(IUCr) New method for determining hexagonal direction indices and their relationship to crystallographic directions

Volume 36
Part 1
Pages 23-28
February 2003

Received 3 June 2002
Accepted 27 September 2002

New method for determining hexagonal direction indices and their relationship to crystallographic directions

D. C. Hofmann ^a and V. A. Lubarda ^a ^*

^aDepartment of Mechanical and Aerospace Engineering, University of California, San Diego, La Jolla, CA 92093-0411,USA
Correspondence e-mail: vlubarda@ucsd.edu

A new method for determining the hexagonal direction indices is introduced, which is more practical than other commonly used methods. The relationship between the crystallographic directions and hexagonal direction indices for face-centered cubic crystals is derived. Different sets of hexagonal indices for equivalent crystallographic directions are discussed.

Keywords: crystallographic directions; hexagonal indices; transformation of indices.

1. Introduction

The hexagonal direction indices are well known and routinely applied to designate the crystallographic directions in hexagonal close packed (h.c.p.) crystals. They can also be used to designate the crystallographic directions in face-centered cubic (f.c.c.) crystals, although these are commonly, and more conveniently, labeled by using the standard crystallographic direction indices (e.g. Askeland, 1994; Smith, 1999). In this paper, we describe a new method to determine the hexagonal indices in f.c.c. and h.c.p. crystals, which is more practical and easier to apply than other commonly used methods. We then elaborate on the relationship between the hexagonal direction indices and the crystallographic directions for f.c.c. crystals, and derive two different but related four-index notations for equivalent crystallographic directions. The results may be of interest in the geometric analysis of the faults in stacking sequences, diffraction patterns, and related problems in the theory of crystalline lattices and their defects (Barrett & Massalski, 1980; McKie & McKie, 1986; Kelly et al., 2000).

2. Geometric preliminaries

Consider a unit cell of an f.c.c. crystal with the lattice parameter a. The stacking sequence ABC... of the close-packed (111) planes of interplanar spacing a/(3^1/2) is shown in Fig. 1. The unit vectors along the edges of the cubic cell are i₁, i₂ and i₃. The crystallographic direction indices for a lattice direction parallel to the vector x = x₁i₁ + x₂i₂ + x₃i₃ are the set of integers [m₁ m₂ m₃], where

$[x_1 = m_1a\,,\quad x_2 = m_2a\,,\quad x_3 = m_3a\,.\eqno (1)]$

The hexagonal atomic packing within the (111) plane through the origin at one of the atoms, A, is shown in Fig. 2. The directions along the cell edges project in the (111) plane along the unit directions e₁, e₂ and e₃. These vectors are at the angle 2 [pi] /3, so that

$[{\bf e}_i\cdot{\bf e}_j = \cases{ 1 & if $i = j$,\cr -1/2 & if $i \neq j$.}\eqno (2)]$

The projected axes [xi] ₁, [xi] ₂ and [xi] ₃ do not pass through the first-nearest neighbors to A, but through the second-nearest neighbors. Denoting the unit vector normal to the octahedral (111) plane by n = (i₁ + i₂ + i₃)/(3^1/2), the direction vector x can be decomposed into the component within the (111) plane, and the orthogonal component $[({\bf x}\cdot{\bf n})\,{\bf n}]$ , such that

$[{\bf x} = \boldxi + { 1 \over{3^{1/2}}}(x_1 + x_2 + x_3){\bf n}.\eqno (3)]$

If a₁, a₂ and a₃ are the components of along the edges of the cubic cell,

$[\boldxi = a_1{\bf i}_1 + a_2{\bf i}_2 + a_3{\bf i}_3,\eqno (4)]$

it readily follows that

$[\left[\matrix{ a_1\cr a_2\cr a_3 }\right] = {1 \over 3} \left(\matrix{ \hfill 2 & -1 & -1\cr -1 & \hfill 2 & -1\cr -1 & -1 & \hfill 2 }\right)\cdot\left[\matrix{ x_1\cr x_2\cr x_3 }\right], \eqno (5)]$

and

$[\left[\matrix{ {\bf e}_1\cr {\bf e}_2\cr {\bf e}_3 }\right] = { 1 \over 6^{1/2}} \left(\matrix{ \hfill 2 & -1 & -1\cr -1 & \hfill 2 & -1\cr -1 & -1 & \hfill 2 }\right)\cdot\left[\matrix{ {\bf i}_1\cr {\bf i}_2\cr {\bf i}_3 }\right].\eqno (6)]$

