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Planck distribution Consider a set of identical harmonic oscillators in thermal equilibrium.
N n1 / N n exp( / k BT ) exp( / )
then
N n N 0 exp( n / )
n
1 exp( / ) 1
<n>
4
3
2
1
0
0
x k BT /
1
2
3
4
The energy of the phonons with wavevector K and polariztion p
U K , p nK , p K , p K , p exp( K , p / ) 1
N0 0 <n>
s
N n N 0 exp( n / )
n
1 d x s s x , and sx x x 1 x dx s 1 x s s
n
s exp( s / ) exp( s / )
s s
1 exp( / ) 1
The average excitation quantum number of an oscillator is
n
sN N
s s
s
s exp( s / ) exp( s / )
s s
s
N
N
s
s
The total number and the total energy inside the square are same as those of the oscillators.
U d D p ( )
p
K , p exp( K , p / ) 1
D p ( ) dN p / d
The lattice heat capacity is
Clat x 2 exp x U k B d D p ( ) 2 T (exp x 1 ) V p
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
(N+1)
u( sa) u( sa L)
N1
N
s=1
2 3 4 . .
us u (0) exp[i ( sKa K , p t )],
with
2 4π 6π Nπ K 0, , , , ... , . L L L L
dN dN dK L d D( ) d dK d dK
1
The density of the state with the periodic boundary condition:
dN dN dK L d D( ) d dK d 2 dK
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)]}
whence
2 4 N K x , K y , K z 0, , , ... , L L L
Therefore there is one allowed value of K per volume (2/L)3 in K space.
D ( ) (VK 2 / 2 2 )( dK / d ) V ( / v) 2 1 V 3 2 2 v 2 2 v 3
The cutoff frequency D:
D (6 2v 3 N he cutoff wavevector KD:
Phonon heat capacity
phonon gas
U the heat capacity at constant volume CV 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:
for / a K / a
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. The density of the state with the fixed boundary condition:
Chapter 5
Key points:
Phonons II. Thermal properties
• Phonon heat capacity • Density of state • Debye model and Einstein model • Thermal expansion • Thermal conductivity
Due to the fixed boundary condition, The wavevector K is
2 3 ( N 1) K , , , ... . L L 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. dN / dK L / , for 0 K / a
The total energy in thermal equilibrium is K , p U U K , p K p K p exp( K , p / ) 1 Suppose that the crystal has Dp( )d modes of a given polarization p in the frequency range to + d . Then the energy is
Method 2: periodic boundary conditions
Method 1: fixed boundary conditions
L
Fixed
s=0 1
Fixed a ui-1 … … …… i1 i i+1 …… N-1 N
us u (0) exp( iK , p t ) sin sKa
. .
.
..
. i+1 ui i
i1
.
..
.
a
For periodic boundary conditions the number of modes per unit range of K is L/2 for /a K /a, otherwise 0.
dN / dK L / 2 ,
U U K , p nK , p K , p
K p K p
where <nK,p> is the thermal equilibrium occupancy of phonons of wavevector K and polarization p, and is given by the Planck distribution function: 1 nK , p exp( / ) 1
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 v
2 3
D
The total number of modes with wavevector less than K for a given polarization and a given branch is:
N ( K ) ( L / 2 )3 (4K 3 / 3)
The density of states for each polarization is
where
x / k BT
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 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 which D() may be calculated.