Chapter 5.Chapter 5. Phonons II. Thermal Properties

Thermal properties of phonons• As mentioned before, we

F t l C T βT3are now going to look at how what we know about phonons will lead us to a

• For a metal, C = γT+βT3

electronic termphonons will lead us to a description of the heat capacity C as a function of C/T

lattice termAt low temperaturesy

temperature. • What we have to do is to

establish the rules we needC/T=γ+βT2

establish the rules we need to count how many phonons are active at a

Slop = β

Y axis intercept =phonons are active at a certain temperature, and then figure out how much energy goes into each

T2

Y-axis intercept = γ

energy goes into each. C=dU/dT

Thermal phononsPhonons : 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 transducerExternal perturbations vibrations or sound transducer

• Scattering of particles – energy transferred into lattice vibrations

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

Thermal phonons : consider a system with energy level En

Probability of occupancy

at temperature T

“B l f ”

EnEn-1

nn

B

EP(E ) expk T

⎛ ⎞∝ −⎜ ⎟

⎝ ⎠“Boltzmann factor”En-3

1⎛ ⎞Phonon number average

Excitation level amplitude (n) for w/. energymode k, ω

( ) sEs e x pP E⎛ ⎞⎜ ⎟∑∑Average of phonons

1n ω2

⎛ ⎞+⎜ ⎟⎝ ⎠

( )

( )s

s Bs

s ss

s e x p s P E k Tn

P E Ee x pk

−⎜ ⎟⎝ ⎠= =

⎛ ⎞−⎜ ⎟

∑∑∑ ∑

g p

ss B

pk T

(s 1 /2 ) ωs e x pk T

⎜ ⎟⎝ ⎠

⎛ ⎞+−⎜ ⎟

⎝ ⎠

∑

∑s Bk T

( s 1 /2 ) ωe x p

k T

⎜ ⎟⎝ ⎠=

⎛ ⎞+−⎜ ⎟

⎝ ⎠

∑

∑

( )

s B

s

k Td e x p s x

ωd x w h e re x

⎝ ⎠

− −= =

∑( ) B

s

w h e re xe x p s x k T

= =−∑

Planck distribution of < n(ω, T) > 1T)n(ω =( , )

average number of phonons excited per mode at ω

1Tkωexp

T),n(ω

B

−⎥⎦

⎤⎢⎣

⎡=

4 0

Thermal energy

3.0

3.5

4.0

High T ( kBT >> ω )k T

2.0

2.5

ω, T

) >

< n(ω, T) > ~ Bk Tω

0 5

1.0

1.5

< n(

ω

Low T ( kBT < ω )

( T) Bk T⎛ ⎞⎜ ⎟

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.00.0

0.5

1

< n(ω, T) > ~ Bexpω

⎛ ⎞−⎜ ⎟⎝ ⎠

x-1 = kBT / ω

mode

i i k,p kpi k

U n ω n ω= =∑ ∑∑Thermal energy: density of modes per polarizationi p k

pp

dω D (ω ) n(ω ) ω

ω

= ∑∫

∫

per polarization

polarizationp

p

B

ω dω D (ω )ωexp 1

k T

=⎛ ⎞

−⎜ ⎟⎝ ⎠

∑∫ thermal equilibriumpolarization

1D density of states D(k)Density of states (modes) : uniform in k-space

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

D(k)dk number of states from k to k+dkA linear chain of length L carries N+1 particles with separation a.

0 s=N

us(t)=u exp[-iωk pt] sin(ska) where k=π 2π 3π 4π (N-1)π, , , , ... L L L L L

Fixed-end boundary conditions : u0(t)=0 and uN(t)=0s=0

us( ) u e p[ ωk,p ] s (s a) e e L L L L L

Why is there no Nπ/L for allowed k? us(t) ∝ =0( )sN asin sin sLπ π⎛ ⎞ =⎜ ⎟

⎝ ⎠

1 mode/mobile atom

πΔkL

= L π for kπ a

⎧ ≤⎪⎪⎨

No motion of any atomOne mode for each interval

Th b f d it f k

L⎝ ⎠

π aD(k)π0 for ka

⎪= ⎨⎪ >⎪⎩

The number of modes per unit range of k

Periodic boundary conditions

Another way for enumerating modes as N b l i th f i dibecomes large is the use of periodic boundary conditions.

