lecture+12+mak +phonon
TRANSCRIPT
-
7/30/2019 Lecture+12+MAK +Phonon
1/116
SOUND WAVES
LATTICE VIBRATIONS OF 1D CRYSTALS
chain of identical atoms
chain of two types of atoms
LATTICE VIBRATIONS OF 3D CRYSTALS
PHONONS
HEAT CAPACITY FROM LATTICE VIBRATIONS
ANHARMONIC EFFECTS
THERMAL CONDUCTION BY PHONONS
-
7/30/2019 Lecture+12+MAK +Phonon
2/116
Crystal Dynamics
Concern with the spectrum of characteristics vibrations of a crystallinesolid.
Leads to;
consideration of the conditions for wave propagation in a periodic
lattice, the energy content,
the specific heat of lattice waves,
the particle aspects of quantized lattice vibrations (phonons)
consequences of an harmonic coupling between atoms.
-
7/30/2019 Lecture+12+MAK +Phonon
3/116
Crystal Dynamics
These introduces us to the concepts of forbidden and permitted frequency ranges, and
electronic spectra of solids
-
7/30/2019 Lecture+12+MAK +Phonon
4/116
Crystal Dynamics In previous chapters we have assumed that the atoms were at
rest at their equilibrium position. This can not be entirely correct(against to the HUP); Atoms vibrate about their equilibrium position atabsolute zero.
The energy they possess as a result of zero point motion is known aszero point energy.
The amplitude of the motion increases as the atoms gain more thermalenergy at higher temperatures.
In this chapter we discuss the nature of atomic motions, sometimes
referred to as lattice vibrations.
In crystal dynamics we will use the harmonic approximation , amplitudeof the lattice vibration is small. At higher amplitude some unharmonic
effects occur.
-
7/30/2019 Lecture+12+MAK +Phonon
5/116
Crystal Dynamics Our calculations will be restricted to lattice vibrations of small
amplitude. Since the solid is then close to a position of stableequilibrium its motion can be calculated by a generalization of themethod used to analyse a simple harmonic oscillator.The smallamplitude limit is known as harmonic limit.
In the linear region (the region of elastic deformation), the restoringforce on each atom is approximately proportional to its displacement(Hookes Law).
There are some effects of nonlinearity or anharmonicity for largeratomic displacements.
Anharmonic effects are important for interactions between phononsand photons.
-
7/30/2019 Lecture+12+MAK +Phonon
6/116
Crystal Dynamics
Atomic motions are governed by the forces exerted on atomswhen they are displaced from theirequilibrium positions.
To calculate the forces it is necessary to determine the
wavefunctions and energies of the electrons within the crystal.
Fortunately many important properties of the atomic motions
can be deduced without doing these calculations.
-
7/30/2019 Lecture+12+MAK +Phonon
7/116
Hooke's Law
One of the properties of elasticity is that it takes about twice asmuch force to stretch a spring twice as far. That linear
dependence of displacement upon stretching force is called
Hooke's law.
xkFspring .
FSpring constant k
It takes twice
as much force
to stretch a
spring twiceas far.
F2
-
7/30/2019 Lecture+12+MAK +Phonon
8/116
The point at which the Elastic Region ends is called the inelasticlimit, or the proportional limit. In actuality, these two points are notquite the same.
The inelastic Limit is the point at which permanent deformationoccurs, that is, after the elastic limit, if the force is taken off the
sample, it will not return to its original size and shape, permanentdeformation has occurred.
The Proportional Limit is the point at which the deformation is nolonger directly proportional to the applied force (Hooke's Law nolonger holds). Although these two points are slightly different, wewill treat them as the same in this course.
Hookes Law
-
7/30/2019 Lecture+12+MAK +Phonon
9/116
SOUND WAVES
Mechanical waves are waves which propagate through amaterial medium (solid, liquid, or gas) at a wave speedwhich depends on the elastic and inertial properties of thatmedium. There are two basic types of wave motion formechanical waves: longitudinal waves and transversewaves.
Longitudinal Waves
Transverse Waves
It corresponds to the atomic vibrations with a long .
Presence of atoms has no significance in this wavelengthlimit, since >>a, so there will no scattering due to thepresence of atoms.
-
7/30/2019 Lecture+12+MAK +Phonon
10/116
SOUND WAVES
Sound waves propagate through solids. This tells us thatwavelike lattice vibrations of wavelength long compared to theinteratomic spacing are possible. The detailed atomic structureis unimportant for these waves and their propagation isgoverned by the macroscopic elastic properties of the crystal.
We discuss sound waves since they must correspond to thelow frequency, long wavelength limit of the more general latticevibrations considered later in this chapter.
At a given frequency and in a given direction in a crystal it ispossible to transmit three sound waves, differing in theirdirection of polarization and in general also in theirvelocity.
-
7/30/2019 Lecture+12+MAK +Phonon
11/116
Elastic Waves
A solid is composed of discrete atoms, however when thewavelength is very long, one may disregard the atomic natureand treat the solid as a continous medium. Such vibrations arereferred to as elastic waves.
At the point x elastic displacement isU(x) and strain e is defined as thechange in length per unit length.
dUe
dx
x x+dx
A
Elastic Wave Propagation (longitudinal) in a bar
-
7/30/2019 Lecture+12+MAK +Phonon
12/116
According to Hookes law stress S (force per unit area) is
proportional to the strain e.
To examine the dynamics of the bar, we choose an arbitrary
segment of length dx as shown above. Using Newtons second
law, we can write for the motion of this segment,
.S C e
2
2( ) ( ) ( )
uAdx S x dx S x A
t
x x+dx
A
C = Young modulus
Mass x Acceleration Net Force resulting from stresses
Elastic Waves
-
7/30/2019 Lecture+12+MAK +Phonon
13/116
Equation of motion
2 2
2 2
u uC
t x
2
2
( ) ( ) ( )u
Adx S x dx S x At
( ) ( )S
S x dx S x dxx
.S C edu
e
dx
2
2
.
.
uS C
x
S uC
x x
2 2
2 2( ) u uAdx C Adxt x
Cancelling common terms of Adx;
Which is the wave eqn. with an offered
soln and velocity of sound waves ;
( )i kx t u Ae k = wave number (2/)
= frequency of the wave
A = wave amplitude /
s
s
v k
v C
Elastic Waves
-
7/30/2019 Lecture+12+MAK +Phonon
14/116
sv k
The relation connecting the frequency and wave number is
known as the dispersion relation.
k
Continuum
Discrete
0
* Slope of the curve gives
the velocity of the wave.
At small k (scattering occurs)
At long k 0 (no scattering)
When k increases velocitydecreases. As k increases further,the scattering becomes greater since
the strength of scattering increasesas the wavelength decreases, andthe velocity decreases even further.
Dispersion Relation
-
7/30/2019 Lecture+12+MAK +Phonon
15/116
Speed of Sound Wave
The speed with which a longitudinal wave moves through aliquid of density is
L C
V
C = Elastic bulk modulus
= Mass density
The velocity of sound is in general a function of the direction
of propagation in crystalline materials.
Solids will sustain the propagation of transverse waves, whichtravel more slowly than longitudinal waves.
