Phonons II. Thermal Properties

Transcription

1 Chapter 5. Phonons II. Thermal Properties

2 Thermal properties of phonons As mentioned before, we are now going to look at how what we know about phonons will lead us to a description of the heat capacity C as a function of temperature. What we have to do is to establish the rules we need to count how many phonons are active at a certain temperature, and then figure out how much energy goes into each. For a metal, C = γt+βtt electronic term lattice term C/T At low temperatures C/T=γ+βT Slop = β Y-axis intercept = γ C=dU/dT T

3 Thermal phonons Phonons : dominate thermal properties of materials and affect the electrical transports of conductors by scatterings of electrons Phonon generations : How are phonons created or excited in a crystal? External perturbations vibrations or sound transducer Scattering of particles energy transferred into lattice vibrations Thermal (K B T) excited at any finite temperature (T 0K) Thermal phonons : consider a system with energy level E n E n E n-1 E n- Probability of occupancy E n P(E at temperature T n) exp kt B Boltzmann factor

4 Phonon number average Excitation level amplitude (n) for mode k, ω 1 w/. energy n + ω Average of phonons E s s P ( E s ) s exp s s k B T n = = P ( E s ) E s s exp s k B T (s + 1/) ω s exp s k B T = (s + 1/) ω exp s k B T d exp ( sx) dx s ω = where x = exp sx k T s ( ) B Planck distribution of < n(ω, T) > 1 n(ω,t) = ω average number of phonons exp k T excited per mode at ω B 1

5 Thermal energy < n(ω ω, T) > High T ( k B T >> ω ) kb T < n(ω, T) > ~ ω Low T ( k B T < ω ) k BT < n(ω, T) > ~ exp ω 1 x -1 = k B T / ω Thermal energy: polarization mode U = n ω = n ω i i k,p kp i p k = dω (ω ) n(ω ) ω = p p p ω dω (ω p ) ω exp 1 kt B density of modes per polarization thermal equilibrium

6 1 density of states (k) ensity of states (modes) : uniform in k-space 1 (k) density of states = number of states per unit k at k (k)dk number of states from k to k+dk A linear chain of length L carries N+1 particles with separation a. s=0 Fixed-end boundary conditions : u 0 (t)=0 and u N (t)=0 s=n π π π 4π (N-1)π u s (t)=u () uexp[-iω ω kp k,p t] ]s sin(ska) where eek=,,,,... L L L L L 1 mode/mobile atom snπ a Why is there no Nπ/L for allowed k? u s (t) sin = sin ( sπ ) =0 π L No motion of any atom One mode for each interval Δk = L L π for k The number of modes per unit range of k π a (k) = π 0 for k > a

7 Periodic boundary conditions u s s( (t)=u exp[ i(ska-ω kp k,p t) )] where k= π One mode for each interval Δk = L The number of modes per unit range of k Another way for enumerating modes as N becomes large is the use of periodic boundary conditions. Unbounded medium but w/. Periodic boundary conditions u(sa)=u(sa+l) ( ) for a large L=Na system π 4π 6π Nπ 0, ±, ±, ±,..., L L L L L π π (k) = for k π a a = 0 otherwise The number of modes is still equal to the number of mobile atoms (N in this case instead of N-1; but as N gets large, this does not matter).

8 ensity of state (ω) L (k)dk = dk and for ± of k π dk (ω )dω = (k)dk = (k) dω dω The number of modes per unit frequency range dk (k) ( ω ) = (k) = = dω dω /dk (k) v g ispersion relation Singularity at v g =0, determined by ω(k)

9 (ω) for 1 monatomic lattice (ω ) 4C ka ω = sin M ( ) (k) Na/π N M ω = = = dω /dk 4C a ka π C ωmax ω cos M (k) L/π (N/π)(M/C) 1/ (ω) max -π/a Total number of modes N = k 0 π/a 0 (4C/M) 1/ π/a max L π (k)dk = = N = ( ω )dω π a = π/a ω 0 4C M 0 ( ω )dω ω

