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    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

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    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.

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    Crystal Dynamics

    These introduces us to the concepts of forbidden and permitted frequency ranges, and

    electronic spectra of solids

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    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.

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    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.

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    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.

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    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

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    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

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    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.

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    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.

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    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

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    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

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    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

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    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

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    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.

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    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

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    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

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    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

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    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

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    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

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    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

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    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

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    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

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    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

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    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

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    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

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    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

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    What is the physical significance of wave numbers

    outside the range of ?2/a

    x

    Monoatomic Chain

    un

    un

    x

    a

    Monoatomic Chain

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    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

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    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

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    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

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    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

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    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

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    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

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    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

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    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

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    ..

    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

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    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

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    Chain of two types of atom

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    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

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    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

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    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

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    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

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    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

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    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

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    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

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    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.

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    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)

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    Transverse optical mode for

    diatomic chain

    Amplitude of vibration is strongly exaggerated!

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    Transverse acoustical mode fordiatomic chain

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    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

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    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.

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    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

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    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

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    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

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    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 .

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    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.

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    1D crystals

    kMultiply by

    k

    Energy of

    phonons

    Crystal momentum

    Phonons are not conserved

    They can be created and destroyed during collisions .

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    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.

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    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.

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    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

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    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

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    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

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    _

    /

    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

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    _

    /

    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.

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    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

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    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

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    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

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    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

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    ,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

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    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

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    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

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    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.

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    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

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    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.

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    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

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    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

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    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

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    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

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    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

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    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

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    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

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    ( )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

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    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

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    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

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    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

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    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

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    +

    +

    + -

    -

    -

    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:

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    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

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    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

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    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

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    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=

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    /

    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

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    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

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    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

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    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:

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    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

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    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

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    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

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    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

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    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

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    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

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    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

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    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

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    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

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    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

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    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

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    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

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    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

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    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

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    Experimental results do tend towards this behaviour at high temperatures as

    sho n in fig re (a)

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    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

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    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

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    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

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    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