The larger the elastic modules and smaller the density, the
more rapidly can sound waves travel.
-
7/30/2019 Lecture+12+MAK +Phonon
16/116
Speed of sound for some typical solids
SolidStructure
Type
NearestNeighbour
Distance
(A)
Density
(kg/m3)
Elastic bulkmodules
Y
(1010 N/m2)
Calculated Wave
Speed
(m/s)
Observedspeed of
sound
(m/s)
Sodium B.C.C 3.71 970 0.52 2320 2250
Copper F.C.C 2.55 8966 13.4 3880 3830
Aluminum F.C.C 2.86 2700 7.35 5200 5110
Lead F.C.C 3.49 11340 4.34 1960 1320
Silicon Diamond 2.35 2330 10.1 6600 9150
Germanium Diamond 2.44 5360 7.9 3830 5400
NaCl Rocksalt 2.82 2170 2.5 3400 4730
VL values are comparable with direct observations of speed of sound.
Sound speeds are of the order of 5000 m/s in typical metallic, covalent
and ionic solids.
Sound Wave Speed
-
7/30/2019 Lecture+12+MAK +Phonon
17/116
They can be characterized by
A propagation velocity, v
Wavelength or wavevector
A frequency or angular frequency =2
An equation of motion for any displacement can be
produced by means of considering the restoring forceson displaced atoms.
A lattice vibrational wave in a crystal is a repetitive andsystematic sequence of atomic displacements of
longitudinal,
transverse, or
some combination of the two
As a result we can generate a dispersion relationship
between frequency and wavelength or between angular
frequency and wavevector.
Sound Wave Speed
-
7/30/2019 Lecture+12+MAK +Phonon
18/116
Lattice vibrations of 1D crystal
Chain of identical atoms
Atoms interact with a potential V(r) which can be written inTaylors series.
2 22
( ) ( ) ...........2
r a
r a d VV r V a
dr
rR
V(R)
0 r0=4
Repulsive
Attractivemin
This equation looks like as the potential energy
associated of a spring with a spring constant :
ardr
VdK
2
2
We should relate K with elastic modulus C:
KaC
a
arCForce
)( arKForce
-
7/30/2019 Lecture+12+MAK +Phonon
19/116
Monoatomic Chain
The simplest crystal is the one dimensional chain of identical atoms. Chain consists of a very large number of identical atoms with identical
masses.
Atoms are separated by a distance of a.
Atoms move only in a direction parallel to the chain.
Only nearest neighbours interact (short-range forces).
a a a a a a
Un-2 Un-1 Un Un+1 Un+2
-
7/30/2019 Lecture+12+MAK +Phonon
20/116
Start with the simplest caseof monoatomic linear chainwith only nearest neighbourinteraction
)2( 11
..
nnnn uuuKum
If one expands the energy near the equilibrium point for the nthatom and use elastic approximation, Newtons equation becomes
a a
Un-1 Un Un+1
Monoatomic Chain
C
-
7/30/2019 Lecture+12+MAK +Phonon
21/116
0)2( 11
..
nnnn uuuKum
The force on the nth atom;
)( 1 nn uuK
The force to the right;
The force to the left;
)( 1 nn uuK
The total force = Force to the right Force to the left
Monoatomic Chain
a a
Un-1 Un Un+1
Eqns of motion of all atoms are of this form, only the
value of n varies
M t i Ch i
-
7/30/2019 Lecture+12+MAK +Phonon
22/116
All atoms oscillate with a same amplitude A and frequency .
Then we can offer a solution;
.
0expnn nduu i A i kx t dt
Monoatomic Chain
0expn nu A i kx t
2..2 2 0
2
expnnn
d uu i A i kx t
dt
..2
nn
u u
naxn
0
nn unax Undisplaced
position
Displaced
position
M t i Ch i Equation of motion for nth atom
-
7/30/2019 Lecture+12+MAK +Phonon
23/116
Monoatomic Chain Equation of motion for nth atom
..
1 1( 2 )n nn nmu K u u u
0 0 0 0
1 12e e 2 e e
n n n ni kx t i kx t i kx t i kx t m K A A AA
kna( 1)k n a
( 1)k n a
kna
2 e e e 2 e e eika ikai kna t i kna t i kna t i kna t
m K A A AA
Cancel Common terms
2 e 2 eika ikam K
2 e e 2 e ei kna t i kna ka t i kna t i kna ka t m K A A AA
M t i Ch i
-
7/30/2019 Lecture+12+MAK +Phonon
24/116
Monoatomic Chain
2 e 2 eika ikam K 2cosix ixe e x
e e 2cosika ika ka
2 2cos 2
2 (1 cos )
m K ka
K ka
21 cos 2sin2
xx
22 4 sin
2
kam K
242 sin
2
K ka
m
4 sin2
K kam
Maximum value of it is 1
max
4K
m
M t i Ch i
-
7/30/2019 Lecture+12+MAK +Phonon
25/116
versus k relation;
max 2
/s
K
m
V k
0 /a 2/a/a k
Monoatomic Chain
Normal mode frequencies of a 1D chain
The points A, B and C correspond to the same frequency, therefore
they all have the same instantaneous atomic displacements.
The dispersion relation is periodic with a period of2/a.
k
C AB
0
Monoatomic Chain
-
7/30/2019 Lecture+12+MAK +Phonon
26/116
Note that:
In above equation n is cancelled out, this means that the eqn. of motionof all atoms leads to the same algebraic eqn. This shows that our trialfunction Un is indeed a solution of the eqn. of motion of n-th atom.
We started from the eqn. of motion of N coupled harmonic oscillators. Ifone atom starts vibrating it does not continue with constant amplitude, buttransfer energy to the others in a complicated way; the vibrations ofindividual atoms are not simple harmonic because of this exchangeenergy among them.
Our wavelike solutions on the other hand are uncoupled oscillationscalled normal modes;each k has a definite w given by above eqn. andoscillates independently of the other modes.
So the number of modes is expected to be the same as the number ofequations N. Lets see whether this is the case;
4sin
2
K ka
m
Monoatomic Chain
-
7/30/2019 Lecture+12+MAK +Phonon
27/116
pa
Nkkp
NapNa
22
Establish which wavenumbers are possible for our one dimensional chain.
Not all values are allowed because nth atom is the same as the (N+n)th as
the chain is joined on itself. This means that the wave eqn. of
must satisfy the periodic boundary condition
which requires that there should be an integral number of wavelengths in
the length of our ring of atoms
Thus, in a range of 2/a of k, there are N allowed values of k.
0expn nu A i kx t
pNa
nNn uu
Monoatomic Chain
-
7/30/2019 Lecture+12+MAK +Phonon
28/116
What is the physical significance of wave numbers
outside the range of ?2/a
x
Monoatomic Chain
un
un
x
a
Monoatomic Chain
-
7/30/2019 Lecture+12+MAK +Phonon
29/116
This value of k corresponds to themaximum frequency; alternate atoms
oscillate in antiphase and the waves at thisvalue of k are standing waves.
