Chapter 5: Phonons (2) Thermal Properties -...
Transcript of Chapter 5: Phonons (2) Thermal Properties -...
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Chapter 5: Phonons (2) Thermal Properties
Phonon heat capacity
Planck distribution
Density of states
Debye model for D.O.S.
Debye T3 law
Einstein model for D.O.S.
Thermal conductivity
Thermal resistivity
Umklapp process
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Phonon Heat Capacity
• Heat capacity at constant volume ~
• Phonons’ contribution to CV ~ the lattice heat capacity (Clat)
• other than phonons? electrons’ contribution to CV (Ch. 6)
• Total energy of the phonons at a temperature T in a crystal ~ the sum of the energies over
all phonon modes
where is the energyV
V
UU
CT
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x <n>
1 0.5
2 1.5
3 2.5
x x - 0.5
~ 진동수 w의 phonon이 온도 T일 때
몇 개인지에 대한 통계적 분포
Wave vector K를 가지면서 진동수 w인 phonon이 평균적으로 몇 개인가?
또는 몇 번째 excited state를 평균적으로 점유하고 있는가?
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Planck Distribution (1)
• Consider a set of identical harmonic oscillators in thermal equilibrium
• Since the number of particles with an energy of Ei is proportional to exp(-Ei/kBT)
(Boltzmann factor), the ratio of the number of oscillators in the (n+1)th state to the number
in the n-th state is
• The fraction of the total number of oscillators in the n-th quantum state is
• • •
T 1
For each harmonic oscillator,
1( )
2n
n n
E n
E E
w
w
1 1exp( / ) exp[ ( ) / ]
exp( / ) exp( / )exp( / ) 1n B n B
n B n B
nB
n
E k T E k T
E k T
Nk T
E kN Tw
w
0
0
2
01 2
0
exp( / )
exp( /(1
))
n
n Bn
B
s
N n k TP
N N
N
Ns k T
NN
w
w
임의의 조화진동자가 n번째 양자상태에 있을 확률
1
/ Bk Tw
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Planck Distribution (2)
• The average excitation quantum number of an oscillator is
• The Planck distribution is given by
• Total energy of phonon at T is
0
0
0
exp( / )
exp( / )
B
ss
sB
s
s s k T
n sP
s k T
w
w
온도 T의 평형 상태에서 진동자의 평균적인 양자상태의 수
1
exp( / ) 1B
nk Tw
,
,
,
,exp( / ) 1
K p
lat
K p K p
K p K p
K p
nUw
w w
x-1
0 2 40
2
4
<n
>
x
0 1 2 3 40
1
2
3
<n
>
x-1
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Normal Model Enumeration
• Total energy of oscillators of frequency w in thermal equilibrium is
• For the calculation, an integral form is preferred.
• When the crystal has Dp(w)dw modes between w and w+dw,
• Lattice heat capacity,
• The central problem is to find D(w), the number of modes per unit frequency range ~ the
density of modes or the density of states
,
,exp( / ) 1
K p
K p K p
Uw
w
,
,
( )exp( / ) 1 exp( / ) 1
K p
p
K p pK p
U d Dw w
w ww w
2
2( ) wher
(e
)
1/
x
B p xp
x ek d D x
eww w
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Density of States in One Dimension – Fixed boundary condition
• Consider vibrations of a 1-dim line of length L carrying N+1 particles at separation a.
• Boundary conditions : the end atoms at s = 0 and N are fixed
• Vibrational mode : standing wave with a displacement, us, of the particle s
N+1 particles in a line
with a length of L = Na
( ) 0 sin( ) 0 ~ trivial solution
( ) sin( ) 0 ~ , / /
2 3 ( 1) , , , ,
i s sKa
ii s N NKa NKa n K n Na n L
NK
L L L L
NK
L a
s = N
~ indistinguishable with other K' < /a
(there are N-1 moving particles)
N half-waves
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Density of States in One Dimension – Periodic boundary condition
• For the unbounded medium, the solutions are periodic over a large distance L
~ u(sa) = u(sa+L) : periodic boundary condition
• The traveling wave solution is
-One mode for the interval of DK = 2/L
s = N
L = Na
0 ,exp[ ( )]s K pu u i sKa tw
~ N independent solutions
~ N particles can move independently
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• In 1-dimensional case,
the number of modes per unit frequency range is D(w) ~ density of states
- the number of modes D(w)dw is given by
where dw/dK is the group velocity, which is obtained from the dispersion relation
• In 2-dimensional caes :
- When a periodic boundary condition is applied over a square of side L for a square lattice
of lattice constant a,
- The interval DK = 2/L for each direction of x and y
- Average area of (2/L)2 per one value of K
- Within the circle of area K2, the number of allowed
points (vibration modes or states) is
1(
( )2 /( )2)
L dKD d d
d
L ddN
d dKw w w
w
w
w
222
222
K LK
L
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Density of States in 3-Dimensions
• Apply periodic boundary conditions over N3 primitive cells within a cube of side L
L L
L = Na
Number of allowed values of K per unit volume of K space
22
23( )
24
8
dN dN dK V dKK
d dK d
K
d
VK dD
d
w w w
w w
How to find dK/dw ? (i) Debye model (ii) Einstein model
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Debye Model for Density of States
K
w
v
2
2( )
2
VK dKD
dw
w
Wavevector larger than KD is not allowed in the Debye model.
