Crystal vibrations

1. Vibrationsof Crystals with
monoatomicbasis
Mrs. P
.Kanmani,
Assistant Professorof Physics,
V.V.VanniaperumalCollegeforWomen,
Virudhunagar

3. C R Y S T A L S
E L A S T I C
C O N S T A N T
WAVE
V E C T O R
B R I L L O U I N
z O N E
Terms to
recall…
Vibrations of Crystalswith monoatomic
basis

4. C R Y S T A L S
Something is crystalline if the
atoms or ions that compose it
are arranged in a regular way
(i.e., a crystal has internal order
due to the periodic arrangement
of atoms in a basis ...)

5. E L A S T I C
C O N S T A N T
subjected
When an elastic
to
body is
stress, a
proportionate amount of strain
is produced. The ratio of the
applied stresses to the strains
generated will always be
constant and is known as elastic
constant.

6. WAVE
V E C T O R
Wavevector gives the direction
of propagation of any wave and
its magnitude is a number called
2𝜋
wavenumber given by . It can
𝜆
be calculated as number of
waves crossing per unit distance
times multiplied by 𝜋.

7. B R I L L O U IN
z O N E
WeignerSeitz cell (locus of all
the points in space closer to that
than any other lattice
point
points)
known
in reciprocal lattice is
as brillouin zone .
Brillouin zone is made by
drawing planes normal to the
segments joining nearest lattice
point to a particular lattice point
through the midpoints of such
segments.

8. Consider elastic vibrations
in crystals with monoatomic
basis
To find frequency in terms of
wavevector and elastic
constants
Solution is simple in
[1 0 0], [1 1 0], and [1 1 1]
propagation directions in
crystals
Crystal Vibrations in monoatomicbasis

9. Crystal Vibrations in monoatomicbasis
• Consider𝑢𝑠, the displacement of planesfrom its equilibrium position.
• Foreachwave vector there arethree modescorrespondingto onelongitudinal
and two transversepolarizations.
• Elastic responseof crystal is alinearfunction of the forces.
• i.e. elastic energyis aquadratic function of the relative displacementof any two
points in thecrystal.
• Linear termswill vanish, cubicand higherorderterms maybeneglected for small
deformations.

10. Crystal Vibrations in monoatomicbasis
• Forceonthe planescausedby the displacement of plane s+pis proportionalto
difference𝑢𝑠+1 − 𝑢𝑠 of theirdisplacements.
• Consideringnearestneighbourinteractions, p =±1
• Thetotal forceon scomesfrom planess =±1
𝐹𝑠 = 𝐶 𝑢𝑠+1 − 𝑢𝑠 + 𝐶(𝑢𝑠−1 − 𝑢𝑠) (1)
expression is linear in displacements and is of the form Hooke’slaw, Cis force
constant.

11. Crystal Vibrations in monoatomicbasis
• Theequation of motion of plane s is,
𝑑𝑡2
2
𝑀 𝑑 𝑢𝑠
= 𝐶 𝑢 − 𝑢 − 2𝑢
𝑠+1 𝑠−1 𝑠 (2)
Solution will beof time dependent form i.e 𝑒𝑥𝑝 −𝑖𝜔𝑡 , then
𝑑2𝑢𝑠
𝑑𝑡2 𝑠
= −𝜔2𝑢 , and (2)becomes,
−𝑀𝜔2𝑢𝑠 = 𝐶 𝑢𝑠+1 − 𝑢𝑠−1 − 2𝑢𝑠 (3)

12. Crystal Vibrations in monoatomicbasis
• The above equn is adifference equation in displacement u and has traveling
wave solutions of the form,
𝑢𝑠±1 = 𝑢𝑒𝑥𝑝 𝑖𝑠𝐾𝑎 𝑒𝑥𝑝 ± 𝑖𝐾𝑎 (4)
• aspacing between planes and K is wavevector. The value to use for awill
depend onK.

13. Crystal Vibrations in monoatomicbasis
+ 𝑒𝑥𝑝 𝑖 𝑠 − 1 𝐾𝑎 − 2exp(𝑖𝑠𝐾𝑎)
• With (4)we have from (3)
−𝜔2𝑀𝑢𝑒𝑥𝑝 𝑖𝑠𝐾𝑎 = 𝐶𝑢 exp 𝑖 𝑠 + 1 𝐾𝑎
• Cancel 𝑢𝑒𝑥𝑝 𝑖𝑠𝐾𝑎 from both sides,
𝜔2𝑀 = −𝐶[exp 𝑖𝐾𝑎 + exp −𝑖𝐾𝑎 − 2]
+ exp −𝑖𝐾𝑎 , wehave thedispersion
2
With identity 2cosKa =exp 𝑖𝐾𝑎
relation connecting 𝜔 and K,
𝜔 = (
2𝐶
𝑀
)(1 − 𝑐𝑜𝑠𝐾𝑎)

14. Crystal Vibrations in monoatomicbasis
• Theboundary of the first Brillouin zone lies at 𝐾 = ±
𝜋
𝑎
.
• Theslope of 𝜔 versusK is zero at the zone boundary,
=
𝑑𝜔2 2𝐶𝑎
𝑑𝐾 𝑀
𝑠𝑖𝑛𝐾𝑎 = 0
At K =0 for sinKa =sin (±Π)=0
2
By trigonometry 𝜔 =
4𝐶
𝑀
2 1 1
𝑠𝑖𝑛 𝐾𝑎; ω = 4𝐶/𝑀2
2
1
𝑠𝑖𝑛 𝐾𝑎
2

15. Crystal Vibrations in monoatomicbasis
• A plot of 𝜔 versus K is shown in figure