WHY STUDY Structures of Metals and Ceramics?
5. If necessary, these three numbers are multiplied or divided by a common factor to reduce them to the smallest integer values
6. The three resulting indices, not separated by commas, are enclosed in square brackets, thus: [uvw]. The u, v, and w integers correspond to the normalized coordinate differences referenced to the x, y, and z axes, respectively.
In summary, the u, v, and w indices may be determined using the following equations:
u=n( x2−x1
a ) (3.11a)
v=n( y2−y1
b ) (3.11b)
w=n( z2−z1
c ) (3.11c)
In these expressions, n is the factor that may be required to reduce u, v, and w to integers.
For each of the three axes, there are both positive and negative coordinates. Thus, negative indices are also possible, which are represented by a bar over the appropriate index. For example, the [111] direction has a component in the −y direction. Also, changing the signs of all indices produces an antiparallel direction; that is, [111] is di- rectly opposite to [111]. If more than one direction (or plane) is to be specified for a particular crystal structure, it is imperative for maintaining consistency that a positive–
negative convention, once established, not be changed.
The [100], [110], and [111] directions are common ones; they are drawn in the unit cell shown in Figure 3.21.
: VMSE Crystallographic
Directions
3.13 CRYSTALLOGRAPHIC DIRECTIONS
Tutorial Video:
Crystallographic Planes and Directions
Figure 3.21 The [100], [110], and [111] directions within a unit cell.
z
y
x
[111]
[110]
[100]
EXAMPLE PROBLEM 3.10
Determination of Directional Indices Determine the indices for the direction shown in the accompanying figure.
Solution
It is first necessary to take note of the vector tail and head coordinates. From the illustration, tail coordinates are as follows:
x1=a y1=0b z1=0c For the head coordinates,
x2=0a y2=b z2=c/2 Now taking point coordinate differences,
x2−x1=0a−a= −a y2−y1=b−0b=b z2−z1=c/2−0c=c/2
It is now possible to use Equations 3.11a through 3.11c to compute values of u, v, and w.
However, because the z2 − z1 difference is a fraction (i.e., c/2), we anticipate that in order to have integer values for the three indices, it is necessary to assign n a value of 2. Thus,
u=n( x2−x1
a )=2(−a
a )= −2 v=n(
y2−y1
b )=2(b b)=2 w=n(
z2−z1
c )=2(c2 c )=1
And, finally enclosure of the −2, 2, and 1 indices in brackets leads to [221] as the direction designation.5
This procedure is summarized as follows:
z
y
x
a
b c O
x2 = 0a y2 = b z2 = c/2
x1 = a y1 = 0b z1 = 0c
x y z
Head coordinates (x2, y2, z2,) 0a b c/2 Tail coordinates (x1, y1, z1,) a 0b 0c
Coordinate differences −a b c/2
Calculated values of u, v, and w u = −2 v = 2 w = 1
Enclosure [221]
5If these u, v, and w values are not integers, it is necessary to choose another value for n.
3.13 Crystallographic Directions • 77
EXAMPLE PROBLEM 3.11
Construction of a Specified Crystallographic Direction Within the following unit cell draw a [110] direction with its tail located at the origin of the coordinate system, point O.
Solution
This problem is solved by reversing the procedure of the preced- ing example. For this [110] direction,
u=1 v= −1 w=0
Because the tail of the direction vector is positioned at the origin, its coordinates are as follows:
x1=0a y1=0b z1=0c
We now want to solve for the coordinates of the vector head—that is, x2, y2, and z2. This is possible using rearranged forms of Equations 3.11a through 3.11c and incorporating the above values for the three direction indices (u, v, and w) and vector tail coordinates. Taking the value of n to be 1 because the three direction indices are all integers leads to
x2=ua+x1=(1) (a)+0a=a y2=vb+y1=(−1) (b)+0b= −b z2=wc+z1=(0) (c)+0c=0c The construction process for this direction vec-
tor is shown in the following figure.
Because the tail of the vector is positioned at the origin, we start at the point labeled O and then move in a stepwise manner to locate the vector head. Because the x head coordinate (x2) is a, we proceed from point O, a units along the x axis to point Q. From point Q, we move b units parallel to the −y axis to point P, because the y head coordinate (y2) is −b. There is no z component to the vector inasmuch as the z head coordinate (z2) is 0c. Finally, the vector corre- sponding to this [110] direction is constructed by drawing a line from point O to point P, as noted in the illustration.
z
y
x
a
b c O
z
x
y a
b O
Q c y2 = –b
x2 = a –y
P [110] Direction
For some crystal structures, several nonparallel directions with different indices are crystallographically equivalent, meaning that the spacing of atoms along each direc- tion is the same. For example, in cubic crystals, all the directions represented by the following indices are equivalent: [100], [100], [010], [010], [001], and [001]. As a con- venience, equivalent directions are grouped together into a family, which is enclosed in Tutorial Video
angle brackets, thus: 〈100〉. Furthermore, directions in cubic crystals having the same indices without regard to order or sign—for example, [123] and [213]—are equivalent.
This is, in general, not true for other crystal systems. For example, for crystals of te- tragonal symmetry, the [100] and [010] directions are equivalent, whereas the [100] and [001] are not.
