chapter 4 (crystal dynamics i) [compatibility mode]

1

SOUND WAVESLATTICE VIBRATIONS OF 1D CRYSTALS

chain of identical atomschain of two types of atoms

LATTICE VIBRATIONS OF 3D CRYSTALSLATTICE VIBRATIONS OF 3D CRYSTALSPHONONSHEAT CAPACITY FROM LATTICE VIBRATIONSANHARMONIC EFFECTSTHERMAL CONDUCTION BY PHONONS

Crystal DynamicsConcern with the spectrum of characteristics vibrations of a crystalline solid.

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.

2

Crystal Dynamics

Th i t d t th t fThese introduces us to the concepts of forbidden and permitted frequency ranges, andelectronic spectra of solids

Crystal DynamicsIn 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, sometimesIn this chapter we discuss the nature of atomic motions, sometimesreferred to as lattice vibrations.

In crystal dynamics we will use the harmonic approximation , amplitudeof the lattice vibration is small. At higher amplitude some unharmoniceffects occur.

3

Crystal DynamicsOur calculations will be restricted to lattice vibrations of smallamplitude. Since the solid is then close to a position of stableequilibrium its motion can be calculated by a generalization of the

h d d l i l h i ill Th llmethod 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(Hooke’s Law).

There are some effects of nonlinearity or ‘anharmonicity’ for largeratomic displacementsatomic displacements.

Anharmonic effects are important for interactions between phononsand photons.

Crystal DynamicsAtomic motions are governed by the forces exerted on atomswhen they are displaced from their equilibrium positions.

To calculate the forces it is necessary to determine thewavefunctions and energies of the electrons within the crystal.Fortunately many important properties of the atomic motionscan be deduced without doing these calculations.

4

Hooke's LawOne of the properties of elasticity is that it takes about twice asmuch force to stretch a spring twice as far. That lineardependence of displacement upon stretching force is calleddepe de ce o d sp ace e upo s e c g o ce s ca edHooke's law.

kFIt takes twiceas much forcet t t hxkFspring .−=

↓FSpring constant k

to stretch aspring twiceas far.

↓F2

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.

Hooke’s Law

The inelastic Limit is the point at which permanent deformationoccurs, that is, after the elastic limit, if the force is taken off thesample, 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 nol h ld ) Alth h th t i t li htl diff tlonger holds). Although these two points are slightly different, wewill treat them as the same in this course.

5

SOUND WAVESMechanical 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 for

• It corresponds to the atomic vibrations with a long λ.

• Presence of atoms has no significance in this wavelengthmedium. There are two basic types of wave motion formechanical waves: longitudinal waves and transversewaves.

Longitudinal Waves

Presence of atoms has no significance in this wavelengthlimit, since λ>>a, so there will no scattering due to thepresence of atoms.

Transverse Waves

SOUND WAVESSound waves propagate through solids. This tells us thatwavelike lattice vibrations of wavelength long compared to theinteratomic spacing are possible The detailed atomic structureinteratomic 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 i f d i i di ti i t l it iAt 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 their velocity.

6

Elastic WavesA 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 are

f d l ireferred to as elastic waves.

• At the point x elastic displacement is U( ) d t i ‘ ’ i d fi d th

A

Elastic Wave Propagation (longitudinal) in a bar

U(x) and strain ‘e’ is defined as the change in length per unit length.

dUedx

=

x x+dx

According to Hooke’s law stress S (force per unit area) isproportional to the strain e.

A

Elastic Waves

To examine the dynamics of the bar, we choose an arbitrarysegment of length dx as shown above. Using Newton’s secondlaw, we can write for the motion of this segment,

.S C e=x x+dx

A

C = Young modulus

[ ]2

2( ) ( ) ( )uAdx S x dx S x At

ρ ∂= + −

∂

Mass x Acceleration Net Force resulting from stresses

7

Equation of motion

[ ]2

2( ) ( ) ( )uAdx S x dx S x At

ρ ∂= + −

∂

[ ]( ) ( ) SS x dx S x dx∂+

.S C e=d ued x

=

2

. uS Cx

∂=

∂

Elastic Waves

2 2

2 2

u uCt x

ρ ∂ ∂=

∂ ∂

[ ]( ) ( )S x dx S x dxx

+ − =∂

2

2.S uCx x

∂ ∂=

∂ ∂2 2

2 2( ) u uAdx C Adxt x

ρ ∂ ∂=

∂ ∂Cancelling common terms of Adx;

Which is the wave eqn. with an offered sol’n and velocity of sound waves ;t x∂ ∂ sol n and velocity of sound waves ;

( )i kx tu Ae ω−=– k = wave number (2π/λ)– ω = frequency of the wave – A = wave amplitude /

s

s

v k

v C

ω

ρ

=

=

v kω =

The relation connecting the frequency and wave number is known as the dispersion relation.

Dispersion Relation

sv kω =

ω Continuum

Discrete

• At small λ k → ∞ (scattering occurs)

• At long λ k → 0 (no scattering)

• When k increases velocitydecreases. As k increases further,the scattering becomes greater sincek

0

* Slope of the curve givesthe velocity of the wave.

the scattering becomes greater sincethe strength of scattering increasesas the wavelength decreases, andthe velocity decreases even further.

8

Speed of Sound Wave The speed with which a longitudinal wave moves through aliquid of density ρ is

L CV λνρ

= =C = Elastic bulk modulusρ = Mass density

• The velocity of sound is in general a function of the directionof propagation in crystalline materials.p p g y

• Solids will sustain the propagation of transverse waves, whichtravel more slowly than longitudinal waves.

• The larger the elastic modules and smaller the density, themore rapidly can sound waves travel.

Speed of sound for some typical solids

Solid StructureType

NearestNeighbourDistance

Densityρ

(k / 3)

Elastic bulkmodules

Y

Calculated WaveSpeed

Observedspeed of

sound

Sound Wave Speed

(A°) (kg/m3) (1010 N/m2) (m/s) (m/s)

Sodium B.C.C 3.71 970 0.52 2320 2250Copper F.C.C 2.55 8966 13.4 3880 3830Aluminum F.C.C 2.86 2700 7.35 5200 5110Lead F.C.C 3.49 11340 4.34 1960 1320Silicon Diamond 2 35 2330 10 1 6600 9150Silicon Diamond 2.35 2330 10.1 6600 9150Germanium Diamond 2.44 5360 7.9 3830 5400NaCl 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, covalentand ionic solids.

9

A lattice vibrational wave in a crystal is a repetitive and systematic sequence of atomic displacements of

longitudinal,transverse or

Sound Wave Speed

• They can be characterized by• An equation of motion for any displacement can be

transverse, orsome combination of the two

• They can be characterized by– A propagation velocity, v– Wavelength λ or wavevector – A frequency ν or angular frequency ω=2πν

produced by means of considering the restoring forceson displaced atoms.

• As a result we can generate a dispersion relationshipbetween frequency and wavelength or between angularfrequency and wavevector.

Lattice vibrations of 1D crystalChain of identical atoms

Atoms interact with a potential V(r) which can be written in Taylor’s series.

