Homework # 4Chapter 4 KittelProb # 1 to 7Phys 175ADr. Ray KwokSJSUProb. 1 – Monatomic linear latticeConsider a longitudinal wave: us= u cos(ωt- sKa)which propagates in a monatomic linear lattice of atoms of mass M, spacing a, and nearest- neighbor interaction C.Victor Chikhani(a) Show that the total energy of the wave is:E= ½ M Σ(dus/dt)2+ ½ C Σ(us- us+1)2 The total energy is equal to the kinetic energy ( ½ Mv2) plus the potential energy ( ½ Cx2) for each atom, summed over all atoms. M, and C are the same for all atoms v=(dus/dt).E= ½ M Σ(dus/dt)2+ ½ C Σ(us-us+1)2 (1)(b) By substitution of u, in this expression, show that the time- average total energy per atom is: ¼ Mω2u2+ ½ C(1- cosKa)u2= ½ Mω2u2Substitution of us=u cos(ωt - sKa) into (1) E = ½ M(ω2u2sin2(ωt-sKa)+ ½ C[u2cos2(ωt-sKa)+u2cos2(ωt-(s+1)Ka)-2u2cos(ωt-sKa)cos(ωt-(s+1)Ka)](b) con’t Integrate (from 0 to 2π/ω) over time to find time- average total energy: <E> = ∫{½ M(ω2u2sin2(ωt- sKa) + ½ C[u2cos2(ωt- sKa) + u2cos2(ωt- (s+1)Ka) - 2u2cos(ωt- sKa)cos(ωt- (s+1)Ka)]}dt Knowing that ∫sin2(ωt- sKa)dt = ∫cos2(ωt- sKa)dt = ∫cos2(ωt-(s+1)Ka)dt = ½ And using the trig. relation that : ∫cos(ωt- sKa)cos(ωt- (s+1)Ka)dt=∫[½ cos[(ωt- sKa) - (ωt- (s+1)Ka)] + ½ cos[(ωt- sKa)- (ωt- (s+1)Ka)]dt = ½ cos(Ka)(b) con’tTerm with ωt will cancel out and the remaining terms become ½ C (1 – cos(Ka))u2<E> = ¼ Mω2u2+ ½ C(1-cosKa)u2From (9) ω2 = (4C/M)sin2(½ Ka) and from the relation sin2(α) = ½ (1-cos2α) we get:(1-cosKa) = 2ω2M/4C Therefore, ¼ Mω2u2+ ½ C(2ω2M/4C)u2 = ½ Mω2u2Show that for long wavelengths the equation of motionreduces to the continuum elastic equation:Prob. 2 – Continum wave equationJason ThorsenProve reduces toSolution: The group velocity is given as: Where the wavevectorFor large wavelengths K << 1 and, therefore, andThe equation of motion can be rewritten as:a is the separation distance between planes so let a = ∆x.And, us+1– usis the change in u over the distance ∆x.Q.E.D.Prove reduces toFor the problem treated by (18) to (26), find the amplitude ratios u/v for the two branches at . Show that at this value of K the two lattices act as if decoupled: one lattice remains at rest while the other lattice moves.aK /maxπ=Prob. 3 – Kohn AnomalyAdam GrayShow Decoupling at K=π/aStarting with Equation 20:CuiKaCvuM 2)]exp(1[12−−+=−ωCviKaCuvM 2]1)[exp(22−+=−ωWe then solve at .aK /maxπ=CuiCvuM 2)]exp(1[12−−+=−πωCviCuvM 2]1)[exp(22−+=−πωThis leaves:CuuM 212−=−ωCvvM 222−=−ωWhich are decoupled with frequencies121M/C2=ω222M/C2=ωAt , the u lattice moves while the v lattice is at rest.Likewise, at , the v lattice moves while the u lattice is at rest.121M/C2=ω222M/C2=ωCuiKaCvuM 2)]exp(1[12−−+=−ωNote: at ω1Requires v = 0 for any K. i.e., only “u” lattice moves.Likewise, at ω2, only “v” lattice moves.4.4 Kohn Anomaly – We suppose that the interplanar force constant Cpbetween planes s and s+p is of the formCp= A (sin(pk0a)/pa)Where A and k0are constants and p runs over all integers. Such a form is expected in metals. Use this and Eq. (16a) to find an expression for ω2and also ∂ω2/∂K is infinite when K=k0. Thus plot ω2versus K or of ω versus K has a vertical tangent at k0 (there is a kink in k0in the phonon dispersion relation ω(K)).Prob. 4 – Kohn AnomalyDaniel WolpertKnow: Cp = A (sin(pk0a)/pa)A and k0are constants p is an integerEq. 