The constraint conditions are

$[a_1 + a_2 + a_3 = 0, \quad{\bf e}_1 + {\bf e}_2 + {\bf e}_3 = 0.\eqno (7)]$

Note that the coefficient 1/(6^1/2) in equation (6) appears in crystalline plasticity as the Schmid factor in f.c.c. crystals for uniaxial loading along the [001] direction. Indeed, the product $[({\bf n}\cdot{\bf i}_3)({\bf m}\cdot{\bf i}_3)]$ , where m is the vector in the close-packed direction $[[0 \bar{1} 1]]$ , is equal to this value.

Figure 1
A unit cell of an f.c.c. crystal with the lattice parameter a. The ABC... stacking sequence of close-packed (111) planes is indicated.

Figure 2
The hexagonal atomic arrangement within a (111) plane. The directions along the cubic cell edges project in this plane along the directions e₁, e₂, e₃. The interatomic spacing is a₀ = a/(2^1/2).

3. New method for determining the hexagonal direction indices

The in-plane hexagonal decomposition of the vector along the three axes e₁, e₂ and e₃,

$[\boldxi = \xi_1{\bf e}_1 + \xi_2{\bf e}_2 + \xi_3{\bf e}_3,\eqno (8)]$

is defined such that [xi] ₁ is the projection of the vector component a₁i₁ onto e₁, and similarly for [xi] ₂ and [xi] ₃, i.e.

$[\xi_1 = a_1({\bf i}_1\cdot{\bf e}_1)\,,\quad \xi_2 = a_2({\bf i}_2\cdot{\bf e}_2)\,,\quad \xi_3 = a_3({\bf i}_3\cdot{\bf e}_3).\eqno (9)]$

Since $[{\bf i}_1\cdot{\bf e}_1]$ = $[{\bf i}_2\cdot{\bf e}_2]$ = $[{\bf i}_3\cdot{\bf e}_3]$ = (2/3)^1/2, it follows that

$[\eqalignno{\left[\matrix{ \xi_1\cr \xi_2\cr \xi_3 }\right] =\hskip.2em& \left({2\over 3}\right)^{1/2}\left[\matrix{ a_1\cr a_2\cr a_3 }\right] \cr =\hskip.2em& \left({2\over 3}\right)^{1/2} {1\over 3} \left(\matrix{ \hfill 2 & -1 & -1\cr -1 & \hfill 2 & -1\cr -1 & -1 & \hfill 2 }\right)\cdot\left[\matrix{ x_1\cr x_2\cr x_3 }\right],&(10)}]$

with the redundancy condition

$[\xi_1 + \xi_2 + \xi_3 = 0.\eqno (11)]$

The following equalities are observed

$[\xi_1 = {{2}\over{3}}\left(\boldxi\cdot{\bf e}_1\right),\quad \xi_2 = {{2}\over{3}}\left(\boldxi\cdot{\bf e}_2\right),\quad \xi_3 = {{2}\over{3}}\left(\boldxi\cdot{\bf e}_3\right). \eqno (12)]$

To prove the first, for example, it is enough to observe that

$[ \boldxi \cdot{\bf e}_1 = \xi_1 - {{1}\over{2}}\xi_2 - {{1}\over{2}}\xi_3 = {{3}\over{2}}\xi_1,\eqno (13)]$

because [xi] ₁ + [xi] ₂ + [xi] ₃ = 0, and similarly for $[\boldxi \cdot {\bf e}_2]$ and $[\boldxi \cdot {\bf e}_3]$ . The formulae (12) have simple but convenient geometric interpretations, shown in Fig. 3. The orthogonal projections of the vector onto the axes e₁, e₂ and e₃, scaled by the factor of 2/3, define the hexagonal components [xi] ₁, [xi] ₂ and [xi] ₃. The simple recipe (12) is probably the most convenient method to determine the hexagonal in-plane indices, although it is surprisingly not mentioned in any of the standard texts on the subject (e.g. Askeland, 1994; Smith, 1999). There, instead, the vector is first decomposed into the directions e₁ and e₂ as