( )2 2

2( ) ( ) ...........2 r a

r a d VV r V adr

=

− ⎛ ⎞= + +⎜ ⎟

⎝ ⎠V(R)

0 r0=4

Repulsive This equation looks like as the potential energy associated of a spring with a spring constant :

VdK ⎟⎟⎞

⎜⎜⎛

=2

r R

0

Attractivemin

ardrK

=⎟⎟⎠

⎜⎜⎝

= 2

We should relate K with elastic modulus C:

KaC =

( )a

arCForce −×= )( arKForce −=

10

Monoatomic ChainThe 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

Start with the simplest caseof monoatomic linear chainwith only nearest neighbour

a a

Monoatomic Chain

with only nearest neighbourinteraction

• If one expands the energy near the equilibrium point for the nth

atom and use elastic approximation Newton’s equation becomes

Un-1 Un Un+1

)2( 11

..

−+ +−= nnnn uuuKum

atom and use elastic approximation, Newton s equation becomes

11

The force on the nth atom;

)( uuK −•The force to the right;

Monoatomic Chain

a a

)( 1 nn uuK −+

•The force to the left;

)( 1−− nn uuK

•The total force = Force to the right – Force to the left

Un-1 Un Un+1

0)2( 11

..=−−+ +− nnnn uuuKum

Eqn’s of motion of all atoms are of this form, only the value of ‘n’ varies

All atoms oscillate with a same amplitude A and frequency ω. Then we can offer a solution;

Monoatomic Chain

( )0expu A i kx tω⎡ ⎤= −⎣ ⎦

( ).

0expnn n

duu i A i kx tdt

ω ω⎡ ⎤= = − −⎣ ⎦

( )expn nu A i kx tω⎡ ⎤⎣ ⎦

( ) ( )2.. 2 2 0e pnd u i A i k tω ω⎡ ⎤⎣ ⎦

..2( ) ( )2 0

2 expnn nu i A i kx t

dtω ω⎡ ⎤= = −⎣ ⎦

2n nu uω= −

nax n =0nn unax +=Undisplaced

position

Displaced position

12

Monoatomic Chain Equation of motion for nth atom ..

1 1( 2 )n nn nmu K u u u+ −= − +

( ) ( ) ( ) ( )( )0 0 0 01 12

2n n n ni kx t i kx t i kx t i kx t

K A A AAω ω ω ω+ −− − − −( ) ( ) ( ) ( )( )e e 2 e em K A A AAω =− − +

kna ( 1)k n a−( 1)k n a+kna

( ) ( ) ( ) ( )( )2e e 2 e e

i kna t i kna ka t i kna t i kna ka tm K A A AA

ω ω ω ωω

− + − − − −− = − +

( ) ( ) ( ) ( )( )2e e e 2 e e e

ika ikai kna t i kna t i kna t i kna tm K A A AA

ω ω ω ωω

−− − − −− = − +

Cancel Common terms

( )2 e 2 eika ikam Kω −− = − +

Monoatomic Chain

( )2 e 2 eika ikam Kω −− = − +2cosix ixe e x−+ =

e e 2cosika ika ka−+ =

( )2 2cos 22 (1 cos )

m K kaK ka

ω− = −

x⎛ ⎞ 2 (1 cos )K ka= − − ( ) 21 cos 2 sin2xx ⎛ ⎞− = ⎜ ⎟

⎝ ⎠22 4 sin2kam Kω ⎛ ⎞= ⎜ ⎟

⎝ ⎠

242 sin2

K kam

ω ⎛ ⎞= ⎜ ⎟⎝ ⎠ Maximum value of it is 1

4 sin2

K kam

ω ⎛ ⎞= ⎜ ⎟⎝ ⎠

m a x4 Km

ω =

13

ω versus k relation;

max 2 Km

ω =

Monoatomic Chain

ω

/sV kω=

0 л/a 2л/aл/a kk

C AB

00 л/a 2л/a–л/a

• Normal mode frequencies of a 1D chainThe 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 of 2π/a.

0

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 trial

4 sin2

K kam

ω =

Monoatomic Chain

of 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. Let’s see whether this is the case;

14

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 asthe 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

( )0expn nu A i kx tω⎡ ⎤= −⎣ ⎦

nNn uu +=

pa

Nkkp

NapNa ππλλ 22=⇒==⇒=

which requires that there should be an integral number of wavelengths in the length of our ring of atoms

Thus, in a range of 2π/a of k, there are N allowed values of k.

λpNa =

What is the physical significance of wave numbers outside the range of ?2π/a

Monoatomic Chain

un

un

x

a

x

15

This value of k corresponds to themaximum frequency; alternate atomsoscillate in antiphase and the waves at thisvalue of k are standing waves.

22 ;a k k aπ πλ λ= = ⇒ =

White line :

Monoatomic Chain

7 2 84 7 1.1474 74

aa k a a aπ π πλ λ= ⇒ = ⇒ = = =

7 2 63 7 0.8573 73

aa k a a aπ π πλ λ= ⇒ = ⇒ = = =

Green line :

u

ω-k relation

nx

un

a

•The points A and C both have same frequency and same atomic displacements

•They are waves moving to the left.

•The green line corresponds to the point Bin dispersion diagram.

Monoatomic Chain

-π/a k

KVmK

s /

2

ω

ω

=

=

k

ω

C AB

0 π/a 2π/ap g

•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 (dω/dk)>0.

•The points A and C are exactlyequivalent; adding any multiple of2π/a to k does not change thefrequency and its group velocity, sopoint A has no physical significance.

•k=±π/a has special significanceu

ω-k relation (dispertion diagram)

•θ=90o

x

2 2nn n nk aπ πλ π= = ⇒

2 2 sin901

a d d a= ⇒ =14243

Bragg reflection can be obtained at

k= ±nπ/a 2 24 sin 2

kam Kω =For the whole range of k (λ)

nx

un

a

16

At the beginning of the chapter, in the long wavelength limit, the velocity of sound waves has been derived as

c Kaρ ρ= 2K am

cVs ρ=

Using elastic properties, let’s seewhether the dispersion relationleads to the same equation in thelong λ limit.

1ka ppIf λ is very long; so sin ka ka≈

maρ =

ρ

KV as k mω= =2 22 4 4

k am Kω =

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 many

Monoatomic Chain

inert-gas solids, its not a good assumption for manysolids.If we use a model in which each atom is attached bysprings 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 ω

i till i di f ti f k ith i d 2 /is still periodic function of k with period 2π/a• Group velocity vanishes at k=(±)π/a• There are still N distinct normal modes• Furthermore the motion at long wavelengths corresponds to

sound waves with a velocity given by (velocity formulü)

17

Chain of two types of atomTwo different types of atoms of masses M and m are connected by identical springs of spring constant K;

(n 2) (n 1) (n) (n+1) (n+2)K K K K

M Mm Mm a)

b)

(n-2) (n-1) (n) (n+1) (n+2)

a

Un-2Un-1 Un Un+1 Un+2

• This is the simplest possible model of an ionic crystal.• Since a is the repeat distance, the nearest neighbors

separations is a/2

We will consider only the first neighbour interaction although itis a poor approximation in ionic crystals because there is along range interaction between the ions.