16a) ω2= (2/M) Σp > 0(Cp(1-cos(pKa))Substitute Cp into 16a : ω2= (2/M) Σp > 0((A (sin(pk0a)/pa))(1-cos(pKa))Find dω2/dK : = (2/M) * A * Σp > 0(sin(pk0a))(sin(pKa))Apply the identity: sin(a) * sin(b) = cos(a-b) + cos(a+b)dω2/dK = Σp > 0½ [cos(p(k0-K)) + cos(p(k0+K))]When K = k0= Σp > 0½ [cos(p(k0-k0)) + cos(p(k0+k0))]= Σp > 0½ [cos(0)) + cos(p(2k0))]When the series increases, the second cos term will oscillate from 1 to -1, the net result will cause that term to average to zero.Σp > 0[½ + ½ cos(p(2k0))]dω2/dK = Σp > 0½ (diverge)As this increments, it will cause the function dω2/dK to go to infinityPlot ω2versus K to show there is a kink at k00 2 4 6 8 10 12 14 16 18 20-1.8-1.6-1.4-1.2-1-0.8-0.6-0.4-0.200.2Statement of the problem5. Diatomic Chain. Consider the normal modes of a linear chain in which the force constants between nearest-neighbor atoms are alternately C and 10C. Let the masses be equal, and let the nearest-neighbor separation be a/2. Find ω(K) at K = 0 and K = π/2. Sketch in the dispersion relation by eye. This problem simulates a crystal of diatomic molecules such as H2.Prob. 5 – Diatomic ChainBrian Jenningsusvsus+1vs+1m m m m mmC C C10C 10C|------- a/2-------|Kvs-1us-1Equations of motionusvsus+1vs+1m m m m mmC C C10C 10C|------- a/2-------|Kvs-1us-1Substitute the travelling wave equationsandto getusvsus+1vs+1m m m m mmC C C10C 10C|------- a/2-------|Kvs-1us-1Solve as a quadratic equationWhich isSet the determinant to zerousvsus+1vs+1m m m m mmC C C10C 10C|------- a/2-------|Kvs-1us-1And the final equation isAt K=0, the radical becomesandAt K= , the radical becomes andusvsus+1vs+1m m m m mmC C C10C 10C|------- a/2-------|Kvs-1us-10KωGiven parameters1- the sodium ion mass is M2- the charge of the ion is “e”3- the number density of ions “conduction electrons is  the displacement of ion from equilibrium is rProb. 6 – Atomic Vibrations in a metalNabel AlkhawaniObjective 1- prove that the frequency is 2- estimate the frequency value for sodium3- estimate the order of magnitude of the velocity of sound in the metal1- the electric force by the electron sea on the displaced ion is where n( r ) is the number of electrons2- n(r)= 3- Plug- in the value of n(r) will yield 4-5-The frequency is given by By plug in the value of C in this equation we will getSecond objectiveR for Na ion is roughly 2* 10-10mM is (4*10-26kg)The frequency is roughly 3*1013HzThird objectiveThe maximum wave vector K should be in the order of 1010m-1Assume the oscillation frequency is associated with the maximum wave vector v= ω/k will yield 3*103mLine of ions of equal mass but alternating in charge ep= e(-1)pas the charge on the pth ion. Inter-atomic potential is the sum of two contributions: (1) a short-range interaction of force constant C = γ, and (2) a coulomb interaction between all