10. Appendix

Solution to Problem 3.2(c)
Solution to Problem 3.2(c) 601
Let us assume that a pair of equal and opposite forces, F and F, are applied
along the [111] direction: F (F/ √ 3)(1, 1, 1). These forces are applied to a
pair of surfaces of area A whose normals are also oriented along the [111]
directions. It is convenient to use a vector parallel to the normal of a surface
to represent both its area and direction. In the case of the surface of a solid
the convention we shall adopt is: a normal pointing away from the solid is
positive. For example, the surface to which the above force F is applied can be
represented by the vector: A (A/ √ 3)(1, 1, 1). Similarly, the force F would
now be applied to a surface, represented by the vector A, on the opposite
side of the solid. The combined effect of F and F would be to stretch the
solid. Since the forces F and F lie along the same axis, the solid is said to be
under a tensile uniaxial stress. The magnitude of the uniaxial stress X is equal
to F/A. However, the stress on a solid cannot be represented, in general, by a
scalar or a vector. Instead, we have to define a second rank tensor Xij, known
as a stress tensor, such that when Xij is contracted with the vector A we obtain
the force F. For the present case of F applied along the [111] direction to the
area represented by A, the component of F along the direction x is Fx
(F/ √ 3). We can decompose A into three mutually perpendicular components:
Ax, Ay and Az. The tensor elements X11, X12 and X13 are defined by: Fx/Ax,
Fx/Ay and Fx/Az, respectively. Since Ax Ay Az for the area A we have
defined above, we expect that X11 X12 X13. Similarly, we find that X21
X22 X23 and X31 X32 X33. Finally, since Fx Fy Fz we find all the
elements of the stress tensor Xij to be equal: X11 X22 X33 ... etc.. Thus,
the second rank tensor Xij is given by:
⎛ ⎞
1 1 1
C ⎝ 1 1 1⎠
1 1 1
where C is a constant to be determined. To obtain C we note that a contraction
between the stress tensor Xij and the vector A would yield the force F
or XijAj Fi. This implies 3C(A/ √ 3) F/ √ 3orC X/3. Thus, the stress
tensor corresponding to a tensile stress applied along the [111] direction is:
X
3
⎛ ⎞
1 1 1
⎝ 1 1 1⎠
1 1 1
Notice that our sign convention for the area vector A leads to the following
sign convention for the stress: a tensile stress is positive.
Solution to Problem 3.3
To determine the non-zero and linearly independent components of the compliance
tensor Sijkl we apply symmetry operations of the zincblende structure
to Sijkl.