Chain of two types of atom

The model is complicated due to the presence of two differenttypes of atoms which move in opposite directions.

Our aim is to obtain ω-k relation for diatomic lattice

Two equations of motion must be written;

One for mass M, andOne for mass m.

18


M m M m M


Equation of motion for mass M (nth):mass x acceleration = restoring force

..

1 1( ) ( )n n n n nM u K u u K u u+ −= − − −

1 1( 2 )n n nK u u u+ −= − +Equation of motion for mass m (n-1)th:

..

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


M m M m M


0 / 2nx na=( )0expn nu A i kx tω⎡ ⎤= −⎣ ⎦

Offer a solution for the mass M

For the mass m;

( )0expu A i kx tα ω⎡ ⎤= −⎣ ⎦

α : complex number which determines the relative amplitude and phase of the vibrational wave.

( )expn nu A i kx tα ω⎡ ⎤= −⎣ ⎦

( )..

2 0expn nu A i kx tω ω⎡ ⎤= − −⎣ ⎦

-1

19

..

1 1( 2 )n n n nM u K u u u+ −= − +

( ) ( )1 12 22 2 22

k n a k n akna knai t i ti t i tMA K A A A

ω ωω ω+ −⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞− −− −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎛ ⎞⎜ ⎟+

For nth atom (M):


2 2 22MAe K Ae Ae Aeω α α⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎜ ⎟− = − +⎜ ⎟⎝ ⎠

2 2 2 2 22 22kna kna kna knaka kai t i t i t i ti i

MAe K Ae e Ae Ae eω ω ω ω

ω α α⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− − − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞− = − +⎜ ⎟⎜ ⎟

⎝ ⎠

Cancel common terms

2 2 1 cos2kaM Kω α⎛ ⎞= −⎜ ⎟

⎝ ⎠

2cosix ixe e x−+ =2 2 22ka kai i

M K e eω α α−⎛ ⎞

− = − +⎜ ⎟⎝ ⎠

..

1 1 2( 2 )n n n nmu K u u u− − −= − +

( ) ( ) ( )1 1 2k n a k n a k n aknai t i t i ti tω ω ωω− − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞− − −−⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟

⎛ ⎞⎜ ⎟

For the (n-1)th atom (m)


22 2 2 2 22 2 22

kna kna kna knaka ka kai t i t i t i ti i imAe e K Ae Ae e Ae e

ω ω ω ωαω α

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− − − −− − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞− = − +⎜⎜

⎝ ⎠

2 2 22 2 2i t

A me K Ae Ae Aeω

α ω α⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎜ ⎟− = − +

⎜ ⎟⎝ ⎠

Cancel common terms

⎛ ⎞

2 2 cos2kam Kαω α⎛ ⎞− = −⎜ ⎟

⎝ ⎠

2cosix ixe e x−+ =

2 2 21 2ka kai i ikame K e eαω α

− − −⎛ ⎞− = − +⎜ ⎟

⎝ ⎠2 2 22

ka kai im K e eαω α

−⎛ ⎞− = − +⎜ ⎟

⎝ ⎠

20


2 kaK ⎛ ⎞⎜ ⎟

2 2 1 cos2kaM Kω α⎛ ⎞= −⎜ ⎟

⎝ ⎠for M

Now we have a pair of algebraic equations for α and ω as afunction of k. α can be found as

22 cos( / 2) 2K ka K Mωα −= =

2 cos2

m Kαω α⎛ ⎞= −⎜ ⎟⎝ ⎠ for m

A quadratic equation for ω2 can be obtained by cross-multiplication

22 2 cos( / 2)K m K kaα

ω−

k

2 2 2 2 44 cos ( ) 4 2 ( )2kaK K K M m Mmω ω= − + +

2

2

2 cos( / 2) 22 2 cos( / 2)K ka K M

K m K kaωα

ω−

= =−


2 2 2 44 (1 cos ( )) 2 ( ) 02kaK K m M Mmω ω− − + + =

24 2 2 sin ( / 2)2 ( ) 4 0m M kaK K

mM mMω ω+

− + =

2 4b b±The two roots are;

22 2 1/ 2( ) 4sin ( / 2)[( ) ]K m M m M kaK

mM mM mMω + +

= −m m

2

1,24

2b b acx

a− ± −

=

21

ω versus k relation for diatomic chain;


ωA

B

N l d f i f h i f t t f t

0 л/a 2л/a–л/a k

BC

• Normal mode frequencies of a chain of two types of atoms. At A, the two atoms are oscillating in antiphase with their centre of mass 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 find 2N normal modes of vibrations and this is the total number of atoms and hence the total number of equations of motion for mass M and m.

As there are two values of ω for each value of k, thedispersion relation is said to have two branches;


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

0 л/a 2л/a–л/a k

ωA

BC

• The dispersion relation is periodic in k with a period2 π /a = 2 π /(unit cell length).

• This result remains valid for a chain of containing anarbitrary number of atoms per unit cell.

0 л/a 2л/a–л/a

22

Let’s examine the limiting solutions at 0, A, B and C.In long wavelength region (ka«1); sin(ka/2)≈ ka/2 in ω-k.

22 2 1/ 2

1,2( ) 4sin ( / 2)[( ) ]K m M m M kaKmM mM mM

ω + += −m


( )1 22 2 2

2 44

K m M m M k aKmM mM mM

ω⎡ ⎤+ +⎛ ⎞≈ ± −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

( )( )

1 2

2 221 1

K m M mM k amM m M

⎡ ⎤⎛ ⎞+ ⎢ ⎥= ± −⎜ ⎟⎜ ⎟⎢ ⎥+⎝ ⎠⎣ ⎦

( )2 2 221 1

2( )K m M mM k a

mM m Mω

⎡ ⎤+ ⎛ ⎞≈ ± −⎢ ⎥⎜ ⎟+⎝ ⎠⎣ ⎦

( )⎝ ⎠⎣ ⎦

Use Taylor expansion: for smal( )1 21 1 2x x− ≈ −


Taking +ve root; sinka«1 (max value of optical branch)

( )2max

2opt

K m MmM

ω+

≈

( ) 2 2 2 22

min . 22( ) 2( )acus

K m M mMk a Kk amM m M m M

ω+ ⎡ ⎤

≈ ≈⎢ ⎥+ +⎣ ⎦

Taking -ve root; (min value of acoustical brach)

By substituting these values of ω in α (relative amplitude)ti d i (k /2) 1 f k 1 fi d th

222 cos( / 2)

K MK ka

ωα −= 1α ≈ M

mα ≈ −

equation and using cos(ka/2) ≈1 for ka«1 we find thecorresponding values of α as;

OR

23

1α ≈


Substitute into relative amplitude α22

2 cos( / 2)K M

K kaωα −

=ac

2 22min

K(k a ) 2(m M)

ω ≈+

ac

2min ω

This solution represents long-wavelength sound waves in theneighborhood of point 0 in the graph; the two types of atomsoscillate with same amplitude and phase, and the velocity ofsound is

ωA

BOptical

1/ 2w K⎛ ⎞

⎜ ⎟

k

BC

Acoustical

2( )sv ak m M

= = ⎜ ⎟+⎝ ⎠

0 π/a 2π/a–π/a


Substitute into relative amplitude we obtain,op

2maxω

This solution corresponds to point A indispersion graph. This value of α shows

Mm

α ≈ −22

2 cos( / 2)K M

K kaωα −

=op

2max

2K(m M) mM

ω +≈

ωA

BC

Optical

p g pthat the two atoms oscillate inantiphase with their center of mass atrest.

