Chapter 5Phonons – Thermal PropertiesPhys 175ADr. Ray KwokSJSUHeat capacity of metal Metal, treated as ideal electronic gas, should carry KE = (3/2)NkT = U (internal energy) C = ∂U/∂T = (3/2)Nk = constant at low T A more advanced model of metal suggested N ~ kT. i.e. the number of electrons is available to carry heat is proportional to kT, restricted by Pauli Exclusion Principle (later chapters).  Then C should be linear with T. Experiments showed a T3behavior. Need to take phonon energy into account.• By heat capacity we mean change in the thermal energy of the system for unit change in temperature for fixed volume,i.e.•The total energy of the phonon bath at a temperature T equals the sum of all branches p and all phonon modes K,where <nK,p> is the occupancy factor of a mode K from branch p..Phonon heat capacity.constVVTUC=∂∂=∑∑∑∑ω==K pp,Kp,KK pp,KnUU hThe Planck distribution was developed during his work of black body radiation by considering a set of identical harmonic oscillators in the (n+1)stand nthstate:which is basically the Maxwell – Boltzmann probability function ~ e−En/kBT= e−nћω/kBTin the nthSHO state.The fraction of the total # of oscillators in the nthquantum state is therefore()TkNNBnn/exp/1ω−=+h()( )∑ω−ω−=sBBnTksTknf/exp/exphhPlanck Distribution FunctionUsing The occupancy factor, i.e. the average excitation number of the oscillator, is then given by (x = ћω/kBT):( )( )1e1e11eeenenfnxx2xxssxnnxnn−=−−−===−−−−−∑∑∑Bose-Einstein Distribution ()21/and11−−=∑−=∑−−−−− xxssxxssxeeseee 11/−=TkBenωhwhich is often referred to as Bose-Einstein Distribution Function, the average occupation number of that state with given ω.•The energy of a collection of oscillators of frequency ωK,pis: (wave vector K on pthbranch)• By conserving the # of states, i.e. going from sum over K to an integral over ω, we can express the lattice heat capacity as (writing Dp(ω) as the density of state in ω-space) for the p-polarization branch( )∑ ∑−ωω=ω∑∑=K pBpKpKpKK ppKTknU1/exp,,,,hhh[ ]2xx2ppBpxpVKp1eex)(Ddk1e)(DdTC)(Dd−ωω=−ωωω∂∂=ωω→∑∫∑∫∑∫hHeat energy kT/xω=hwithDensity of States KK+dKωω+dωD(ω)dω# of states with wave vector K and frequency ω is dN = D(ω)dωor, the density of states D(ω) = dN/dω. To count # of quantum states, one uses Fixed or Periodic Boundary Conditionsor Vanishing Boundary Conditions.Consider a 1D line of length L that carries (N+1) particles. The particles at s = 0 and at s = N are held fixed. From the harmonic approximationand the condition that uN=0, we get:i.e. there are (N-1) allowed independent non-zero values of K.Note: the phase is from 0 to π so K > 0 is sufficient. Hence, the number of modes per unit range of K is L/π.For each polarization (1 longitudinal, 2 transverse)()()KsatiUupKssinexp,0ω−=LNLLKnKNaππππ)1(2,,,,0−=→= KFixed Boundary Conditions (easy to see)()()00Nu t u t= =dK/dN = π/LdN/dK = L/π( )LD Kπ=∫∫ππ=π=⋅a/0a/0NdKL)K(DdK()()00Nu t u t= =()(),, , , 0 sinK pi tsu K p t u p e sKaω−=withK mLπ=( ),,0 sinK pi tsmu p eNωπ−=mNaπ=1D Fixed Boundarye.g. N = 9L = Na# of allowed modes = N = 9 (m = Nπ/L = 9 would be the same as m = 0) m = 0,1, 2, 3…. 