Chapter 5.
Phonons II. Thermal Properties
Thermal properties of phonons
• As mentioned before, we
are now going to look at
how what we know about
phonons will lead us to a
description of the heat
capacityy C as a function of
temperature.
• What we have to do is to
establish the rules we need
to count how many
phonons are active at a
certain temperature, and
then figure out how much
energy goes into each
each.
• For
F a metal,
t l C = γT+βT
T βT3
electronic term
C/T
lattice term
At low temperatures
C/T=γ+βT2
Slop = β
Y axis intercept = γ
Y-axis
T2
C=dU/dT
Thermal phonons
Phonons : 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
Probability of occupancy
En
En-1
En-3
⎛ E ⎞
P(En ) ∝ exp ⎜ − n ⎟
⎝ kBT ⎠
“B l
“Boltzmann
ffactor””
at temperature T
Phonon number average
G
Excitation level amplitude (n) for mode k, ω
Average
g of p
phonons
n
=
∑ s P (E )
∑ P (E )
s
s
s
s
⎛
⎝
⎛
Es ⎞
∑s s e x p ⎜ − k T ⎟
B
⎝
⎠
=
⎛
Es ⎞
∑s e x p ⎜ − k T ⎟
B
⎝
⎠
⎛ (s + 1 /2 )= ω ⎞
∑s s e x p ⎜ −
⎟
k
T
B
⎝
⎠
=
⎛ (s + 1 /2 )= ω ⎞
∑s e x p ⎜ −
⎟
k
T
B
⎝
⎠
d
−
∑ ex p (− sx )
dx s
=
∑ ex p (− sx )
s
Planck distribution of < n((ω, T)) >
average number of phonons
excited per mode at ω
1⎞
2⎠
w/. energy ⎜ n + ⎟ =ω
n(ω , T) =
w h e re x =
1
⎡ =ω ⎤
exp ⎢
−1
⎥
⎣ k BT ⎦
=ω
k BT
Thermal energy
40
4.0
3.5
High T ( kBT >> =ω )
k BT
< n(ω, T) > ~
=ω
< n(ω, T) >
3.0
2.5
2.0
1.5
Low T ( kBT < =ω )
⎛ k BT ⎞
< n((ω, T) > ~ exp ⎜ −
⎟
⎝ =ω ⎠
1.0
05
0.5
0.0
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
-1
1
x = kBT / = ω
Thermal energy:
U =
mode
∑
i
n i =ωi = ∑∑ n k,p =ωkp
p
k
= ∑ ∫ dω D p (ω ) n(ω ) =ω
density of modes
per polarization
p
polarization
=
∑ ∫ dω D (ω )
p
p
=ω
⎛ =ω ⎞
exp ⎜
⎟ −1
k
T
⎝ B ⎠
thermal equilibrium
1D density of states D(k)
Density of states (modes) : uniform in k-space
1D
( )
y of states = number of states p
per unit k at k
D(k)≡density
D(k)dk number of states from k to k+dk
A linear chain of length L carries N+1 particles with separation a.
s=0
0
Fixed-end boundary conditions : u0(t)=0 and uN(t)=0
us((t)=u
) ue
exp[-iω
p[ ωk,p
s (s a)
k pt]] sin(ska)
s=N
π 2π 3π 4π
(N-1)π
,
,
,
,
...
where
e e k=
L L L L
L
1 mode/mobile atom
⎛ sN π a ⎞
Why is there no Nπ/L for allowed k? us(t) ∝ sin ⎜
⎟ = sin ( sπ ) =0
⎝ L No
⎠ motion of any atom
π
One mode for each interval Δk =
π
⎧L
L
≤
for
k
⎪⎪ π
a
Th number
The
b off modes
d per unit
it range off k D(k) = ⎨
⎪ 0 for k > π
⎪⎩
a
Periodic boundary conditions
Another way for enumerating modes as N
b
becomes
llarge iis th
the use off periodic
i di
boundary conditions.
Unbounded medium but w/.