10 In two dimensions : periodic boundary condition, N primitive cells within a square of side L exp[ i(k x x+k y y) ] = exp[ i( k x (x+l) + k y (y+l) ) ] π 4π 6π Nπ L L L L 1 π L = = Δk xδk y L 4π whence k x, k y = 0, ±, ±, ±,..., One mode per unit area in k-space Number of modes with wavevector from k to k+dk in k-space 1 L (k)d k = dk xdk y = π k dk Δk xδky 4π The number of modes per unit frequency range k y k k X (k) () ( ω ) = = dω /dk A 4π π k 1 v g In three dimensions : (k) ( ω ) = = dω /dk V 8π 4π k 1 v g

11 ensity of states for continuum wave (ω)dω = (k)d k for each polarization complicated! -- must map out dispersion relation and count all k-values with each frequency Continuum waves: ω = v g k depending only on amplitude of k ( ω )d dω = (k) d k V = 4π kdk 8π V ω dω = π v g vg V ω = dω π v g The number of modes per unit frequency range for each polarization V ω ( ω ) = a quadratic dependence! π vg

12 ebye model The actual density of state can be (k) V obtained from dispersion curve ( ω ) = = 4π k dω /dk 8π but it is usually very complicated Vω Peter ebye made two approximations:, ω ω continuum elastic phonon mode (ω ) = π vg (low energy) ω = v g k 0, ω> ω, only up to some cutoff frequency ω Number of phonon mode for each polarization is equal to N ω ω ebye frequency 1 v ω ω V ω V ω N= ( ω )dω = d 0 = 0 π v 6π v ω g g 6π v g N = V v 1/ g ω 6π N ebye wavevector k = = vg V 0 k k ebye temperature ω Θ = ω k B 1/ g

13 ensity of states of the ebye model ebye solid Quadratic at low ω ω = v g k Peaks at high ω -- cutoff of ω max in different k direction ω Actual crystal ω ebye solid N primitive cells in the crystal, A total number of acoustic phonon mode is N for each polarization ω ω V ω V ω Cutoff frequency Cutoff wavevector ω k N= ( ω )dω = dω = π v 6π v 0 0 1/ 6π vg N = V 1/ ω 6π N = = vg V g g ( ω ) = V ω π vg

14 Thermal energy of the ebye model For each polarization: U = dω (ω ) n(ω ) ω = ω 0 Vω dω π v g ω ω exp 1 kt B There are three polarizations : transverse + 1 longitudinal Assume all polarization have the same energy dependence U = V π v g 0 ω ω exp 1 kt B ω dω 4 x V kt B x ω = dx where x = π vg exp 0 ( x) 1 kbt

15 efine The ebye temperature Θ 1/ 1/ ω 6π vn g vg 6π N Θ = = = kb k B V kb V Therefore x = ω /k B T= Θ /T The total phonon energy Θ /T T x U 9Nk T dx = B ( ) Θ exp x 1 0 In classical model : equipartition theorem (k B T/ for each excitation mode) translational + vibrational modes : six degrees of freedom U = N 6 (k B T/) = N k B T for N atoms in the crystal C v = Nk B ulong and Petit Law

16 C V Heat capacity C V ω U V ω = = dω T V π vg T exp 0 ( ω/kbt) 1 ( ω/kbt) ( ) ω 4 V 1 ω exp = dω π vg kb T 0 exp ω/k T 1 x 4 x T x e = 9Nk dx Θ 0 e 1 ( ) B x ( B ) At high T limit, x=θ Θ /T <<1 dx ( x ) = i.e., For T >> Θ, U Nk B T x x x x e Θ T x xe 1 ( ) 1 Θ x dx x = e 1 T ( ) C V Nk B ulong and Petit value

17 The Einstein model classical model high T C V g V Nk B Einstein model (1907) : N identical oscillators of C V (N Nk B ) Einstein i model 0. low T Nk B e - ω/kt C V frequency ω 0 The Einstein model is often used to approximate the optical phonon part of the phonon spectrum. Einstein model x -1 = k B T / ω ( ω)= N δω ( - ω ); U = N n ω C 0 0 ( ω 0 /kb T ) ( k ) U ω ep exp 0 = = Nk T k T exp ω / T 1 N ω0 = exp ω /k T 1 ( ) V B V B 0 B ( ) 0 B