7 2 84 7 1.1474 7
4
aa ka a a
7 2 63 7 0.8573 7
3
aa ka a a
22 ;a k ka
White line :
Green line :
-k relation
Monoatomic Chain
u
n
x
u
n
a
Monoatomic Chain
-
7/30/2019 Lecture+12+MAK +Phonon
30/116
The pointsA and C both have same
frequency and same atomic displacements
They are waves moving to the left.
The green line corresponds to the point B
in dispersion diagram.
The point B has the same frequency and
displacement with that of the points A and C
with a difference.
The point B represents a wave moving to
the right since its group velocity (d/dk)>0.
The points A and C are exactly
equivalent; adding any multiple of2/a to k does not change the
frequency and its group velocity, so
point A has no physical significance.
k=/a has special significance
=90o
x
2 2nn n nk a
2 2 sin90
1
a d d a
Bragg reflection can be obtained at
k= n/a
2 24 sin2
kam K
For the whole range of k ()
Monoatomic Chain
u
n
x
u
n
a
-/a k
KV
m
K
s/
2
k
C AB
0
-k relation (dispertion diagram)
/a 2/a
At th b i i f th h t
-
7/30/2019 Lecture+12+MAK +Phonon
31/116
At the beginning of the chapter,
in the long wavelength limit, the
velocity of sound waves has
been derived as
Using elastic properties, lets see
whether the dispersion relationleads to the same equation in the
long limit.
1kaIf is very long; so sin ka ka
c Ka
2Kam
m
a
KV as k m
2 22 44
k am K
cVs
Monoatomic Chain
-
7/30/2019 Lecture+12+MAK +Phonon
32/116
Since there is only one possible propagation directionand one polarization direction, the 1D crystal has onlyone sound velocity.
In this calculation we only take nearest neighborinteraction although this is a good approximation for theinert-gas solids, its not a good assumption for manysolids.
If we use a model in which each atom is attached by
springs of different spring constant to neighbors atdifferent distances many of the features in abovecalculation are preserved. Wave equation solution still satisfies.
The detailed form of the dispersion relation is changed but is still periodic function of k with period 2/a
Group velocity vanishes at k=()/a There are still N distinct normal modes
Furthermore the motion at long wavelengths corresponds tosound waves with a velocity given by (velocity formul)
Monoatomic Chain
-
7/30/2019 Lecture+12+MAK +Phonon
33/116
Chain of two types of atom Two different types of atoms of masses M and m are
connected by identical springs of spring constant K;
Un-2 Un-1 Un Un+1 Un+2
K K K K
M Mm Mm a)
b)
(n-2) (n-1) (n) (n+1) (n+2)
a
This is the simplest possible model of an ionic crystal.
Since a is the repeat distance, the nearest neighbors
separations is a/2
Chain of two types of atom
-
7/30/2019 Lecture+12+MAK +Phonon
34/116
We will consider only the first neighbour interaction although it
is a poor approximation in ionic crystals because there is along range interaction between the ions.
The model is complicated due to the presence of two differenttypes of atoms which move in opposite directions.
Our aim is to obtain -k relation for diatomic lattice
Chain of two types of atom
Two equations of motion must be written;
One for mass M, and
One for mass m.
Chain of two types of atom
-
7/30/2019 Lecture+12+MAK +Phonon
35/116
Chain of two types of atom
M m M m M
Un-2Un-1 Un Un+1 Un+2
Equation of motionfor mass M (nth):
mass x acceleration = restoring force
Equation of motion for mass m (n-1)th:
..
1 1( ) ( )n n n n nu K u u K u u
1 1( 2 )n n nK u u u
..
1 1 2( ) ( )n n n n nmu K u u K u u
..
1 2( 2 )n n n nmu K u u u
-1
-1
Chain of two types of atom
-
7/30/2019 Lecture+12+MAK +Phonon
36/116
Chain of two types of atom
M m M m M
Un-2Un-1 Un Un+1 Un+2
0
/ 2nx na
0
expn nu A i kx t
Offer a solution for the mass M
For the mass m;
: complex number which determines the relative amplitudeand phase of the vibrational wave.
0expn nu A i kx t
..
2 0expn nu A i kx t
-1
Chain of two types of atom
-
7/30/2019 Lecture+12+MAK +Phonon
37/116
..
1 1( 2 )n n n nu K u u u
1 12 22 2 22
k n a k n akna knai t i t i t i t
MAe K Ae Ae Ae
For nth atom (M):
2 2 2 2 22 22kna kna kna knaka kai t i t i t i t i i
MAe K Ae e Ae Ae e
Cancel common terms
22 1 cos
2
kaM K
2cosix ix
e e x 2 2 22ka kai i
M K e e
Chain of two types of atom
Chain of two types of atom
-
7/30/2019 Lecture+12+MAK +Phonon
38/116
2
2 2 2 2 22 2 22
kna kna kna knaka ka kai t i t i t i t i i i
mAe e K Ae Ae e Ae e
..
1 1 2( 2 )n n n nmu K u u u
1 1 22 2 22 2 2
k n a k n a k n aknai t i t i ti t
A me K Ae Ae Ae
Cancel common terms
For the (n-1)th atom (m)
Chain of two types of atom
2 2 cos2
kam K
2cosix ixe e x
2 2 21 2
ka kai i
ikame K e e
2 2 22ka ka
i i
m K e e
-
7/30/2019 Lecture+12+MAK +Phonon
39/116
Chain of two types of atom
-
7/30/2019 Lecture+12+MAK +Phonon
40/116
The two roots are;
2 2 2 44 (1 cos ( )) 2 ( ) 02
kaK K m M Mm
2 2 2 2 44 cos ( ) 4 2 ( )2
kaK K K M m Mm
24 2 2 sin ( / 2)2 ( ) 4 0
m M kaK K
mM mM
22 2 1/ 2( ) 4sin ( / 2)[( ) ]
K m M m M kaK
mM mM mM
2
2
2 cos( / 2) 2
2 2 cos( / 2)
K ka K M
K m K ka
yp
2
1,2
4
2
b b ac
x a
Chain of two types of atom
-
7/30/2019 Lecture+12+MAK +Phonon
41/116
versus k relation for diatomic chain;
yp
Normal mode frequencies of a chain of two types of atoms.
At A, the two atoms are oscillating in antiphase with their centre ofmass at rest;
at B, the lighter mass m is oscillating and M is at rest;
at C, M is oscillating and m is at rest.
If the crystal contains N unit cells we would expect to find2N normal modes of vibrations and this is the total number ofatoms and hence the total number of equations of motion for
mass M and m.
0 /a 2/a/a k
A
B
C
Chain of two types of atom
-
7/30/2019 Lecture+12+MAK +Phonon
42/116
As there are two values of for each value of k, the
dispersion relation is said to have two branches;
yp
Upper branch is due to the
+ve sign of the root.
Lower branch is due to the
-ve sign of the root.
Optical Branch
Acoustical Branch
The dispersion relation is periodic in k with a period
2 /a = 2 /(unit cell length).
This result remains valid for a chain of containing an
arbitrary number of atoms per unit cell.
0 /a 2/a/a k
A
B
C
Chain of two types of atom
-
7/30/2019 Lecture+12+MAK +Phonon
43/116
Lets examine the limiting solutions at 0, A, B and C.