Periodic boundary conditions
u(sa)=u(sa+L)
( ) (
)
for a large
g L=Na system
y
us((t)=u
) exp[
p[ i(ska-ω
(
)]
k pt)
k,p
where k=
0, ±
2π
4π
6π
Nπ
, ± , ± , ... ,
L
L
L
L
2π
L
L
D(k)
=
The number of modes per unit range of k
One mode for each interval Δk =
2π
= 0
π
π
≤k≤
a
a
otherwise
for −
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)
L
D(k)dk =
dk
2π
and 2 for ± of k
dk
D(ω )dω = 2D(k)dk = 2D(k)
dω
dω
The number of modes per unit frequency range
D( ω ) = 2D(k)
dk 2D(k) 2D(k)
=
=
dω dω /dk
vg
Dispersion relation
Singularity at vg=0, determined by ω(k)
D(ω) for 1D monatomic lattice
4C
⎛ ka ⎞
sin ⎜ ⎟
M
⎝ 2 ⎠
ω=
2D(k)
D(ω ) =
=
dω /dk
D(k)
2 ( Na/2π )
N M
=
π C
4C a
⎛ ka ⎞
cos ⎜ ⎟
M 2
⎝ 2 ⎠
D(ω)
ω2max
2
ωmax
− ω2
L/2π
(N/π)(M/C)1/2
-π/a
0
π/a
k
0
Total number of modes
π/a
L 22π
N = ∫ D(k)dk =
=N=
2π a
− π/a
ω
(4C/M)1/2
max
4C
M
0
0
∫ D( ω )dω = ∫ D( ω )dω
ω
In two dimensions :
• periodic boundary condition
condition, N2 primitive cells within a square of side L
exp[ i(kxx+kyy) ] = exp[ i( kx(x+L) + ky(y+L) ) ]
2π
4π
6π
Nπ
whence kx, ky = 0,
0 ± , ± , ± , ... ,
L
L
L
L
−2
1
L2
⎛ 2π ⎞
• One mode per unit area in k-space
=⎜ ⎟ = 2
Δk x Δk y ⎝ L ⎠
4π
• Number of modes with wavevector
from k to k+dk in k-space
1
L2
2
D(k)d k =
dk x dk y = 2 2π k dk
Δk x Δk y
4π
• The number of modes per unit
frequency range
D( ω ) =
ky
k
D(k)
( )
A
1
= 2 22π k
dω /dk 4π
vg
In three dimensions :
D(k)
V
2 1
D( ω ) =
= 3 4π k
dω /dk 8π
vg
kX
Density of states for continuum wave
D(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 k
D( ω ) ddω = D(k) d 3 k
=
V
2
4π
k
dk
3
8
8π
V ⎛ω
= 2⎜
2π ⎜⎝ v g
2
⎞ dω
⎟⎟
⎠ vg
V ω2
= 2 3 dω
2π v g
The number of modes per unit frequency range for each polarization
V ω2
D( ω ) = 2 3
2π v g
a quadratic dependence !
Debye model
• 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
(low
energy)) ω = vgk
– only up to some cutoff frequency ωD
D( ω ) =
D(k)
V
1
= 3 4π k 2
dω /dk 8π
vg
⎧ Vω2
⎪ 2 3 , ω ≤ ωD
D(ω ) = ⎨ 2π vg
⎪0
, ω > ωD ,
⎩
2
ωD
ωD V ω
V ω3D
•N
Number
b off phonon
h
mode
d ffor each
h
N= ∫ D( ω )dω = ∫
dω = 2 3
0
0 2π 2 v 3
polarization is equal to N
6π v g
g
ω
ω
D
0
⎛ 66π v N ⎞
=⎜
⎟⎟
⎜
V
⎝
⎠
2
Debye frequency
vg
kD
ωD
3
g
1/ 3
ω D ⎛ 6π 2 N ⎞
Debye wavevector k D =
=⎜
⎟
vg
V
⎠
⎝
