UNIVERSITY OF SOUTH AFRICA
College of Science, Engineering and Technology
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
PHY3707: Solid State Physics
Assignment 3 — Lattice Dynamics and Phonons
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
PHY3707
Module Code:
Solid State Physics
Module Name:
Lattice Dynamics and Phonons
Assignment Topic:
Assignment 3
Assignment Number:
197029
Unique Number:
19 June 2026
Due Date:
Submitted in partial fulfilment of the requirements for Solid State Physics — UNISA
,UNISA | PHY3707 Assignment 3 — Lattice Dynamics and Phonons
Question 1: Dispersion Relation for a Chain with Alternating Force Constants
Question
In a linear chain all the atoms have equal mass but are connected alternately by springs
of force constants f1 and f2 . Derive the frequency-wavevector relation for this chain. Are
there still two branches? Explain.
Solution. A linear chain in which every atom has the same mass but the bonds alternate be-
tween two force constants is a useful variant of the standard diatomic chain, because it iso-
lates the effect of bond alternation from the effect of mass alternation. The derivation below
follows the standard 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:
Page 2 of 11
,UNISA | PHY3707 Assignment 3 — Lattice Dynamics and Phonons
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 :
(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
ω = 2
± 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
Page 3 of 11
,UNISA | PHY3707 Assignment 3 — Lattice Dynamics and Phonons
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):
s
f1 + f2 1 ka
2
ωoptical = + (f1 + f2 )2 2
− 4f1 f2 sin (10)
M M 2
s
f1 + f2 1 ka
2
ωacoustic = − (f1 + f2 )2 2
− 4f1 f2 sin (11)
M M 2
At the zone centre (k = 0) the acoustic branch vanishes, ωacoustic = 0, corresponding to a
rigid translation of the whole chain, while the optical branch reaches its maximum, ωoptical
2 =
2(f1 + f2 )/M . At the zone boundary (ka = π) the two branches separate into ω 2 = 2f1 /M and
ω 2 = 2f2 /M , opening a frequency gap whenever f1 ̸= f2 .
ω (arbitrary units)
2
1
Acoustic branch
Optical branch
0 π
−π −π/2 0 π/2
ka
Figure 2: Acoustic and optical branches for the alternating-spring chain, plotted for f1 = 2f2
(illustrative units, M = 1, f2 = 1). A frequency gap opens at the zone boundary ka = π be-
cause f1 ̸= f2 .
Key Distinction
The two branches survive even though every atom has the same mass, because what
creates two branches is the size of the repeating unit, not which physical parameter
alternates. Whether it is the mass that alternates (the standard diatomic chain) or the
force constant that alternates (this problem), the repeat unit still contains two atoms
and spans 2a. This halves the first Brillouin zone compared with a simple monatomic
chain of spacing a, and folding the single monatomic branch back into the smaller zone
produces the familiar acoustic and optical pair. Two branches is therefore a conse-
quence of the two-atom basis, not of the masses being unequal.
Page 4 of 11
, UNISA | PHY3707 Assignment 3 — Lattice Dynamics and Phonons
Question 2: Reststrahlen Wavelength of NaCl
Question
The lattice constant of NaCl is 5.6 Å and its Young’s modulus in the [100] direction is
5 × 1010 N m−2 . Calculate the wavelength at which the electromagnetic 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.)
Solution. Ionic crystals such as NaCl reflect electromagnetic radiation almost completely over
a narrow band in the far infrared, known as the reststrahlen band, because the transverse opti-
cal phonon at the zone centre couples strongly to the electromagnetic field (Kittel, 2005). This
question estimates the wavelength of that reflection band from elastic data using the diatomic-
chain model of Question 1.
