Page 63 - Semiconductor For Micro- and Nanotechnology An Introduction For Engineers

The Crystal Lattice System
Crystal
h
Energy in We return our attention to equation (2.25), where our goal is to rewrite
the harmonic crystal energy U
in terms of the crystal’s strain. Clearly,
Terms of h
U should not be dependent on our choice of coordinate axes, and is
Strain
therefore invariant with respect to rigid-body rotations. We now choose a
deformation field u = θ × R which simply rotates the crystal atoms
θ
from their lattice positions by a constant defined angle . The gradient of
this field is u∇ = ∇ θ × R + θ × ∇ R = 0 , so that the energy associated
u
with pure rotations is clearly zero. The gradient of can always be writ-
ten as the sum of a symmetric and an anti-symmetric part
T
T
⁄
∇ u = ∇ u + ∇ u = [ ∇ u + ( ∇ u) ] 2 + [ ∇ u ( ∇ u) ] 2
⁄
–
s a
(2.32)
= ( ε + κ)
κ
ε
Note that ε = ε T and κ = – κ T . We substitute and into (2.25) to
obtain
h 1 T T T T
--- (
d
U = – 2∫ ε :E:ε + ε :E:κ + κ :E:ε + κ :E:κ) V (2.33)
Ω
for the harmonic energy of the crystal.
Independent The elastic tensor E has inherent symmetries that can be exploited to
Elastic simplify (2.33). We first write an expression for the components of the
Constants
tensor E, centered at site i, in cartesian coordinates
N
(
1   ∂ ∂E R ) 
i 
ER = ------- ∑ ( R ) -------- ----------------- ( R ) (2.34)
()
i αβγµ  ij α   ij γ 
2V ∂x β ∂x µ
j = 1
Consider the argument of the sum. Clearly, E is symmetric with respect
µ
β
to the indices and , since the order of differentiation of the energy
with respect to a spatial coordinate is arbitrary. Furthermore, swapping
T
T
α and also has no effect. Now look at the terms ε :E:κ and κ :E:ε
γ
in (2.33). They vanish because, due the above symmetries of E ,
T T
κ
ε :E:κ = – κ :E:ε . Finally, because the anti-symmetric strain repre-
T
sents a pure rotation, the last term κ :E:κ in (2.33) must also vanish. As
60 Semiconductors for Micro and Nanosystem Technology