스핀전자물성연구실
도중회
Ch. 5 Phonons II : Thermal properties
스핀전자물성연구실
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• 열용량(C :heat capacity) :
계의 온도를 1.0 ºC 높이는데 필요한 열에너지
C
dQ
dT
계의 열에너지
dQ  dU  dW  dU  PdV
뒬롱-프티 법칙 : 대부분의 고체들 : ~25 J/mol K
•
에너지 등분배의 원리 : 자유도가 계에 기여하는 에너지의 양은
1
k BT
2
만큼씩이며
자유도에는 병진운동, 회전 운동, 분자의 진동에 의한 것들이 포함된다.
병진운동
z
1
2
회전운동
mv 2
z
1
2
진동운동
Iw 2
z
1
2
kx 2
(예) 이원자 분자
y
x
자유도 : 3개 =
vx vy vz
3
k BT
2
y
x
y
x
자유도 : 2개 =
Ix Iz
k BT
자유도 : 2개 =
k BT
2
대칭모드, 비대칭모드
스핀전자물성연구실
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Phonon heat capacity
 U 
CV  

 T V
Heat capacity at constant volume
From experiments,
CV  3Nk B
at room T.
CV  T (metal) or CV  T 3 (insulator) at low T
2

p
1
Classical calculation of specific heat
 mw02 r 2
E
2m 2
From statistical mechanics, the equipartition theorem states that the mean
value of each independent quadratic term in the energy is equal to ½ kBT.
In 3D, the total energy for a crystal with NA molecules
 p x2  p y2  p z2 1

 1

E  NA
 mw02 ( x 2  y 2  z 2 )  N A 6  k BT   3 N A k BT
2m
2
 2



The heat capacity is
C
dE
 3 N A k B  3R  25 J/mol  deg ;Dulong & Petit’s law
dT
스핀전자물성연구실
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Plank distribution
The total energy of the phonons at a temperature T in a crystal is
U lat  
K

P
K : wavevector, p : polarization index
<nK,p> : thermal equalibrium occupancy
of phonons of K and p .
nK , p wK , p
Ratio of the number of harmonic oscillators in the (n+1) state to the number
of in the n state is given by
Boltzmann distribution ;

N n1 N n  exp  kBwT

where,  
1
k BT
Average excitation quantum number of an harmonic oscillator
n( w)
 s exp(sw)  sx




exp(

)
s
w

x
s 0
s
S
s
s 0
S
where, x  exp( w)
x
S
s
 1 x  x2  x3  
1
for x  1
1 x
d  1 
d
x  xs x 

x
1
dx  1  x 
dx S
; Planck distribution




s
1
1

x
exp(


w
)

1
x
S
1 x
스핀전자물성연구실
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Phonon heat capacity
Then, the total phonon energy is
U lat  
K
wK,p
 exp(w
p
K,p
/ k BT )  1
   dwD p ( w)
P
w
exp(w / k BT )  1
Dp(w) : density of state
= number of modes per unit frequency range for a given polarization
The phonon heat capacity is
w
 w
e k BT
 U 
CV  
    dwD p ( w)
k B T 2  w k B T  2
 T V
 1
e


k B x 2e x
   dwD p ( w)
where, x  w
2
x
k BT
e 1
2

2

스핀전자물성연구실
Density of state in 1D
Real space
K-space
L = sa
a
/L
Standing wave solution for fixed ends
u s  u (0) sin wt sin kx
boundary condition
For un-fixed ends,
sin kL  0 kL  n
n
k 
L
u s  u (0)e  iwt e iKsa
In a solid, periodic boundary condition is
u ( sa )  u ( sa  L) standing wave for circular ring
KL  2n  K  0,
/a
2 4 N
,
 
L
L
L
도중회
스핀전자물성연구실
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Density of state in 1D
u s  u (0)e  iwt e iKsa
For un-fixed ends,
In a solid, periodic boundary condition is
u ( sa )  u ( sa  L)
standing wave for circular ring
2 4 N
,
 
L
L
L
KL  2n  K  0,
Density in K-space = 1 / unit mode length in K-space
nK 
1
1

2
K
 L

L
2
Total number of modes N = Density in K-space  length in K-space
N
L
K
2
D ( w)  dN
Density of state is
dw
L
 dK 
2  dw 
스핀전자물성연구실
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Density of state in 3D
Density of states in 3-dimensions
L
exp iK x x  K y y  K z z  exp iK x ( x  L)  K y ( y  L)  K z ( z  L)
L
Kx , Ky , Kz  0, 
L
N
2
4
,  ,  
L
L
L
/a
Density in K-space = 1 / unit mode volume in K-space
nK 
1
1

K 3 2 3
L
 

V
8 3
Total number of modes N
= Density in K-space  total volume in K-space
2/L
4K 3
V
N
2 3

3
Density of state is
D ( w)  dN
2
dw
 VK
2
 dK 


2  dw 
2
 L 
2
In 2D, N  
  K
 2 
AK dK
where A  L2

D ( w)  dN
dw 2 dw
스핀전자물성연구실
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Debye model : Phonon heat capacity
Debye approximation : dispersion relation of acoustic phonon c =  / k
3
3
Vw3D
 L  4 ( w / c)
N 