In long wavelength region (ka1); sin(ka/2) ka/2 in -k.
2 2 22
1 12( )
K m M mMk a
mM m M
22 2 1/ 2
1,2( ) 4sin ( / 2)[( ) ]K m M m M kaK
mM mM mM
1 2
2 2 22 4
4
K m M m M k aK
mM mM mM
1 2
2 2
21 1
K m M mMk a
mM m M
Use Taylor expansion: for small x 1 21 1 2x x
Chain of two types of atom
-
7/30/2019 Lecture+12+MAK +Phonon
44/116
Taking +ve root; sinka1
22
2 cos( / 2)
K M
K ka
2 2 2 22min . 22( ) 2( )
acus
K m M mMk a Kk a
mM m M m M
1 M
m
(max value of optical branch)
Taking -ve root; (min value of acoustical brach)
By substituting these values of in (relative amplitude)
equation and using cos(ka/2) 1 for ka1 we find the
corresponding values of as;
OR
2max2
opt
K m M
mM
Chain of two types of atom
-
7/30/2019 Lecture+12+MAK +Phonon
45/116
1
Substitute into relative amplitude
2
22 cos( / 2)
K MK ka
ac
2 22
minK(k a )
2(m M)
This solution represents long-wavelength sound waves in the
neighborhood of point 0 in the graph; the two types of atoms
oscillate with same amplitude and phase, and the velocity ofsound is
k
A
B
C
Optical
Acoustical
1/ 2
2( )s
w Kv a
k m M
ac
2
min
0 /a 2/a/a
Chain of two types of atom
-
7/30/2019 Lecture+12+MAK +Phonon
46/116
This solution corresponds to point A in
dispersion graph. This value of shows
that the two atoms oscillate inantiphase with their center of mass at
rest.
M
m
22
2 cos( / 2)
K M
K ka
op
2
max
2K(m M)
mM
Substitute into relative amplitude we obtain,op2
max
0 /a 2/a/ak
A
B
C
Optical
Acoustical
Chain of two types of atom
-
7/30/2019 Lecture+12+MAK +Phonon
47/116
The other limiting solutions of equation 2 are for ka= ,
i.e sin(ka/2)=1. In this case
1/ 22
2
max
( ) 4ac
K m M M mK
Mm Mm Mm
( ) ( )K m M K M m
Mm
2
max
2
ac
K
M OR 2
min
2
op
K
m (C) (B)
At max.acoustical point C, M oscillates and m is at rest.
At min.optical point B, m oscillates and M is at rest.
-
7/30/2019 Lecture+12+MAK +Phonon
48/116
Acoustic/Optical Branches
The acoustic branch has this name because it gives rise tolong wavelength vibrations - speed of sound.
The optical branch is a higher energy vibration (the frequencyis higher, and you need a certain amount of energy to excitethis mode). The term optical comes from how these werediscovered - notice that if atom 1 is +ve and atom 2 is -ve, thatthe charges are moving in opposite directions. You can excitethese modes with electromagnetic radiation (ie. The oscillating
electric fields generated by EM radiation)
-
7/30/2019 Lecture+12+MAK +Phonon
49/116
Transverse optical mode for
diatomic chain
Amplitude of vibration is strongly exaggerated!
-
7/30/2019 Lecture+12+MAK +Phonon
50/116
Transverse acoustical mode fordiatomic chain
-
7/30/2019 Lecture+12+MAK +Phonon
51/116
What is phonon? Consider the regular lattice of atoms in a uniform solid
material.
There should be energy associated with the vibrations of theseatoms.
But they are tied together with bonds, so they can't vibrate
independently. The vibrations take the form of collective modes which
propagate through the material.
Such propagating lattice vibrations can be considered to besound waves.
And their propagation speed is the speed of sound in thematerial.
Phonon
-
7/30/2019 Lecture+12+MAK +Phonon
52/116
The vibrational energies of molecules are quantized and
treated as quantum harmonic oscillators. Quantum harmonic oscillators have equally spaced
energy levels with separationE = h.
So the oscillators can accept or lose energy only indiscrete units of energy h.
The evidence on the behaviour of vibrational energy inperiodic solids is that the collective vibrational modes canaccept energy only in discrete amounts, and thesequanta of energy have been labelled "phonons".
Like the photons of electromagnetic energy, they obeyBose-Einstein statistics.
-
7/30/2019 Lecture+12+MAK +Phonon
53/116
-
7/30/2019 Lecture+12+MAK +Phonon
54/116
sphonon
hE
PHONONS
Quanta of lattice vibrations
Energies of phonons are
quantized
~a0=10-10m
phonon
hp
PHOTONS
Quanta of electromagneticradiation
Energies of photons arequantized as well
photon
hcE
~10-6m
photon
hp
-
7/30/2019 Lecture+12+MAK +Phonon
55/116
Energy of harmonic oscillator
Obtained by in a classical way of considering the normal modes
that we have found are independent and harmonic.
2
1nn
Make a transition to Q.M.
Represents equally spaced
energy levels
Energy, E
Energy levels of atoms
vibrating at a single
frequency
-
7/30/2019 Lecture+12+MAK +Phonon
56/116
It is possible to consider as constructed by adding n excitation
quanta each of energy to the ground state.
n
2
10
A transition from a lower energy level to a higher energy level.
2
1
2
112 nn
2 1unity
n n
absorption of phonon
-
7/30/2019 Lecture+12+MAK +Phonon
57/116
The converse transition results an emission of phonon
with an energy .
Phonons are quanta of lattice vibrations with an
angular frequency of .
Phonons are not localized particles.
Its momentum is exact, but position can not be
determined because of the uncertainity princible.
However, a slightly localized wavepacket can be
considered by combining modes of slightly different
and .
-
7/30/2019 Lecture+12+MAK +Phonon
58/116
Assume waves with a spread of k of ; so this wavepacket will
be localized within 10 unit cells.
a10
This wavepacket will represent a fairly localized phonon moving with
group velocity .dk
d
Phonons can be treated as localized particles within some limits.
-
7/30/2019 Lecture+12+MAK +Phonon
59/116
1D crystals
kMultiply by
k
Energy of
phonons
Crystal momentum
Phonons are not conserved
They can be created and destroyed during collisions .
-
7/30/2019 Lecture+12+MAK +Phonon
60/116
Thermal energy and lattice vibrations
Atoms vibrate about their equilibrium position.
They produce vibrational waves.
This motion is increased as the temperature is
raised.
In a solid, the energy associated with this vibration and perhaps also with
the rotation of atoms and molecules is called as thermal energy.
Note: In a gas, the translational motion of atoms and molecules
contribute to this energy.
-
7/30/2019 Lecture+12+MAK +Phonon
61/116
Therefore, the concept of thermal energy is fundamental to an
understanding many of the basic properties of solids. We would like to
know:
What is the value of this thermal energy?
How much is available to scatter a conduction electron in a metal;
since this scattering gives rise to electrical resistance. The energy can be used to activate a crystallographic or a
magnetic transition.
How the vibrational energy changes with temperature since this
gives a measure of the heat energy which is necessary to raise thetemperature of the material.
