UNIVERSITY OF SOUTH AFRICA
College of Science, Engineering and Technology
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
PHY3707: Solid State Physics
Lattice Dynamics and Phonons — Tutorial Assignment
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
PHY3707
Module Code:
Solid State Physics
Module Name:
Lattice Dynamics and Phonons
Assignment Topic:
Assignment Number 3
Assignment Number:
197029
Unique Number:
19 June 206
Due Date:
Submitted in partial fulfilment of the requirements for Solid State Physics — UNISA
,Assignment 3
Unique number: 197029
Please note that the Assignment problems are taken from Chapters 4 of the prescribed textbook,
Elements of Solid State Physics (4th edition, 2014) by JP Srivastava.
1) In a linear chain all the atoms have equal mass but are connected alternatively by
springs of force constants, f1 and f2. Derive the frequency-wavevector relation for this
chain. Are there still two branches? Explain. [34]
2) The lattice constant of NaCl is 5.6 Å and its Young’s modulus in the [100] direction is
5 x 1010 N m-2. Calculate the wavelength at which the e.m.radiation is strongly reflected
from a NaCl crystal. State the assumptions under which your calculation is valid. Take
the atomic weights of Na and Cl as 23 and 37 respectively. (Hint: Young’s modulus =
f/a).
[10]
3) Consider a longitudinal wave
which propagates in a monatomic linear lattice of atoms of mass M, spacing a and the
force constant f. Show that the total energy of the wave is
where n runs over all atoms.
[6]
Total [50 marks]
2
,UNISA | [Module Code] Lattice Dynamics and Phonons
Question 1: Dispersion Relation for a Chain with Alternating Force Constants
A linear chain in which every atom has the same mass but the bonds alternate between two
force constants is a useful variant of the standard diatomic chain, because it isolates the effect
of bond alternation from the effect of mass alternation. The derivation below follows the stan-
dard travelling-wave method for lattice dynamics set out in Kittel (2005).
1.1 Setting Up the Equations of Motion
Let every atom have mass M , and let the chain be built from a repeating two-atom unit in
which atom displacements are written un and vn . The springs alternate along the chain: a
bond of force constant f1 joins each un to its neighbouring vn , and a bond of force constant f2
joins each vn to the next un+1 .
f1 f2 f1 f2
un−1 vn−1 un vn un+1
Figure 1: Identical atoms of mass M joined by springs that alternate between force constants
f1 and f2 . The repeating unit contains two atoms and has length 2a.
Newton’s second law applied to each atom in the basis gives
M ün = −f1 (un − vn ) − f2 (un − vn−1 ) (1)
M v̈n = −f1 (vn − un ) − f2 (vn − un+1 ) (2)
1.2 Travelling-Wave Trial Solutions
As for any periodic lattice, both atoms in the basis oscillate at the same frequency ω and the
same wavevector k, with a fixed amplitude and phase relationship between them:
un = U ei(nka−ωt) , vn = V ei(nka−ωt) (3)
Substituting these into Equations (1) and (2) and cancelling the common exponential factor
gives a pair of linear equations in U and V :
Page 1 of 10
, UNISA | [Module Code] Lattice Dynamics and Phonons
(f1 + f2 ) − M ω 2 U − f1 + f2 e−ika V = 0 (4)
− f1 + f2 eika U + (f1 + f2 ) − M ω 2 V = 0 (5)
1.3 The Dispersion Relation
A non-trivial solution for U and V exists only when the determinant of the coefficient matrix
vanishes:
(f1 + f2 ) − M ω 2 − f1 + f2 e−ika
=0 (6)
− f1 + f2 eika (f1 + f2 ) − M ω 2
Expanding the determinant, and using f1 + f2 e−ika f1 + f2 eika = f12 + f22 + 2f1 f2 cos ka,
gives
2
(f1 + f2 ) − M ω 2 = f12 + f22 + 2f1 f2 cos ka (7)
so that
f1 + f2 1
q
ω2 = ± f12 + f22 + 2f1 f2 cos ka (8)
M M
It is conventional to rewrite the term under the root using the half-angle identity cos ka =
1 − 2 sin2 (ka/2), which puts the dispersion relation in its more familiar form:
s
f1 + f2 1 ka
ω2 = ± (f1 + f2 )2 2
− 4f1 f2 sin (9)
M M 2
1.4 Are There Still Two Branches?
Yes. The ± sign produces exactly two solutions for every value of k, exactly as it does for the
conventional diatomic chain of two different masses on identical springs (Kittel, 2005):
Page 2 of 10
College of Science, Engineering and Technology
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
PHY3707: Solid State Physics
Lattice Dynamics and Phonons — Tutorial Assignment
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
PHY3707
Module Code:
Solid State Physics
Module Name:
Lattice Dynamics and Phonons
Assignment Topic:
Assignment Number 3
Assignment Number:
197029
Unique Number:
19 June 206
Due Date:
Submitted in partial fulfilment of the requirements for Solid State Physics — UNISA
,Assignment 3
Unique number: 197029
Please note that the Assignment problems are taken from Chapters 4 of the prescribed textbook,
Elements of Solid State Physics (4th edition, 2014) by JP Srivastava.