0 π/a 2π/a–π/a

k

C

Acoustical

24


The other limiting solutions of equation ω2 are for ka= π ,i.e sin(ka/2)=1. In this case

1/ 22( ) 4K M M⎡ ⎤+ +⎛ ⎞2max

( ) 4ac

K m M M mKMm Mm Mm

ω⎡ ⎤+ +⎛ ⎞= −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

m

( ) ( )K m M K M mMm

+ −=

m

2 2K 2 2K2max

2ac

KM

ω = OR 2min

2op

Km

ω =(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.

Acoustic/Optical BranchesThe 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 exciteth d ith l t ti di ti (i Th ill tithese modes with electromagnetic radiation (ie. The oscillatingelectric fields generated by EM radiation)

25

Transverse optical mode for diatomic chain

Amplitude of vibration is strongly exaggerated!

Transverse acoustical mode for diatomic chain

26

What is phonon?Consider the regular lattice of atoms in a uniform solid material.There should be energy associated with the vibrations of theseThere should be energy associated with the vibrations of these atoms.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 be sound wavessound waves. And their propagation speed is the speed of sound in the material.

The vibrational energies of molecules are quantized and treated as quantum harmonic oscillators. Quantum harmonic oscillators have equally spaced

l l ith ti ∆E h

Phonon

energy levels with separation ∆E = hν. So the oscillators can accept or lose energy only in discrete units of energy hν.The evidence on the behaviour of vibrational energy in periodic solids is that the collective vibrational modes can accept energy only in discrete amounts, and these quanta of energy have been labelled "phonons".quanta of energy have been labelled phonons . Like the photons of electromagnetic energy, they obey Bose-Einstein statistics.

27

PHONONS• Quanta of lattice vibrations• Energies of phonons are

PHOTONS• Quanta of electromagnetic

radiation

sphonon

hE νλ

=

• Energies of phonons are quantized

~a =10-10m

• Energies of photons arequantized as well

photonhcEλ

=a0=10 m

phononhpλ

=

~10-6m

photonhpλ

=

28

Energy of harmonic oscillatorObtained by in a classical way of considering the normal modesthat we have found are independent and harmonic.

⎞⎛ ωε h⎟⎠⎞

⎜⎝⎛ +=

21nn

• Make a transition to Q.M.

• Represents equally spacedenergy levels

ωh

Energy, E

ωhωhωhωh

Energy levels of atoms vibrating at a single

frequency ω

It is possible to consider as constructed by adding n excitationquanta each of energy to the ground state.

nε

ωh

ωε h21

0 = 20

A transition from a lower energy level to a higher energy level.

ωωε hh ⎟⎠⎞

⎜⎝⎛ +−⎟

⎠⎞

⎜⎝⎛ +=Δ

21

21

12 nn

( )2 1

unity

n nε ω ε ωΔ = − ⇒ Δ =h h14243

absorption of phonon

29

The converse transition results an emission of phononwith an energy .Phonons are quanta of lattice vibrations with an

ωhPhonons are quanta of lattice vibrations with anangular frequency of .Phonons are not localized particles.Its momentum is exact, but position can not bedetermined because of the uncertainity princible.However, a slightly localized wavepacket can be

ω

considered by combining modes of slightly different and . ω

λ

Assume waves with a spread of k of ; so this wavepacket will be localized within 10 unit cells. a10

π

This wavepacket will represent a fairly localized phonon moving with group velocity .

dkdω

Phonons can be treated as localized particles within some limits.

30

1D crystalsω ωh

Energy of phonons

k Multiply by h kh

Crystal momentumCrystal momentum

•Phonons are not conserved

•They can be created and destroyed during collisions .

Thermal energy and lattice vibrations•Atoms vibrate about their equilibrium position.

•The prod ce ibrational a es•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.gy

Note: In a gas, the translational motion of atoms and molecules contribute to this energy.

31

Therefore, the concept of thermal energy is fundamental to anunderstanding many of the basic properties of solids. We would like toknow:

Wh i h l f hi h l ?• 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 amagnetic transition.

• How the vibrational energy changes with temperature since thisHow the vibrational energy changes with temperature since thisgives 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 energywhich is required to raise the temperature of unit mass or 1gmoleby one Kelvin.

The energy given to lattice vibrations is the dominantcontribution to the heat capacity in most solids. In non-magnetic

Heat capacity from Lattice vibrations

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 zeroo c v b o s e ds o b d o o ode eque c es o e oup 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.

32

∑_

Energy and heat capacity of a harmonic oscillator, Einstein Model

nn

nP εε ∑=Avarage energy of a harmonicoscillator and hence of a latticemode of angular frequency attemperature T Energy of oscillator

12n nε ω⎛ ⎞= +⎜ ⎟

⎝ ⎠hThe probability of the oscillator being in this

level as given by the Boltzman factor

exp( / )n Bk Tε−

_0

1 1exp /2 2

1e p /

Bn

n n k T

k T

ω ωε

ω

∞

=∞

⎡ ⎤⎛ ⎞ ⎛ ⎞+ − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦=⎡ ⎤⎛ ⎞+⎜ ⎟⎢ ⎥

∑

∑

h h

h

(*)n

nnP εε ∑=

_

0

/ 2 3 / 2 5 / 2

1exp[ ( ) ]2

.....B B B

n Bk T k T k T

z nk T

z e e eω ω ω

ω∞

=

− − −

= − +

= + + +

∑h h h

h

0

exp /2 B

n

n k Tω=

− +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦∑ h

/ 2 / 2 /

/ 2 / 1

(1 .....(1 )

B B B

B B

k T k T k T

k T k T

z e e ez e e

ω ω ω

ω ω

− − −

− − −

= + + +

= −

h h h

h h

According to the Binomial expansion for x«1 where / Bx k Tω= −h

33

Eqn (*) can be written_

2 2

/ 2_2

/

1 (ln )

ln1

B

B

B B

k T

B k T

zk T k T zz T T

ek TT e

ω

ω

ε

ε−

−

∂ ∂= =

∂ ∂⎛ ⎞∂

= ⎜ ⎟∂ −⎝ ⎠

h

h

'

(ln ) xxx x∂

=∂

( )

( )