Unbounded medium but w/.Periodic boundary conditionsu(sa)=u(sa+L) for a large L=Na system( ) ( ) g y

us(t)=u exp[ i(ska-ωk pt) ] where k= 2π 4π 6π N0, , , , ... ,L L L L

π± ± ±s( ) p[ ( k,p ) ] L L L L

2πΔkL

=One mode for each intervalL π πL π πD(k) for k2π a a

0 otherwise

= − ≤ ≤

=

The number of modes per unit range of k

The number of modes is still equal to the number of mobile atoms (N in this case instead of N-1; but as N gets large, this does not matter).

Density of state D(ω)LD(k)dk dk and 2 for of k2π

dk

= ±

dkD(ω )dω 2D(k)dk 2D(k) dωdω

= =

2D(k)2D(k)dk

The number of modes per unit frequency range

gv2D(k)

/dkdω2D(k)

dωdkD(k)2) ω D( ===

Dispersion relation

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

D(ω) for 1D monatomic lattice 4C kaω sinM 2

⎛ ⎞= ⎜ ⎟⎝ ⎠

( ) 2max

2 2

2 Na/2π ω2D(k) N MD(ω ) dω /dk π C ω ω4C a ka

M 2

⎝ ⎠

= = =⎛ ⎞ maxdω /dk π C ω ω4C a kacos

M 2 2−⎛ ⎞

⎜ ⎟⎝ ⎠

D(ω)D(k) D(ω)D(k)

L/2π

ωk

(N/π)(M/C)1/2

k-π/a π/a0 (4C/M)1/20

Total number of modes

2 M4C

ωπ/a

)dω ω D()dω ω D(Na

2π2πLdkD(k)N

M

0

ω

0

π/a

π/a

max

∫∫∫ =====−

In two dimensions :

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

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

whence k k = 2π 4π 6π Nπ0 ± ± ±whence kx, ky = 0, , , , ... ,L L L L

± ± ±

• One mode per unit area in k-space2 2

2x y

1 2π LΔk Δk L 4π

−⎛ ⎞= =⎜ ⎟⎝ ⎠x y ⎝ ⎠

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

21 L

ky

22

x y 2x y

1 LD(k)d k dk dk 2π k dkΔk Δk 4π

= =

• The number of modes per unitk

The number of modes per unit frequency range

1k2AD(k))D(

kX

g2 v

k2π4π /dkdω

( )) ωD( ==

1VD(k)

g

23 v

1k 4π8πV

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

Density of states for continuum waveD(ω)dω = D(k)d3k for each polarization

complicated ! -- must map out dispersion relation and count all k-values with each frequency

Continuum waves: ω = vgk depending only on amplitude of k3D( ) d D(k) d k3

23

D( ω ) dω D(k) d kV 4π k dk

8

=

= 3

2

2

8π

V ω dω ⎛ ⎞

= ⎜ ⎟⎜ ⎟2g g

2

2π v v

V ω dω

⎜ ⎟⎜ ⎟⎝ ⎠

= 2 3g

dω2π v

=

2VThe number of modes per unit frequency range for each polarization

3g

2

2 vω

2πV) ω D( = a quadratic dependence !

Debye model1VD(k)The actual density of state can be

⎧

g

23 v

1k 4π8πV

/dkdωD(k)) ω D( ==

• The actual density of state can be obtained from dispersion curve but it is usually very complicated

• Peter Debye made two approximations:– continuum elastic phonon mode

(lo energ ) = k

2

D2 3g

Vω , ω ω2π vD(ω )

