Homework # 4 Chapter 4 Kittel Prob # 1 to 7 Phys 175A Dr. Ray Kwok SJSU
Prob. 1 – Monatomic linear lattice Victor Chikhani
Consider a longitudinal wave: us = u cos(ωt- sKa) which propagates in a monatomic linear lattice of atoms of mass M, spacing a, and nearest- e nighbor interaction C.
(a)
Show that the total energy of the wave is: E= ½ M Σ(dus/dt)2 + ½ C Σ(u-s s+1 u )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 Mω2u2
Substitution of us=u cos(ωt - sKa) into (1) E = ½ M(ω2u2sin2(ωt-sKa)+ ½ C[u2cos2(ωtsKa)+u2cos2(ωt-(s+1)Ka)-2u2cos(ωtsKa)cos(ωt-(s+1)Ka)]
(b) con’t Integrate (from 0 to 2π/ω) over time to find time - average total energy: = ∫{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’t Term with ωt will cancel out and the remaining terms become ½ C (1 – cos(Ka))u2 = ¼ Mω2u2 + ½ C(1-cosKa)u2 From (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 Mω2u2
Prob. 2 – Continum wave equation Jason Thorsen
Show that for long wavelengths the equation of motion
reduces to the continuum elastic equation:
Prove
reduces to
Solution: The group velocity is given as:
Where the wavevector For large wavelengths K << 1 and, therefore, and
Prove
reduces to
The equation of motion can be rewritten as: a is the separation distance between planes so let a = ∆x. And, us+1 – us is the change in u over the distance ∆x.
Q.E.D.
Prob. 3 – Kohn Anomaly Adam Gray
For the problem treated by (18) to (26), find the amplitude ratios u/v for the two branches at Kmax = π / a . Show that at this value of K the two lattices act as if decoupled: one lattice remains at rest while the other lattice moves.
Show Decoupling at K=π/a Starting with Equation 20: 2
− ω M1u = Cv[1 + exp(−iKa)] − 2Cu 2
− ω M 2 v = Cu[exp(iKa ) + 1] − 2Cv We then solve at Kmax =π / a . 2
− ω M1u = Cv[1 + exp(−iπ )] − 2Cu 2 − ω M 2 v = Cu[exp( iπ ) + 1] − 2Cv
This leaves:
2
− ω M1u = −2Cu 2 − ω M2v = −2Cv
Which are decoupled with frequencies 2
ω1 = 2C / M1 2
ω2 = 2C / M 2
At
2
ω1 = 2C / M1 , the u lattice moves while the v lattice is at rest. 2 Likewise, at ω2 = 2C / M 2 , the v lattice moves while the u lattice is at rest. Note: at ω1
− ω 2 M 1u = Cv[1 + exp( −iKa )] − 2Cu
Requires v = 0 for any K. i.e., only “u” lattice moves. Likewise, at ω2, only “v” lattice moves.
Prob. 4 – Kohn Anomaly Daniel Wolpert
4.4 Kohn Anomaly – We suppose that the interplanar force constant Cp between planes s and s+p is of the form Cp = A (sin(pk0a)/pa) Where A and k0 are constants and p runs over all integers. Such a form is expected in metals. Use this and Eq. (16a) to find an expression for ω2 and also ∂ω2/∂K is infinite when K=k0. Thus plot ω2 versus K or of ω versus K has a vertical tangent at k0 (there is a kink in k0 in the phonon dispersion relation ω(K)).
Know:
Cp = A (sin(pk0a)/pa) A and k0 are constants
Eq. 16a) ω2 = (2/M) Σp > 0(Cp (1-cos(pKa))
p is an integer
Substitute Cp into 16a : Find dω2/dK : Apply the identity:
ω2 = (2/M) Σp > 0((A (sin(pk0a)/pa))(1-cos(pKa))
= (2/M) * A * Σp > 0(sin(pk0a))(sin(pKa)) sin(a) * sin(b) = cos(a-b) + cos(a+b) dω2/dK = Σp > 0 ½ [cos(p(k0-K)) + cos(p(k0+K))]
When K = k0
= Σp > 0 ½ [cos(p(k0-k0)) + cos(p(k0+k0))] = Σp > 0 ½ [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 > 0 ½
(diverge)
As this increments, it will cause the function dω2/dK to go to infinity
Plot ω2 versus K to show there is a kink at k0 0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-1.2
-1.4
-1.6
-1.8
0
2
4
6
8
10
12
14
16
18
20
Prob. 5 – Diatomic Chain Brian Jennings
Statement of the problem 5. 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.