Recall that the specific heat or heat capacity is the thermal energy
which is required to raise the temperature of unit mass or 1gmole
by one Kelvin.
-
7/30/2019 Lecture+12+MAK +Phonon
62/116
The energy given to lattice vibrations is the dominantcontribution to the heat capacity in most solids. In non-magnetic
insulators, it is the only contribution.
Other contributions;
In metals from the conduction electrons.
n magnetic materials from magneting ordering.
Atomic vibrations leads to band of normal mode frequencies from zero
up to some maximum value. Calculation of the lattice energy and heatcapacity of a solid therefore falls into two parts:
i) the evaluation of the contribution of a single mode, and
ii) the summation over the frequency distribution of the modes.
Heat capacity from Lattice vibrations
-
7/30/2019 Lecture+12+MAK +Phonon
63/116
n
n
nP _
Avarage energy of a harmonicoscillator and hence of a lattice
mode of angular frequency at
temperature T Energy of oscillator
1
2n n
The probability of the oscillator being in this
level as given by the Boltzman factor
exp( / )n Bk T
Energy and heat capacity of a harmonicoscillator, Einstein Model
-
7/30/2019 Lecture+12+MAK +Phonon
64/116
0
/ 2 3 / 2 5 / 2
/ 2 / 2 /
/ 2 / 1
1exp[ ( ) ]2
.....
(1 .....
(1 )
B B B
B B B
B B
n B
k T k T k T
k T k T k T
k T k T
z nk T
z e e e
z e e e
z e e
_ 0
0
1 1exp /
2 21
exp /2
B
n
B
n
n n k T
n k T
(*)
According to the Binomial expansion for x1 where / Bx k T
n
n
nP _
Eqn (*) can be written
-
7/30/2019 Lecture+12+MAK +Phonon
65/116
_
/
1
2 1Bk T
e
'
(ln )x
xx x
_2 2
/ 2_
2/
_/ 2 /2
_ /2
/
2 2_2
2 2 /
1(ln )
ln1
ln ln 1
ln 12
2
4 1
B
B
B B
B
B
B
B B
k T
B k T
k T k T
B
k TB
B
k TB
B B
B k TB
zk T k T z
z T T
ek TT e
k T e eT
k T eT k T T
ke
k k Tk T
k T e
/
/
1
2 1
B
B
k T
k T
e
e
-
7/30/2019 Lecture+12+MAK +Phonon
66/116
_
/
1
2 1Bk T
e
This is the mean energy of phonons.The first term in the above
equation is the zero-point energy. As we have mentioned before even
at 0K atoms vibrate in the crystal and have zero-point energy. This is
the minimum energy of the system.
The avarage number of phonons is given by Bose-Einstein
distribution as
1
1)(
TBken
(number of phonons) x (energy of phonon)=(second term in )_
The second term in the mean energy is the contribution of
phonons to the energy.
-
7/30/2019 Lecture+12+MAK +Phonon
67/116
Mean energy of a
harmonic oscillatoras a function of T
low temperature limit
T
2
1
TkB
TkB
12
1_
TBke
2
1_ Zero point energy
Since exponential term
gets bigger
-
7/30/2019 Lecture+12+MAK +Phonon
68/116
is independent of frequency ofoscillation.
This is the classical limit because the
energy steps are now small compared with
the energy of the harmonic oscillator.
So that is the thermal energy of the
classical 1D harmonic oscillator.
..........!2
12
x
xex
Tke B
TBk
1
112
1_
TkB
_
12
Bk T
_
Bk T
high temp erature l imi t
T
2
1
TkB Mean energy of a
harmonic oscillator as
a function of T
Bk T
-
7/30/2019 Lecture+12+MAK +Phonon
69/116
Heat Capacity C
Heat capacity C can be found by differentiating the average energy ofphonons of
12
1_
TBke
2
2
1
k TB
k TB
B
B
v
ke
k TdC
dTe
2
2 2
1
k TB
k TB
v B
B
eC k
k Te
k
2
2
1
T
T
v B
eC k
Te
Let
-
7/30/2019 Lecture+12+MAK +Phonon
70/116
2
2
1
T
T
v B
eC k
Te
Area= 2
T
Bk
B
k
Specific heat vanishes
exponentially at low Ts and tends to
classical value at high temperatures.
The features are common to allquantum systems; the energy tends
to the zero-point-energy at low Ts
and to the classical value of
Boltzmann constant at high Ts.
vC
vC
k
where
Plot of as a function of T
-
7/30/2019 Lecture+12+MAK +Phonon
71/116
,T K
3R
This range usually includes RT.From the figure it is seen that Cv is
equal to 3R at high temperatures
regardless of the substance. This fact
is known as Dulong-Petit law. This law
states that specific heat of a givennumber of atoms of any solid is
independent of temperature and is the
same forall materials!
vC
Specific heat at constant volume depends on temperature as shown infigure below. At high temperatures the value of Cv is close to 3R,
where R is the universal gas constant. Since R is approximately 2
cal/K-mole, at high temperatures Cv is app. 6 cal/K-mole.
Plot of as a function of TvC
Classical theory of
-
7/30/2019 Lecture+12+MAK +Phonon
72/116
Classical theory of
heat capacity of solids
The solid is one in which each atom is bound to its side bya harmonic force. When the solid is heated, the atoms vibrate
around their sites like a set of harmonic oscillators. The
average energy for a 1D oscillator is kT. Therefore, the
averaga energy per atom, regarded as a 3D oscillator, is 3kT,and consequently the energy per mole is
=
where N is Avagadros number, kB is Boltzmann constant and
R is the gas constant. The differentiation wrt temperaturegives;
3 3BNk T RT
23 233 3 6.02 10 ( / ) 1.38 10 ( / )vC R atoms mole J K
24.9 ;1 0.2388 6
( ) ( )
J CalCv J Cal Cv
K mole K mole
v
dC
dT
Einstein heat capacity of solids
-
7/30/2019 Lecture+12+MAK +Phonon
73/116
p y The theory explained by Einstein is the first quantum theory of solids.
He made the simplifying assumption that all 3N vibrational modes of
a 3D solid of N atoms had the same frequency, so that the whole solidhad a heat capacity 3N times
In this model, the atoms are treated as independent oscillators, but
the energy of the oscillators are taken quantum mechanically as
This refers to an isolated oscillator, but the atomic oscillators in a solid
are not isolated.They are continually exchanging their energy with
their surrounding atoms.
Even this crude model gave the correct limit at high temperatures, a
heat capacity of
Dulong-Petit law where R is universal gas constant.
2
2
1
T
T
v B
eC k
Te
3 3BNk R
-
7/30/2019 Lecture+12+MAK +Phonon
74/116
At high temperatures, all crystalline solids have a specific heat of
6 cal/K per mole; they require 6 calories per mole to raise their
temperature 1 K.
This arrangement between observation and classical theorybreak
down if the temperature is not high.
Observations show that at room temperatures and belowthe
specific heat of crystalline solids is not a universal constant.
6cal
Kmol
Bk
vC
T
3vC R
In all of these materials
(Pb,Al, Si,and Diamond)
specific heat approaches
constant value asymptoticallyat high Ts. But at low Ts, the
specific heat decreases
towards zero which is in a
complete contradiction with
the above classical result.