0

⎧≤⎪= ⎨

⎪⎩(low energy) ω = vgk

– only up to some cutoff frequency ωD

D0 , ω ω ,⎪ >⎩

N b f h d f h 32V V• Number of phonon mode for each polarization is equal to N

1 / 32 36 N⎛ ⎞

D32D

2 3 2 30 0g g

ωV ω VN= D( ω )d d2π v 6π v

Dω ωω ω= =∫ ∫

1 / 3

2 3g

D

6π v NV

ω

⎛ ⎞

⎛ ⎞= ⎜ ⎟⎜ ⎟

⎝ ⎠Debye frequencyω

ωD v 1 / 32

DD

g

ω 6π Nkv V

⎛ ⎞= = ⎜ ⎟

⎝ ⎠Debye wavevector

D vg

DD

Bkω

Θ =Debye temperaturekkD0

Density of states of the Debye modelPeaks at high ω-- cutoff of ωmax in

Debye solid

Actual crystal

max different k direction

Quadratic

solid

Debyeat low ωω = vgk

Debye solid

ω2

ωD

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

1 / 32 3⎛ ⎞

D32D

2 3 2 30 0g g

ωV ω VN= D( ω )d d2π v 6π v

Dω ωω ω= =∫ ∫

1 / 32 3g

D

6π v Nω

V⎛ ⎞

= ⎜ ⎟⎜ ⎟⎝ ⎠

Cutoff frequency

3

2

2

ω2V)ωD( =

1 / 32D

Dg

ω 6π Nkv V

⎛ ⎞= = ⎜ ⎟

⎝ ⎠Cutoff wavevector

3g

2 v2π)(

∫Thermal energy of the Debye model

U dω D(ω ) n(ω ) ω=

⎛ ⎞⎜ ⎟

∫For each polarization:

Dω 2

2 3g0

Vω ω dω 2π v ωexp 1

⎜ ⎟⎜ ⎟= ⎜ ⎟⎛ ⎞

−⎜ ⎟⎜ ⎟⎜ ⎟

∫B

exp 1k T

⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

• There are three polarizations : 2 transverse + 1 longitudinalThere are three polarizations : 2 transverse 1 longitudinal• Assume all polarization have the same energy dependence

⎛ ⎞⎜ ⎟D 3ω

2 3g 0

3V ωU dω 2π v ω 1

⎜ ⎟⎜ ⎟= ⎜ ⎟⎛ ⎞⎜ ⎟⎜ ⎟

∫

D 3

g 0

B

4 x

ωexp 1k T

k T3V x ω

⎛ ⎞−⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞ ∫ ( )3

B2 3

g B0

k T3V x ω dx where x2π v exp x 1 k T

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠∫

The Debye temperature ΘD1/3 1/3⎛ ⎞1/3 1/32 3 2

g gDD

B B B

6π v N vω 6π Nk k V k V

⎛ ⎞ ⎛ ⎞Θ = = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

Define

Therefore xD= ωD /kBT= ΘD/T

D 33 Θ /T

BT xU 9Nk T dx

⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟ ⎜ ⎟∫

The total phonon energy

( )BD 0

U 9Nk T dx exp x 1⎜ ⎟⎜ ⎟ ⎜ ⎟Θ −⎝ ⎠ ⎝ ⎠

∫

In classical model : equipartition theorem (kBT/2 for each excitation mode)

3 translational + 3 vibrational modes : six degrees of freedom

U = N 6 (kBT/2) = 3 N kBT for N atoms in the crystal

Cv = 3NkB Dulong and Petit Law

Heat capacity CV

( )Dω 3

2 3g B0

VV

3V ωdω2π v T exp ω/k T 1

UC T

⎛ ⎞ ⎡ ⎤∂= ⎜ ⎟ ⎢ ⎥⎜ ⎟ ∂ −⎢ ⎥⎣ ⎦⎝ ⎠

∂=

∂ ∫

( )( )( )

Dω 4B

22 3 2B 0

ω exp ω/k T3V 1 dω 2π v k T exp ω/k T 1

⎣ ⎦⎝ ⎠⎛ ⎞ ⎛ ⎞⎛ ⎞= − −⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠ −⎝ ⎠⎝ ⎠

∫ ( )( )

( )D

g B

3 x 4 x

B 2

B02π v k T exp ω/k T 1

T x e 9Nk dx

⎝ ⎠ −

⎛ ⎞⎜ ⎟

⎝ ⎠⎝ ⎠

= ∫ ( )B 2xD 0 e 1

⎜ ⎟Θ⎝ ⎠ −∫

( )3

D3x 11xd

3D

⎟⎞

⎜⎛ Θ

→⎟⎟⎞

⎜⎜⎛

∫At high T limit x=ΘD /T <<1 ( ) DDx

0 T3x

3

1e dx ⎟

⎠⎞

⎜⎝⎛=→⎟⎟

⎠⎜⎜⎝ −∫

( )D 4

3x x3 Dx e 1 1d Θ⎛ ⎞

⎜ ⎟∫

At high T limit, x ΘD /T <<1

i.e., For T >> ΘD, U 3NkBT( )

( )3 DD2x

0

x e 1 1dx x3 3 Te 1

Θ⎛ ⎞→ = ⎜ ⎟⎝ ⎠−

∫

CV 3NkB Dulong and Petit value

The Einstein model

1.0

high T CV 3NkBclassical model • Einstein model (1907) :

0.6

0.8

Nk B)

g V B

Ei t i d l

( )N identical oscillators of frequency ω0

• The Einstein model is

0.4

CV

(3N Einstein model The Einstein model is

often used to approximate the optical phonon part of the

0.0

0.2 low T CV 3NkBe- ω/kT

phonon part of the phonon spectrum.