_/ 2 /2

_/2

/2 2

ln ln 1

ln 12

2

B B

B

B

k T k TB

k TB

B

k TB

k T e eT

k T eT k T T

k ek k T

ω ω

ω

ω

ε

ωε

ωω

− −

−

−

∂ ⎡ ⎤= − −⎣ ⎦∂⎡ ⎤⎛ ⎞∂ ∂

= − − −⎢ ⎥⎜ ⎟∂ ∂⎝ ⎠⎣ ⎦⎡ ⎤⎢

h h

h

h

h

h

h /1 Bk Te ωω −⎥ hh

_

/

12 1Bk Te ω

ωε ω= +−h

hh

( )_

22 2 /

24 1 B

B BB k T

B

k k Tk Tk T e ω

ωε−

⎢⎢= +

−⎢⎢⎣ ⎦

h

h

( )/

12 1

B

Bk T

ee ω

ωω−

⎥⎥ = +

−⎥⎥

h

hh

_

/

12 1Bk Te ω

ωε ω= +−h

hh

This is the mean energy of phonons.The first term in the abovegy pequation is the zero-point energy. As we have mentioned before evenat 0ºK atoms vibrate in the crystal and have zero-point energy. This isthe minimum energy of the system.

The avarage number of phonons is given by Bose-Einsteindistribution as

( b f h ) ( f h ) ( d t i )_

1

1)(−

=TBke

n ωω h

(number of phonons) x (energy of phonon)=(second term in )ε

The second term in the mean energy is the contribution of phonons to the energy.

34

ε Mean energy of a harmonic oscillator as a function of T

TkB

low temperature limit

T

ωh21

TkBffhω B

121_

−+=

TBkeω

ωωε h

hh

ωε h21_

= Zero point energy

Since exponential term gets bigger

12xx

ε

1

TkB Mean energy of a harmonic oscillator as

a function of T

• is independent of frequency ofoscillationε

..........!2

1 +++= xex

Tke

B

TBk ωω hh

+=1

1121_

−++=

Tkωωωεh

hh

high temperature limit

T

ωh21

Bk Tωh

oscillation.

•This is the classical limit because theenergy steps are now small compared withthe energy of the harmonic oscillator.

•So that is the thermal energy of theclassical 1D harmonic oscillator.

TkB_ 1

2 Bk Tε ω= +h

_

Bk Tε ≈

35

Heat Capacity CHeat capacity C can be found by differentiating the average energy ofphonons ofp

121_

−+=

TBkeω

ωωε h

hh

( )

( )2

2

1

k TB

k TB

B

Bv

k ek TdC

dT e

ω

ω

ωωε

−−

= =

h

h

hh ( )

( ) ( )2

2 2

1

k TB

k TB

v BB

eC kk T e

ω

ω

ω=

h

h

h

( )1Be − ( ) ( )1k TBB e −

kωθ =h

( )2

2

1

T

T

v BeC k

T e

θ

θ

θ⎛ ⎞= ⎜ ⎟⎝ ⎠ −

Let

2TeC k

θ

θ⎛ ⎞⎜ ⎟

vC

ωhh

Plot of as a function of T

( )2

1T

v BC kT e

θ= ⎜ ⎟

⎝ ⎠ −

Bk

Specific heat vanishesexponentially at low T’s and tends toclassical value at high temperatures.

vC

kωθ =hwhere

Area=2ωh

T

Bkωh

The features are common to allquantum systems; the energy tendsto the zero-point-energy at low T’sand to the classical value ofBoltzmann constant at high T’s.

36

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

Plot of as a function of TvC

3R

This range usually includes RT.From the figure it is seen that Cv isequal to 3R at high temperaturesregardless of the substance. This facti k D l P tit l Thi l

vC

where R is the universal gas constant. Since R is approximately 2cal/K-mole, at high temperatures Cv is app. 6 cal/K-mole.

,T K

is known as Dulong-Petit law. This lawstates that specific heat of a givennumber of atoms of any solid isindependent of temperature and is thesame for all materials!

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 vibratearound their sites like a set of harmonic oscillators. Theaverage energy for a 1D oscillator is kT. Therefore, theaveraga energy per atom, regarded as a 3D oscillator, is 3kT,and consequently the energy per mole is

=where N is Avagadro’s number, kB is Boltzmann constant and

3 3BNk T RT=εg , B

R is the gas constant. The differentiation wrt temperaturegives;

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 CvK mole K mole

= = ⇒− −

vdCdTε

=

37

Einstein heat capacity of solidsThe theory explained by Einstein is the first quantum theory of solids.He made the simplifying assumption that all 3N vibrational modes ofa 3D solid of N atoms had the same frequency, so that the whole solidhad a heat capacity 3N times

2 θ

⎛ ⎞

In this model, the atoms are treated as independent oscillators, butthe energy of the oscillators are taken quantum mechanically as

This refers to an isolated oscillator, but the atomic oscillators in a solidt i l t d Th ti ll h i th i ith

( )2

2

1

T

T

v BeC k

T eθ

θ⎛ ⎞= ⎜ ⎟⎝ ⎠ −

ωh

are not isolated.They are continually exchanging their energy withtheir surrounding atoms.Even this crude model gave the correct limit at high temperatures, aheat capacity of

Dulong-Petit law where R is universal gas constant.

3 3BNk R=

• At high temperatures, all crystalline solids have a specific heat of6 cal/K per mole; they require 6 calories per mole to raise theirtemperature 1 K.

•This arrangement between observation and classical theory breakdown if the temperature is not high.

•Observations show that at room temperatures and below thespecific heat of crystalline solids is not a universal constant.

6 calvC

In all of these materials (Pb,Al, Si,and Diamond)

f6Kmol

Bkωh

T

3vC R=

specific heat approaches constant value asymptotically at high T’s. But at low T’s, the

specific heat decreases towards zero which is in a

complete contradiction with the above classical result.

38

The Discrepancy of Einstein model

Einstein model also gave correctly a specific heat tending tozero at absolute zero, but the temperature dependence near T=0did not agree with experiment.

Taking into account the actual distribution of vibrationfrequencies in a solid this discrepancy can be accounted usingone dimensional model of monoatomic lattice

Density of StatesAccording to Quantum Mechanics if a particle is constrained;

the energy of particle can only have special discrete energyvalues.it cannot increase infinitely from one value to another.it has to go up in steps.

39

These steps can be so small depending on the system that theenergy can be considered as continuous.This is the case of classical mechanics.But on atomic scale the energy can only jump by a discreteamount from one value to another.