k Debye temperature Θ = = ω D
D
kB
1/ 3
Density of states of the Debye model
Peaks at high ω
-- cutoff of ωmax in
different k direction
Debye
solid
Actual
crystal
Quadratic
at low ω
ω = vgk
Debye
solid
ω2
ωD
N primitive cells in the crystal
crystal, A total number of acoustic phonon mode is
2
ωD
ωD V ω
V ω3D
N for each polarization
N= ∫ D( ω )dω = ∫
0
Cutoff frequency
ωD
Cutoff wavevector k D
⎛ 6π 2 v 3g N
=⎜
⎜
V
⎝
0
⎞
⎟⎟
⎠
1/ 3
ω D ⎛ 6π 2 N ⎞
=
=⎜
⎟
vg
V
⎝
⎠
1/ 3
2
2π v
3
g
dω =
6π 2 v3g
V ω2
D(( ω ) = 2 3
2 vg
2π
Thermal energy of the Debye model
U = ∫ dω D(ω ) n(ω ) =ω
For each polarization:
⎛
⎞
⎟
ωD
2 ⎜
Vω ⎜
=ω
⎟
= ∫ dω
2 3 ⎜
2π
vg
⎛ =ω ⎞ ⎟
0
⎜⎜ exp ⎜
⎟ − 1 ⎟⎟
⎝ k BT ⎠ ⎠
⎝
• There are three polarizations : 2 transverse + 1 longitudinal
• Assume all polarization have the same energy dependence
⎛
⎞
⎜
⎟
ωD
3
3V=
ω
⎜
⎟
U = 2 3 ∫ dω
⎜
2π v g 0
⎛ =ω ⎞ ⎟
⎜⎜ exp ⎜
⎟ − 1 ⎟⎟
⎝ k BT ⎠ ⎠
⎝
3V= ⎛ k T ⎞
= 2 3⎜ B ⎟
2π v g ⎝ = ⎠
4
⎛
⎞
x3
∫0 dx ⎜⎜ exp ( x ) − 1 ⎟⎟
⎝
⎠
xD
where x =
=ω
k BT
The Debye temperature ΘD
Define
1/ 3
=ωD
= ⎛ 6π v N ⎞
ΘD =
=
⎜⎜
⎟⎟
kB
kB ⎝ V ⎠
2
3
g
1/ 3
=v g ⎛ 6π 2 N ⎞
=
⎜
⎟
kB ⎝ V ⎠
Therefore xD= =ωD /kBT= ΘD/T
The total phonon energy
⎛ T ⎞
U = 9Nk BT ⎜
⎟
Θ
⎝ D⎠
3
ΘD /T
∫
0
⎛
⎞
x3
dx ⎜⎜
⎟⎟
⎝ exp ( x ) − 1 ⎠
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
Cv = 3NkB
for N atoms in the crystal
Dulong and Petit Law
Heat capacity CV
⎛ 3V=
∂U
C V=
=⎜ 2 3
∂T V ⎜⎝ 2π v g
⎛ 3V=
=⎜ 2 3
⎜ 2π v
g
⎝
⎞ ∂ ⎡ ωD
⎤
ω3
⎟⎟
⎢ ∫ dω
⎥
exp ( =ω/k BT ) − 1 ⎦⎥
⎠ ∂T ⎢⎣ 0
⎞ ⎛ = ⎞ ⎛ 1 ⎞ ωD
ω4 exp ( =ω/k BT )
⎟⎟ ⎜ − ⎟ ⎜ − 2 ⎟ ∫ dω
2
k
T
⎝
⎠
exp
=
ω/k
T
−
1
0
( ( B ) )
⎠ ⎝ B⎠
⎛ T ⎞
= 9Nk B ⎜
⎟
Θ
⎝ D⎠
3 x
D
∫ dx
0
x 4ex
(e
x
− 1)
At high T limit,
limit x
x=ΘD /T <<1
2
xD
⎛ x3 ⎞
1
1 ⎛ ΘD ⎞
3
⎜
⎟
(
)
x
d
dx
→
=
∫0 ⎜⎝ e x − 1 ⎟⎠ 3 D 3 ⎜⎝ T ⎟⎠
xD
d
∫ dx
0
i.e., For T >> ΘD, U
CV
3NkBT
3NkB
x e
4
(e
x
x
− 1)
2
→
3
1
1 ⎛ ΘD ⎞
3
x
=
( D)
⎜
⎟
3
3⎝ T ⎠
Dulong and Petit value
3
The Einstein model
1.0
classical model
high
g T CV
CV (3N
NkB)
0.8
0.6
3NkB
Ei t i model
Einstein
d l
0.4
0.2
low T
CV 3NkBe-=ω/kT
0.0
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
• Einstein model ((1907)) :
N identical oscillators of
frequency ω0
• The Einstein model is
often used to
approximate the optical
phonon part of the
phonon spectrum.