18 The Einstein model Experimental values of imond high T C V Nk B classical l model Einstein i model The Einstein model is often used to approximate the optical phonon part of the phonon spectrum. low T C V Nk B e - ω/kt Einstein model (1907) : N identical oscillators of frequency ω At high T, C V Nk B same as the ulong and Petit value At low T, C V Nk B exp(- ω/k B T) Experimental data for other materials show T dependence of C V instead

19 Comparison between the Einstein i and ebye models ebye Einstein Copper At low temperatures, t the ebye model gives the phonon contribution to the heat capacity its experimental T dependence form. Actual density of states (ω) for silicon ispersion curve for two atoms per primitive basis - π/a ω M 1 >M π/a C M C M 1 k

20 ebye T model ω, Θ depend on v g, n=n/v, ~ v g n 1/ High for stiff, light materials Kittel : Table 1 in ch.5 (P.116) material A Cu Ag Au Pb Θ (K) x V kt x B U = dx x π vg e 1 0 x 4 x T x e C = 9Nk dx Θ x 0 e 1 ( ) V B x dx = 0 π 4 x e 1 15 At very low temperature, T<<Θ, x = Θ /T U BT 1π Nk BT and CV 4Nk B 5Θ π Nk 5Θ T ebye T Θ approximation

21 ebye temperature θ 1/ 1/ ω 6π vn g vg 6π N = = = k B k B V kb V Usually, a harder material has a higher ebye temperature

22 ebye T dependence T observed in most insulators for T<0.1Θ solid Ar w/. Θ =9K Why T at low temperatures? Only long wave length acoustic modes are thermally excited. These modes can be treated as an elastic continuum. The energy of short wavelength modes is too high for them to be populated significantly at low temperatures.

23 C/T=γ+βT C ule Mole -1 K - V /T (Jou ) Slop = β Y-axis intercept = γ KCl Cu Phys. Rev. 91, 154 (195) Low Temperature Solid State Physics (196) T (K )

24 Simple reasoning for T dependence kb Total phonon mode : ω ω (or k k = Θ ) v g k y Excited phonon mode k B ω k B T/ (or k ω/v g = T = k T ) k x Others are frozen out k T Fraction excited at T : Each mode has energy k B T k v Thermal wavevector g k T T = k Θ of the total volume in k-space U ~ T N Θ Θ k T B T 1Nk B Θ C V ~ T V too small but correct T dependence

25 Success of ebye theory NaCl high T C V Nk B diamond low T C V 0 C V may appear different in the above figure, But after rescaling the temperature by the ebye temperature θ, which differs from material to material, C V of many materials can fit on the same cure, i.e., a universal behavior emerges.

26 (ω )dω General result for (ω) L = π shell dk the number of states between ω and ω+dω ds ω : an element of area on the surface in K space of selected constant frequency ω. dk d k = dsωdk shell ω k dk dω = ω ω+dω dω dω ds L dsω ωdk = dsω dsω ω = (ω )dω = dω v k g k z π v g (ω) V ds ω = ( ) π vg k x ds w k y

27 Van Hove singularities ebye solid Quadratic at tlow ω ω = v g k Peaks at high ω -- cutoff of ω max in different k direction ω Actual crystal ebye solid ω Cutoff frequency Cutoff wavevector ω k 1/ 6π v g N V ds = V ω 6π N = = vg V 1/ (ω) = ( ) π v Critical points at which v g =0 produce singularities know as Van Hove singularities). iscontinuities develop at singular points g ω

28 ispersion and density of states of Cu Experimental results Phys. Rev. 155, 619 (1967) Phys. Rev. B7, 9 (1967) Cu (ω) V = ( ) π ds v g ω Solid line Numerical calculation based on experimental data ω(k) ashed line -- Numerical fit w/. ω 45 = rad./sec d/ Θ = 44K