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

x-1 = kBT / ω

3N ω( )

( )2

0B

00

0

e p ω /k TU

D( )= N ( - 3N ω U 3Nω ); n ωexp ω T 1k/

ω δ ω

⎛ ⎞∂

=−

=

Einstein model( )

( )( )0 B0

V B 2V B 0 B

exp ω /k TωUC 3NkT k T exp ω / T 1k

⎛ ⎞∂= = ⎜ ⎟∂ −⎠⎝

The Einstein model

high T CV 3NkB

l i l d l

Experimental valuesof Dimond

classical model

Ei t i d lThe Einstein model is often used to approximate the optical

Einstein model

approximate the optical phonon part of the phonon spectrum.

low T CV 3NkBe- ω/kT

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

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

At low T, CV → 3NkBexp(- ω/kBT)At low T, CV → 3NkBexp( ω/kBT)

Experimental data for other materials show T3 dependence of CV instead

Comparison between the Ei t i d D b d lEinstein and Debye models

At l t t thAt low temperatures, the Debye model gives the phonon contribution to the

CDebyeEinstein

heat capacity its experimentalT3 dependence form.

Copper

Actual density of states D( ) for silicon

2Cω

D(ω) for silicon

Dispersion curve for two

1M2C

2M2C

M1 >M2

curve for two atoms per primitive b i

π/a- π/a kbasis

Debye T3 modelωD, ΘD depend on vg, n=N/V, ~ vgn1/3

High for stiff light materialsKittel : Table 1 in ch.5 (P.116)

High for stiff, light materials

material A Cu Ag Au Pb

ΘD(K) 428 343 225 165 105D 3

4 xBk T3V xU dx

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟ ∫B

2 3 xg 0

U dx 2π v e 1

= ⎜ ⎟⎜ ⎟ −⎝ ⎠ ⎝ ⎠∫

D3 x 4 xT x e⎛ ⎞

3 4

x0

x πdx e 1 15

∞ ⎛ ⎞=⎜ ⎟−⎝ ⎠

∫

( )V B 2xD 0

T x eC 9Nk dx e

1

⎛ ⎞= ⎜ ⎟Θ⎝ ⎠ −

∫0 ⎝ ⎠

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

TTNk12πTNk3π33444 ⎞⎛ Debye T3

T234Nk5Θ

TNk12πC and 5Θ

TNk3πUD

B3D

BV3

D

B⎟⎟⎠

⎞⎜⎜⎝

⎛Θ

≅≅≅Debye T

approximation

Debye temperature1/3 1/32 3 2

g gDD

6π v N vω 6π Nθk k V k V

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠B B Bk k V k V⎜ ⎟ ⎝ ⎠⎝ ⎠

Usually, a harder material has a higher Debye temperature

Debye T3 dependence

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.

C/T=γ+βT2

e-1 K

-2) KCl Phys. Rev. 91, 1354

(1953) Slop = βY-axis intercept = γ

ule

Mol

eC

V/T

(Jou

Low Temperature

C

CuSolid State Physics (1963)

T2 (K2)

k

Simple reasoning for T3 dependenceTotal phonon mode : ω ≤ ωD (or k ≤ kD= )

kyE it d h d

BD

g

k Θv Thermal wavevector

kD

k

Excited phonon mode

ω ≤ kBT/ (or k ≤ ω/vg = = kT )B

g

k Tv

kT

kx Others are frozen out

Fraction excited at T :

g

3 3

T

D D

k Tk Θ

⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠of the total volume in k-space

D D⎝ ⎠ ⎝ ⎠

Each mode has energy kBTEach mode has energy kBT

U ~3

BT3N k T

⎛ ⎞⎜ ⎟Θ⎝ ⎠

CV ~ ∝ T3

DΘ⎝ ⎠3

BT12Nk

⎛ ⎞⎜ ⎟Θ⎝ ⎠

V

too small but correct T3 dependence

BD

⎜ ⎟Θ⎝ ⎠

Success of Debye theory

high T CV 3NkB

NaCl

diamond

low T CV 0

• CV may appear different in the V y ppabove figure,

• But after rescaling the temperature by the Debyetemperature by the Debye temperature θD, which differs from material to material, CVof many materials can fit onof many materials can fit on the same cure, i.e., a universal behavior emerges.