-
7/30/2019 Lecture+12+MAK +Phonon
75/116
The Discrepancy of Einstein model
Einstein model also gave correctly a specific heat tending to
zero at absolute zero, but the temperature dependence near T=0
did not agree with experiment.
Taking into account the actual distribution of vibration
frequencies in a solid this discrepancy can be accounted using
one dimensional model of monoatomic lattice
-
7/30/2019 Lecture+12+MAK +Phonon
76/116
Density of States
According to Quantum Mechanics if a particle is constrained; the energy of particle can only have special discrete energy
values.
it cannot increase infinitely from one value to another.
it has to go up in steps.
-
7/30/2019 Lecture+12+MAK +Phonon
77/116
These steps can be so small depending on the system that the
energy can be considered as continuous. This is the case of classical mechanics.
But on atomic scale the energy can only jump by a discrete
amount from one value to another.
Definite energy levels Steps get small Energy is continuous
-
7/30/2019 Lecture+12+MAK +Phonon
78/116
In some cases, each particular energy level can be
associated with more than one different state (orwavefunction )
This energy level is said to be degenerate.
The density of states is the number of discrete states
per unit energy interval, and so that the number of states
between and will be .
( )
( )d d
There are two sets of waves for solution;
Running waves
-
7/30/2019 Lecture+12+MAK +Phonon
79/116
Running waves
Standing waves
0 2
L
2
L
4
L
4
L
6
L
k
These allowed k wavenumbers corresponds to the running
waves; all positive and negative values of k are allowed. By
means ofperiodic boundary condition
2 2 2NaL Na p k p k p
p k Na L
an integer
Length of
the 1D
chain
Running waves:
These allowed wavenumbers are uniformly distibuted in k at a
density of between k and k+dk. R k
running waves
2RL
k dk dk
Standing waves: 45 3
-
7/30/2019 Lecture+12+MAK +Phonon
80/116
In some cases it is more suitable to use standing waves,i.e. chain
with fixed ends. Therefore we will have an integral number of half
wavelengths in the chain;
Standing waves:
0L
2
L
6
L
LL
2 2;
2 2
n n nL k k k
L L
3
L
7
L
k0
L
2
L
3
L
These are the allowed wavenumbers for standing waves; onlypositive values are allowed.
2k p
L
for
running wavesk p
L
for
standing waves
These allowed ks are uniformly distributed between k and k+dk
-
7/30/2019 Lecture+12+MAK +Phonon
81/116
at a density of
( )SL
k dk dk
2
R
Lk dk dk
DOS of standing wave
DOS of running wave
( )S k
The density of standing wave states is twice that of the running waves.
However in the case of standing waves only positive values are
allowed
Then the total number of states forboth running and standing waveswill be the same in a range dk of the magnitude k
The standing waves have the same dispersion relation as running
waves, and fora chain containing N atoms there are exactly N distinct
states with k values in the range 0 to ./a
-
7/30/2019 Lecture+12+MAK +Phonon
82/116
modes with frequency from to +d corresponds
modes with wavenumber from kto k+dk
The density of states per unit frequency range g():
The number of modes with frequencies and +d will be
g()d.
g() can be written in terms ofS(k) and R(k).
dn
dR
;( ) ( )Rdn k dk g d ( ) ( )Sdn k dk g d
-
7/30/2019 Lecture+12+MAK +Phonon
83/116
Choose standing waves to obtain ( )g
Lets remember dispertion relation for 1D monoatomic lattice
2 24 sin2
K ka
m 2 sin
2
K ka
m
dkddkd
d
dk
2cos2 2
a K ka
m cos 2K ka
a m 1cos
2
K kaa
m
1 1cos
2
mkaa K
R
( ) ( )Sg k
( ) ( )Sg k
( ) ( )Sg k 1 1
cos / 2
m
a K ka
-
7/30/2019 Lecture+12+MAK +Phonon
84/116
( )gNa
cos / 2a K ka
2cos 1 sin
2 2
ka ka
2 2 2sin cos 1 cos 1 sinx x x x
( ) ( )Sg k 2
1 1 4
41 sin
2
m
a K ka
Multibly and divide
( ) ( )Sg k 2
1 2
4 4sin
2
a K K ka
m m
( )SL
k dk dk
Lets remember:
( )gL
2 2
max
2 1
a
L Na
2
max
4K
m
2 24 sin2
K ka
m
( )2
gN
1/ 2
2 2
max
True dens ity o f states
1/ 2
2 22( )N
g
( )g
-
7/30/2019 Lecture+12+MAK +Phonon
85/116
constant density of states
N mK
max 2
K
m
K
m
True density of states by
means of above equation
max( )g
True DOS(density of states) tends to infinity at ,since the group velocity goes to zero at this value of .
Constant density of states can be obtained by ignoring the
dispersion of sound at wavelengths comparable to atomic spacing.
max 2
K
m /d dk
The energy of lattice vibrations will then be found by
-
7/30/2019 Lecture+12+MAK +Phonon
86/116
gy y
integrating the energy of single oscillator over the distribution
of vibration frequencies. Thus
/0
1
2 1kTg d
e
1/ 2
2 2max2N
Mean energy of a harmonic
oscillator
One can obtain same expression of by means of using
running waves.
for 1D
It should be better to find 3D DOS in order to compare the
results with experiment.
( )g
3D DOS
-
7/30/2019 Lecture+12+MAK +Phonon
87/116
3D DOS
Lets do it first for 2D
Then for 3D.
Consider a crystal in the shape of 2D box with crystal lengths of L.
+
+
+ -
-
-
L0
L
y
x
L
Standing wave pattern for a
2D box
Configuration in k-space
xk
yk
L
Lets calculate the number of modes within a range of
-
7/30/2019 Lecture+12+MAK +Phonon
88/116
wavevector k.
Standing waves are choosen but running waves will lead
same expressions.
Standing waves will be of the form
Assuming the boundary conditions of
Vibration amplitude should vanish at edges of
Choosing
0 sin sinx yU U k x k y
0; 0; ;x y x L y L
;x yp q
k kL L
positive integer
y yk
-
7/30/2019 Lecture+12+MAK +Phonon
89/116
+
+
+ -
-
-
L0
L
x
The allowed k values lie on a square lattice of side inthe positive quadrant ofk-space.
These values will so be distributed uniformly with a density
of per unit area.
This result can be extended to 3D.
Standing wave pattern for
a 2D box
L
L
Configuration in k-space
/L
2/L
xk
LOctant of the crystal:
-
7/30/2019 Lecture+12+MAK +Phonon
90/116
L
L
y
kx,ky,kz(all have positive values)
The number of standing waves;
3
3 3 3
3s
L Vk d k d k d k
/L
k
dk
zk
yk
xk
21 48
k dk
3 231
48
s
Vk d k k dk
2
3
22
sVkk d k dk
2
22S
Vkk
2Vk
-
7/30/2019 Lecture+12+MAK +Phonon
91/116
is a new density of states defined as the
number of states per unit magnitude of in 3D.This eqn
can be obtained by using running waves as well.
