Lecture 12

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Lecture 12
Crystal Vibrations
From aclassical standpoint:
The vibrationalmodes of the atoms are considered
sound waves
From a modern standpoint:
The vibrational modes of the atoms are considered
phonons
The speed with which a longitudinal wave
propagationthrough a medium of elastic material(for
example solid) of density 𝜌 given by:
𝑣0 =πœ†πœ = (
Β𝑠 ½
)
𝜌
…… 1
Where Β𝑠 is adiabatic elastic bulk modulus, or
stiffness coefficient.
4-1Vibrations of crystals with monatomic basis
Consider a one-dimensional (linear) lattice with a
period (π‘Ž) and with identical atoms of mass m,
vibrating around each lattice point(see Fig.26).Each
atom is indexed by an integer (𝑛), and its
displacement from its equilibrium position is denoted
(𝑒). The atoms are taken to oscillate in the same
direction as the lattice.
We want to find the angular frequency(w) of an
elastic wave in terms of wavevector (k).
In this one-dimension case we will take into account
only the interaction between nearest neighbors.
The force acted on the nth atom:
𝐹 n=𝑐[𝑒n+1+𝑒n-1-2𝑒n]
In a homogeneous solid the transmission of a plane
wave in the x-direction can be represented by the
displacement equation:
𝑒=𝐴exp[i(π‘˜π‘₯ − 𝑀𝑑)],and𝑒n=𝐴exp[i(π‘˜π‘›π‘Ž-𝑀𝑑)]
where π‘₯ = π‘›π‘Ž
𝑑2𝑒𝑛
𝑑𝑑 2
=-𝑀2 𝐴 exp[i(π‘˜π‘›π‘Ž-𝑀𝑑)] ,
𝑒n=-𝑀2𝑒𝑛
∴ Fn=-m𝑀2𝑒𝑛
-m𝑀2𝑒𝑛 = C [𝑒n+1+ 𝑒n-1-2𝑒n]
𝑐
𝑒𝑛+1
π‘š
𝑒𝑛
𝑀2= [2-
-
𝑒𝑛−1
𝑒𝑛
]
𝑐
𝑐
π‘š
π‘š
𝑀2= [2-𝑒 π‘–π‘˜π‘Ž - 𝑒 −π‘–π‘˜π‘Ž ]= 2 [1- π‘π‘œπ‘ π‘˜π‘Ž]
or
𝑐
π‘˜π‘Ž
π‘š
2
= 4 sin2 (
)
𝒄
π’Œπ’‚
π’Ž
𝟐
𝑾=±πŸ( )½ π’”π’Šπ’ (
𝑐
)=±π’˜m π’”π’Šπ’ (
π’Œπ’‚
)
where
𝟐 ……………. 2
wm =±2( )½
π‘š
This relation is called the phonon dispersion
relationbetween w and k for allowed longitudinal
wave in a linear monatomic chain, and is plotted in
Fig. 27.
Note: the plus and minus signs in Eq. 2 denote
waves traveling either to the right or to the left. The
motion in any point is periodic in time.
k
Fig. 27
2ndBrillouin zone2nd Brillouin zone
k
in the case of propertiesπ‘˜π‘Ž β‰ͺ 1(i.e. in the long
wavelength limit ), 𝑠𝑖𝑛 (
π‘˜π‘Ž
2
)becomes (
π‘˜π‘Ž
2
)
and Eq. ( 2 ) becomes:
𝑀 ⋍ 𝑣0 π‘˜
𝑐
𝑣0 = π‘Ž( )½ …. 3 for π‘˜π‘Ž β‰ͺ 1(non-dispetion)
π‘š
𝑣0 :is the rapid spread of acoustic wave in a medium
of elastic material, and It is approximately fixed amount
πœ‹
In the case of propertiesπ‘˜ = ± (i.e. in the short
π‘Ž
wavelength limit, asπ‘˜ increases, the slope of 𝑀
decreases and becomes flat at the zone boundaries
πœ‹
π‘˜=± ).
π‘Ž
We have π‘˜ =
2πœ‹
πœ†
,thenπœ† = 2π‘Ž , andwm =2
𝑣0
π‘Ž
phase and group velocities are in general given
by:
𝑀
πœ—π‘β„Ž =
πœ•π‘€
πœ—π‘” =
πœ•π‘˜
π‘˜
=
2
π‘˜
(
𝑐 ½
)
π‘š
= 𝑣0 π‘π‘œπ‘  (
π‘˜π‘Ž
2
𝑠𝑖𝑛 (
π‘˜π‘Ž
2
) = 𝑣0 [
π‘˜π‘Ž
)
2
π‘˜π‘Ž
2
sin ⁑
(
]
)
Note:in the case of π‘˜π‘Ž β‰ͺ 1 (or πœ† ≫ π‘Ž) the
atoms moving in one direction and the same phase
and this explains the fact that𝑀 = 0 when π‘˜ = π‘œ.
Problems
𝑐
1- Use Eq.𝑣0 = π‘Ž( )½ to find the relation
π‘š
between 𝑐 and young model’s Y.?
The answer
(𝑐 = π‘Œπ‘Ž).
2- in the case of properties π‘˜π‘Ž β‰ͺ 1, show that
πœ—π‘” =𝑣0 , πœ—π‘β„Ž =𝑣0 and
πœ—π‘” =πœ—π‘β„Ž
πœ‹
3- In the case of properties π‘˜ = ± , show that
π‘Ž
πœ—π‘” =0 and
2𝑣0
πœ—π‘β„Ž =
πœ‹
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