General result for D(ω)the number of states between ω and ω+dω

33

shell

LD(ω )dω d k2π

⎛ ⎞= ⎜ ⎟⎝ ⎠ ∫

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

3d k dS dk∫ ∫dk⊥3

ωshell

d k dS dk⊥=∫ ∫ω+dω

⊥

ω dk dω∇ =ωkω dk dω⊥∇ =

dω dωdS dk dS dS= = dωdSL)dωD(ω ω3

∫⎟⎞

⎜⎛=ω ω ω

k g

dS dk dS dSω v⊥ = =

∇dω

v2π)dωD(ω

g∫⎟

⎠⎜⎝

=

kz

∫= ωdSV)D(ωdSw

( ) ∫=g

3 v2π) D(ω

kxky

Van Hove singularitiesPeaks at high ω-- cutoff of ωmax in diff t k di ti

Debye solid

Actual crystal

different k direction

Quadratic t l Debyeat low ω

ω = vgk

Debye solid

ω2

ωD

1 / 32 36π v N⎛ ⎞∫

dSVgD

1 / 32

6π v Nω

V

6 N

⎛ ⎞= ⎜ ⎟⎜ ⎟

⎝ ⎠

⎛ ⎞

Cutoff frequency( ) ∫=

g

ω3 v

dS2πV)D(ω

• Critical points at which vg=0 produce singularities know as Van Hove singularities)

2D

Dg

ω 6π Nkv V

⎛ ⎞= = ⎜ ⎟

⎝ ⎠Cutoff wavevector

as Van Hove singularities).• Discontinuities develop at

singular points

Dispersion and density of states of Cu

Experimental results

Phys. Rev. 155, 619 (1967)

Phys. Rev. B7, 2393 (1967)

dSV

C( ) ∫=

g

ω3 v

dS2πV) D(ω

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

Cu

based on experimental data ω(k)

Dashed line -- Numerical fit w/. 4 5 1013 d /ωD = 4.5× 1013 rad./sec

ΘD = 344K

Thermal properties (Review)Lattice vibrations : mode (k,ω )

k is in BZ, discrete,ω(k) dispersion relationD(ω ) density of statesE(ω ) = (n+1/2) ωE(ω ) = (n+1/2) ω

Phonons : number n energy = ω1Phonons : number n energy = ω

crystal momentum kBω/k Tn

e 1=

−

Thermal properties (equilibrium)

ω) n(ω ) D(ω dωU ∫=thermal energy

heat capacity dUheat capacity

dTdUCV =

Transport propertiesConduction of sound and heat through the crystal (non-equilibrium)

vibration energygy

Ultrasonic attenuation Thermal conductionexcite single phonon mode apply temperature gradient measure decay of amplitude measure heat current by phonons

Phonon thermal conductivity

Apply temperature gradient ∇T → determine heat current density jUpp y p g y jU

TH TL∇T

jU

UdTj κdx

= −The flux of the thermal energythe energy transmitted across unit area per unit time

κ : thermal conductivity coefficient

Anharmonic effects• In the theory we have covering so far for phonons,

we have assumed that the potential energy kx2/2we have assumed that the potential energy kx2/2 (harmonic, Hook’s law)

• However real systems have anharmonic effectsHowever, real systems have anharmonic effects.• In terms of physical phenomena, what assumes is

that :– Two lattice waves do not interact– A single wave does not decay or change form with time

Th i th l i f th l tti– There is no thermal expansion of the lattice– The force (elastic) constants between the atoms do not

change as a function of temperature or pressurechange as a function of temperature or pressure• In reality, all of these effects are real and they can

modify our dispersion curves and how we look at hphonons.

