1

Chapter Five Phonons II. Thermal PropertiesPhonons : dominate thermal properties of materials and affect the

electrical transports of conductors by scatterings of electrons

Phonon generations : How are phonons created or excited in a crystal?

• External perturbations – vibrations or sound transducer

• Scattering of particles – energy transferred into lattice vibrations

• Thermal (KBT) – excited at any finite temperature (T≠0K)

Thermal phonons : consider a system with energy level En

En

En-3

En-1

Probability of occupancy

at temperature T

“Boltzmann factor”

−∝

TkEexp)P(E

B

nn

2

Excitation level amplitude (n) w/. energy

ω ,k moder

( )

( )

( )

( ) Tkω xwhere

sxexp

sxexpdxd

Tkω1/2)(sexp

Tkω1/2)(sexp s

TkEexp

Tk

Eexp s

EP

EP sn

Bs

s

s B

s B

s B

s

s B

s

ss

ss

h

h

h

=−

−−=

+−

+−

=

−

−

==

∑∑

∑

∑

∑

∑∑∑

Planck distribution of < n(ω, T) >

average # of phonons excited per mode at ω

1Tkωexp

1 T), n(ω

B

−

=

h

ω21n h

+

Average of phonons

3

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.00.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

< n(ω

, T) >

x-1 = kBT / hω

High T ( kBT >> hω )

< n(ω, T) > ~

Low T ( kBT < hω )

< n(ω, T) > ~

1Tkωexp

ω) D(ω dω

ω) n(ω ) D(ω dω

ωnU

B

i

mode

ii

−

=

=

=

∫

∫

∑

h

h

h

hThermal energy

thermal equilibrium

density of modes

ωhTk B

−

ωhTkexp B

4

Density of states (modes) : uniform in k-space

1D D(k)≡density of states = number of states per unit k at k

D(k)dk number of states from k to k+dk

A linear chain of length L carries N+1 particles with separation a.

Boundary condition : u0(t)=0 and uN(t)=0 fixed pointss=0 s=N

L1)π-(N ... ,

L4π ,

L3π ,

L2π ,

Lπ

No motion at all.

πL

asN

us(t)=u exp[-iωk,pt] sin(ska) where k=

Why is there no Nπ/L for allowed k? us(t) ∝ =0( )πssinsin =

Lπ∆k =One mode for each interval

>

≤=

aπkfor 0aπkfor

πL

D(k)The number of modes per unit range of k

5

Unbounded medium but w/. periodic solution over the distance L

Periodic boundary conditions u(sa)=u(sa+L) for a large system

us(t)=u exp[ i(ska-ωk,pt) ] where k= ... ,L6π ,

L4π ,

L2π ,0 ±±±

L2π∆k =One mode for each interval

otherwise 0 aπk

aπfor

2πLD(k)

=

≤≤−=The number of modes per unit range of k

dωdωdkD(k)2D(k)dk2)dω D(ω

k of for 2 anddk 2πLD(k)dk

==

±=

The number of modes per unit frequency range

gv2D(k)

/dkdω2D(k)

dωdkD(k)2) ω D( ===

Singularity at vg=0, determined by ω(k)

Dispersion relation

6

One dimensional monatomic lattice

( )22

max

2max

ωωω

CM

πN

2kacos

2a

M4C

Na/2π2 /dkdω

2D(k) ) D(ω

2kasin

M4Cω

−=

==

=

D(ω)

ωk

D(k)

L/2π

-π/a π/a0

(N/π)(M/C)1/2

(4C/M)1/20

Total number of modes

)dω ω D()dω ω D(Na

2π2πLdkD(k)N

M4C

0

ω

0

π/a

π/a

max

∫∫∫ =====−

7

In two dimensions :

periodic boundary condition, N2 primitive cells within a square of side L

exp[ i(kxx+kyy) ] = exp[ i( kx(x+L) + ky(y+L) ) ]

whence kx, ky = ... ,L6π ,

L4π ,

L2π ,0 ±±±

2

22

yx 4πL

L2π

∆k∆k1

=

=

−

One mode per unit area in k-space

Number of modes with wavevector from k to k+dk in k-space

dkk 2π4πLdkdk

∆k∆k1D(k)dk 2

2

yxyx

==

The number of modes per unit frequency range

g2 v

1k 2π4πA

/dkdωD(k)) ω D( ==

g

23 v

1k 4π8πV

/dkdωD(k)) ω D( ==In three dimensions :