$[ \boldxi = \hat{\xi}_1{\bf e}_1 + \hat{\xi}_2{\bf e}_2,\eqno (14)]$

so that comparison with equation (8) establishes the connections

$[\hat{\xi}_1 = \xi_1-\xi_3 = 2\xi_1 + \xi_2, \quad \hat{\xi}_2 = \xi_2 - \xi_3 = 2\xi_2 + \xi_1,\eqno (15)]$

with the inverse relationships

$[\xi_1 = {{1}\over{3}}\left(2\hat{\xi}_1 - \hat{\xi}_2\right),\quad \xi_2 = {{1}\over{3}}\left(2\hat{\xi}_2 - \hat{\xi}_1\right).\eqno (16)]$

These formulae are clearly not needed if the simple recipe (12) is used.

Figure 3
Geometric interpretation of the vector decomposition = [xi]

₁e₁ +

₂e₂ +

₃e₃, subject to the constraint [xi]

₁ +

₂ +

₃ = 0. The components [xi]

₁,

₂,

₃ are defined by the orthogonal projections on the axes e₁, e₂, e₃, with the appropriate sign and the scaling factor of 2/3.

As an illustration, consider a unit direction ( $[|\boldxi|]$ = 1) at an angle of 30° to the [xi] ₁ axis. From Fig. 4 it immediately follows that (3/2) [xi] ₁ = cos30° = 3^1/2/2 and [xi] ₂ = 0 (because is orthogonal to [xi] ₂). Thus, the hexagonal indices are [ $[10 \bar{1} 0]]$ . On the other hand, if the classical method is used, we first need to calculate the components $[\hat{\xi}_1]$ = 2/(3^1/2) and $[\hat{\xi}_2]$ = 1/(3^1/2), then apply equations (16) to obtain [xi] ₁ = 1/(3^1/2) and [xi] ₂ = 0, which finally yields $[[1 0 \bar{1} 0]]$ . The advantage of the former method is clear.

Figure 4
The unit direction at an angle of 30° relative to the [xi]

₁ axis. Two types of projections on the coordinate directions are shown, both of which can be used to determine the corresponding hexagonal indices.

3.1. Trigonometric relations

Several useful trigonometric expressions are derived as follows. From Fig. 3 it is recognized that

$[\boldxi \cdot {\bf e}_1\!= \!\xi \cos \left({ \pi \over 3} - \theta\right),\,\, \boldxi \cdot {\bf e}_2\! = \!\xi \cos \left({ \pi \over 3} + \theta\right),\,\, \boldxi \cdot {\bf e}_3\! = \!-\xi \cos \theta,\eqno (17)]$

where [xi] = $[|\boldxi |]$ is the length of the vector . Consequently, the following relationships hold:

$[\cos\theta = - { 3 \over 2} {{\xi_3}\over \xi},\quad \cos\left({ \pi \over 3} - \theta\right) = { 3 \over 2} {{\xi_1}\over \xi},\quad \cos\left({ \pi \over 3} + \theta\right) = { 3 \over 2} {\xi_2 \over \xi}.\eqno (18)]$

and

$[\sin \left({{\pi}\over 3} - \theta \right) = { 3^{1/2}\over 2} {\xi_2 - \xi_3 \over \xi} ,\quad \sin \left( {\pi \over 3} + \theta \right) = { 3^{1/2} \over 2 }{ \xi_1 - \xi_3 \over \xi} .\eqno (20)]$

Since $[\cos 3\theta]$ = $[4 \cos^3\theta]$ - $[3 \cos\theta]$ , and $[\sin 3\theta]$ = $[ 3 \sin \theta ]$ - $[4 \sin^3\theta]$ , we also have

$[\cos 3\theta = - {{27}\over{2}} {{\xi_1\xi_2\xi_3}\over{\xi^3}},]$

$[ \sin 3\theta = - {{3(3^{1/2})}\over{2}}{{(\xi_1 - \xi_2)(\xi_2 - \xi_3)(\xi_3 - \xi_1)}\over{\xi^3}}.\eqno (21)]$

The length of the vector can be calculated from

$[\xi^2 = {{3}\over{2}}\left(\xi_1^2 + \xi_2^2 + \xi_3^2\right) = {{1}\over{2}} \left[(\xi_1 - \xi_2)^2 + (\xi_2 - \xi_3)^2 + (\xi_3-\xi_1)^2\right].\eqno (22)]$