2.0
-1
x = kBT / = ω
Einstein model
3N=ω0
D(ω )= Nδ (ω - ω0 ); U = 3N n =ω0 =
exp ( =ω0 /k B T ) − 1
e p ( =ω0 /k BT )
exp
⎛ =ω0 ⎞
∂U
CV =
= 3Nk B ⎜
⎟
2
∂T V
k
T
⎝ B ⎠ ( exp ( =ω0 /k B T ) − 1)
2
The Einstein model
Experimental values
of Dimond
Ei t i model
Einstein
d l
low T CV
high T CV 3NkB
classical
l
i l model
d l
The Einstein model is
often used to
approximate the optical
phonon part of the
phonon spectrum.
3NkBe-=ω/kT
Einstein model ((1907)) : N identical oscillators of frequency
q
yω
At high T, CV → 3NkB same as the Dulong and Petit value
At low T, CV → 3NkBexp(
exp(-=ω/k
=ω/kBT)
Experimental data for other materials show T3 dependence of CV instead
Comparison between the
Ei t i and
Einstein
d Debye
D b models
d l
C
Copper
Debye
Einstein
At low
l
t
temperatures,
t
the
th
Debye model gives the
phonon contribution to the
heat capacity its experimental
T3 dependence form.
Actual density of states
D( ) for silicon
D(ω)
Dispersion
curve for two
atoms per
primitive
b i
basis
- π/a
ω
M1 >M2
2C
M2
2C
M1
π/a k
Debye T3 model
ωD, ΘD depend on vg, n=N/V,
~ vgn1/3
Kittel : Table 1 in ch.5 (P.116)
High for stiff,
stiff light materials
material
AA
Cu
Ag
Au
Pb
ΘD(K)
428
343
225
165
105
3V= ⎛ k T ⎞
U = 2 3⎜ B ⎟
2π v g ⎝ = ⎠
⎛ T ⎞
C V = 9Nk B ⎜
⎟
Θ
⎝ D⎠
4
xD
⎛ x3 ⎞
∫0 dx ⎝⎜ ex − 1 ⎠⎟
3 x
D
∫ dx
0
3π Nk BT
U≅
5Θ 3D
4
(e
x
− 1)
2
xD= ΘD/T → ∞
⎛ T ⎞
12π Nk BT
⎟⎟
≅ 234Nk B ⎜⎜
CV ≅
3
5Θ D
⎝ ΘD ⎠
4
and
⎛ x3 ⎞
π4
∫0 dx ⎜⎝ ex − 1 ⎟⎠ = 15
x 4ex
At very low temperature, T<<ΘD,
4
∞
3
3
Debye T3
approximation
Debye temperature
1/ 3
2 3
=ω D
= ⎛ 6π v g N ⎞
=
θD =
⎜
⎟
kB
k B ⎜⎝ V ⎟⎠
1/ 3
=v g ⎛ 6π 2 N ⎞
=
⎜
⎟
kB ⎝ 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.
CV/T (Jou
ule Mole
e-1K-2)
C/T=γ+βT2
Slop = β
Y-axis intercept = γ
KCl
Cu
T2 (K2)
Phys. Rev. 91, 1354
(1953)
Low Temperature
Solid State Physics
(1963)
Simple reasoning for T3 dependence
kB
ΘD )
Total phonon mode : ω ≤ ωD (or k ≤ kD=
=v g
Thermal wavevector
ky
E it d phonon
Excited
h
mode
d
kD
k
ω ≤ kBT/= (or k ≤ ω/vg = B T = kT )
kx
kT
Others are frozen out
Fraction excited at T :
=v g
3
⎛ kT ⎞ ⎛ T ⎞
⎜ ⎟ =⎜
⎟
k
Θ
⎝ D⎠ ⎝ D⎠
3
of the total volume in k-space
Each mode has energy kBT
3
⎛ T ⎞
3N
U~
⎜
⎟ k BT
⎝ ΘD ⎠
⎛ T ⎞
12Nk
CV ~
⎟
B⎜
Θ
⎝ D⎠
3
∝ T3
too small but correct T3 dependence
Success of Debye theory
NaCl
high T CV
3NkB
diamond
low T CV
0
• CV may
y appear
pp
different in the
above figure,
• But after rescaling the
temperature by the Debye
temperature θD, which differs
from material to material, CV
of many materials can fit on
the same cure, i.e., a
universal behavior emerges.