8

D(ω)dω = D(k)d3kcomplicated ! -- must map out dispersion relation and count all

k-values with each frequency

for each polarization

Continuum waves : ω = vgk depending only on amplitude of k

dωvω

2πV

vdω

vω

2πV

dkk 4π8πV

kd D(k)dω ) ω D(

3g

2

2

g

2

g2

23

3

=

=

=

=

kX

ky

k

The number of modes per unit frequency range for each polarization

3g

2

2 vω

2πV) ω D( = a quadratic dependence !

9

Quadratic at low ω

ωD

Peaks at high ω-- cutoff of ωmax in different k direction

N primitive cells in the crystal,A total number of acoustic phonon mode is N for each polarization

3/12

g

DD

3/13g

2

D

VN6π

vωk

VNv6π

ω

==

=Cutoff frequency

Cutoff wavevector

10

Thermal energy

−

=

=

∫

∫

1Tkωexp

ωv2π

Vω dω

ω) n(ω) D(ω dωU

B

3g

2

2ω

0

D

h

h

h

each polarization

There are three polarizations : 2 transverse + 1 longitudinal

( ) Tkω xwhere

1xexpx dx Tk

v2π3V

1Tkωexp

ω dωv2π

3V U

B

x

0

4B

3g

2

B

ω

03g

2

3D

3D

h

h

h

h

h

=

−

=

−

=

∫

∫

11

Defining the Debye temperature ΘD

3/12

B

g

3/13g

2

BB

DD V

N6πkv

VNv6π

kkω

=

==Θ

hhh

( ) 1xexp

x dx TT9Nk U3D/TΘ

0

3

DB

−

Θ

= ∫Peter Debye, 1884-19661936 Nobel prize winner

in chemistry

Therefore xD= hωD /kBT= ΘD/T

The total phonon energy

In classical model : equipartition theorem (0.5kBT for each excitation mode)

3 translational + 3 vibrational modes : six degrees of freedom

U = N 6 (0.5kBT) = 3 N kBT for N atoms in the crystal

Cv = 3NkB Dulong and Petit Law

12

( )

( )( )( )

( ) 1e

exdx T9Nk

1Tω/kexp

Tω/kexpωdωT1

k

v2π3V

1Tω/kexpωdω

Tv2π3V

TU C

D

D

D

x

02x

x43

DB

2B

B4ω

02

B3g

2

ω

0 B

3

3g

2V

V

∫

∫

∫

−

Θ

=

−

−

−

=

−∂∂

=

∂∂

=

h

hhh

h

h

( )3

D3Dx

x

0 T31x

31

1ex dx

3D

Θ=→

−∫

x= ΘD /T <<1

At T >> ΘD, U 3NkBT

CV 3NkB

( )( )

( )( )2B

B

2

BB

VV

B

1Tω/kexpTω/kexp

TkωNk3

TUC

1Tω/kexp

ω3NωnN3 U

−

=

∂∂

=

−==

h

hh

h

hh

Einstein model

( ) ( )3

D3D2x

xx

0 T31x

31

1eexdx

4D

Θ=→

−∫

13

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.00.0

0.2

0.4

0.6

0.8

1.0

CV

(3N

k B)

x-1 = kBT / hω

high T CV 3NkB

low T CV 3NkB (hω/kBT)2 e-hω/kT

Einstein model

classical model

Einstein model (1907) : N identical oscillators of frequency ω

At high T, CV → 3NkB same as the Dulong and Petit value

At low T, CV → 3NkB (hω/kBT)2 exp(-hω/kBT)

( )( )( )2B

B

2

BBV 1Tω/kexp

Tω/kexpTkωNk3C

−

=

h

hh

14

T/ΘD

C (c

al/m

ole-

K) Diamond,Experimental data – red points

Einstein’s model – blue curvew/. ΘD=1320K

Ann. Physik 22, 180 (1907)

“If Planck’s theory of radiation has hit upon the heart of the matter, thenwe must also expect to find contradictions between the present kinetic molecular theory and practical experience in other area of heat theory, contradiction can be removed in the same way.”