Propagations of phononsBallistic No interaction/scattering

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

independent plane waves and standing waves d

gdωv vdk

= =

Diffusion Phonons scatter, random walk through crystal

Phonons scatter in real crystalsPhonons scatter in real crystals.

Scattering processes : boundary scattering

defect scattering

phonon-phonon scattering

gdωv vdk

<< =

Thermal conductivityThe flux of thermal energy is based on that

the process of thermal energy transfer is a random process.p gy pie. the energy diffuses through the crystal, suffering frequent collisions.

J ΔT∝Ballistic : across the whole sampleU

U

J ΔTdTJ Tdx

∝

∝ ∇ =

Ballistic :

Diffusive :

across the whole sample

localdx

j

For diffusion, thermal conductivity is defined byphonon properties

( )UjκT

≡−∇

p p pscatteringcrystal quality (size, defect)temperature

the energy transmitted across unit area per unit timejU [Watt/m2]:

κ [(Watt/m2)/(K/m)] = [Watt/m/K]

p

consider phonons as gases contained in a crystal volumeKinetic theory of gases (phonons)consider phonons as gases contained in a crystal volumecalculate diffusion in the presence of temperature gradient

TH TL∇T

jH LjUn Fick’s law

dTdndn

xdxdT

dTdn

dxdn

=

t ti f l ln : concentration of moleculesC : heat capacity per unit volume = ncvg : phonon velocity

h f th ( di tx x

dT dTΔT v τdx dx

= =: phonon mean free path (average distance

that a lattice vibration travels before it collide with another phonon=vgτ

dx dx

dT 1 dTp g2 2

U xdT 1 dTj n v cτ n v cτdx 3 dx

= − = −U xj |v | c Tn= − Δ

2U g

1 dT 1 dTj nv cτ Cv3 dx 3 dxg= − = −g

1κ Cv3

=Thermal

conductivity

Anharmonic effects• In the theory we have covering so far for phonons,

we have assumed that the potential energy kx2/2we have assumed that the potential energy kx2/2 (harmonic, Hook’s law)

• However real systems have anharmonic effectsHowever, real systems have anharmonic effects.• In terms of physical phenomena, what assumes is

that :– Two lattice waves do not interact– A single wave does not decay or change form with time

Th i th l i f th l tti– There is no thermal expansion of the lattice– The force (elastic) constants between the atoms do not

change as a function of temperature or pressurechange as a function of temperature or pressure• In reality, all of these effects are real and they can

modify our dispersion curves and how we look at hphonons.

Anharmonic terms in binding potentialpotential

U

xo

The general shape applies for any type of binding

xo

( ) ( ) ( )2 3

2 3o o o o2 3

U 1 U 1 UU(x) U(x ) x x x x x x ...x 2 x 6 x

∂ ∂ ∂= + − + − + − +

∂ ∂ ∂o o ox x xx 2 x 6 x∂ ∂ ∂

( ) ( )2 3

2 3o o o2 3

x x

1 U 1 UU(x) U(x)-U(x ) x x x x ...2 x 6 x

∂ ∂Δ = = − + − +

∂ ∂o ox x

Reset the equilibrium, let displacement x-xo → x2 3 4U(x) cx gx fx=

harmonic term anharmonic term

U(x) cx gx fx ...= − −

Thermal expansion As a solid is heated up, the lattice expands (due to the average displacement of the atoms increasing with thermal energy which causes fluctuation of x from xo.o.

2 3 4 2 4cx gxdx exp[ ]k T k T

U(x) cx gx fxdx x exp[ ] dx x exp[ ]k T k T

∞ ∞ ∞−−

−− −∫ ∫ ∫

B B B2

B

B B

2 3 4

B

k T k Tcxdx

k T k Tx

U(x) cx gx fxdx exp[ ] dx exp[ ]k T k

exp[TT

]k

−∞ −∞∞ ∞

∞ ∞

−∞∞

∞

≈−

= ≅− −

− −

∫ ∫ ∫

∫ ∫ ∫B BB−∞ −∞ −∞

2 3 4 2 3 4cx gx fx cx gx fxexp exp expk T k T k T k T

⎛ ⎞ ⎛ ⎞ ⎛ ⎞− −− = − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠anharmonic term gives the net change of <x>B B B B