General result for D(ω)
⎛ L⎞
D(ω )dω = ⎜ ⎟
⎝ 2π ⎠
3
∫
d3k
the number of states between ω and ω+dω
shell
dSω : an element of area on the surface in K space of selected
constant frequency ω.
dk⊥
3
d
k
=
dS
dk
∫
∫ ω ⊥
shell
∇ k ω dk ⊥ = dω
dω
dω
dSω dk ⊥ = dSω
= dSω
∇k ω
vg
ω+dω
ω
⎛ L⎞
D(ω )dω = ⎜ ⎟
⎝ 2π ⎠
kz
V dSω
D(ω ) =
(2π)3 ∫ vg
dSw
kx
ky
3
dSω
∫ vg dω
Van Hove singularities
Actual
crystal
Peaks at high ω
-- cutoff of ωmax in
diff
different
t k direction
di ti
Debye
solid
Quadratic
att low
l
ω
ω = vgk
Debye
solid
ω2
ωD
⎛ 6π v N ⎞
=⎜
⎟⎟
⎜
V
⎝
⎠
2
Cutoff frequency
ωD
Cutoff wavevector k D =
3
g
1/ 3
ω D ⎛ 6π
6 N⎞
=⎜
⎟
vg
V
⎝
⎠
2
V dSω
D(ω ) =
(2π)3 ∫ vg
1/ 3
• Critical points at which vg=0
produce singularities know
as Van Hove singularities)
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)
V dSω
D(ω ) =
3 ∫
(2π) vg
C
Cu
Solid line – Numerical calculation
based on experimental data ω(k)
Dashed line -- Numerical fit w/.
ωD = 4.5×
4 5 1013 rad./sec
d/
ΘD = 344K
Thermal properties (Review)
Lattice vibrations : mode (k,ω )
k is in BZ,, discrete
ω(k) dispersion relation
D(ω ) density of states
E(ω ) = (n+1/2) =ω
Phonons : number n
n =
1
e
=ω/k B T
energy = =ω
−1
crystal momentum =k
Thermal properties (equilibrium)
thermal energy
U = ∫ dω D(ω ) n(ω ) =ω
heat capacity
dU
CV =
dT
Transport properties
Conduction of sound and heat through the crystal (non-equilibrium)
vibration
energy
gy
Ultrasonic attenuation
excite single phonon mode
measure decay of amplitude
Thermal conduction
apply temperature gradient
measure heat current by phonons
Phonon thermal conductivity
Apply
pp y temperature
p
g
gradient ∇T → determine heat current density
y jU
TH
∇T
jU
The flux of the thermal energy
jU = − κ
TL
dT
dx
the 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 kx2/2
(harmonic, Hook’s law)
• However,
However real systems have anharmonic effects
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
There
is
i no thermal
th
l expansion
i off th
the llattice
tti
The force (elastic) constants between the atoms do not
change 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
phonons.
h
Propagations of phonons
Ballistic
No interaction/scattering
In harmonic approximation in perfect, infinite crystal,
Expect no scattering → phonon modes are uncoupled,
independent plane waves and standing waves
dω
d
v = vg =
dk
Diffusion
Phonons scatter, random walk through crystal
Phonons scatter in real crystals.
crystals
Scattering processes :
9
9
9
boundary scattering
defect scattering
phonon-phonon scattering
dω
v << v g =
dk
Thermal conductivity
The flux of thermal energy is based on that
process of thermal energy
gy transfer is a random p
process.
the p
ie. the energy diffuses through the crystal, suffering frequent collisions.
