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1、固体物理ppt 5 Four short words sum up what has lifted most successful Four short words sum up what has lifted most successful individuals above the crowd: a little bit more. individuals above the crowd: a little bit more. -author -author -date-datePhonon heat capacityphonon gasthe heat capacity at const
2、ant volumeVVTUCthe lattice heat capacity Clat The total energy of the phonons at a temperature ( kBT) in a crystal is the sum of the energies over all phonon modes:KppKpKKppKnUU,where is the thermal equilibrium occupancy of phonons of wavevector K and polarization p, and is given by the Planck distr
3、ibution function:1)/exp(1,pKnPlanck distributionConsider a set of identical harmonic oscillators in thermal equilibrium.)/exp()/exp(/1TkNNBnnthen )/exp(0nNNnThe average excitation quantum number of an oscillator is sssssssssNsNn)/exp()/exp(ssN)/exp(0nNNnn0NN0The total number and the total energy ins
4、ide the square are same as those of the oscillators.xxxdxdxsxxxssssss1 and ,11 1)/exp(1)/exp()/exp(sssssn0123401234/TkxB1)/exp(1nThe energy of the phonons with wavevector K and polariztion p1)/exp(,pKpKpKpKpKnUThe total energy in thermal equilibrium isKppKpKKppKUU1)/exp(,Suppose that the crystal has
5、 Dp( )d modes of a given polarization p in the frequency range to + d . Then the energy isppKpKpDdU1)/exp()( ,ddNDpp/)(The lattice heat capacity isppBVlatxxxDdkTUC22) 1(expexp)( where TkxB/The central problem is to find D(), the number of modes per unit frquency range.D() is called the density of mo
6、des or density of states (DOS).The best practical way to obtain the density of state is to measure the dispersion relation vs K in selected crystal directions by inelastic neutron scattering and then to make a theoretical analytic fit to give the dispersion relation in general direction, from which
7、D() may be calculated.Density of states in one dimensionThere are two equivalent methods for enumerating the number of the mode. Method 1: fixed boundary conditionsMethod 2: periodic boundary conditionsConsider vibrations of a 1d line of length L carrying N+1 particles at separation a. Method 1: fix
8、ed boundary conditionss = 0 1 i1 i i+1 N-1 NLaui-1FixedFixedsKatiuupKssin)exp()0(,Due to the fixed boundary condition, The wavevector K is.) 1( . ,3 ,2 ,LNLLLKThe number of mode is equal to the number of particles allowed to move. There is one mode for ach interval K = /L, so that the number of mode
9、s per unit range of K is L/ for K /a, and 0 for K /a.Method 2: periodic boundary conditionsauis = 1NN1i+1i1i234.)()(Lsausau(N+1),(exp)0(,tsKaiuupKswith . , . ,6 ,4 ,2 , 0LNLLLKFor periodic boundary conditions the number of modes per unit range of K is L/2 for /a K /a, and 0 otherwise./for ,2/aKaLdKd
10、NThere are three polarizations p for each value of K: 2 transverse modes, 1 longitudinal mode.The density of the state:dKddLdddKLdD/)(The number of allowed K values in a single branch for the 1st Brillouin zone equals to the number of the primitive cells.The number of modes equals to the total degre
11、es of the freedom.Density of states in two dimensionsWithin the circle of area K2 the smoothed number of allowed points is K2(L/2)2.Density of states in three dimensionsApply periodic boundary conditions over N3 primitive cells within a cube of side L.)()()(exp )(expLzKLyKLxKizKyKxKizyxzyx)(exp)0(,t
12、rKiuupKwhenceLNLLKKKzyx , . ,4 ,2 , 0,Therefore there is one allowed value of K per volume (2/L)3 in K space.The total number of modes with wavevector less than K for a given polarization and a given branch is:)3/4()2/()(33KLKNThe density of states for each polarization is)/)(2/(/)(22ddKVKddNDDebye
13、model for density of statesDebye approximation: the velocity of waves is constant for each polarization type.vK i.e.3232222212)/()/)(2/()(vVvvVddKVKDThe density of states:The cutoff frequency D:3/132)/6(VNvDThe cutoff wavevector KD:3/12)/6(VNKDThere is NO wavevector larger than KD in Debye model. Th
14、e thermal energy for each polarization:DpppppppdvVnDdU03321)/exp( 2 )()( Assume that the phonon velocity is independent of the polarization.DDxxBppppppexdxvTVkdvVUUU033324403321 23 1)/exp( 233The total thermal energy:where TTkxTkxBDDB/ and ,/Then the total thermal energy isDxxBexdxTTNkU0331 9The hea
15、t capacity:DxxxBVeexdxTNkTUC0243) 1( 9The Debye temperature 3/126VNkvBDiscussion:1. Dulong-Petit lawAt high temperature T , i.e. xD 1xex1At high temperature the heat capacity approaches to the classical value of 3NkB from the Dulong-Petit law.BxBxxxBVNkxxdxTNkeexdxTNkCDD3 9 ) 1( 9024302432. Debye T3
16、 lawAt very low temperature T .The potential energy has been limited to terms quadratic in the interatomic displacements.221CuUAnharmonic crystal interactionsIn real crystal the harmonic theory is not satisfied accurately. The deviations may be attributed to the neglect of anharmonic (higher than qu
17、adratic) terms in the interatomic displacements.Three-phonon processesExperiments: the interaction of two phonons to produce a third phonon at frequency 3 = 1 + 2.Theory: due to the third-order terms in the lattice potential energy, e.g. U3 = Aexxeyyezz.The presence of one phonon causes a periodic e
18、lastic stain (through the anharmonic interaction) which modulates the elastic constant. A second phonon perceives the modulation of the elastic constant and thereupon is scattered to produce a third phonon.Thermal expansionTake the higher order terms of the potential energy into account.432)(fxgxcxx