2 3 4

k T k T k T k T

cx gx fx exp 1k T k T k T

⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞⎛ ⎞

≅ − + +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⋅⋅⋅++++=

xxx132

xehigh T limit

the net change of <x>

B B Bk T k T k T⎝ ⎠⎝ ⎠ !3!21/2 5/2 3/2

BB1/2 1/2 2

(3π /4)(g/c )(k T) 3gx k T= = linear dependence of TB1/2 1/2 2B(π/c) (k T) 4c

ea depe de ce o

Coefficient of linear expansion

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

scatter other phonons

Normal processes : all ks are in BZthree phonon process

p

2k3k

1 2 3k k k+ =

T t l t f th h1st BZ ink-space

1k

3k Total momentum of the phonon gas is conserved by such collision

Umklapp processes : k3* is outside BZ “Folding over”

1st BZ in

2k*k

G 3

*1 2k k k+ = The only meaningful

phonon K’s lie in the 1st BZ ink-space

1k3

k3k

3

*3k G k+ = 1st BZ, so any

longer K must be brought back into the

outside BZ1 2 3k k k G+ = +

crystal momentum is not conserved

g1st BZ by adding a G.

Normal and umklapp processes

• Total momentum of

N-process

the phonon gas is conserved by such collisioncollision

• No thermal resistance.q3qq4

U-process

31 2 4qqq q G+ += =

A collision of two phonons both with a positive value of qx can by an U-process,by an U process, create a phonon with negative qx.

Umklapp processesU-processes occur at high temperature : require large k (ie. large ω)

How large ?kz

D b h D1k ~ kHow large ?

k

Debye spherek≤kD

D

D

21ω ~ ω2

kx

ky

D B D

21 1E ~ ω ~ k Θ2 2

Phonon-phonon scattering : rate τ-1∝ # of phonons involvedU process : 1 N k T exp( Θ /2T)U-process : τ-1∝ NU ~ kBT exp(-ΘD/2T)

(Boltzmann factor of phonons w/. large k only)at intermediate (low) temperatures

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

Phonon mean free pathLog-log plot

Very low T ,=vgτ =constant→ L (sample’s size)

Phonon mean free path Exponential Intermediate T ,

(1/T) (1/T)p

( ∝ τ ) =vgτ ∝ (1/T)exp(1/T)dominated by U process

Slope: -1Slope: -1 High T ,=vgτ ∝ T-1

∝ (number of phonons)-1

T (K)

No distinction between N and U processes

T-dependent thermal conductivityBelow 5K, enriched Ge74 shows

T3 dependence of κdue to boundary scatteringIsotope due to boundary scattering

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

Isotope effect

(sample s size)Phonon propagation ~ ballisticκ =(1/3)vg CV ~ vgLCVg g

κ ∝ CV ∝ T3 DebyeLog-log plot of κ(T)

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

2τCV

V BC constant 3Nk= =

/TΘ

B

DeTk

1~κ

U-processes

DΘ /T

B

1τ~ ek T

T-dependent thermal conductivityImpurity scatterings

break periodicityOther effectsDefect scatterings

break periodicity

Log-log plotSlope: 3

κ(Watt/m/K) Exponential

Slope: -1

T (K)

Thermal conductivity of LiF crystal barbar

Different cross sectional area

(a)1.33mm × 0.91mm

(b)7.55mm × 6.97mm

m-1

K-1

)

Data show

κ(W

att m 1. Below 10K, κ ∝ T3

2. As temperature increases, κ increases

κ and reaches a maximum around 18K.

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

4. Cross sectional area influences κbelow 20K Bigger area crystal has

T(K)

below 20K. Bigger area crystal has, larger κ it has.

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”)slow liquid )

Crystalline order is defined by the regular positions of the nucleicr stal str ct re lattice + basiscrystal structure = lattice + basis

Lattice and reciprocal latticeDiffraction and experimental studiesBrillouin zone

Crystal bindingType of bindingEl ti t t d l tiElastic constants and elastic waves

Summary of part (I)y p ( )Vibrations of atoms

Harmonic approximation

Quantization of vibrationsQuantization of vibrations phonons act like particles-- can be created or destroyed by inelastic scatteringsy y g

Thermal properties F d t l l f b biliti (B lt f t )Fundamental law of probabilities (Boltzmann factor)Planck distribution for phonons

• Heat capacity : Cp yLow T, C ∝ T3 and High T, C ~ constant

• Thermal conductivity : κmaximum; as function of Tmaximum; as function of T