~ D.O.S. (Debye model)
3 3
3 2
4
8 3 6
V K VKN
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2
2 3( )
2
VD
v
ww
1/326
D
Nv
V
w
N
, BB
k Tk Tx d dxw w
3 4
22 3 2 3 2
0 0
3 3 exp( / )
2 exp( / ) 1 2 exp( / ) 1
D D
V B
U V Vd d
T v T v k T
w ww w w
w w w w
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3T
3
BNk
for T
(Debye theory)
Experiment
3T
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Debye T3 Law (1)
• At very low temperature, xD = /T >> 1
2 2 3
1
10
11 1 1
1
0
0 0 0
1 1
( ) :
(1 ) exp( )1 (1 )
, Let
1 1
x x x x x x
x x xs
n sx
s
nn sx u n u
ns
t
ns
z
s
e e e e e e sxe e e
dx x e u sx
udx x e d
z dt t e Gamma fun
u e du u es s s
ction
0
11
4 43
10
1
( 1)
1( 1) ( 1) ( 1)
(4) (4) 690 1
( ) :
5
n u
ns
sx
s
z
n
z n Riem
du u e n
n n ns
ann zeta functio
dx x e
n
3 43
41 10 0
1 exp( ) 6
1 15xs
xdx dx x sx
e s
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Debye T3 Law (2) 3 4
01 15x
xdx
e
• Low temperature heat capacity of solid argon, plotted
against T3. Here, = 92 K
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Einstein Model of the Density of States
• • •
T
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General Result for D(w)
• How to obtain a general expression for D(w), the number of states per unit frequency range, from
the phonon dispersion relation w(K).
Element of area dSw on a constant
frequency surface in K space.
~ Perpendicular distance between two
constant frequency surfaces in K space,
one at w and the other at frequency w+dw
d
dK ddK
ww
K공간상의 w면에서 적분을 수행하고
여기에 w+dw 면까지의 거리를 곱함
~ Integral taken over the area of the surface w constant in K space
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• Debye solid : D(w) is proportional to w2
• Actual crystal structure :
- Kinks or discontinuities develop at singular points where group velocity vg = 0.
- Van Hove singularities
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~ O(nm)
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• From the kinetic theory of gases, the thermal conductivity is given by
where C is the heat capacity per unit volume, v is the average particle velocity, and is the mfp of a
particle between collisions.
• Thermal conductivity in dielectric solids ~ C as the heat capacity of the phonons, v the phonon velocity,
and the phonon mean free path.
• Elementary kinetic theory :
1
3K Cv
T T+DT x
- Flux of particles in the x direction : where n is the concentration of molecules
(there is a flux of equal magnitude in the opposite direction)
- Particle moving from T+DT to T gives up energy cDT, where c is the heat capacity of a particle.
- Temperature difference during a free path of the particle is given by
- Net flux of energy :
- Since v is constant for phonons (~ Debye approximation)
2
2
1 1( ) ( )
2 2
1
3
U x x x x x x
dT dTj n v c T n v c T n v c T n v c v n v c
dx dx
dTn v c
dx
D D D
, : average time between collisionsx
dT dTT v
dx dx D
2 1=
3
1 1 where , , and
3 3U
dT dTj Cv K
dx dx
dTnv c v C nc K Cv
dx
(생략 가능)
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~ K-1
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Normal Collision Process
• Normal phonon collision (or three-phonon) process : K1 + K2 = K3
- Since DK = K3 – (K1 + K2) = 0, the phonon momentum is conserved
- Energy is conserved ~ w1 w2 w3
- Phonon flux is unchanged (no effect on thermal resistivity)
~ Kn unchanged after collisions K
J K
1st B.Z.
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Umklapp (“flip-over”) Process
• When the resulting wavevector K1+K2 reaches out of the first Brillouin zone,
we can define K3 satisfying K1 + K2 = K3 + G, where G is a reciprocal
lattice vector.
• Energy is conserved (w1 w2 w3) but momentum (phonon wavevector) is
not conserved
• Always possible in periodic lattices ~ the wavevector in the first Brillouin
zone is the only meaningful phonon
• A collision of two phonons both with a positive value of Kx can create a
phonon with negative Kx ~ “flip-over” or umklapp process
• K1 and K2 should be large enough to generate K3 exceeding the FBZ ~ the U
process requires high temperature to generate such high valued phonons.
• At high temperature T > , substantial proportion of all phonon collisions
are U processes ~ decrease of the mean-free path ~ finite thermal resistivity
http://upload.wikimedia.org/wikipedia/en/7/7b/Phonon_k_3k.gif
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Size effect (l ~ D)