(frequency) space can be related to k-space:
2
22
Vkk
g d k dk dk
g kd
Lets find C at low and high temperature by means of
using the expression of . g
High and Low Temperature Limits
-
7/30/2019 Lecture+12+MAK +Phonon
92/116
B
Tk
High and Low Temperature Limits
This result is true only if
At low Ts only lattice modes having low frequencies
can be excited from their ground states;
3 BNk T Each of the 3N lattice
modes of a crystal
containing N atomsdCdT
3 BC Nk
k
a0
Low frequency long
sound waves
sv k sv k
2
Vk dk g
1 1k dkv
and
-
7/30/2019 Lecture+12+MAK +Phonon
93/116
depends on the direction and there are two transverse,one longitudinal acoustic branch:
22
gd
s
s s
vk v d v
and
2
2
2
1
2
s
s
Vv
gv
at low Ts
sv
2 2
2 3 2 3 3
1 1 2
2 2s L T
V Vg g
v v v
Velocities of sound in
longitudinal and
transverse direction
1
g d
Zero point energy=
-
7/30/2019 Lecture+12+MAK +Phonon
94/116
/
02 1kT
g de
3
3
3
/
0 01 1
B
B
kT x
k Tx
k Td dx
e e
Zero point energy= z
2
/ 2 3 3
0
1 1 22 1 2kT L T
V de v v
3
2 3 3 /
0
1 2
2 1z kT
L T
Vd
v v e
B
xk T
Bk Tx
Bk Td dx
4
43 3
/ 3
0 0
15
1 1
B
kT x
k T xd dx
e e
4 4
2 3 3 3
1 2
2 15Bz
L T
k TV
v v
24 3
3 3 3
1 24
30v B
L T
d VC k T
dT v v
3
2
3 3
2 1 2
15
Bv B
L T
k TdC V k
dT v v
at low temperatures
How good is the Debye approximation
-
7/30/2019 Lecture+12+MAK +Phonon
95/116
g y pp
at low T?
3T
32
3 3
2 1 2
15
B
v B
L T
k TdC V k
dT v v
The lattice heat capacity of solids thus
varies as at low temperatures; this is
referred to as the Debye law.
Figure illustrates the excellent aggrement
of this prediction with experiment for a
non-magnetic insulator. The heat
capacity vanishes more slowly than theexponential behaviour of a single
harmonic oscillator because the vibration
spectrum extends down to zero
frequency.
3T
The Debye interpolation scheme
-
7/30/2019 Lecture+12+MAK +Phonon
96/116
The Debye interpolation scheme
The calculation of is a very heavy calculation for 3D, so itmust be calculated numerically.
Debye obtained a good approximation to the resulting heatcapacity by neglecting the dispersion of the acoustic waves, i.e.assuming
for arbitrary wavenumber. In a one dimensional crystal this isequivalent to taking as given by the broken line of density ofstates figure rather than full curve. Debyes approximation gives the
correct answer in either the high and low temperature limits, and thelanguage associated with it is still widely used today.
( )g
s k
( )g
-
7/30/2019 Lecture+12+MAK +Phonon
97/116
1. Approximate the dispersion relation of any branch by a linearextrapolation of the small k behaviour:
Einstein
approximationto thedispersion
Debye
approximationto thedispersion
vk
The Debye approximation has two main steps:
-
7/30/2019 Lecture+12+MAK +Phonon
98/116
2
3
9( )
D
Ng
2 3 3 3 31 2 3 9( ) 32 L T D DV N N
v v
Debye cut-off frequency
2. Ensure the correct number of modes by imposing a cut-offfrequency , above which there are no modes. The cut-offfreqency is chosen to make the total number of lattice modescorrect. Since there are 3N lattice vibration modes in a crystal
having N atoms, we choose so that
0
( ) 3D
g d N
2
2 3 3
1 2( ) ( )
2 L T
Vg
v v
2
2 3 3
0
1 2( ) 3
2
D
L T
Vd N
v v
3
2 3 3
1 2( ) 36
D
L T
V Nv v
D
D
D
2( ) /g
-
7/30/2019 Lecture+12+MAK +Phonon
99/116
The lattice vibration energy of
becomes
and,
First term is the estimate of the zero point energy, and all T dependence isin the second term. The heat capacity is obtained by differentiating above eqnwrt temperature.
/
0
1( ) ( )2 1Bk T
E g de
3 32
/ /3 30 0 0
9 1 9( )
2 1 2 1
D D D
B Bk T k T D D
N NE d d d
e e
3
/3
0
9 9
8 1
D
BD k T
D
N dE N
e
-
7/30/2019 Lecture+12+MAK +Phonon
100/116
dE
CdT
/2 4
23 2/
0
9
1
DB
B
k T
Dk T
D B
dE N eC ddT k T e
3
/3
0
9 98 1
D
BD k T
D
N dE Ne
Lets convert this complicated integral into an expression for
the specific heat changing variables to
and define the Debye temperature
x
D
B
xk T
DD
Bk
The heat capacity is
d kT
dx
kTx
-
7/30/2019 Lecture+12+MAK +Phonon
101/116
4 /2 4
23 2
0
9
1
D T x
B BD
xD B
k T k T dE N x eC dx
dT k T e
3
/ 4
2
0
91
DT
x
D Bx
D
T x eC Nk dx
e
The Debye prediction for lattice specific heat
DD
Bk
where
-
7/30/2019 Lecture+12+MAK +Phonon
102/116
3 /
2
0
9 3
D T
D D B B
D
TT C Nk x dx Nk
How does limit at high and low temperatures?
High temperature
DC
DT
2 3
12! 3!
x x xe x
4 4 4 2
2 2 2(1 ) (1 )
1 11
x
x
x e x x x x xxxe
X is always small
-
7/30/2019 Lecture+12+MAK +Phonon
103/116
DT
3 / 4
2
0
91
D T x
D D Bx
D
T x eT C Nk dx
e
3412
5B
D
D
Nk TC
How does limit at high and low temperatures?
Low temperature
44 /15
DC
For low temperature the upper limit of the integral is infinite; the
integral is then a known integral of .
We obtain the Debye law in the form3T
Lattice heat capacity due to Debye interpolation
scheme
-
7/30/2019 Lecture+12+MAK +Phonon
104/116
scheme
Figure shows the heat capacity
between the two limits of high and lowT as predicted by the Debyeinterpolation formula.
3 B
C
Nk T
Because it is exact in both high and low Tlimits the Debye formula gives quite a good
representation of the heat capacity of most solids,
even though the actual phonon-density of states
curve may differ appreciably from the Debye
assumption.Debye frequency and Debye temperature scale with the velocity of sound in
the solid. So solids with low densities and large elastic moduli have high . Values offor various solids is given in table. Debye energy can be used to estimate
the maximum phonon energy in a solid.
Lattice heat capacity of a solid as
predicted by the Debye interpolation
scheme
3 / 4
2
0
91
D T x
D Bx
D
T x eC Nk dx
e
/ DT
1
1
Solid Ar Na Cs Fe Cu Pb C KCl
93 158 38 457 343 105 2230 235( )D K
D
D
D
Anharmonic Effects
-
7/30/2019 Lecture+12+MAK +Phonon
105/116
Anharmonic Effects
Any real crystal resists compression to a smaller volume than its equilibrium valuemore strongly than expansion due to a larger volume.
