固体物理ppt 5

Chapter 5 Phonons II. Thermal
properties
Key points:
• Phonon heat capacity
• Density of state
• Debye model
• Thermal conductivity
Phonon heat capacity
phonon gas
The central problem is to find D(), the number
of modes per unit frquency range.
D() is called the density of modes or density
of states (DOS).
The best practical way to obtain the density of state is to measure the dispersion relation
nK , p
1
exp( /
)
1
Planck distribution Consider a set of identical harmonic oscillators in thermal equilibrium.
Nn1 / Nn exp( / kBT ) exp( / )
then
Nn N0 exp( n / )
Method 1: fixed boundary conditions
Fixed s=0 1
L
… ui-1
a …
Fixed
…… i1 i i+1 …… N-1 N
us u(0) exp( iK , pt) sin sKa
Due to the fixed boundary condition, The wavevector K is
Density of states in three dimensions
Apply periodic boundary conditions over N3
primitive
cells
within
a
cube
of
side
L.
u u(0) exp[i(K r K,pt)]
exp[i(Kx x K y y Kz z)] exp{i[Kx (x L) K y ( y L) Kz (z L)]}
the heat capacity at constant volume
CV
U T
V
the lattice heat capacity Clat
The total energy of the phonons at a temperature
( kBT) in a crystal is the sum of the energies
over all phonon modes:
U
UK,p
nK, p K, p
Kp
Kp
where <nK,p> is the thermal equilibrium occupancy of phonons of wavevector K and polarization p, and is
given by the Planck distribution function:
i+1
i1
i
ui a
For periodic boundary conditions the number of
modes per unit range of K is L/2 for /a K /a, and 0 otherwise.
dN / dK L / 2 , for a / K a /
Method 2: periodic boundary conditions
(N+1)
u(sa) u(sa L)
s=1
N
2
us u(0) exp[ i(sKa K, pt)],
with
K 0, 2 , 4π , 6π , ..., Nπ .
LLL
L
N1
3
4
..... .
. .....
The cutoff wavevector KD:KD (6 2 N /V )1/3
There is NO wavevector larger than KD in Debye
model.
The thermal energy for each polarization:
U p d D(p ) n(p ) p
V
2 2v3
D 0
d
p
3 p
exp(p / ) 1
Assume that the phonon velocity is independent of the po
The total thermal energy:
U
p
Up
3U p
3V
2 2v3
D 0
d p
3 p
exp(p / ) 1
3VkB4T 4
to + d .
Then
the
energy U
p
is
d
Dp
()
K , exp(K, p
p
/
)
1
Dp () dN p / d
The lattice heat capacity is
Clat
U T V
kB
p
d
Dp
()
x2 exp x (exp x 1)2
where x / kBT
9
Nk
B
T
3
xD dx
0
x4 x2
3NkB
At high temperature the heat capacity approaches
to the classical value of 3NkB from the Dulong-
Petit law.
2. Debye T3 law
At very low temperature T << ,
Einstein model for density of states
There are three polarizations p for each value of K: 2 transverse modes, 1 longitudinal mode.
The number of allowed K values in a single branch
for the 1st Brillouin zone equals to the number of the primitive cells. The number of modes equals to the total degrees of the freedom.
N0 Nn N0 exp( n / )
0 <n>
n
xs 1 , and sxs x d xs x
s
1 x
s
dx s
1 x
n
s exp( s / )
s
exp( s / )
1
exp( /
)
1
s
n
exp(
1
/
)
1
4
< n>
3
2
1
0
0
1
2
3
4
x kBT /
The energy of the phonons with wavevector K and polarizti
0
dx
x3 ex 1
4
15
Thus
U
3 4 NkB 5 3
T
4
Then
CV
12 4 Nk 5 3
B
T3
T3
i.e. xD
Debye T3 law matches
quite good experimental results for isolators at sufficiently low temperature.
constant for
each
polarization type. i.e. vK
The density of states:
D() (VK2 / 2 2 )(dK / d)
V ( / v)2 1 V3 2 2 v 2 2v3
The cutoff frequency D: D (6 2v3N /V )1/3
UK,p
nK, p
K , p
K , p exp(K,p / ) 1
The total energy in thermal equilibrium is
U
K
UK,p
p
K
K , p p exp(K, p / ) 1
Suppose that the crystal has Dp( )d modes of a given polarization p in the frequency range
2 2v33
xD dx x3 0 ex 1
where x / kBT , and xD D / kBT / T
The
Debye
temperature
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v kB
6 2
V
N
1/ 3
Then the total thermal energy is
U
9
Nk
BT
T
3
xD dx
0
x3 ex 1
The heat capacity:
CV
U T
9
Nk
B
T
3
xD dx
0
x4ex (ex 1)2
Discussion: 1. Dulong-Petit law
At high temperature T >> ,
i.e. xD << 1
ex 1 x
CV
9
Nk
B
T
3
xD dx
0
x4ex (ex 1)2
N (K ) (L / 2 )3(4K 3 / 3)
The density of states for each polarization is
D() dN / d (VK 2 / 2 2 )(dK / d)
Debye model for density of states
Debye approximation: the velocity of waves is
Density of states in one dimension
Consider vibrations of a 1d line of length L carrying N+1 particles at separation a.
There are two equivalent methods for enumerating the number of the mode. Method 1: fixed boundary conditions Method 2: periodic boundary conditions
The density of the state:
D()d L dK d L d d d / dK
Density of states in two dimensions
Within the circle of area K2 the smoothed number of allowed points is K2(L/2)2.
whence
Kx , K y , Kz 0,
2 ,
L
4 ,
L
..., N
L
Therefore there is one allowed value of K per volume (2/L)3 in K space.
The total number of modes with wavevector less than K for a given polarization and a given branch is:
vs K in selected crystal directions by
inelastic neutron scattering and then to make a theoretical analytic fit to give the dispersion relation in general direction, from
K , 2 , 3 , ... (N 1) .
LL L
L
The number of mode is equal to the number of
particles allowed to move. There is one mode for
ach interval K = /L, so that the number of modes per unit range of K is L/ for K /a, and 0 for K /a.
The average excitation quantum number of an
oscillator is
sNs s exp( s / )
n
s
s
Ns exp( s / )
s
s
N
Ns
s
The total number and the total energy inside the square are same as those of the oscillators.