Ballistic :
Diffusive :
JJG
across the whole sample
J U ∝ ΔT
JJG
dT
J U ∝ ∇T =
local
dx
For diffusion, thermal conductivity is defined by
JJG
jU
κ≡ K
( −∇T )
jU [Watt/m2]:
p
phonon
p
properties
p
scattering
crystal quality (size, defect)
temperature
the energy transmitted
per unit time
across unit area p
κ [(Watt/m2)/(K/m)] = [Watt/m/K]
Kinetic theory of gases (phonons)
consider phonons as gases contained in a crystal volume
calculate diffusion in the presence of temperature gradient
∇T
TL
TH
jU
n
Fick’s law
x
dn dn dT
=
dx dT dx
n : concentration
t ti off molecules
l
l
C : heat capacity per unit volume = nc
dT
dT
vg : phonon velocity
Ax =
ΔT =
vx τ
dx
dx
A : phonon
h
mean ffree path
th (average
(
distance
di t
that a lattice vibration travels before it
phonon=vgτ
collide with another p
jU = −n |v x | cΔT
Thermal
conductivity
1
κ = Cv g A
3
dT
1
dT
2
jU = − n v cτ
= − n v cτ
dx
3
dx
2
x
1
dT
1
dT
= − Cv g A
jU = − nv 2g cτ
3
dx
3
dx
Anharmonic effects
• In the theory we have covering so far for phonons,
we have assumed that the potential energy kx2/2
(harmonic, Hook’s law)
• However,
However real systems have anharmonic effects
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
There
is
i no thermal
th
l expansion
i off th
the llattice
tti
The force (elastic) constants between the atoms do not
change 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
phonons.
h
Anharmonic terms in binding
potential
U
The general shape applies for
any type of binding
xo
∂U
U(x) = U(x o ) +
∂x
x
1 ∂2U
1 ∂3U
2
3
−
+
−
+
−
x
x
x
x
x
x
(
(
(
o)
o)
o ) + ...
2
3
2 ∂x x
6 ∂x x
xo
ΔU(x) = U(x)-U(x o ) =
o
o
Reset the equilibrium, let displacement x-xo
U(x) = cx 2 − gx 3 − fx 4 ...
harmonic term
o
1∂ U
1∂ U
2
3
x
−
x
+
x
−
x
+ ...
(
)
(
)
o
o
2
3
2 ∂x x
6 ∂x x
2
anharmonic term
3
o
→ x
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.
∞
x =
U(x)
dx
x
exp[
−
]
∫−∞
k BT
∞
U(x)
dx
exp[
−
]
∫−∞∞
k BT
∞
≅
cx 2 − gx 3 − fx 4
]
∫−∞ dx x exp[ −
k BT
∞
cx − gx − fx
dx
exp[
−
]
∫−∞∞
k BT
2
3
4
⎛ cx 2 − gx 3 − fx 4 ⎞
⎛ cx 2 ⎞
⎛ gx 3 fx 4 ⎞
exp ⎜ −
+
⎟ = exp ⎜ −
⎟ exp ⎜
⎟
k
T
k
T
k
T
k
T
⎝
B
⎠
⎝ B ⎠
⎝ B
B ⎠
⎛ cx 2 ⎞ ⎛
gx 3 fx 4 ⎞
high T limit
≅ exp ⎜ −
+
⎟ ⎜1 +
⎟
k
T
k
T
k
T
⎝ B ⎠⎝
B
B ⎠
(3π1/2 /4)(g/c5/2 )(k BT)3/2 3g
x =
= 2 k BT
1/2
1/2
(π/c) (k BT)
4c
∞
≈
cx 2 gx 4
∫−∞ dx exp[− k BT ] k BT
∞
cx 2
∫−∞∞ dx exp[− k BT ]
anharmonic term gives
the net change of <x>
x2 x3
e = 1+ x +
+ + ⋅⋅⋅
2! 3!