29 Thermal properties (Review) Lattice vibrations : mode (k,ω ) k is in BZ, discrete ω(k) dispersion relation (ω ) density of states E(ω )=(n+1/) ω 1 ω/k B e T 1 Phonons : number n energy = ω n = Thermal properties (equilibrium) crystal momentum k thermal energy U = dω (ω ) n(ω ) ω heat capacity C V = du dt

30 Transport properties Conduction of sound and heat through the crystal (non-equilibrium) vibration energy Ultrasonic attenuation Thermal conduction excite single phonon mode apply temperature gradient measure decay of amplitude measure heat current by phonons Phonon thermal conductivity Apply temperature gradient T determine heat current density j U T H j U T T L The flux of the thermal energy j U dt = κ dx the energy transmitted across unit area per unit time κ : thermal conductivity coefficient

31 Anharmonic effects In the theory we have covering so far for phonons, we have assumed that the potential energy kx / (harmonic, Hook s law) However, real systems have anharmonic effects. In terms of physical phenomena, what assumes is that : Two lattice waves do not interact A single wave does not decay or change form with time There is no thermal expansion of the lattice The force (elastic) constants between the atoms do not change as a function of temperature or pressure In reality, all of these effects are real and they can modify our dispersion curves and how we look at phonons.

32 Propagations of phonons Ballistic No interaction/scattering In harmonic approximation in perfect, infinite crystal, Expect no scattering phonon modes are uncoupled, independent plane waves and standing waves v = v = g dω dk iffusion Phonons scatter, random walk through crystal Phonons scatter in real crystals. Scattering processes : boundary scattering defect scattering phonon-phonon scattering v<< v = g dω dk

33 Thermal conductivity The flux of thermal energy is based on that the process of thermal energy transfer is a random process. ie. the energy diffuses through the crystal, suffering frequent collisions. Ballistic : iffusive : j U T JU ΔT dt JU T= dx across the whole sample local For diffusion, thermal conductivity is defined by κ ( ) phonon propertiesp scattering crystal quality (size, defect) temperature j U [Watt/m ]: the energy transmitted across unit area per unit time κ [(Watt/m )/(K/m)] = [Watt/m/K]

34 Kinetic theory of gases (phonons) consider phonons as gases contained in a crystal volume calculate diffusion in the presence of temperature gradient T T T H j L U n Fick s law x n : concentration ti of molecules l C : heat capacity per unit volume = nc v g : phonon velocity : phonon mean free path (average distance that a lattice vibration travels before it collide with another phonon=v g τ j U = n v cδt x g dn = dx dn dt dt dx dt dt ΔT = x = vxτ dx dx dt 1 dt ju = n vx cτ = n v cτ dx dx Thermal conductivity κ 1 = Cvg 1 dt 1 dt = = dx dx ju nvgcτ Cvg

35 Anharmonic effects In the theory we have covering so far for phonons, we have assumed that the potential energy kx / (harmonic, Hook s law) However, real systems have anharmonic effects. In terms of physical phenomena, what assumes is that : Two lattice waves do not interact A single wave does not decay or change form with time There is no thermal expansion of the lattice The force (elastic) constants between the atoms do not change as a function of temperature or pressure In reality, all of these effects are real and they can modify our dispersion curves and how we look at phonons.

36 Anharmonic terms in binding potential U The general shape applies for any type of binding x o x U 1 U 1 U U(x) = U(x ) + x x + x x + x x +... ( ) ( ) ( ) o o o o x x x 6 x o xo xo 1 U 1 U Δ U(x) = U(x)-U(x ) = x x + x x +... ( ) ( ) o o o x 6 x x x Reset the equilibrium, let displacement x-x o x 4 U(x) = cx gx fx... o o harmonic term anharmonic term

