m
∂^2 zp,q,r
∂t^2=C 1 (zp,q,r− 1 − 2 zp,q,r+zp,q,r+1)Once again, we assume plane wave solutions like
rp,q,r=rkei(pka^1 +qka^2 +rka^3 −ωt)(a 1 ,a 2 anda 3 are the primitive lattice vectors of the simple cubic lattice) and obtain the equations
−ωi^2 m= 2C 1 (1−cos(kia))withi {x,y,z}. So in this case, where we only considered the next nearest neighbors, the oscillations
in thex-,y- andz-direction are independent from each other and the energy is just the sum over the
oscillations’ energies in all directions:
Ek=~(ωx+ωy+ωz) (43)Now we also want to take a look at the second-nearest neighbors, but we are limiting ourselves to a 2
dimensional problem, because otherwise the equations would get too complicated. Hence, Newton’s
laws read as follows:
m∂^2 xp,q
∂t^2
=C 1 (xp− 1 ,q− 2 xp,q+xp+1,q) +C 2
√
2
(xp− 1 ,q− 1 +xp− 1 ,q+1+xp+1,q− 1 +xp+1,q+1− 4 xp,q)+
C 2
√
2
(yp− 1 ,q− 1 +yp− 1 ,q+1+yp+1,q− 1 +yp+1,q+1− 4 yp,q)m
∂^2 yp,q
∂t^2
=C 1 (yp− 1 ,q− 2 yp,q+yp+1,q) +C 2
√
2
(yp− 1 ,q− 1 +yp− 1 ,q+1+yp+1,q− 1 +yp+1,q+1− 4 yp,q)+
C 2
√
2
(xp− 1 ,q− 1 +xp− 1 ,q+1+xp+1,q− 1 +xp+1,q+1− 4 xp,q)The factor √^12 has only been introduced to indicate the direction and could also be involved in a
constantC ̃ 2 =√C^22. By assuming plane waves like
rp,q,r=rkei(pka^1 +qka^2 −ωt)we get the following matrix equation:
(
axx axy
ayx ayy
)(
xk
yk)
=mω^2(
xk
yk)axx= 2C 1 (1−cos (kxa)) +4 C 2
√
2
(1−cos (kxa) cos (kya))axy=ayx=−4 C 2
√
2
sin (kxa) sin (kya)ayy= 2C 1 (1−cos (kya)) +4 C 2
√
2
(1−cos (kxa) cos (kya))