19、UThe average displacement at finite temperature in classical statisticsTcgkcxdxcxfxgxxdxcfxcgxcxdxfxgxcxxdxxUdxxUxdxxB222432243243)exp( )exp()1 ( )/1 (exp )exp( )(exp )(exp Thermal conductivityThe thermal conductivity coefficient K of a solid is defined with respect to the steady-state flow of heat
20、down a long rod with a temperature gradient,dxdTKjUwhere jU is the flux of thermal energy. The process of thermal energy transfer is a random process. The energy diffuses through the specimen, suffering frequent collisions. The random nature of the conductivity process brings the temperature gradien
21、t.The mean free path of a particle between collisions.The kinetic theory of gasesConsider a system with temperature gradient along the x direction. The energy transfer is due to the collisions between particles. The flux of the particles in the x direction isIf c is the heat capacity of a particle,
22、then in moving from a region at local temperature T + T to a region at local temperature T a particle will give up energyU = c TT the ends of a free path of the particle isnwhere n is the concentration of the particles, and is the average velocity along the x direction.xxvdxdTldxdTTthe flux of therm
23、al energydxdTcvndxdTcvnvdxdTcvnUvnjxxxxU2231 dxdTCvljU31If v is constant as for phononwhere l v is the mean free path of the particle between collisions and C nc is the heat capacity of the particles.Thus the thermal conductivity coefficient CvlK31Thermal resistivity of phonon gasThe phonon mean fre
24、e path l is determined by geometrical scattering (scattering with crystal boundary) scattering with lattice imperfections scattering with other phononsFor harmonic interaction, l is limited only by collision of a phonon with the crystal boundary, and by lattice imperfections.For anharmonic interacti
25、on, the coupling between different phonons limits l.At high temperature (T D)l 1/TTkTknBB1)/exp(1n/1Tvl/1TK/1 thusTo define a thermal conductivity there must exist mechanisms in the crystal whereby the distribution of phonons may be brought locally into thermal equilibrium.Flow of gas molecules in a
26、 state of drifting equilibrium. Elastic collision processes do not change the momentum and energy flux of the gas. The energy is transported by mass flow without being driven temperature gradient. The thermal resistivity is zero and the thermal conductivity is infinite.The usual definition of therma
27、l conductivity in gas refers to a situation where no mass flow is permitted. With a temperature gradient, the colliding pairs with above-average center mass velocities will tend to be directed to the right. A slight concentration gradient, high on the right, will be set up to allow a net energy tran
28、sport without net mass transport.It is not sufficient to have only a way of limiting the mean free path, but there must also be a way of establishing a local thermal equilibrium distribution of phonons.Phonon collisions with a static imperfection or a crystal boundary will not by themselves establis
29、h thermal equilibrium, because such collisions do not change the energy of individual phonons.A three-phonon collision process K1 + K2 = K3 will also not establish thermal equilibrium, because the total momentum of the phonon gas is not changed by such a collision. Similar like the gas flow in a sta
30、te of drifting equilibrium.Unklapp processesNormal (N) processes: K1 + K2 = K3Unklapp (U) processes: K1 + K2 = K3 + GIn N processes, K is conserved. However, due to the periodic lattices U processes are possible which changes the momentum of phonons in collisions. For all processes, N or U, energy m
31、ust be conserved, so that 1 + 2 = 3.In N processes, the phonon flux is unchanged in momentum on collision and some phonon flux will persist down the length of the crystal. The thermal resistivity is zero.In U processes there is a large net change in phonon momentum in each collision event. An initia
32、l net phonon flow will rapidly decay as we move to the right. Net energy transport under a temperature gradient occurs.At high temperature (T ), all phonon modes are exited because kBT max. A substantial proportion of all phonon collisions will be U processes. A lattice thermal resistivity T.If both
33、 phonons have low K, there is no way to have an U process. The energy of phonones K1, K2 suitable for umklapp is of the order of kB/2. At low temperature the number of suitable phonons of the high energy kB/2 required may be expected to vary roughly as exp(/2T). Therefore the thermal resistivity is
34、roughly proportional to exp(/2T).Note the phonon mean free path for thermal conductivity is the mean free path for umklapp collisions between phonons and not for all collisions between phonons.ImperfectionsGeometrical effects may also be important in limiting the mean free path.The size effect:When
35、at low temperatures the mean free path l becomes comparable with the width of the specimen, the value of l is limited by the width, and the thermal conductivity becomes a function of the dimensions of the specimen. The abrupt decrease in thermal conductivity of pure crystals at low temperatures is c
36、aused by the size effect.At low temperatures the umklapp process becomes ineffective in limiting the thermal conductivity, and the size effect becomes dominant.l is limited by the width of the specimen.Dl 3TCvDCvlKDielectric crystals may have thermal conductivities as high as metals.The distribution of isotopes of the chemical elements often provides an important mechanism for phonon scattering.