Einstein theory shows correct trends with temperature.

For simple harmonic oscillator of spring constant C and mass M, .MC

=ω

15

C (c

al/m

ole-

K)

T (K)

Ag

Einstein’s model

At low T, there are systematic deviations between data and Einstein model.

Einstein realized that the oscillations of a solid where complex, far from single frequency.

Key point is that however low the temperature, there are always some modes with low enough frequencies to be excited.

By Walther Nernst

Einstein model : At low T, CV → 3NkB (hω/kBT)2 exp(-hω/kBT)

Experimental data show T3 dependence of CV instead

16

Debye and Einstein modelsC

(J/m

ole-

K)

T/ΘD

Red points : Experimental data of Ag

ΘD=225K

In the Einstein model, C decreases too rapidly at low temperatures.

Debye model gives correct T3 dependence of C at low T.

17

Debye T3 model

Assume continuum elastic phonon mode only up to some cutoff frequency ωD

,ωω , 0

ωω ,v2π

Vω ) D(ω

D

D3g

2

2

>

≤=

Number of phonon mode for each polarization is equal to N

v2π

Vω dω) D(ω dωNDD ω

03g

2

2ω

0∫∫ ==

B

DD

3/12

g

DD

3/13g

2

D

kω

VN6π

vωk

VNv6π

ω

h=Θ

==

=Debye frequency

Debye wavevector

Debye temperature

ωD

0

ω

kD

vg

k

18

ωD, ΘD depend on vg, n, ~ vgn1/3

High for stiff, light materials

105165225343428ΘD(K)

PbAuAgCuAlmaterialKittel : Table 1 in ch.5 (P.116)

1e

x dx Tkv2π

3V U x

x

0

4B

3g

2

3D

−

= ∫h

h

( ) 1e

exdxT9Nk CDx

02x

x43

DBV ∫ −

Θ

=

At very low temperature, T<<ΘD, xD= ΘD/T → ∞

15π

1ex dx

4

x0

3

=

−∫

∞

T234Nk5Θ

TNk12πC and 5Θ

TNk3πU3

DB3

D

3B

4

V3D

4B

4

Θ

≅≅≅Debye T3

approximation

19

T3 observed in most insulators for T<0.1ΘD

solid Ar w/. ΘD=92K

Why T3 at low temperatures ?

Only long wave length acoustic modes are thermally excited.

These modes can be treated as an elastic continuum.

The energy of short wavelength modes is too high for them to be populated significantly at low temperatures.

20

CV/T

(Jou

le M

ole-1

K-2

)

T2 (K2)

KCl

Cu

Phys. Rev. 91, 1354 (1953)

Low Temperature Solid State Physics (1963)

21

Other simple idea to understand T3 dependence :

Total phonon mode : ω ≤ ωD (or k ≤ kD= )

kD

kT

kx

kyExcited phonon mode

ω ≤ kBT/h (or k ≤ ω/vg = = kT )

Others are frozen out

Fraction excited at T :

of the total volume in k-space

Tv

k

g

B

h

Dg

B Θv

kh

3

D

3

D

T

ΘT

kk

=

Thermal wavevector

Each mode has energy kBT

U ~

CV ~ ∝ T3B

T2Nk1

TkTN3 B

3

D

Θ

3

D

Θ

too small but correct T3 dependence

22

General result for D(ω) : the number of states per unit frequency

∫

=

shell

33

kd2πL )dω D(ω

dSω : an element of area on the surface in K space of selectedconstant frequency ω.