The identity

$[x_1 - x_2 = a_1 - a_2 = \left({ 3 \over 2}\right)^{1/2} (\xi_1 - \xi_2),\eqno (23)]$

with similar identities for [xi] ₂ - [xi] ₃ and [xi] ₃ - [xi] ₁ are noted, in addition to

$[\xi_1^2 + \xi_2^2 + \xi_3^2 = - 2\left(\xi_1\xi_2 + \xi_2\xi_3 + \xi_3\xi_1\right).\eqno(24)]$

Analogous expressions in terms of the principal deviatoric stress components are well known from the theory of plasticity and geometric representation of the yield locus in the deviatoric [pi] plane (e.g. Kachanov, 1971; Lubarda, 2002).

4. Transformation rules under rotation within the (111) plane

For the derivation of the relationship between the crystallographic directions and hexagonal direction indices for an f.c.c. crystal, it is useful first to derive the general formulae for the transformation of coordinates ( $[\xi_1,\xi_2,\xi_3]$ ) under the rotation of the coordinate system within the (111) plane. If [varphi] is the angle of counterclockwise rotation (Fig. 5), it can be readily verified that

$[\xi_1 = {{2}\over{3}}\xi \cos \left({{\pi}\over{3}} - \theta \right),\quad \xi_2 = {{2}\over{3}} \xi \cos \left({{\pi}\over{3}} + \theta \right),]$

$[ \xi_3 = - {{2}\over{3}} \xi \cos \theta,\eqno (25)]$

and

$[\xi^*_1 = {{2}\over{3}}\xi \cos \left({{\pi}\over{3}} - \theta^*\right),\quad \xi^*_2 = {{2}\over{3}} \xi \cos \left({{\pi}\over{3}} + \theta^*\right),]$

$[\xi^*_3 = - {{2}\over{3}} \xi \cos \theta^*.\eqno (26)]$

where $[\theta^*]$ = [theta] + [varphi] . By expanding the right-hand sides of equation (26) and by using equations (18), (19) and (20), it follows that

$[\eqalignno{&\left[\matrix{ \xi _1^*\cr \xi _2^*\cr \xi _3^* }\right] = \cr&\quad{{2}\over{3^{1/2}}} \left(\matrix{ \sin\left({\displaystyle{\pi \over 3}} - \varphi\right) & 0 & -\sin\varphi\cr -\sin\varphi & \sin\left({\displaystyle{\pi \over 3}}-\varphi\right) & 0 \cr 0 & -\sin\varphi & \sin\left({\displaystyle{\pi \over 3}}-\varphi\right) }\right)\cdot\left[\matrix{\xi _1\cr \xi _2\cr \xi _3 }\right].\cr&& (27)}]$

The same transformation relates the unit direction vectors $[{\bf e}_1^*]$ , $[{\bf e}_2^*]$ , $[{\bf e}_3^*]$ and $[{\bf e}_1]$ , $[{\bf e}_2]$ , $[{\bf e}_3]$ . A text by Sands (1982) can be consulted for further analysis of the transformation rules for vectors and tensors in crystallography.

Figure 5
The coordinate system ( $[\xi_1^*,\xi_2^*,\xi_3^*]$ ) is obtained from the coordinate system ( $[\xi_1,\xi_2,\xi_3]$ ) through a counterclockwise rotation by an angle [varphi]

. The vector makes an angle [theta]

with the negative [xi]

₃ axis.

5. Relationship between crystallographic and hexagonal direction indices

If the angle of rotation discussed in the previous section is [varphi] = [pi] /6, we obtain the hexagonal coordinates $[\xi_1^{o}]$ , $[\xi_2^{o}]$ , $[\xi_3^{o}]$ within the (111) plane, which are shown in Fig. 6. The transformation rules for the unit vectors in this case reduce to