 
3
6 2 c 3
 2 
w  6 c N
3
D
2 3
kD 
V
wD
Phonon energy is given by
WD
U 3 D  3 dwD( w)n( w)w  3
0
1L+2 T modes
 T 

 9 Nk BT 

 D
3

xD
0
x3
dx x
e 1
Then, the heat capacity is
If phonons are filled up to wD only
Ky

 6 N
c
2
V

1
3
KD
 Vw2   w 
dw 2 3  w
 2 c   e k BT 1 
KT
w
wd Debye approx.
 x dw  dx kBT

c


w
D  kD
 D
xd  D
kBT
T
kB
w
 T 
dx dU
 dU 

 9 Nk B 
CV  
 
 dT V dT dx
 D 
k BT
3

xD
0
dx
x 4e x
e
x
Actual
Kd

1
Kx
2
스핀전자물성연구실
K
도중회
Debye model vs. Einstein model
•Debye approximation at a low T
w

x3
xD  D  
dx x
k BT
0
e 1

CV 
 T 
dU 12π


Nk B 
dT
5
 ΘD 
4
at a high T, xD 

4
3
 T  4
 
U  3 Nk BT 
5
 D 
15
3
wE  1
e wE  1  wE
wD
0
k BT
U  3 Nk BT
 CV  3 Nk B
• Einstein model
Consider N oscillators of the same frequency wE
U 3 D  3 N n wE 
3 NwE
e
wE
k BT 1
e  wE
 dU 
2


CV  
Nk
w



3

B
E
2
 dT V
e  wE  1


at high T CV  3 Nk B
at low T  wE  1
• In a real system, total heat capacity at low T is
CV  e

wE
k BT
스핀전자물성연구실
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Anharmonic crystal interaction
The consequences of harmonic theory are :
• Two lattice waves do not interact ; a single wave does not decay or change
form with time
• No thermal expansion.
• Adiabatic & isothermal elastic constants are equal.
• The elastic constants are independent of pressure and temperature.
• The heat capacity becomes constant at high T＞ 
No real crystal satisfies above conditions.
In a real crystal, anharmonic effect should be considered.
- third order or higher order terms in the potential energy
U ( x)  cx 2  gx 3  fx 4 
In the experiments, the interaction of two phonons produces a third phonon
at a frequency 3=1+2
스핀전자물성연구실
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Thermal expansion
• Thermal expansion of a solid : an increase of the length with the temperature.
We may understand thermal expansion by considering for a classical oscillator
the effect of anharmonic term in the potential energy.
Let’s take a potential energy
g: asymmetry of the mutual repulsion of the atoms
f :softening of the vibration at large amplitude
U ( x)  cx 2  gx 3  fx 4
x : a displacement from the equilibrium separation at T=0
Then, the average displacement at T is given by





x 
xe  U dx



e  U dx





xe  cx e  gx
2



 fx 4


e  cx dx

e  cx e  gx
2
e  cx dx

xe  cx 1  gx 3  fx 4 dx
2
dx
2


3
2
 

3
 fx 4

 e  cx 1  gx 3  fx 4 
2

3g
k BT ; Length of a solid  temperature.
4c 2
1  3 (n  1)(2a)  ( n 1) / 2 ( / 2)1/ 2 for n  0 & even
n  ax
0 x e dx   2  4 (n  1)(2a) ( n1) / 2 ( / 2)1/ 2 for n  1 & odd
2
The thermal expansion does not involve the symmetric term -fx4, but only the
asymmetric term -gx3
스핀전자물성연구실
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Thermal conductivity
Q: In insulators, how the energy is transferred from hot region to cold region?
A : Energy can diffuse through phonon collisions in a solid.
dT
T  T ( x   )  T ( x ) 

Energy transmitted across unit area per unit time depends on temperature dx
gradient
dT
T 
 
dx
dT
v x
dx
x
T
l : mean free path of a particle between collision,
 : collision time
x
T+∆T
The net flux of energy = Energy transmitted per unit area per unit time
ju
   nvx cΔT 
 n v x2 c
ju
 K
n : density, c : heat capacity per particle
dT
1
dT
1
dT
  v 2 nc
  v 2 C
where, C  nc
3
3
dx
dx
dx
dT
dx
1
 K  Cv
3
스핀전자물성연구실
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Thermal resistivity of phonon gas
The phonon mean free path l is determined by geometrical scattering and
scattering by other phonons.
i) If purely harmonic, the l is limited by collisions with the crystal boundary
and lattice imperfections.
ii) If anharmonic, the l is limited by coupling with other phonons and it is
proportional to 1/T at high temperature because the number of the
excited phonons is proportional to T.
• Normal phonon collision
source
sink



K1  K 2  K 3
• Umklapp phonon collision
hot
cold




K1  K 2  K 3  G
Phonon flux is unchanged in momentum.
Thermal resistivity = 0
Energy conservation : 3=1+2
  
K1 , K 2 , K 3 : phonons

G : reciprocal lattice vector
Energy conservation : 3=1+2
Phonon flux is decreased as they moves to the right.
Thermal resistivity ≠ 0, Energy transfer ≠ 0