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A crystal in this system contains three diads, which must be at right angles to one another. In Figure 2.6 they are the crystal axes. From Table 1.3 and Figures 1.19d, e, f and g, the lattice parameters may be all unequal.

The point group containing just three diad axes at right angles to one another is shown in Figure 2.6 and is designated 222. In general, a pole in a crystal belonging to this point group is repeated four times. If the indices of the initial pole are (hkl), where there is no special relationship between h, k and l, the operation of all of the symmetry elements of the point group on this one initial pole produces three other poles. These are , and (Figure 2.7). The assemblage of crystal faces produced by repetition of an initial crystal face with indices (hkl) is called the form hkl and is given the symbol {hkl}.³ If the assemblage of faces encloses space, the form is said to be closed; otherwise it is open. In this case, {hkl} is closed. The symbol {hkl} with curly brackets means all faces of the form hkl. Here, for the point group 222, this means (hkl), , and . The form {hkl} is said to show a multiplicity of four. Then {hkl} would be said to be a general form; that is, a form that bears no special relationship to the symmetry elements of the point group.

Figure 2.7 A stereogram of an orthorhombic crystal of point group 222 centred on 001. The diad axes are parallel to the x‐, y‐ and z‐axes

Special forms in this crystal class would be {100}, {010} and {001}; each of these forms gives just two faces: for {100}, these would be the faces (100) and (00) (Figure 2.7). These forms are all open ones. They are easily recognized as special forms since their multiplicity is less than that of the general form. Forms such as {hk0}, {h0l} and {0kl} which have one index zero and no special relationship between the other two would also be described as special, even though, as is evident from Figure 2.7, the multiplicity of each is four, as for the general form in this crystal class. One reason for this is that these poles lie normal to a diad axis. This is a special position with respect to this axis and has the result that if a crystal grew with faces parallel only to the planes of indices {hk0}, {h0l} and {0kl}, it would appear to show mirror symmetry as well as the three diad axes. A second reason is that each of the {hk0}, {h0l} and {0kl} forms is open, while the general {hkl} form is closed. Special forms usually correspond to poles lying normal to or on an axis of symmetry, and normal to or on mirror planes, and sometimes to poles lying midway between two axes of symmetry. However, the best definition of a special form is as follows: a form is special if the development of the complete form shows a symmetry of arrangement of the poles which is higher than the one the crystal actually possesses. Special forms in all of the crystal classes are listed later, in Table 2.1.

Table 2.1 Special forms in the crystal classes

2	32	6mm
{010}, {00}, {h0l}	{0001}, , , , {hki0}, , , ,	{0001}, , , , {hki0}, ,
m and 2/m	4	m2
{010}, {h0l}	{001}, , {hk0}	{0001}, , , , {hki0}, , ,
mm2	and 4/m	622 and 6/mmm
{001}, , {100}, {010}, {hk0}, {h0l}, {0kl}	{001}, {hk0}	{0001}, , , {hki0}, ,
222 and mmm	4mm	23
{100}, {010}, {001}, {hk0}, {h0l}, {0kl}	{001}, , {100}, {110}, {hk0}, {h0l}, {hhl}	{100}, {110}, {hk0}, {kh0}, {111}, {11}, {hll}, , {hhl},
3	2m
{0001}, , {hki0}	{001}, {100}, {110}, {hk0}, {h0l}, {hhl},	{100}, {110}, {hk0}, {kh0}, {111}, {hll}, {hhl}
	42 and 4/mmm	432
{0001}, {hki0}	{001}, {100}, {110}, {hk0}, {h0l}, {hhl}	{100}, {110}, {hk0}, {111}, {hll}, {hhl}
3m	6	3m
{0001}, , , , , {hki0}, , ,	{0001}, , {hki0}	{100}, {110}, {hk0}, {111}, {11}, {hll}, {hl}, {hhl},
m	and 6/m	m m
{0001}, , , {hki0}, , ,	{0001}, {hki0}	{100}, {110}, {hk0}, {111}, {hll}, {hhl}

The orthorhombic system also contains the classes 2mm and mmm (Figure 2.6). The former contains two mirror planes, which must be at right angles. Two mirror planes at right angles automatically show diad symmetry along the line of intersection (Figure 2.6). Since m ≡ , this group could be designated 2 , and this is why it appears in the orthorhombic system, which is defined as possessing three diad axes. This crystal class could simply be designated mm since the diad is automatically present. However, mm2 is usually used for the later development of space groups (see Section 2.12). A crystal containing three diad axes can also contain mirrors normal to all of these without an axis of higher symmetry. Such a point group is designated mmm, or could be designated 2/mm. As can be seen from Figure 2.6, the multiplicity of the general form is now eight. The special forms {hk0}, {h0l} and {0kl} now show a lower multiplicity than the general one. The point group mmm shows the highest symmetry in the orthorhombic system. The point group showing the highest symmetry in a particular crystal system is said to be the holosymmetric class.