$[\eqalignno{\left[\matrix{ {\bf e}_1^{o}\cr {\bf e}_2^{o}\cr {\bf e}_3^{o} }\right] =\hskip.2em& { 1 \over 3^{1/2} } \left(\matrix{ \hfill 1&\hfill 0& \hfill -1\cr \hfill - 1&\hfill 1&\hfill 0 \cr \hfill 0&\hfill -1&\hfill 1 }\right) \cdot \left[\matrix{ {\bf e}_1\cr {\bf e}_2\cr {\bf e}_3 }\right] \cr =\hskip.2em& { 1 \over 2^{1/2}} \left(\matrix{ \hfill 1&\hfill 0&\hfill -1\cr \hfill - 1&\hfill 1&\hfill 0 \cr \hfill 0&\hfill -1&\hfill 1 }\right) \cdot \left[\matrix{ {\bf i}_1\cr {\bf i}_2 \cr {\bf i}_3 }\right],& (28)}]$

and similarly for the two sets of coordinates $[\xi_1^0]$ , $[\xi_2^0]$ , $[\xi_3^0]$ and $[\xi_1]$ , $[\xi_2]$ , $[\xi_3]$ . By using the expressions (10) for the components [xi] ₁, [xi] ₂, [xi] ₃ in terms of the rectangular components x₁ = m₁a, x₂ = m₂a, x₃ = m₃a, it follows that

$[\left[\matrix{ \xi_1^{o}\cr \xi_2^{o}\cr \xi_3^{o} }\right] = { 2a_0 \over 3} \left(\matrix{ \hfill 1&\hfill 0&\hfill -1\cr \hfill -1&\hfill 1&\hfill 0 \cr \hfill 0&\hfill -1&\hfill 1 }\right)\cdot\left[\matrix{ m_1\cr m_2\cr m_3 }\right],\eqno (29)]$

where a₀ = a/(2^1/2) is the atomic distance between the first neighbors in the close-packed (111) plane. Thus, the projected vector can be written as

$[\boldxi = { 2 \over 3} a_0\left[(m_1-m_3){\bf e}_1^{o}+(m_2-m_1){\bf e}_2^{o}+(m_3-m_2){\bf e}_3^{o}\right].\eqno (30)]$

On the other hand, the component of the vector x orthogonal to is

$[{ 1 \over 3^{1/2}} (x_1+x_2 + x_3){\bf n} = {{1}\over{3}}(m_1 + m_2 + m_3)c_0{\bf n},\eqno (31)]$

where c₀ = 3^1/2a = 6^1/2a₀ represents the interplanar spacing between the A-type planes. This scale in the n direction is chosen so that the spacing between the A, B and C planes is equal to c₀/3. Therefore, the direction vector x can be decomposed as

$[\eqalignno{{\bf x} =\hskip.2em& { 2 \over 3}a_0 [(m_1-m_3){\bf e}_1^{o} +(m_2-m_1){\bf e}_2^{o}+(m_3-m_2){\bf e}_3^{o} ]\cr &\! + { 1 \over 3} (m_1 + m_2 + m_3)c_0{\bf n}.&(32)}]$

This representation defines the hexagonal indices [h_o k_o i_o l_o] for the considered direction, which are

$[\eqalign{&h_{o} = 2(m_1 - m_3),\quad k_{o} = 2(m_2-m_1),\cr& i_{o} = 2(m_3-m_2),\quad l_{o} = m_1 + m_2 + m_3\semi }\eqno(33)]$

then

$[3{\bf x} = a_0\left(h_{o}{\bf e}_1^{o} + k_{o}{\bf e}_2^{o}+i_{o}{\bf e}_3^{o}\right) + l_{o}c_0{\bf n}.\eqno (34)]$

The reduction to lowest integers usually needs to be made in equation (33). In the matrix notation, the transformation between the two sets of indices is

$[\left[\matrix{ h_{o}\cr k_{o}\cr i_{o}\cr l_{o} }\right] = \left(\matrix{ \hfill 2&\hfill 0&\hfill-2\cr \hfill-2&\hfill 2&\hfill 0\cr \hfill 0&\hfill -2&\hfill 2\cr \hfill 1&\hfill 1&\hfill 1 }\right)\cdot\left[\matrix{ m_1\cr m_2\cr m_3 }\right]. \eqno (35)]$

The inverse relations are

$[m_1 = h_{o} - k_{o} + 2l_{o},\,\, m_2 = k_{o} - i_{o} + 2l_{o},\,\, m_3 = i_{o} - h_{o} + 2l_{o}\eqno (36)]$

with the matrix counterpart

$[\left[\matrix{ m_1\cr m_2\cr m_3 }\right] = \left(\matrix{ \hfill 1&\hfill -1&\hfill 0&\hfill 2\cr \hfill 0&\hfill 1&\hfill -1&\hfill 2\cr \hfill -1&\hfill 0&\hfill 1&\hfill 2\cr }\right)\cdot\left[\matrix{ h_{o}\cr k_{o}\cr i_{o}\cr l_{o} }\right].\eqno (37)]$

The clearing of fractions was used in the above derivation, so that in this case m₁ + m₂ + m₃ = 6l_o. In general, an additional reduction to lowest integers is needed.