Directions in Hexagonal Crystals
A problem arises for crystals having hexagonal symmetry in that some equivalent crys- tallographic directions do not have the same set of indices. This situation is addressed using a four-axis, or Miller–Bravais, coordinate system, which is shown in Figure 3.22a.
The three a1, a2, and a3 axes are all contained within a single plane (called the basal plane) and are at 120° angles to one another. The z axis is perpendicular to this basal plane. Directional indices, which are obtained as described earlier, are denoted by four indices, as [uvtw]; by convention, the u, v, and t relate to vector coordinate differences referenced to the respective a1, a2, and a3 axes in the basal plane; the fourth index per- tains to the z axis.
Conversion from the three-index system (using the a1–a2–z coordinate axes of Figure 3.22b) to the four-index system as
[UVW]→[uvtw] is accomplished using the following formulas6:
u=1
3(2U−V) (3.12a)
v=1
3(2V−U) (3.12b)
t= −(u+v) (3.12c)
w=W (3.12d)
Here, uppercase U, V, and W indices are associated with the three-index scheme (instead of u, v, and w as previously), whereas lowercase u, v, t, and w correlate with the Miller–Bravais four-index system. For example, using these equations, the [010]
direction becomes [1210]; furthermore, [1210] is also equivalent to the following: [1210], [1210], [1210].
6Reduction to the lowest set of integers may be necessary, as discussed earlier.
Figure 3.22 Coordinate axis systems for a hexagonal unit cell: (a) four-axis Miller–Bravais; (b) three-axis.
a1 a2
a3
z
120°
(a)
a1 a2 z
120°
(b)
3.13 Crystallographic Directions • 79
Several directions have been drawn in the hexagonal unit cell of Figure 3.23.
Determination of directional indices is carried out using a procedure similar to the one used for other crystal systems—by the subtraction of vector tail point coordi- nates from head point coordinates. To simplify the demonstration of this procedure, we first determine the U, V, and W indices using the three-axis a1–a2–z coordinate system of Figure 3.22b and then convert to the u, v, t, and w indices using Equations 3.12a–3.12d.
The designation scheme for the three sets of head and tail coordinates is as follows:
Figure 3.23 For the hexagonal crystal system, the [0001], [1100], and [1120] directions.
[0001]
[1120]
[1100]
a1 a2
a3
z
Head Tail Axis Coordinate Coordinate
a1 a″1 a′1
a2 a″2 a′2
ÿ
z z″ z′
Using this scheme, the U, V, and W hexagonal index equivalents of Equations 3.11a through 3.11c are as follows:
U=n( a″1−a′1
a ) (3.13a)
V=n( a″2−a′2
a ) (3.13b)
W=n(z″ −z′
c ) (3.13c)
In these expressions, the parameter n is included to facilitate, if necessary, reduction of the U, V, and W to integer values.
EXAMPLE PROBLEM 3.12
Determination of Directional Indices for a Hexagonal Unit Cell For the direction shown in the accompanying figure, do the following:
(a) Determine the directional indices referenced to the three-axis coordinate system of Figure 3.22b.
(b) Convert these indices into an index set referenced to the four-axis scheme (Figure 3.22a).
Solution
The first thing we need to do is determine U, V, and W indices for the vector referenced to the three-axis scheme represented in the sketch; this is possible us- ing Equations 3.13a through 3.13c. Because the vector passes through the origin, a′1=a′2=0a and z′ =0c.
Furthermore, from the sketch, coordinates for the vector head are as follows:
a″1=0a a″2= −a z″ =c
2
Because the denominator in z″ is 2, we assume that n = 2. Therefore,
U=n( a″1−a′1
ÿ
a )=2(0a−0a a )=0 V=n(
a″2−a′2
ÿ
a )=2(−a−0a
a )= −2 W=n(z″ −zÿ′
c )=2(c/2−0c c )=1
This direction is represented by enclosing the above indices in brackets—namely, [021].
(b) To convert these indices into an index set referenced to the four-axis scheme requires the use of Equations 3.12a–3.12d. For this [021] direction
U=0 V= −2 W=1 and
u=1
3(2U−V)=1
3[(2) (0)−(−2)]=2 3 v=1
3(2V−U)=1
3[(2) (−2) −0]= −4 3 t= −(u+v)= −(2
3−4 3)=2
3 w=W=1
Multiplication of the preceding indices by 3 reduces them to the lowest set, which yields values for u, v, t, and w of 2, −4, 2, and 3, respectively. Hence, the direction vector shown in the figure is [2423].
a1 a2
z
a a
a″1 = 0a c a″2 = –a z″ = c/2
The procedure used to plot direction vectors in crystals having hexagonal symmetry given their sets of indices is relatively complicated; therefore, we have elected to omit a description of this procedure.
3.14 Crystallographic Planes • 81
The orientations of planes for a crystal structure are represented in a similar manner.
Again, the unit cell is the basis, with the three-axis coordinate system as represented in Figure 3.19. In all but the hexagonal crystal system, crystallographic planes are specified by three Miller indices as (hkl). Any two planes parallel to each other are equivalent and have identical indices. The procedure used to determine the h, k, and l index numbers is as follows:
1. If the plane passes through the selected origin, either another parallel plane must