chapter 5: phonons (2) thermal properties -...
TRANSCRIPT
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
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
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?
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
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
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
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
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
• 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
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
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
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
3T
3
BNk
for T
(Debye theory)
Experiment
3T
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
Debye T3 Law (2) 3 4
01 15x
xdx
e
• Low temperature heat capacity of solid argon, plotted
against T3. Here, = 92 K
Einstein Model of the Density of States
• • •
T
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
• 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
~ O(nm)
• 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)
~ K-1
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.
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
Size effect (l ~ D)