Figure 6
The coordinate system ( $[\xi_1^{o},\xi_2^{o},\xi_3^{o}]$ ) is obtained from the coordinate system ( $[\xi_1,\xi_2,\xi_3]$ ) through a counterclockwise rotation by an angle [pi]

/6. The new axes $[\xi_1^{o}]$ , $[\xi_2^{o}]$ , and $[\xi_3^{o}]$ are aligned along the close-packed atomic directions of interatomic spacing a₀.

As an example which illustrates the use of the transformation formulae (35) and (37), consider the direction $[[1 2 \bar{1}]]$ in the crystallographic direction notation. From equation (35) it follows that h_o = 4, k_o = 2, i_o = -6 and l_o = 2, so that upon the reduction to lowest integers the hexagonal indices become $[[2 1 \bar{3} 1]]$ . On the other hand, for the direction with the hexagonal indices $[[2 \bar{2} 0 1]]$ , equation (37) gives m₁ = 6, m₂ = m₃ = 0, which yield the crystallographic direction indices [100].

5.1. Transformation to three-axis hexagonal indices

Returning to equation (32), the substitution of $[{\bf e}_3^{o}]$ = $[-({\bf e}_1^{o}]$ + $[{\bf e}_2^{o})]$ yields

$[\eqalignno{3{\bf x} =\hskip.2em& 2a_0 [ (m_1 + m_2-2m_3){\bf e}_1^{o} + (-m_1 + 2m_2-m_3){\bf e}_2^{o}]\cr &\! + (m_1 + m_2 + m_3)c_0{\bf n}.&(38)}]$

Thus, the transformation rule between the crystallographic direction indices [m₁ m₂ m₃] and the three-axis hexagonal direction indices $[[h'_{o}\,k'_{o}\,l'_{o}]]$ is

$[\left[\matrix{ h'_{o}\cr k'_{o}\cr l'_{o} }\right] = \left(\matrix{ \hfill 2&\hfill 2&\hfill -4\cr \hfill -2&\hfill 4&\hfill -2\cr \hfill 1&\hfill 1&\hfill 1 }\right)\cdot\left[\matrix{ m_1\cr m_2\cr m_3 }\right].\eqno (39)]$

The inverse transformation (after clearing the fractions) is

$[\left[\matrix{ m_1\cr m_2\cr m_3 }\right] = \left(\matrix{ \hfill 1&\hfill -1&\hfill 2\cr \hfill 0&\hfill 1&\hfill 2\cr \hfill -1&\hfill 0&\hfill 2 }\right)\cdot\left[\matrix{ h'_{o}\cr k'_{o}\cr l'_{o} }\right].\eqno (40)]$

The well known connections between the two sets of indices $[[h'_{o}\,k'_{o}\,l'_{o}]]$ and [h_o k_o i_o l_o] are

$[h'_{o} = 2h_{o} + k_{o},\quad k'_{o} = h_{o} + 2k_{o},\eqno (41)]$

$[h_{o} = {{1}\over{3}}(2h'_{o}-k'_{o}),\quad k_{o} = {{1}\over{3}}(2k'_{o}-h'_{o}),\eqno (42)]$

$[i_{o} = - (h_{o} + k_{o}) = - {{1}\over{3}}(h'_{o} + k'_{o}),\quad l_{o} = l'_{o}.\eqno (43)]$