Definite energy levels Steps get small Energy is continuous

In some cases, each particular energy level can beassociated 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( )ρ εper unit energy interval, and so that the number of statesbetween and will be .( )dρ ε εdε ε+ε

40

There are two sets of waves for solution;Running wavesStanding waves

0 2π2π 4π4π 6πk

Running waves:

0 2Lπ2

Lπ

−4Lπ4

Lπ

−6Lπ

These allowed k wavenumbers corresponds to the runningwaves; all positive and negative values of k are allowed. Bymeans of periodic boundary condition

2 2 2NaL Na p k p k pp k Na L

π π πλ λ= = ⇒ = = ⇒ = ⇒ =

an integer

Length of the 1D chain

pchain

These allowed wavenumbers are uniformly distibuted in k at a density of between k and k+dk.( )R kρ

running waves ( )2RLk dk dkρπ

=

Standing waves:

0Lπ

2Lπ

6Lπ

4Lπ5

Lπ 3

Lπ

7Lπ

k0

Lπ 2

Lπ 3

Lπ

In some cases it is more suitable to use standing waves,i.e. chainwith fixed ends. Therefore we will have an integral number of halfwavelengths in the chain;

L

2 2;2 2

n n nL k k kL L

λ π π πλ

= = ⇒ = ⇒ =

L

These are the allowed wavenumbers for standing waves; onlypositive values are allowed.

2k pLπ

= for

running wavesk p

Lπ

= for

standing waves

41

These allowed k’s are uniformly distributed between k and k+dk at a density of

( )SLk dk dkρπ

= DOS of standing wave

( )S kρ

( )2RLk dk dkρπ

= DOS of running wave

•The density of standing wave states is twice that of the running waves.

•However in the case of standing waves only positive values areallowed

•Then the total number of states for both running and standing waveswill be the same in a range dk of the magnitude k

•The standing waves have the same dispersion relation as runningwaves, and for a chain containing N atoms there are exactly N distinctstates with k values in the range 0 to ./aπ

The density of states per unit frequency range g(ω):

The number of modes with frequencies ω and ω+dω will be ( )d

modes with frequency from ω to ω+dω corresponds

g(ω)dω.g(ω) can be written in terms of ρS(k) and ρR(k).

dR

modes with wavenumber from k to k+dkdn

42

Choose standing waves to obtain ( )g ωdkdωdkdω

;( ) ( )Rdn k dk g dρ ω ω= = ( ) ( )Sdn k dk g dρ ω ω= =

( ) ( )Sg kω ρ=

Let’s remember dispertion relation for 1D monoatomic lattice

2 24 sin2

K kam

ω = 2 sin2

K kam

ω =

ddkω 2 cos

2 2a K ka

m= cos

2K kaam

= 1

cos2

K kaam

1 1

cos2

mkaa K ⎛ ⎞

⎜ ⎟⎝ ⎠

( ) ( )Sg kω ρ=

( ) ( )Sg kω ρ=( )

1 1cos / 2

ma K ka

2cos 1 sin2 2ka ka⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠2 2 2sin cos 1 cos 1 sinx x x x+ = ⇒ = −

1 1 4( ) ( )Sg kω ρ=2

1 1 44

1 sin2

ma K ka⎛ ⎞− ⎜ ⎟

⎝ ⎠

Multibly and divide

( ) ( )Sg kω ρ=2

1 24 4 i

a K K ka⎛ ⎞⎜ ⎟

( )SLk dk dkρπ

=

Let’s remember:

( )g Naωπ

=

2sin2m m

⎛ ⎞− ⎜ ⎟⎝ ⎠

π

( )g Lωπ

=2 2max

2 1a ω ω−

L Na=

2max

4Km

ω =

2 24 sin2

K kam

ω ⎛ ⎞= ⎜ ⎟⎝ ⎠( ) 2g Nω

π= ( ) 1/ 22 2

maxω ω−

−

True density of states

43

N mKπ

True density of states by means of above equation

( ) 1/ 22 2max

2( ) Ng ω ω ωπ

−= −

( )g ω

constant density of states

ωπ

max 2 Km

ω =Km

π

e s o bove equ o

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 thedispersion of sound at wavelengths comparable to atomic spacing.

max 2 Km

ω =

ω/d dkω

The energy of lattice vibrations will then be found byintegrating the energy of single oscillator over the distributionof vibration frequencies. Thus

( )/

12 1kT g dω

ωε ω ω ω∞⎛ ⎞= + ×⎜ ⎟⎝ ⎠∫ h

hh ( )/

0 2 1kT ge ω⎜ ⎟−⎝ ⎠∫ h

( ) 1/ 22 2max

2N ω ωπ

−−

Mean energy of a harmonic oscillator

for 1D

One can obtain same expression of by means of using running waves.

It should be better to find 3D DOS in order to compare the results with experiment.

( )g ω

44

3D DOS

Let’s do it first for 2DThen for 3DThen for 3D.Consider a crystal in the shape of 2D box with crystal lengths of L.

+ -

y

Lπ

yk

Lπ

+

+ -

-

L0

L

x

L

Standing wave pattern for a 2D box

Configuration in k-space

xk

L

•Let’s calculate the number of modes within a range of wavevector k.

•Standing waves are choosen but running waves will lead same expressions.

St di ill b f th f•Standing waves will be of the form

• Assuming the boundary conditions of

•Vibration amplitude should vanish at edges of

( ) ( )0 sin sinx yU U k x k y=

Vibration amplitude should vanish at edges of

Choosing

0; 0; ;x y x L y L= = = =

;x yp qk kL Lπ π

= =

positive integer

45

+

+

+ -

-

-

L0

L

y

x

Lπ

Lπ

yk

xk

•The allowed k values lie on a square lattice of side in the positive quadrant of k-space.

•These values will so be distributed uniformly with a density

Standing wave pattern for a 2D box

Configuration in k-space

/ Lπ

( )2of per unit area.

• This result can be extended to 3D.

( )2/L π

L Octant of the crystal:

kx,ky,kz(all have positive values)

The number of standing waves;L

L( )

33 3 3

3sL Vk d k d k d kρπ π⎛ ⎞= =⎜ ⎟⎝ ⎠

/L π

zk

21 48

k dkπ×

( ) 3 23

1 48s

Vk d k k dkρ π= ×

k

dk

yk

xk

( ) 3 8π

( )2

322s

Vkk d k dkρπ

=

( )2

22SVkkρπ

=

46

• is a new density of states defined as the

number of states per unit magnitude of in 3D.This eqn

( )2

22Vkkρπ

=

can be obtained by using running waves as well.

