y plane (q). To do this we recognise that the figure abpq is a simple trapezoid with ap parallel to bq and that d can be calculated from the cosine rule if we know the length e; e is calculated using pythagarus' theorum.






In the above pair of equations we can eliminate e2 and solve for cos(d) since all the other quantities are known.



The x,y and z co-ordinates of the end of the crystal vector b are simply



Hence we can define the mapping matrix M as



Let's see how this works in practice by finding the cartesian vector [xyz] that is equivalent to the crystal vector [111] in the monoclinic, tetragonal and cubic forms of ZrO2. The crystal constants for zirconia are (dimensions in Å).

Crystal a b c a b g d, calc
Monoclinic 5.22 5.27 5.38 90 99.46 90 90
Tetragonal 3.64 3.64 5.27 90 90 90 90
Cubic 5.065 5.065 5.065 90 90 90 90


In the cubic form [111] becomes [111] in cartesian space, in the tetragonal form [111] becomes [1,1,1.448] while in the monoclinic form [111] becomes [1.148,1.177,1] in cartesian space. The poles are plotted below on a standard [001] stereographic projection




Now lets' find the normal to any plane (hkl) in any crystal system.