6. An alternative set of hexagonal indices

The hexagonal direction indices [h_o k_o i_o l_o] were derived in the previous section by using the coordinate axes $[\xi_1^{o}]$ , $[\xi_2^{o}]$ , $[\xi_3^{o}]$ that pass through the first-nearest neighbors relative to an atom A at the origin. Another set of four hexagonal indices [h k i l] for the same crystallographic direction can be obtained by using the coordinate axes [xi] ₁, [xi] ₂, [xi] ₃ that pass through the second-nearest neighbors (Fig. 7). The direction [xi] ₂ is a reciprocal direction to $[\xi_1^0]$ , in the sense that $[{\bf e}_2\cdot{\bf e}_1^0]$ = 0. The distance between the second-nearest neighbors is $[\hat{a}_0]$ = 3^1/2a₀ = (3/2)^1/2a. Thus, considering a direction x with the crystallographic direction indices [m₁ m₂ m₃], such that x = m₁ai₁ + m₂ai₂ + m₃ai₃, from equation (10) it follows that

$[\left[\matrix{ \xi_1\cr \xi_2\cr \xi_3 }\right] = \left({2\over 3}\right)^{1/2}{a\over 3}\left(\matrix{ \hfill 2 & \hfill -1 & \hfill -1\cr \hfill -1 & \hfill 2 & \hfill -1\cr \hfill -1 & \hfill -1 & \hfill 2 }\right)\cdot\left[\matrix{ m_1\cr m_2\cr m_3 }\right] \eqno (44)]$

and

$[\eqalignno{{\bf x} =\hskip.2em& { 2 \over 9} \hat{a}_0 [(2m_1-m_2-m_3){\bf e}_1 + (-m_1+2m_2-m_3){\bf e}_2 \cr &\! + (-m_1-m_2+2m_3){\bf e}_3 ] + {{1}\over{3}}(m_1 + m_2 + m_3)c_0{\bf n}.\cr&&\hfill\llap{(45)}}]$

This defines the four indices

$[\eqalign{&h = 2(2m_1-m_2-m_3) ,\quad k = 2(-m_1+2m_2-m_3) ,\cr &i = 2(-m_1-m_2+2m_3) ,\quad l = 3(m_1+m_2+m_3) , }\eqno (46)]$

i.e.

$[\left[\matrix{ h\cr k\cr i\cr l }\right] = \left(\matrix{ \hfill 4&\hfill -2&\hfill -2\cr \hfill -2&\hfill 4&\hfill -2\cr \hfill -2&\hfill -2&\hfill 4\cr \hfill 3&\hfill 3&\hfill 3 }\right)\cdot\left[\matrix{ m_1\cr m_2\cr m_3 }\right],\eqno (47)]$

such that

$[9{\bf x} = \hat{a}_0\left(h{\bf e}_1+k{\bf e}_2+i{\bf e}_3\right) + lc_0{\bf n}.\eqno (48)]$

On the other hand, if the direction indices [h k i l] are known, the corresponding crystallographic direction indices are determined from

$[m_1 = 3h + 2l ,\quad m_2 = 3k + 2l ,\quad m_3 = 3i + 2l,\eqno (49)]$

with the matrix counterpart

$[\left[\matrix{ m_1\cr m_2\cr m_3 }\right] = \left(\matrix{ 3&0&0&2\cr 0&3&0&2\cr 0&0&3&2\cr }\right)\cdot\left[\matrix{ h\cr k\cr i\cr l }\right].\eqno (50)]$

For example, for the previously considered direction $[[1 2 \bar{1}]]$ in the crystallographic direction notation, from equation (47) we have h = 2, k = 8, i = -10, l = 6, so that upon reducing to lowest integers the set of four indices becomes $[[1 4 \bar{5} 3]]$ . Similarly, for the direction with the indices $[[h\, k\, i\, l]\simeq [2 \bar{2} 0 1]]$ , from equation (50) we have m₁ = 8, m₂ = -4, m₃ = 2, which is equivalent to the $[[4 \bar{2} 1]]$ crystallographic direction. It can also be easily verified that the crystallographic direction [100] corresponds to $[[h\, k\, i\, l] \simeq [4 \bar{2} \bar{2} 3]]$ .

Figure 7
The coordinate axes $[\xi_1^{o}]$ , $[\xi_2^{o}]$ , $[\xi_3^{o}]$ are aligned along directions of the first-nearest atoms to the atom at the origin A. The coordinate axes [xi]

₁,

₂,

₃ are along directions of the second-nearest atoms to the origin at A. The corresponding six atoms form the hexagon (indicated by dashed lines) of interatomic spacing $[\hat{a}_0]$ = 3^1/2a₀, where a₀ is the interatomic distance between the first-nearest neighbors.