⊥∫∫ = dkdSkd ωshell

3

ωω+dω

dk⊥

dωdkωk =∇ ⊥

gω

kωω v

dωdSω

dωdSdkdS =∇

=⊥ dωv

dS2πL )dω D(ω

g

ω3

∫

=

kxky

kz

dSw( ) ∫=g

ω3 v

dS2πV) D(ω

23

Lattice vibrations : mode (k,ω )k is in BZ, discreteω(k) dispersion relationD(ω ) density of statesE(ω ) = (n+1/2) hω

Phonons : number n energy = hω

crystal momentum hk1e1n Tω/kB −

=h

Thermal properties (equilibrium)

ω) n(ω ) D(ω dωU h∫=thermal energy

heat capacity

dTdUCV =

24

Transport properties (non-equilibrium)

Conduction of sound and heat through the crystalvibration energy

Ultrasonic attenuation Thermal conductionexcite single phonon mode measure decay of amplitude

apply temperature gradient measure heat current by phonons

Phonon thermal conductivity

Apply temperature gradient ∇T → determine heat current density jU

TH TL∇T

jU

dxdTκjU −= the energy transmitted

across unit area per unit timeThe flux of the thermal energy

κ : thermal conductivity coefficient

25

In solids, heat is transported by phonon and free electrons.

For metals, it is electronic contribution that dominates the thermal conductivity.

This does not mean that insulators are necessarily poor thermal conductors.

26

Propagations of phononsNo interaction/scatteringBallistic

In harmonic approximation in perfect, infinite crystal,Expect no scattering → phonon modes are uncoupled,

independent plane waves and standing waves

dkdωvv g ==

Diffusion Phonons scatter, random walk through crystal

Phonons scatter in real crystals.

Scattering processes : boundary scattering

defect scattering

phonon-phonon scattering

dkdωvv g =<<

27

The flux of thermal energy is based on that the process of thermal energy transfer is a random process.

ie. the energy diffuses through the crystal, suffering frequent collisions.

dxdTTJ

∆TJ

U

U

=∇∝

∝ across the whole sampleBallistic :

local Diffusive :

For diffusion, thermal conductivity is defined byphonon propertiesscatteringcrystal quality (size, defect)temperature( )T

jκ U

∇−≡ v

jU [Watt/m2], κ [(Watt/m2)/(K/m)] = [Watt/m/K]

28

Kinetic theory of gases:consider phonons as gases contained in a crystal volumecalculate diffusion in the presence of temperature gradient

TH TL∇T

jUFick’s law

dxdT

dTdn

dxdn

=

n

xn : concentration of moleculesC : heat capacity per unit volume = ncvg : phonon velocityl : phonon mean free path =vgτ

dxdTcτvn

31

dxdTcτvnj 22

xU −=−=τvdxdT

dxdT∆T xx == l

lgCv31κ =

dxdTCv

31

dxdTcτnv

31j g

2gU l−=−=

29

From anharmonic terms in binding potentialU

xxo

The general shape applies for any type of binding

( ) ( ) ( ) ...xxxU

61xx

xU

21xx

xU)U(xU(x) 3

ox

3

32

ox

2

2

ox

o

ooo

+−∂∂

+−∂∂

+−∂∂

+=

( ) ( ) ...xxxU

61xx

xU

21)U(x-U(x)U(x) 3

ox

3

32

ox

2

2

o

oo

+−∂∂

+−∂∂

==∆

Reset the equilibrium, let displacement x-xo x

harmonic term anharmonic term

...fxgx cxU(x) 432 −−=

30

Thermal expansion – thermal energy causes fluctuation of x from xo

anharmonic term gives the net change of <x>

∫

∫

∫

∫∞

∞−

∞

∞−∞

∞−

∞

∞−

−−−

−−−

≅−

−=

]Tk

fxgxcxexp[dx

]Tk

fxgxcxexp[dx x

]Tk

U(x)exp[dx

]Tk

U(x)exp[dx x x

B

432B

432

B

B

++

−≅

+

−=

−−−

Tkfx

Tkgx1

Tkcxexp

Tkfx

Tkgxexp

Tkcxexp

Tkfxgxcxexp

B

4

B

3

B

2

B

4

B

3

B

2

B

432

⋅⋅⋅++++= !3

x!2

xx132

xe

Tk4C3gx B2=

linear dependence of Thigh T limit

Coefficient of linear expansion

31

Phonon-phonon scatteringphonon displaces atom which changes the force constant C (anharmonic terms)