2.1 Force Constant from Young’s Modulus
Using the hint Y = f /a, with Y = 5 × 1010 N m−2 and lattice constant a = 5.6 × 10−10 m,
f = Y a = (5 × 1010 )(5.6 × 10−10 ) (12)
f = 28 N m−1 (13)
2.2 Reduced Mass of the Na–Cl Ion Pair
With atomic masses mNa = 23u and mCl = 37u, where u = 1.66 × 10−27 kg, the reduced mass
of the ion pair is
mNa mCl 23 × 37
µ= = u = 14.183 u (14)
mNa + mCl 60
µ = 2.354 × 10−26 kg (15)
Page 5 of 11
College of Science, Engineering and Technology
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
PHY3707: Solid State Physics
Assignment 3 — Lattice Dynamics and Phonons
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
PHY3707
Module Code:
Solid State Physics
Module Name:
Lattice Dynamics and Phonons
Assignment Topic:
Assignment 3
Assignment Number:
197029
Unique Number:
19 June 2026
Due Date:
Submitted in partial fulfilment of the requirements for Solid State Physics — UNISA
,UNISA | PHY3707 Assignment 3 — Lattice Dynamics and Phonons
Question 1: Dispersion Relation for a Chain with Alternating Force Constants
Question
In a linear chain all the atoms have equal mass but are connected alternately by springs
of force constants f1 and f2 . Derive the frequency-wavevector relation for this chain. Are
there still two branches? Explain.
Solution. A linear chain in which every atom has the same mass but the bonds alternate be-
tween two force constants is a useful variant of the standard diatomic chain, because it iso-
lates the effect of bond alternation from the effect of mass alternation. The derivation below
follows the standard 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:
Page 2 of 11
,UNISA | PHY3707 Assignment 3 — Lattice Dynamics and Phonons
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 :
(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
ω = 2
± 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
Page 3 of 11
,UNISA | PHY3707 Assignment 3 — Lattice Dynamics and Phonons
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):
s
f1 + f2 1 ka
2
ωoptical = + (f1 + f2 )2 2
− 4f1 f2 sin (10)
M M 2
s
f1 + f2 1 ka
2
ωacoustic = − (f1 + f2 )2 2
− 4f1 f2 sin (11)
M M 2
At the zone centre (k = 0) the acoustic branch vanishes, ωacoustic = 0, corresponding to a
rigid translation of the whole chain, while the optical branch reaches its maximum, ωoptical
2 =
2(f1 + f2 )/M . At the zone boundary (ka = π) the two branches separate into ω 2 = 2f1 /M and
ω 2 = 2f2 /M , opening a frequency gap whenever f1 ̸= f2 .
ω (arbitrary units)
2
1
Acoustic branch
Optical branch
0 π
−π −π/2 0 π/2
ka
Figure 2: Acoustic and optical branches for the alternating-spring chain, plotted for f1 = 2f2
(illustrative units, M = 1, f2 = 1). A frequency gap opens at the zone boundary ka = π be-
cause f1 ̸= f2 .
Key Distinction
The two branches survive even though every atom has the same mass, because what
creates two branches is the size of the repeating unit, not which physical parameter
alternates. Whether it is the mass that alternates (the standard diatomic chain) or the
force constant that alternates (this problem), the repeat unit still contains two atoms
and spans 2a. This halves the first Brillouin zone compared with a simple monatomic
chain of spacing a, and folding the single monatomic branch back into the smaller zone
produces the familiar acoustic and optical pair. Two branches is therefore a conse-
quence of the two-atom basis, not of the masses being unequal.
Page 4 of 11
, UNISA | PHY3707 Assignment 3 — Lattice Dynamics and Phonons
Question 2: Reststrahlen Wavelength of NaCl
Question
The lattice constant of NaCl is 5.6 Å and its Young’s modulus in the [100] direction is
5 × 1010 N m−2 . Calculate the wavelength at which the electromagnetic 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.)
Solution. Ionic crystals such as NaCl reflect electromagnetic radiation almost completely over
a narrow band in the far infrared, known as the reststrahlen band, because the transverse opti-
cal phonon at the zone centre couples strongly to the electromagnetic field (Kittel, 2005). This
question estimates the wavelength of that reflection band from elastic data using the diatomic-
chain model of Question 1.
2.1 Force Constant from Young’s Modulus
Using the hint Y = f /a, with Y = 5 × 1010 N m−2 and lattice constant a = 5.6 × 10−10 m,
f = Y a = (5 × 1010 )(5.6 × 10−10 ) (12)
f = 28 N m−1 (13)
2.2 Reduced Mass of the Na–Cl Ion Pair
With atomic masses mNa = 23u and mCl = 37u, where u = 1.66 × 10−27 kg, the reduced mass
of the ion pair is
mNa mCl 23 × 37
µ= = u = 14.183 u (14)
mNa + mCl 60
µ = 2.354 × 10−26 kg (15)
Page 5 of 11