7. Discussion

It is of interest to establish a transformation rule between the two sets of indices [h k i l] and [h_o k_o i_o l_o]. This can be accomplished by equating the expressions for x from equations (34) and (48), i.e.

$[{{a_0}\over{3}}\left(h_{o}{\bf e}_1^{o} + k_{o}{\bf e}_2^{o}+i_{o}{\bf e}_3^{o}\right) + {{c_0}\over{3}}l_{o}{\bf n} = {{\hat{a}_0}\over{9}}\left(h{\bf e}_1 + k{\bf e}_2 + i{\bf e}_3\right) + {{c_0}\over{9}}l{\bf n}.\eqno (51)]$

In view of $[\hat{a}_0]$ = 3^1/2a₀ and the relationship (28) between the two sets of unit vectors, it follows that

$[\left[\matrix{ h\cr k\cr i\cr l }\right] = \left(\matrix{ \hfill 1&\hfill -1&\hfill 0&\hfill 0\cr \hfill 0&\hfill 1&\hfill -1&\hfill 0\cr \hfill -1&\hfill 0&\hfill 1&\hfill 0\cr \hfill 0&\hfill 0&\hfill 0&\hfill 3 }\right)\cdot\left[\matrix{ h_{o}\cr k_{o}\cr i_{o}\cr l_{o} }\right].\eqno (52)]$

The inverse relations are

$[\left[\matrix{ h_{o}\cr k_{o}\cr i_{o}\cr l_{o} }\right] = {1\over 3} \left(\matrix{ \hfill 1&\hfill 0&\hfill -1&\hfill 0\cr \hfill -1&\hfill 1&\hfill 0&\hfill 0\cr \hfill 0&\hfill -1&\hfill 1&\hfill 0\cr \hfill 0&\hfill 0&\hfill 0&\hfill 1 }\right)\cdot\left[\matrix{ h\cr k\cr i\cr l }\right].\eqno (53)]$

For instance, the direction [h_o k_o i_o l_o] $[\simeq]$ $[[2 1 \bar{3} 1]]$ is equivalent to [h k i l] $[\simeq]$ $[[1 4 \bar{5} 3]]$ , while [h_o k_o i_o l_o] $[\simeq]$ $[[2 \bar{2} 0 1]]$ is equivalent to [h k i l] $[\simeq]$ $[[4 \bar{2} \bar{2} 3]]$ , in agreement with earlier calculations. The fact that the equivalent directions have different indices is a natural consequence of different hexagonal axes used for two sets of indices. It is well known that innumerable different sets of axes can be chosen for any crystal, with the corresponding direction indices related by appropriate transformation matrices [e.g. rhombohedral versus hexagonal axes for hexagonal crystals (see Barrett & Massalski, 1980; Hahn, 1995; Kelly et al., 2000)].

References

Askeland, D. R. (1994). The Science and Engineering of Materials, 3rd ed. Boston: PWS.
Barrett, C. S. & Massalski, T. B. (1980). Structure of Metals: Crystallographic Methods, Principles, and Data, 3rd revised ed. Oxford: Pergamon Press.
Hahn, Th. (1995). Editor. International Tables for Crystallography, 4th revised ed. Dordrecht: Kluwer Academic Publishers.
Kachanov, L. M. (1971). Foundations of Theory of Plasticity. Amsterdam: North-Holland.
Kelly, A., Groves, G. W. & Kidd, P. (2000). Crystallography and Crystal Defects, revised ed. New York: John Wiley.
Lubarda, V. A. (2002). Elastoplasticity Theory. Boca Raton, Florida: CRC Press.
McKie, D. & McKie, C. (1986). Essentials of Crystallography. Oxford: Blackwell Scientific.
Sands, D. E. (1982). Vectors and Tensors in Crystallography. Reading, Massachusetts: Addison-Wesley.
Smith, W. F. (1999). Principles of Materials Science and Engineering, 3rd ed. New York: McGraw-Hill.

research papers

New method for determining hexagonal direction indices and their relationship to crystallographic directions

1. Introduction

2. Geometric preliminaries

3. New method for determining the hexagonal direction indices

3.1. Trigonometric relations

4. Transformation rules under rotation within the (111) plane

5. Relationship between crystallographic and hexagonal direction indices

5.1. Transformation to three-axis hexagonal indices

6. An alternative set of hexagonal indices

7. Discussion

References