To plot a stereogram of an orthorhombic crystal centred on 001 if we are given the lattice parameters a, b and c, we proceed as shown in Figure 2.8. In Figure 2.8a the poles of the (001), (010) and (100) planes are immediately inserted at the centre of the primitive and where the x‐ and y‐axes cut the primitive, respectively. A pole (hk0) can be inserted at an angle θ along the primitive to (100), as shown in Figure 2.8a, by noting from Figure 2.8b (which is a section of the crystal parallel to (001) showing the intersection of the plane (hk0) with the crystal axes) that the pole of (100) lies along OM, that (hk0) makes intercepts on the crystal axes of a/h along the x‐axis and b/k along the y‐axis, and that θ is the angle between the normal to (hk0) and the normal to (100). From Figure 2.8b we have:

That is:

(2.1a)

where (100)^∧(hk0) means the angle between the (100) and (hk0) planes; that is, the angle between the pole of (100) and the pole of (hk0). Similarly, if the lattice parameters are given, we can locate (0kl) and (h0l), since:

(2.1b)

and:

(2.1c)

Figure 2.8 (a) General location of a hk0 pole on the stereogram of an orthorhombic crystal, (b) geometry to determine the angle θ between the 100 pole and the hk0 pole, (c) location of the hkl pole given the locations of the hk0, h0l and 0kl poles, (d) the location of the 311 pole given the locations of the 100, 010, 001, 011 and 310 poles

These relations can be seen immediately by drawing diagrams similar to Figure 2.8b but looking along the x‐ and y‐axes, respectively. When poles such as (h0l), (0kl) and (hk0) have been inserted on the stereogram, we can insert a pole such as (hkl) immediately by use of the zone addition rule, given in Section 1.3, Eq. (1.20).

If two poles (h₁k₁l₁), (h₂k₂l₂) both lie in the same zone with the indices [uvw] then so does the pole of the plane (mh₁ + nh₂, mk₁ + nk₂, ml₁ + nl₂); that is, any plane whose indices can be formed by taking linear combinations of the indices of two planes in a given zone can provide the indices of a further plane in that same zone (Section 1.3). In general, m and n can be positive or negative, but if m and n are both positive then the pole of the plane under consideration must lie on the great circle between the poles of (h₁k₁l₁) and (h₂k₂l₂): this simple result can be extremely useful when locating poles on stereographic projections such as Figure 2.7.

In Figure 2.8c, after (001), (010), (100) and (hk0), (h0l), (0kl) are plotted, then to plot, say, (hkl), we note that (hkl) must lie in the zone containing (001) and (hk0), since if we multiply (001) by the number l and add the indices (00l) and (hk0) we obtain (hkl). It then follows that (hkl) lies somewhere on the great circle between (001) and (hk0). Similarly, (hkl) lies in the zone containing (0kl) and (100), since h times (100) gives (h00) and this added to (0kl) gives (hkl). Again, it is true that (hkl) lies somewhere on the great circle between (0kl) and (100), but we can now rationalize that (hkl) must lie at the intersection of the two great circles we have considered. We then draw the great circle (or zone) containing (001) and (hk0) and that containing (001) and (0kl) and we know that (hkl) is situated where these intersect.

A particular example may make the procedure clear. Suppose we wish to locate (311) after plotting (001), (010) and (100) (Figure 2.8d). One way to proceed would be to locate (011) using Eq. (2.1b), setting k = 1 and l = 1 and using the known lattice parameters. We then find (310), on the primitive, by finding the angle between (100) and (310) from Eq. (2.1a), setting h = 3 and k = 1. Finally, we note that (311) lies in the zone containing (001) and (310), since (001) plus (310) yields (311). Also, (311) lies in the zone containing (100) and (011) since three times (100) plus (011) yields (311). The pole of (311) is then immediately located by drawing the great circle through (001) and (310) and that through (011) and (100); (311) is located where these great circles meet.

Using the above procedure for locating poles is usually the quickest way to draw an accurate stereogram when key poles have been located either by calculation or through the use of a computer software package. It must be strongly emphasized that, although we have chosen the orthorhombic system as an example, the use of Eq. (1.20) to locate poles applies to any crystal system and does not depend on the crystal axes being at any particular angle to one another. The utility of Eq. (1.20) is one of the great advantages of the Miller index for denoting crystal planes, and arises naturally from the properties of a space lattice.

Equations such as those in Eq. (2.1) can of course be used to find the ratio of the lattice parameters – the axial ratios – from measurements of the angles between poles.