• ω(frequency) space can be related to k-space:

( ) ( )g d k dkω ω ρ= ( ) ( ) dkg kd

ω ρω

=

Let’s find C at low and high temperature by means of

using the expression of .( )g ω

High and Low Temperature Limits

3 BNk Tε =Each of the 3N latticemodes of a crystalcontaining N atoms

dCdTε

= 3 BC Nk=

B

TkωhThis result is true only if

At low T’s only lattice modes having low frequencies can be excited from their ground states;

θ

ω Low frequency l λ

ka

π0

Low frequency long λ

sound waves

sv kω = svkω

=

47

( )2

22Vk dkg

dω

π ω=1 1

ss s

k dkvk v d vω

ω ω= ⇒ = ⇒ = and

( )

2

2 1s

Vv

g

ω

ω

⎛ ⎞⎜ ⎟⎝ ⎠ t l T’

depends on the direction and there are two transverse, one longitudinal acoustic branch:

( ) 22 s

gv

ωπ⎝ ⎠= at low T’s

sv

( ) ( )2 2

2 3 2 3 3

1 1 2V Vg gω ωω ω⎛ ⎞

= ⇒ = +⎜ ⎟( ) ( )2 3 2 3 32 2s L T

g gv v vπ π ⎜ ⎟

⎝ ⎠

Velocities of sound in longitudinal and transverse direction

( )/0

12 1kT g d

e ω

ωε ω ω ω∞⎛ ⎞= + ×⎜ ⎟−⎝ ⎠∫ h

hh Zero point energy= zε

2

/ 2 3 30

1 1 22 1 2kT

L T

V de v vω

ω ωε ω ωπ

∞ ⎛ ⎞⎛ ⎞= + × +⎜ ⎟⎜ ⎟−⎝ ⎠ ⎝ ⎠∫ h

hh

33

3

/ 1 1

B

BkT x

k T xk Td dx

e eω

ω ω∞ ∞

⎛ ⎞⎜ ⎟⎝ ⎠=∫ ∫h

hh h

h

0 ⎝ ⎠

( )3

2 3 3 /0

1 22 1z kT

L T

V dv v e ω

ωε ε ωπ

∞⎛ ⎞⎡ ⎤⎛ ⎞⎜ ⎟= + +⎢ ⎥⎜ ⎟ ⎜ ⎟−⎝ ⎠⎣ ⎦ ⎝ ⎠∫ h

h

B

xk Tω

=h

Bk T xω =h

k T0 01 1e e− −∫ ∫ h Bk Td dxω =

h( )

4

43 3

/ 30 0

15

1 1B

kT x

k T xd dxe eω

π

ω ω∞ ∞

=− −∫ ∫h

h

h14243

( )4 4

2 3 3 3

1 22 15

Bz

L T

k TVv v

πε επ

⎛ ⎞= + +⎜ ⎟

⎝ ⎠ h

24 3

3 3 3

1 2 430v B

L T

d VC k TdT v vε π ⎛ ⎞

= = +⎜ ⎟⎝ ⎠h

32

3 3

2 1 215

Bv B

L T

k TdC V kdT v vε π

⎛ ⎞⎛ ⎞= = +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ h

at low temperatures

48

How good is the Debye approximation at low T?

32

3 3

2 1 215

Bv B

L T

k TdC V kdT v vε π

⎛ ⎞ ⎛ ⎞= = +⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ h

3T

L T ⎝ ⎠⎝ ⎠

The lattice heat capacity of solids thusvaries as at low temperatures; this isreferred to as the Debye law.Figure illustrates the excellent aggrementof this prediction with experiment for anon-magnetic insulator The heat

3T

non-magnetic insulator. The heatcapacity vanishes more slowlythan the exponential behaviour ofa single harmonic oscillatorbecause the vibration spectrumextends down to zero frequency.

The Debye interpolation schemeThe calculation of is a very heavy calculation for 3D, so it

must be calculated numerically.D b b i d d i i h l i h

( )g ω

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 of

s kω υ=

( )g ωequivalent to taking as given by the broken line of density ofstates figure rather than full curve. Debye’s approximation gives thecorrect answer in either the high and low temperature limits, and thelanguage associated with it is still widely used today.

( )g ω

49

1. Approximate the dispersion relation of any branch by a linearextrapolation of the small k behaviour:

The Debye approximation has two main steps:

extrapolation of the small k behaviour:

Einstein approximation to the dispersion

Debye approximation to the dispersion

vkω =

Debye cut-off frequency2. Ensure the correct number of modes by imposing a cut-off

frequency , above which there are no modes. The cut-offf i h t k th t t l b f l tti d

Dω

Dω

freqency is chosen to make the total number of lattice modescorrect. Since there are 3N lattice vibration modes in a crystalhaving N atoms, we choose so that

0

( ) 3D

g d Nω

ω ω =∫2

2 3 3

1 2( ) ( )2Vg

v vωωπ

= +2

2 3 30

1 2( ) 32

D

L T

V d Nv v

ω

ω ωπ

+ =∫

Dω

23

9( )D

Ng ω ωω

=

2 3 3 3 3

1 2 3 9( ) 32 L T D D

V N Nv vπ ω ω

+ = =

2 L Tv vπ 0L T

32 3 3

1 2( ) 36 D

L T

V Nv v

ωπ

+ =

2( ) /g ω ω

50

The lattice vibration energy of

1( ) ( )E g dωω ω ω∞

+∫h

h

becomes

and,

/0

( ) ( )2 1Bk TE g d

e ωω ω ω= +−∫ h

h

3 32

/ /3 30 0 0

9 1 9( )2 1 2 1

D D D

B Bk T k TD D

N NE d d de e

ω ω ω

ω ω

ω ω ωω ω ω ω ωω ω

⎡ ⎤= + = +⎢ ⎥

− −⎢ ⎥⎣ ⎦∫ ∫ ∫h h

h h hh

3

/3

9 9 D

D k T

N dE Nω ω ωω= + ∫ h

hh

First term is the estimate of the zero point energy, and all T dependence is in the second term. The heat capacity is obtained by differentiating above eqn wrt temperature.

/308 1BD k T

D e ωω −∫ h

dECdT

=

( )/2 4

23 2 /

9

1

D B

B

k T

D k T

dE N eC ddT k T

ω ω

ω

ω ωω

= = ∫h

h

h3

/3

9 98 1

D

BD k T

N dE Ne

ω

ω

ω ωωω

= + ∫ h

hh

The heat capacity is

( )/0 1Bk T

D BdT k T e ωω −∫ h

08 1BD eω −∫

Let’s convert this complicated integral into an expression for the specific heat changing variables to x

xk Tω

=h d k T

d xω

=h

kT xω =h

and define the Debye temperature DΘ

Bk T

DD

Bkω

Θ =h

d x h h

51

4 /2 4

23 2

9 D T xB B

Dk T k TdE N x eC dx

Θ⎛ ⎞⎛ ⎞= = ⎜ ⎟⎜ ⎟ ∫h

The Debye prediction for lattice specific heat

( )23 20 1

D xD B

C dxdT k T eω ⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠ −∫h h

( )

3 / 4

20

91

D T x

D B xD

T x eC N k dxe

Θ⎛ ⎞= ⎜ ⎟Θ⎝ ⎠ −

∫

DD

Bkω

Θ =h

where

How does limit at high and low temperatures?

High temperature

DC

DT Θ

3

2 3

12 ! 3 !

x x xe x= + + + +

( ) ( )

4 4 42

2 2 2

(1 ) (1 )1 11

x

x

x e x x x x xxxe

+ += = =

+ −−

X is always small

3 /2

0

9 3D T

D D B BD

TT C Nk x dx NkΘ⎛ ⎞

Θ ⇒ ≅ =⎜ ⎟Θ⎝ ⎠∫

52

DT Θ

How does limit at high and low temperatures?

Low temperature

DC

( )

3 / 4

20

91

D T x

D D B xD

T x eT C Nk dxe

Θ⎛ ⎞Θ ⇒ ≅ ⎜ ⎟Θ⎝ ⎠ −

∫

44 / 15π

For low temperature the upper limit of the integral is infinite; theintegral is then a known integral of .