8Another way of looking at this problem is to assume that the medium is unbound, but the solutions have to be periodic over a large distance L, so that:Again, for harmonic displacements this leads to:This method gives the same number of modes, one per atom, but we have + or – values and ∆K=2π/L between successive values of K. D(K) = L/2π.Note: K’ = K − G would be the same physical state.i.e. K = π/a and –π/a are the same state on the BZ boundary.Note also: the phase is from 0 to 2π so K can be negative.)()(Lsausauss+=LNLiKLLnKeπππ±±==→= ,,,0212KPeriodic Boundary Conditions (easy in k)i s K a i tsu ueω−=()()s N su t u t+=→()(),, , ,0K pi ti s K asu K p t u p e eω−=with2K mLπ=10, 1, , ,2 2N Nm−= ± ±L()(),2 /, , ,0K pi ti s m Nsu K p t u p e eωπ−=Number of allowed K is NKaπ≤are the sameKaπ = ±  2mN aπ=→= Number of mobile atoms( )2LD Kπ=( )//aadK D K Nππ−=∫1D Periodic Boundarye.g. N = 8L = NaExample : N=3aiKsaosKnKLKNaeuuPeriodic32,02ππ±====π/areciprocal space−π/a0aperiodic-1.5-1-0.500.511.50 0.5 1 1.5 2 2.5 3k1=0k2=2*pi/3k3=-2*pi/3k4=4*pi/3aaosKnKLKNaKsauuFixed323,,0sinπππ====acrystal spaceL = 3aπ/areciprocal space0fixed-1.5-1-0.500.511.50 0.5 1 1.5 2 2.5 3k1=0k2=pi/3k3=2*pi/3k4=4*pi/3Are they the same? discrete match at atoms-1.5-1-0.500.511.50 0.5 1 1.5 2 2.5 3sumk1=0k2=pi/3k3=2*pi/3sum = 0.58sin(k1x+π/2)+0.38sin(k2x-π/2)+0.2sin(k3x-π/2)The DoS function in 1D and in analogy, in 3D, is given by:( )dK/d12VK2VK232LvdLdK/ddL2L2222g)(DdKdKK4d)(D :D3dK2dK)K(Dd)(D :D1ωπππωπωωππ=ω→=π=ωω====ωωDoS (density of state) in energy (ωωωω) spaceK from 0 to KmaxK from -Kmaxto KmaxK from 0 to KmaxD(ω)dω = D(K)d3KD(ω) = L/πvg(1D)= (L/2π)34πK2/vg(3D)Debye ModelωK0()vKω=Kv = velocity of sound (for a given polarization)( )2348V dKD Kdω ππ ω=222V Kvπ=22 32Vvωπ=For a crystal of N primitive cells:( )0DN D dωω ω=∫22 302DVdvωω ωπ=∫32 36DVvωπ=→1/32 36DNvVω π =  1/326DDNKv Vωπ = =  Debye frequencyωDKDBZdoes not have to lie on the BZ boundaryDebye wavevectorThermal (vibrational) energy ( ) ( )0DU d D nωω ω ω ω=∫h2/2 302 1DBk TVU dv eωωωω ωπ=−∫hh4 4 32 3 302 1DxBxV k Txdxv eπ=−∫hBxk Tω=hDDBxk T Tωθ= =hwhereθ= Debye temperature1/326DB Bv Nk k Vωπ = =  hh33031DxBxT xU Nk T dxeθ = − ∫for each acoustic branch3/2 302 1DBVk TVC dv T eωωωωπ ∂= ∂ − ∫hh h( )3/22 3 2/021DBBk Tk TBVd ev k Teωωωω ωωπ=−∫hhh h h( )342031DxxBxT x eNk dxeθ =  −∫Phonon Heat Capacity1 cal/mol-K 4.186 J/mol-K=Heat Capacity examplesfrom KittelDebye T3Law3310 01s xxsxdx dx x ee∞ ∞∞−==−∑∫ ∫For low T, xD→ ∞ :416ss∞−==∑()6 4ζ=→33031DxBxT xU Nk T dxeθ = − ∫345BTNk Tπθ →  for each acoustic branch3445V BTC Nkπθ →  378BTNkθ   3234BTNkθ   To account for all 3 acoustic branches, we set34125V BTC Nkπθ →