Acoustic branch (− sign): ω→0 as q→0. Atoms in the unit cell move in phase.
Optical branch (+ sign): ω=0 at q=0. Atoms in the unit cell move out of phase. Can interact with light (hence the name).
At q=0The optical frequency is ω0=2K(1/m1+1/m2) and the acoustic branch Has ω=vsq with vs=a2K/(m1+m2).
4.3 Quantisation: Phonons
Lattice vibrations are quantised. Each normal mode of wave vector q and branch s has Energy:
Eqs=(nqs+21)ℏωqs
Where nqs is the phonon occupation number. Phonons are bosons obeying Bose-Einstein Statistics:
⟨nqs⟩=eβℏωqs−11
In three dimensions, there are 3 acoustic branches (1 longitudinal, 2 transverse) and 3p−3 Optical branches for a crystal with p atoms per primitive cell.
4.4 Debye Model
The Debye model approximates the phonon spectrum as linear (ω=vsq) up to a cutoff frequency ωD (the Debye frequency):
ωD=vs(V6π2N)1/3
The Debye temperature:ΘD=ℏωD/kB.
Derivation of the phonon density of states. The number of modes with wave vector ∣q∣≤q In 3D is N(q)=3⋅(2π)3V⋅34πq3 (factor of 3 for polarisations). Differentiating: g(q)dq=dN/dqdq=(Vq2/π2)dq. Converting to frequency with ω=vsq:
g(ω)dω=π2Vq2dωdqdω=π2vs3Vω2dω
Since there are 3N total modes, the cutoff is determined by ∫0ωDg(ω)dω=3NGiving g(ω)=2π2vs33Vω2 For 0≤ω≤ωD. ■
Low-temperature limit (T≪ΘE): CV≈3NkB(ΘE/T)2e−ΘE/T Which is exponentially suppressed. This disagrees with the Debye T3 law (and experiment).
4.6 Phonon Thermal Conductivity
Phonons carry heat through the lattice. By the kinetic theory formula:
κph=31CVvsℓph
Where ℓph is the phonon mean free path.
Scattering mechanisms that limit ℓph:
Phonon—phonon scattering: At high T, ℓph∝1/T (Umklapp processes dominate, where the total phonon momentum is not conserved). At low TOnly normal processes (N-processes, conserving momentum) contribute, and ℓph grows exponentially.
Boundary scattering: At very low T, ℓph is limited by the sample size L.
Defect scattering: Point defects, dislocations, and grain boundaries scatter phonons, reducing κph.
Temperature dependence:
Low T (T≪ΘD): κph∝T3 (from CV∝T3With ℓph limited by boundaries).
Intermediate T: κph peaks.
High T (T≳ΘD): κph∝1/T (from ℓph∝1/T and CV≈const).
4.7 Specific Heat: Debye vs Einstein vs Experiment
Feature
Debye
Einstein
Experiment
High T
3NkB
3NkB
3NkB
Low T
∝T3
∝e−ΘE/T
∝T3
Single parameter
ΘD
ΘE
---
Physical basis
Acoustic phonons
Optical phonons
Both contribute
The Debye model captures the correct low-T behaviour because long-wavelength acoustic phonons Dominate the specific heat at low temperatures. The Einstein model is more appropriate for Describing the optical branch contribution, which is nearly flat (dispersionless) and hence well Approximated by a single frequency.
For crystals with optical branches (e.g., NaCl, SiO2), a combined model using Debye for Acoustic modes and Einstein for optical modes gives better agreement with experiment across all Temperatures.
Worked Example: Debye Temperature of Copper
Copper has molar mass M=63.55 g/mol, density ρ=8.96g/cm3And measured Speed of sound vs=3810 m/s (average of longitudinal and transverse).
Number density: n=MρNA=63.558.96×6.022×1023=8.49×1028m−3.
The accepted experimental value is ΘD=343 K. The discrepancy arises because the Debye Model uses a single average sound velocity, while the real phonon spectrum is anisotropic.
:::caution Common Pitfall The Debye and Einstein models describe the lattice contribution to specific heat. At low Temperatures, the electronic specific heat Ce=γT also contributes and can dominate over The lattice T3 term in metals. The total low-T specific heat of a metal is CV=γT+βT3Where β is related to ΘD. A plot of CV/T versus T2 yields γ (intercept) and β (slope).
Worked Example: Comparing Debye and Einstein Specific Heats
For aluminium: ΘD=428 K. Fit an Einstein temperature to match the Debye model at T=ΘD/2=214 K.
Numerical evaluation gives CV/(3NkB)≈0.825 at T/ΘD=0.5.
For the Einstein model, ΘE≈0.75ΘD gives a good match at intermediate Temperatures. The Einstein model overestimates CV at low T and is less accurate overall, but It is simpler to evaluate analytically.
At T=300 K: T/ΘD=0.70And both models give CV≈0.94×3NkB Approaching the Dulong—Petit limit.
Worked Example: Phonon Mean Free Path in Copper
For copper at 300 K: ΘD=343 K, vs=3810 m/s, thermal conductivity κph=401 W/(m⋅K), and CV≈3NkB=3×8.49×1028×1.381×10−23=3.52×106 J/(m3⋅K).
From κ=31CVvsℓ:
ℓ=CVvs3κ=3.52×106×38103×401=8.97×10−8m≈90nm
This is much shorter than the sample size, confirming that phonon—phonon (Umklapp) scattering Dominates at room temperature. At 10 K, the mean free path would be limited by sample boundaries.
4.8 Neutron Scattering
Neutrons are an ideal probe of phonons because their de Broglie wavelength (∼1 Å) matches Lattice spacings, and their energy (∼10—100 meV) matches phonon energies. In an inelastic Neutron scattering experiment, the energy and momentum transfer are measured:
ℏω=Ei−Ef,q=ki−kf
The scattering cross-section is proportional to the dynamical structure factor S(q,ω) Which has peaks when ℏω=ℏωqs (phonon creation) or ℏω=−ℏωqs (phonon annihilation). This allows direct measurement of The full phonon dispersion relation ω(q).
Time-of-flight and triple-axis spectrometers are the primary instruments used. Neutron scattering Has provided definitive measurements of phonon dispersions in virtually all important crystals.