Homework # 5Chapter 5 KittelPhys 175ADr. Ray KwokSJSU(a) From the dispersion relation derived in Chapter 4 for a monatomiclinear lattice of N atoms with nearest-neighbor interactions, show thatthe density of states is where ωmis the maximum frequency.Dispersion relation:Therefore, And,Dω( )=2Nπ      1ωm2−ω2( )12Dω( )=Naπ      dKdωωK( )=ωmsinKa2      Kω( )=2a      sin−1ωωm      dKdω=2a      1ωm2−ω2( )12Dω( )=2Nπ      1ωm2−ω2( )12Prob. 1 – Singularity in density of statesVictor Chikhani(b) Suppose that an optical phonon branch has the form ω(K)= ω0-AK2, near K=0 in three dimensions. Show thatfor ω<ω0and D(ω)=0for ω>ω0. Here the density of modes is discontinuous.From the dispersion relation:we get and For 3-D we knowTherefore,D(ω) =L2π      32πA32      ωo2−ω2( )12ωK()=ωo− AK2Kω( )=1A12ω0−ω( )12dKdω=12A121ωo−ω( )12        Dω( )=L2π      34πK2( )dKdωDω( )=L2π      34π ω0−ω( )A      1A12      12ωo−ω( )12        =L2π      32πA32      ωo−ω( )12(a) Estimate for 300 K the RMS thermal dilation ∆V/V for a primitive cell of sodium. Take the bulk modulus as 7x1010erg cm-3. Note that the Debye temperature 158 K is less than 300 K, so that the thermal energy is of the order kBT.½ kBT = ½ B a3 (∆V/V)**∆V/V = kBT/Ba3∆V/V = .125T = 300KB = 7x1010 erg cm-3a = 4.225 x10-10mkB= 1.38062x10-16erg/K(b) Use this result to estimate the RMS thermal fluctuation ∆a/a of the lattice parameterV = ½ a3∆V/V = 3∆a/a∆a/a = .125/3 = .04Prob. 2 – RMS thermal dilation of crystal cellDaniel Wolpert** B = P/(∆V/V)U = ∫PdV = W (Q=0, adiabatic, no external heat)or U = PVaverage U = ½ kT (longitudinal only – bulk)V = ½ a3(bcc, 2 atoms per cubic cell)A) In the Debye approximation, show that the mean square displacement of an atom at absolute zero is <R2> = 3ħωd2/ 8πρv3 , where v is the velocity of sound. Start from the result (4.29) summed over the independent lattice modes: <R2> = (ħ / 2ρV) Σω-1. We have included a factor of ½ to go from mean square amplitude to mean square displacement.Prob. 3 – Zero point lattice displacement and strainAdam Gray<R2> = (ħ / 2ρV) Σω-1Σω-1 can be expressed as:Where ωdis the cutoff frequency, and D(ω) is the density of states.Using the following two equations in Kittel:(5.20) D(ω) = (VK2/2π2)(dK/dω)(5.21) ω = vKWe end up with D(ω) = (VK2/2π2v)Plugging this into the integral above results in:Σω-1= Vωd2/ 4π2v3Plugging this into the above expression for <R2> :<R2> = (ħ / 2ρV) (Vωd2/ 4π2v3)<R2> = (ħ ωd2/ 8π2ρv3) The final missing factor of 3 comes from the phonon velocity being independent of the polarization according to Kittel. ∫−dDdωωωω01)(B) Show that Σω-1 and <R2> diverge for a one-dimensional lattice, but the mean square strain is finite. Consider <(dR/dx)2)> = (1/4) ΣK2u02 as the mean square strain, and show that it is equal to ħωd2L/4πMNv3for a line of N atoms each of mass M, counting longitudinal modes only. The divergence of R2is not significant for any physical measurement.I think the equation for mean strain should be ¼ instead of ½ Consider R ~ uocos(ωt-kx), dR/dx would give you k·R. Averaging would give ½ for each time average and spatial average.To show the divergence, Equation (5.15) gives thatD(ω) = L/πvIf we use this to evaluate the same integral for Σω-1 from earlier, the result isΣω-1= (L/πv)(Ln(ωd)-Ln(0))Which diverges at low values, since Ln(0) goes to ∞.Starting with <(dR/dx)2)> = (1/4) ΣK2u02 Similar to part a) we can replace the sum by an integral over the frequencies with K2u02 multiplied by D(ω):dωD(ω)K2u02(1/4)From Kittel we have(4.29) u02= 4(n+1/2)ħ/ρVω(5.15) D(ω) = L/πv(5.21)K = ω/vPlugging these in to the above integral yieldsdω(L/πv)(ω/v)2(2ħ/ρVω)(1/4)(ħL/πρVv3) dω ωWhich results in: (ħLωd2/4πρVv3)But, ρV=NM for the given problem.So: <(dR/dx)2)> = (1/4) ΣK2u02 = (ħLωd2/4πΝΜv3)∫dω0∫dω0∫dω0Prob. 4 – Heat capacity of layer latticeGregory KaminskyHeat capacity of layer lattice. (a) Consider a dielectric crystal made up of layers of atoms, with rigid coupling between layers, so that the motion of the atoms is restricted to the plane of the layer. Show that the phonon heat capacity in the Debye approximation in the low temperature limit is proportional to T2.(b) Adjacent layers are now weakly bound, what form would the phonon heat capacity approach at extremely low temperatures.UwwK pK ppK=−∑∑hh,,exp( / )τ1Energy is:U dwD wwwpp=−∫∑( )exp( / )hhτ1D(w) is the number of modes per unit frequency range , also called the Density of states.Apply periodic boundary conditions over N2 primitive cells within a square of side LK is determined byexp[ ( )] exp[ ( ( ) ( )]i K x K y i K x L K y Lx y x y+=+++Whence K Kx y, ;=0±2πL;±4πL;±NLπ;Therefore there is one allowed value of K per area (2π/L)^2 in K space or area per K is (L/ 2π)2= A/4π2The total number of modes with wavevector less then K is (L/2π)2times the area of the sphere of radius K. N = (L/ 2π)2K2πThe density of states for each polarization is D(w) = dN/dw = 2*(L/ 2π)2K(dK/dw) π = (KA/2π)*dK/dwDensity of states = number of modes per unit frequency rangew = vKD(w) = wA/(2πv2)If there are N primitive cells in the specimen, the total number of acoustic phonons is N. A cutoff frequency wd is determined by using equation for N. N = (L/ 2π)2K2π = (L/ 2πv)2w2πwd = (2πv/L)*(N/ π)1/2 For each polarization type, energy is given by:U dwD www=−∫( )exp( / )hhτ1U dwwAvwwwD=−∫0221πτhhexp( / )Assuming that phonon velocity is independent of the polarization type, we multiply by the factor of 2. UAvdwwwwD=−∫hhπτ2021exp( / )With xwk TB=hxwk TTDDB= =hθ/And CUTlat=∂∂UAk TvdwxxBxD=−∫3 32 2021πhexp( )CUTAk TvdwxxlatBxD= =−∫∂∂π313 22 202hexp( )At low temperature limit xD→∞xwk TTDDB= =hθ/dwxx021∞∫−exp( )This is some finite constant.Thus phonon heat capacity is proportional to T2.Adjacent layers are now weakly bound, what form would the phonon heat capacity approach at extremely low