1) In a linear chain all the atoms have equal mass but are connected alternatively by
springs of force constants, f1 and f2. Derive the frequency-wavevector relation for this
chain. Are there still two branches? Explain. [34]
2) The lattice constant of NaCl is 5.6 Å and its Young’s modulus in the [100] direction is
5 x 1010 N m-2. Calculate the wavelength at which the e.m.radiation is strongly reflected
from a NaCl crystal. State the assumptions under which your calculation is valid. Take
the atomic weights of Na and Cl as 23 and 37 respectively. (Hint: Young’s modulus =
f/a).
[10]
3) Consider a longitudinal wave
which propagates in a monatomic linear lattice of atoms of mass M, spacing a and the
force constant f. Show that the total energy of the wave is
where n runs over all atoms.
[6]
Total [50 marks]
2
,UNISA | [Module Code] Lattice Dynamics and Phonons
Question 1: Dispersion Relation for a Chain with Alternating Force Constants
A linear chain in which every atom has the same mass but the bonds alternate between two
force constants is a useful variant of the standard diatomic chain, because it isolates the effect
of bond alternation from the effect of mass alternation. The derivation below follows the stan-
dard travelling-wave method for lattice dynamics set out in Kittel (2005).
1.1 Setting Up the Equations of Motion
Let every atom have mass M , and let the chain be built from a repeating two-atom unit in
which atom displacements are written un and vn . The springs alternate along the chain: a
bond of force constant f1 joins each un to its neighbouring vn , and a bond of force constant f2
joins each vn to the next un+1 .
f1 f2 f1 f2
un−1 vn−1 un vn un+1
Figure 1: Identical atoms of mass M joined by springs that alternate between force constants
f1 and f2 . The repeating unit contains two atoms and has length 2a.
Newton’s second law applied to each atom in the basis gives
M ün = −f1 (un − vn ) − f2 (un − vn−1 ) (1)
M v̈n = −f1 (vn − un ) − f2 (vn − un+1 ) (2)
1.2 Travelling-Wave Trial Solutions
As for any periodic lattice, both atoms in the basis oscillate at the same frequency ω and the
same wavevector k, with a fixed amplitude and phase relationship between them:
un = U ei(nka−ωt) , vn = V ei(nka−ωt) (3)
Substituting these into Equations (1) and (2) and cancelling the common exponential factor
gives a pair of linear equations in U and V :
Page 1 of 10
, UNISA | [Module Code] Lattice Dynamics and Phonons
(f1 + f2 ) − M ω 2 U − f1 + f2 e−ika V = 0 (4)
− f1 + f2 eika U + (f1 + f2 ) − M ω 2 V = 0 (5)
1.3 The Dispersion Relation
A non-trivial solution for U and V exists only when the determinant of the coefficient matrix
vanishes:
(f1 + f2 ) − M ω 2 − f1 + f2 e−ika
=0 (6)
− f1 + f2 eika (f1 + f2 ) − M ω 2
Expanding the determinant, and using f1 + f2 e−ika f1 + f2 eika = f12 + f22 + 2f1 f2 cos ka,
gives
2
(f1 + f2 ) − M ω 2 = f12 + f22 + 2f1 f2 cos ka (7)
so that
f1 + f2 1
q
ω2 = ± f12 + f22 + 2f1 f2 cos ka (8)
M M
It is conventional to rewrite the term under the root using the half-angle identity cos ka =
1 − 2 sin2 (ka/2), which puts the dispersion relation in its more familiar form:
s
f1 + f2 1 ka
ω2 = ± (f1 + f2 )2 2
− 4f1 f2 sin (9)
M M 2
1.4 Are There Still Two Branches?
Yes. The ± sign produces exactly two solutions for every value of k, exactly as it does for the
conventional diatomic chain of two different masses on identical springs (Kittel, 2005):
Page 2 of 10