( )0 1D e⎝ ⎠

34125

BD

D

Nk TC π ⎛ ⎞≅ ⎜ ⎟Θ⎝ ⎠

We obtain the Debye law in the form3T

Lattice heat capacity due to Debye interpolation scheme

Figure shows the heat capacitybetween the two limits of high and lowT as predicted by the Debyei t l ti f l

3 B

CNk T

1interpolation formula.

Because it is exact in both high and low Tlimits the Debye formula gives quite a goodrepresentation of the heat capacity of most solids,even though the actual phonon-density of statescurve may differ appreciably from the Debyeassumption

Lattice heat capacity of a solid as predicted by the Debye interpolation scheme

( )

3 / 4

20

91

D T x

D B xD

T x eC Nk dxe

Θ⎛ ⎞= ⎜ ⎟Θ −⎝ ⎠

∫

/ DT Θ1

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.

sc e e

Solid Ar Na Cs Fe Cu Pb C KCl

93 158 38 457 343 105 2230 235( )D KΘ

DΘDΘ

Dωh

53

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 due to the shape of the interatomic potential curve.This is a departure from Hooke’s 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.

Th l i i l t th h i ff t

( )2 2

2( ) ( ) ....................2 r a

r a d VV r V adr =

− ⎛ ⎞= + +⎜ ⎟

⎝ ⎠

Thermal expansion is an example to the anharmonic effect.In harmonic approximation phonons do not interact with each other, in the absenceof 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.

Phonon-phonon collisionsThe coupling of normal modes by the unharmonic terms in the

interatomic forces can be pictured as collisions between the phonons associated with the modes. A typical collision process of

phonon11 1,kω 3 3, kω

After collision another phonon is produced

phonon2 2 2,kω3 1 2k k k= +

3 1 2k k k= +h h h

3 1 2ω ω ω= +

3 1 2ω ω ω= +h h h

and

conservation of energy

conservation of momentum

54

Phonons are represented by wavenumbers with

ka aπ π

− ≤ ≤

If lies outside this range add a suitable multible of to bring3k 2aπ

1 2

ε3 ' 31 2

ε 3Longitudinal

Transverse

it back within the range of . Then, becomes

3 1 22nk k kaπ

± = +

where , , and are all in the above range.

3 1 2k k k= +ka aπ π

− ≤ ≤

2k 3k1k

This phonon is indistinguishable from a phonon with wavevector 3k

Phonon3 has kaπ

p ; Phonon3 has and Phonon3=Phonon3’kaπ

f

2k

aπ

− 0aπ

Umklapp process

(due to anharmonic effects)

2k

aπ

− 0aπ

Normal process

Transverse

0n = ⇒ 0n ≠ ⇒

Thermal conduction by phononsA flow of heat takes place from a hotter region to acooler region when there is a temperature gradient in ag p gsolid.

The most important contribution to thermal conductioncomes from the flow of phonons in an electricallyinsulating solid.

Transport property is an example of thermal conduction.

Transport property is the process in which the flow ofsome quantity occurs.

Thermal conductivity is a transport coefficient and itdescribes the flow.

The thermal conductivity of a phonon gas in a solid willbe calculated by means of the elementary kinetic theoryof the transport coefficients of gases.

55

Kinetic theoryIn the elementary kinetic theory of gases, the steady state flux of a propertyin the z direction is

P

1 dP_13

dPflux ldz

υ=

Mean free path

Angular averageConstant average speed for molecules

In the simplest case where is the number density of particles the transportcoefficient obtained from above eqn. is the diffusion coefficient .

If is the energy density then the flux W is the heat flow per unit area so that

P_1

3D l υ=

P EIf is the energy density then the flux W is the heat flow per unit area so that_ _1 1

3 3dE dE dTW l ldz dT dz

υ υ= =

Now is the specific heat per unit volume, so that the thermal conductivity;

_13

K l Cυ=

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

conduction in a phonon and real gas;

Phonon gas Real gas

•Speed is approximately constant •No flow of particlesSpeed is approximately constant.

•Both the number density and energydensity is greater at the hot end.

•Heat flow is primarily due to phononflow with phonons being created at thehot end and destroyed at the cold end

No flow of particles

•Average velocity and kinetic energy perparticle are greater at the hot end, but thenumber density is greater at the cold end,and the energy density is uniform due to theuniform pressure.

•Heat flow is solely by transfer of kineticenergy from one particle to another incollisions which is a minor effect in phononcase.hot cold

hot cold

56

Temperature dependence of thermal conductivity K_1

3K l Cυ= Approximately equal to

l it f d d3 velocity of sound and sotemperature independent.Vanishes exponentially at

low T’s and tends to classicalvalue at high T’sBk

?

•Temperature dependence of phonon mean free length is determined byphonon-phonon collisions at low temperatures

•Since the heat flow is associated with a flow of phonons, the most effectivecollisions for limiting the flow are those in which the phonon group velocityis reversed. It is the Umklapp processes that have this property, and these areimportant in limiting the thermal conductivity

Conduction at high temperaturesAt temperatures much greater then the Debye temperature the heat capacity is given by temperature-independent classical result of

DΘ

The rate of collisions of two phonons phonon density.

If collisions involving larger number of phonons are important, however, then the scattering rate will increase more rapidly than this with phonon density.

At high temperatures the average phonon density is constant and

3 BC Nk=∝

the total lattice energy T ; phonon number T , so

Scattering rate T and mean free length

Then the thermal conductivity of .

1T −

∝ ∝∝ ∝

_13

K l Cυ= ∝ 1T −

57

Experimental results do tend towards this behaviour at high temperatures as shown in figure (a).

10

1T

10

0

10-1

0

10-1 3T

5 10 20 50 100

10

( )T K2 5 10 20 50 100

( )T K

(a)Thermal conductivity of a quartz crystal

(b)Thermal conductivity of artificial sapphire rods of different diameters

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?Dθalthough 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 is expected to vary roughly as

when

3 /k aπ>

Dkθ

/D b Tθ− T θ

∝

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 b Te θDT θ

/D bTl eθ∝3T

58

Conduction at low temperatures

for phonon-phonon collisions becomes very long at low T’s and eventually exceeds the size of the solid, because

l

number of high energy phonons necessary for Umklapp processes decay exponentially as

is then limited by collisions with the specimen surface, i.e.

Specimen diameter

/D bTe θ−

l ∝

l

T dependence of K comes from which obeys law in this region

Temperature dependence of dominates.

3TvC

34125

BD

D

Nk TC π ⎛ ⎞≅ ⎜ ⎟Θ⎝ ⎠

vC

Size effectWhen 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.ff

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 T’s the dominant phonon wavelength becomes so long that these imperfectionsare not effective scatterers, so;

the thermal conductivity has a dependence at these temperatures.

The maximum conductivity between and region is controlled byimperfections.

3T

3T/D bTeθ

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.

chapter 4 (crystal dynamics i) [compatibility mode]

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