This is due to the shape of the interatomic potential curve.
This is a departure from Hookes law, since harmonic application does not producethis property.
This is an anharmonic effect due to the higher order terms in potential which areignored in harmonic approximation.
Thermal expansion is an example to theanharmonic effect. In harmonic approximation phonons do not interact with each other, in the absence
of boundaries, lattice defects and impurities (which also scatter the phonons), thethermal conductivity is infinite.
In anharmonic effect phonons collide with each other and these collisions limitthermal conductivity which is due to the flow of phonons.
2
2
2( ) ( ) ....................
2r a
r a d VV r V a
dr
Phonon phonon collisions
-
7/30/2019 Lecture+12+MAK +Phonon
106/116
Phonon-phonon collisions
The coupling of normal modes by the unharmonic terms in theinteratomic forces can be pictured as collisions between the phonons
associated with the modes. A typical collision process of
phonon1
phonon2
1 1, k
2 2, k
3 3, k
After collision another phonon is
produced
3 1 2k k k
3 1 2k k k
3 1 2
3 1 2
and
conservation of energy
conservation of momentum
Phonons are represented by wavenumbers with
-
7/30/2019 Lecture+12+MAK +Phonon
107/116
Phonon3 has k
a
; Phonon3 has andPhonon3=Phonon3k
a
12
k
a 0
a
3 '
Umklapp process
(due to anharmonic effects)
3
12
k
a 0
a
Normal process
3Longitudinal
Transverse
0n 0n
ka a
If lies outside this range add a suitable multible of to bring
it back within the range of . Then, becomes
3k
3 1 2
2nk k k
a
where , , and are all in the above range.
2
a
3 1 2k k k ka a
2k 3k1k
This phonon is indistinguishable
from a phonon with wavevector 3k
Thermal conduction by phonons
-
7/30/2019 Lecture+12+MAK +Phonon
108/116
Thermal conduction by phonons
A flow of heat takes place from a hotter region to a cooler regionwhen there is a temperature gradient in a solid.
The most important contribution to thermal conduction comes fromthe flow of phonons in an electrically insulating solid.
Transport property is an example of thermal conduction. Transport property is the process in which the flow of some quantity
occurs.
Thermal conductivity is a transport coefficient and it describes theflow.
The thermal conductivity of a phonon gas in a solid will becalculated by means of the elementary kinetic theory of the transportcoefficients of gases.
Kinetic theory
-
7/30/2019 Lecture+12+MAK +Phonon
109/116
Kinetic theory
In the elementary kinetic theory of gases, the steady state flux of a propertyin the z direction is
P_1
3
dPflux l
dz
Mean free path
Angular average
Constant average speed for molecules
In the simplest case where is the number density of particles the transport
coefficient obtained from above eqn. is the diffusion coeff icient .
If is the energy density then the flux W is the heat flow per unit area so that_ _
1 13 3
dE dE dT W l ldz dT dz
Now is the specific heat per unit volume, so that the thermal
conductivity;_1
3K l C
P_1
3D l
P E
/dE dT C
Works well for a phonon gas
Heat conduction in a phonon and real gasThe essential differences between the processes of heat
-
7/30/2019 Lecture+12+MAK +Phonon
110/116
conduction in a phonon and real gas;
Phonon gas Real gas
Speed is approximately constant.
Both the number density and energy
density is greater at the hot end.
Heat flow is primarily due to phononflow with phonons being created at the
hot end and destroyedat the cold end
No flow of particles
Average velocity and kinetic energy per
particle are greater at the hot end, but the
number density is greater at the cold end,
and the energy density is uniform due to theuniform pressure.
Heat flow is solely by transfer of kinetic
energy from one particle to another in
collisions which is a minor effect in phonon
case.hot cold
hot cold
-
7/30/2019 Lecture+12+MAK +Phonon
111/116
Temperature dependence of thermal conductivity K
_1
3K l C Approximately equal to
velocity of sound and so
temperature independent.Vanishes exponentially at
low Ts and tends to classical
value at high TsB
k
?
Temperature dependence of phonon mean free length is determined by
phonon-phonon collisions at low temperatures
Since the heat flow is associated with a flow of phonons, the most effective
collisions for limiting the flow are those in which the phonon group velocity
is reversed. It is the Umklapp processes that have this property, and these are
important in limiting the thermal conductivity
-
7/30/2019 Lecture+12+MAK +Phonon
112/116
Experimental results do tend towards this behaviour at high temperatures as
sho n in fig re (a)
-
7/30/2019 Lecture+12+MAK +Phonon
113/116
shown in figure (a).
1
T
5 10 20 50 100
10
0
10-1
( )T K
2 5 10 20 50 100
( )T K
10
0
10-1 3T
(a)Thermal conductivity of a quartz
crystal
(b)Thermal conductivity of artificial
sapphire rods of different diameters
Conduction at intermediate temperatures
-
7/30/2019 Lecture+12+MAK +Phonon
114/116
Conduction at intermediate temperatures
Referring figure aAt T< ; the conductivity rises more steeply with falling temperature,although the heat capacity is falling in this region. Why?
This is due to the fact that Umklapp processes which will only occur if there arephonons of sufficient energy to create a phonon with . So
Energy of phonon must be the Debye energy ( )
The energy of relevant phonons is thus not sharply defined but their number isexpected to vary roughly as
when ,
where b is a number of order unity 2 or 3. Then
This exponential factor dominates any low power of T in thermal conductivity,
such as a factor of from the heat capacity.
D
3/k a
Dk
/D bTe DT
/D bTl e3T
Conduction at low temperatures
-
7/30/2019 Lecture+12+MAK +Phonon
115/116
p
for phonon-phonon collisions becomes very long at low Ts and eventuallyexceeds the size of the solid, because
number of high energy phonons necessary for Umklapp processes decayexponentially as
is then limited by collisions with the specimen surface, i.e.
Specimen diameter
T dependence of K comes from which obeys law in this region
Temperature dependence of dominates.
l
/D bTe
3
T
l
l
vC
3412
5B
D
D
Nk TC
vC
Size effect
-
7/30/2019 Lecture+12+MAK +Phonon
116/116
When the mean free path becomes comparable to the dimensions of the sample,transport coefficient depends on the shape and size of the crystal. This is known as asize effect.
If the specimen is not a perfect crystal and contains imperfections such as dislocations,grain boundaries and impurities, then these will also scatter phonons. At the verylowest Ts the dominant phonon wavelength becomes so long that these imperfections
are not effective scatterers, so;the thermal conductivity has a dependence at these temperatures.
The maximum conductivity between and region is controlled byimperfections.
For an impure or polycrystalline specimen the maximum can be broad and low[figure (a) on pg 59], whereas for a carefully prepared single crystal, as illustrated infigure(b) on pg 59, the maximum is quite sharp and conductivity reaches a very highvalue, of the order that of the metallic copper in which the conductivity ispredominantly due to conduction electrons
3T
3T
/D bTe