x
linear
ea dependence
depe de ce o
of T
Coefficient of linear expansion
Phonon-phonon scattering
phonon displaces atom which changes the force constant C (anharmonic terms)
scatter other phonons
three phonon process
Normal p
processes : all ks are in BZ
k2
1st BZ in
k-space
JJG JJG JJG
k1 + k 2 = k 3
k3
k1
Total
T
t l momentum
t
off the
th phonon
h
gas is conserved by such collision
Umklapp processes : k3* is outside BZ
G
1st BZ in
k-space
k3
k2
k1
k *3
outside BZ
JJG JJG JJG*
k1 + k2 = k3
JJG JG JJG*
k3 + G = k3
JJG JJG JJG JG
k1 + k2 = k3 + G
“Folding over”
The only meaningful
phonon K’s lie in the
1st BZ, so any
longer K must be
g back into the
brought
1st BZ by adding a G.
crystal momentum is not conserved
Normal and umklapp processes
N-process
• Total momentum of
the phonon gas is
conserved by such
collision
• No thermal resistance.
q3
q4
U-process
JJG JJG JJG JJG JG
q1 + q 2 = q 3 = q 4 + G
A collision of two
phonons both with a
positive value of qx can
by an U-process,
U process,
create a phonon with
negative qx.
Umklapp processes
U-processes occur at high temperature : require large k (ie. large ω)
How large ?
kz
D b sphere
Debye
h
k≤kD
ky
kx
1
k ~ kD
2
1
ω ~ ωD
2
1
1
E ~ =ωD ~ k BΘ D
2
2
Phonon-phonon scattering : rate τ-1∝ # of phonons involved
U process : τ-11∝ NU ~ kBT exp(-Θ
U-process
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 mode
U process can not occur
Phonon mean free path
Log-log plot
Phonon
mean free
path A
p
( ∝τ )
Exponential
Slope: -1
T (K)
Very low T ,
A=vgτ =constant
A → L (sample’s size)
Intermediate T ,
A=vgτ ∝ (1/T)exp(1/T)
(1/T)
(1/T)
dominated by U process
High T ,
A=vgτ ∝ T-1
∝ (number of phonons)-1
No distinction between
N and U processes
T-dependent thermal conductivity
Below 5K, enriched Ge74 shows
T3 dependence of κ
due to boundary scattering
Isotope
effect
At low temperatures, A → L
(sample’s
(sample
s size)
Phonon propagation ~ ballistic
κ =(1/3)vgACV ~ vgLCV
κ ∝ CV ∝ T3 Debye
Log-log plot of κ(T)
At intermediate temperatures,
κ=(1/3)vgACV =(1/3)vg2τCV
U-processes CV = constant = 3Nk B
1 ΘD /T
τ~
e
k BT
κ~
1 Θ D /T
e
k BT
T-dependent thermal conductivity
Other effects
Impurity scatterings
Defect scatterings
Log-log plot
Slope: 3
κ
(Watt/m/K)
break periodicity
Exponential
Slope: -1
T (K)
Thermal conductivity of LiF crystal
bar
Different cross sectional area
(a) 1.33mm × 0.91mm
κ (Watt m-1K-1)
(b) 7.55mm × 6.97mm
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
t
temperature
t
and
d follows
f ll
that
th t exp(1/T).
(1/T)
T(K)
4. Cross sectional area influences κ
below 20K.
20K Bigger area crystal has
has,
larger κ it has.
Summary of part (I)
¾ Solids are defined by their capacity to be solid –
to resist shear stress
¾ A crystal is truly solid (as opposed to a glass which is just a
“slow
slow liquid”)
liquid )
¾ Crystalline order is defined by the regular positions of the nuclei
cr stal structure
crystal
str ct re = lattice + basis
¾ Lattice and reciprocal lattice
Diffraction and experimental studies
Brillouin zone
¾ Crystal binding
Type of binding
El ti constants
Elastic
t t and
d elastic
l ti waves
Summary
y of p
part (I)
()
¾ Vibrations of atoms
Harmonic approximation
¾ Quantization of vibrations
phonons act like particles
-- can be created or destroyed
y byy inelastic scatterings
g
¾ Thermal properties
•
•
Fundamental
F
d
t l law
l
off probabilities
b biliti (B
(Boltzmann
lt
ffactor)
t )
Planck distribution for phonons
Heat capacity
p
y:C
Low T, C ∝ T3
and High T, C ~ constant
Thermal conductivity : κ
maximum; as function of T