scatter other phononsthree phonon process

Normal processes : all ks are in BZ

1st BZ ink-space

1k

2k3k

321 kkk =+

crystal momentum is conserved

Umklapp processes : k3 is outside BZ

1st BZ ink-space

1k

2k*3

k3k

G*

21 3kkk =+

*3 3

kGk =+

Gkkk 321 +=+crystal momentum is not conserved

“Folding over”

R. Peierls, Ann. Physik 3, 1055 (1929)

outside BZ

32

U-processes occur at high temperatures : require large k (ie. large ω)

How large ?

DBD

D

D

Θk21~ω

21~E

ω21~ω

k21~k

hkx

ky

kzDebye sphere

k≤kD

Phonon-phonon scattering : rate τ-1∝ # of phonons involvedU-process : τ-1∝ NU ~ exp(-ΘD/2T) (phonons w/. large k only)

at intermediate temperatures

At very low temperatures, phonons are populated at low k modeU process can not occur

33

Phonon mean free path l

( ∝ τ )

T (K)

Log-log plot

Exponential

Slope -1

Very low T ,l=vgτ =constant

Intermediate T ,l=vgτ ∝ (1/T)exp(1/T)dominated by U process

High T ,l=vgτ ∝ T-1

∝(number of phonons)-1

No distinction between N and U processes

34

Log-log plot of κ(T)

Below 5K, enriched Ge74 shows T3 dependence of κ

due to boundary scattering

At low temperatures, l → L (sample’s size)

Phonon propagation ~ ballisticκ =(1/3)vglCV ~ vgLCV

κ ∝ CV ∝ T3 Debye

At intermediate temperatures,κ=(1/3)vglCV =(1/3)vg

2τCV

/TΘ

BV

De~τ

3NkconstantC ==

/TΘDe~κU-processes

35

Thermal conductivity of LiF crystal bar w/. different cross sectional areasκ

(Wat

t m-1

K-1

)

Data show

1. Below 10K, κ ∝ T3

2. As temperature increases, κ increases and reaches a maximum around 18K.

3. Above 18K, κ decreases w/. increasing temperature and follows that exp(1/T).

4. Cross sectional area influences κbelow 20K. Bigger area crystal has, larger κ it has.

T(K)

36

Impurity scatterings

Defect scatteringsbreak periodicityOther effects

Log-log plot

Slope -1

Exponential

Slope 3

κ(Watt/m/K)

T (K)

37

Cu

Experimental results

Phys. Rev. 155, 619 (1967)

Phys. Rev. B7, 2393 (1967)

Solid line – Numerical calculationbased on experimental data ω(k)

Dashed line -- Numerical fit w/. ωD = 4.5× 1013 rad./secΘD = 344K

38

Summary of part (I)

Solids are defined by their capacity to be solid –to resist shear stress

A crystal is truly solid (as opposed to a glass which is just a “slow liquid”)

Crystalline order is defined by the regular positions of the nucleicrystal structure = lattice + basis

Lattice and reciprocal lattice Diffraction and experimental studiesBrillouin zone

Crystal binding Type of bindingElastic constants and elastic waves

39

Summary of part (I)

Vibrations of atoms Harmonic approximation

Quantization of vibrations phonons act like particles-- can be created or destroyed by inelastic scatterings

Thermal properties Fundamental law of probabilitiesPlanck distribution for phonons

Heat capacity : CLow T, C ∝ T3 and High T, C ~ constant

Thermal conductivity : κmaximum as function of T