37 Thermal expansion As a solid is heated up, the lattice expands (due to the average displacement of the atoms increasing with thermal energy which causes fluctuation of x from x o. x 4 4 U(x) cx gx fx dx x exp[ ] dx x exp[ ] kt kt B = 4 U(x) cx gx fx dx exp[ ] dx exp[ ] kt kt cx gx dx exp[ ] kt kt B B B B B cx dx exp[ ] k T B 4 4 cx gx fx cx gx fx exp = exp exp + kt B kt B kt B kt B high T limit 4 anharmonic term gives the net change of <x> cx gx fx exp 1+ + x x x e = 1+ x kt B kt B kt B!! x (π /4)(g/c )(k T) g = = k T linear dependence of T 1/ 5/ / B 1/ 1/ (π/c) (k BT) 4c B Coefficient of linear expansion

38 Phonon-phonon scattering phonon displaces atom which changes the force constant C (anharmonic terms) scatter other phonons Normal processes : all ks are in BZ 1st BZ in k-space k k 1 k three phonon process k + k = k 1 Ttl Total momentum of fthe phonon gas is conserved by such collision Umklapp processes : k * is outside BZ 1st BZ in k-space k G 1 k k 1 outside BZ k * k * k + k = k * k + G= k k + k = k + G 1 Folding over The only meaningful phonon K s lie in the 1st BZ, so any longer K must be brought back into the 1st BZ by adding a G. crystal momentum is not conserved

39 Normal and umklapp processes N-process Total momentum of the phonon gas is conserved by such collision No thermal resistance. q q 4 U-process q + q = q = q + G 1 4 A collision of two phonons both with a positive value of q x can by an U-process, create a phonon with negative q x.

40 Umklapp processes U-processes occur at high temperature : require large k (ie. large ω) k z 1 How large? k ~ k ebye sphere k k 1 ω ~ ω k y k x 1 1 E ~ ω ~ kbθ Phonon-phonon scattering : rate τ -1 # of phonons involved U-process : τ -1 N U ~ k B Texp(Θ exp(-θ /T) (Boltzmann factor of phonons w/. large k only) at intermediate (low) temperatures At very low temperatures, phonons are populated at low k mode U process can not occur

41 Phonon mean free path Log-log plot Very low T, =v g τ =constant L (sample s size) Phonon mean free path ( τ ) Exponential Intermediate T, =v g τ (1/T)exp(1/T) dominated by U process Slope: -1 High T, =v g τ T -1 T (K) (number of phonons) -1 No distinction between N and U processes

42 T-dependent thermal conductivity Isotope effect Log-log plot of κ(t) Below 5K, enriched Ge 74 shows T dependence of κ due to boundary scattering At low temperatures, L (sample s s size) Phonon propagation ~ ballistic κ =(1/)v g C V ~ v g LC V κ C V T ebye At intermediate temperatures, κ=(1/)v g C V =(1/)v g τc V U-processes C = constant = Nk V 1 τ~ e kt B Θ /T B κ ~ 1 k T B e Θ /T

43 T-dependent thermal conductivity Other effects Impurity scatterings efect scatterings break periodicity κ (Watt/m/K) Slope: Log-log plot Exponential Slope: -1 T (K)

44 Thermal conductivity of LiF crystal bar ifferent cross sectional area (a) 1.mm 0.91mm (b) 7.55mm 6.97mm κ (Watt m -1 K -1 ) ata show 1. Below 10K, κ T. As temperature increases, κ increases and reaches a maximum around 18K.. Above 18K, κ decreases w/. increasing temperature t and follows that t exp(1/t). T(K) 4. Cross sectional area influences κ below 0K. Bigger area crystal has, larger κ it has.

45 Summary of part (I) Solids are defined by their capacity to be solid to resist shear stress A crystal is truly solid (as opposed to a glass which is just a slow liquid ) Crystalline order is defined by the regular positions of the nuclei crystal structure re = lattice + basis Lattice and reciprocal lattice iffraction and experimental studies Brillouin zone Crystal binding Type of binding Elastic constants t and elastic waves

46 Summary of part (I) Vibrations of atoms Harmonic approximation Quantization of vibrations phonons act like particles -- can be created or destroyed by inelastic scatterings Thermal properties Fundamental law of probabilities biliti (Boltzmann factor) Planck distribution for phonons Heat capacity : C Low T, C T and High T, C ~ constant Thermal conductivity : κ maximum; as function of T