us-1 m
C
vs-1
10C
us
m m |------- a/2-------|
C
vs m
10C
us+1
C
m
vs+1 m
K
Equations of motion
us-1 m
C
vs-1
10C
us
C
m m |------- a/2-------|
vs
10C
m
us+1
C
m
m K
Substitute the travelling wave equations and to get
vs+1
us-1 m
C
vs-1
10C
us
C
m m |------- a/2-------|
vs
10C
m
us+1
C
m
m K
Set the determinant to zero
Which is
Solve as a quadratic equation
vs+1
us-1 m
C
vs-1
10C
us
C
m m |------- a/2-------|
vs
10C
m
At K= , the radical becomes
C
m
and
and
vs+1 m
K
And the final equation is
At K=0, the radical becomes
us+1
us-1 m
C
vs-1
10C
us
m m |------- a/2-------|
C
vs m
10C
us+1 m
K
vs+1 m
K
ω
0
C
Prob. 6 – Atomic Vibrations in a metal Nabel Alkhawani
Given 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 r
Objective 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 metal
- the electric force by the electron sea on the displaced 1 ion is where n( r ) is the number of electrons - n(r)= 2 - Plug 3 - 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 get
Second objective R for Na ion is roughly 2* 10-10 m M is (4*10-26 kg) The frequency is roughly 3*1013 Hz
Third objective The maximum wave vector K should be in the order of 1010 m-1 Assume the oscillation frequency is associated with the maximum wave vector v= ω/k will yield 3*103 m
Prob. 7 – Soft Phonon modes Gregory Kaminsky
Line of ions of equal mass but alternating in charge ep = e(-1)p as 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 ions.
Show that the contribution of the coulomb interaction to the atomic force constants is C pC
2( − 1) p e 2 = p 3a 3
Well ion feels a force due to all other ions. dF I expanded the force using the Taylor expansion F ( x ) = F0 + x * ( )0 dx and a bunch of other terms that I am ignoring. I assume that x is very small so other terms with x2, x3 are nearly zero. The constant term plays no role so only the restoring force, second term matters. F = kx. The second term is the k (the force constant).
( − 1) p * e 2 F= 4πε * ( pa + x ) 2
− 2 * ( − 1) p * e 2 the F = 4πε * ( pa ) 3
Taking the derivative of force at x = 0. Now the minus can be ignored since we know it is the restoring force. What happened to the 4πε I don’t know it seems to have mysteriously vanished from the answer in the book. [ note: The Coulomb constant = 1 in cgs units ]
[=F’(0)]
From 16a show that the dispersion relation may be written as ∞ 1 w / w = sin Ka + σ ∑ ( − 1) p (1 − cos pKa ) p − 3 2 p=1 2
2 0
with → w02 = 4γ / M
2
e2 and ⇒ σ = γ *a3
∞ 1 w = (4C / M ) sin Ka + (2 / M ) ∑ C pC (1 − cos pKa ) 2 p=1 2
2
Well considering that Dividing w2 by w02 we get this equation after canceling out all the terms. You can try it yourself.
Show that w2 is negative (unstable mode) at the zone boundary Ka = π if σ > 0.475 or 4/7ζ(3) ∞
w 2 / w02 = 1 − σ ∑ (1 − ( − 1) p ) p − 3 = 0 p=1
1 1 1 σ * 2 * (1 + 3 + 3 + 3 ...) = 1 3 5 7
Using a calculator I summed this up approximately to the seventh term (I got lazy afterwards) and yeah if σ = 0.475 all works out and if σ > 0.475 then w2 is negative.
The sum of ζ(3) – even terms ζ = 1.0505307 2*0.475* 1.0505307 = 1 The sum of ζ(3) = 1.202, if σ > 4/7 ζ(3) it also works out since 4/7 ζ(3) = 0.475.
Show that the speed of sound at small Ka is imaginary if σ > (2ln2)-1
Let’s do several approximations. We got:
∞ 1 w 2 / w02 = sin 2 Ka + σ ∑ ( − 1) p (1 − cos pKa ) p − 3 2 p=1
sin 2
1 1 1 Ka = (1 − cos Ka ) = K 2 a 2 2 2 4
2 2 K a 3 Then (1 – cos pKa)/p = 2p
To get imaginary speed, w2 < 0. 2 2 2 2 1 σ ( Ka ) ( Ka ) ( Ka ) ( Ka ) ( Ka ) 2 − (( Ka ) 2 − + − + ...) = 0 4 2 2 3 4 5
Canceling all out, going waco on the formula 1 σ 1 1 1 1 − (1 − + − + ...) = 0 4 2 2 3 4 5
This series is the alternating harmonic, it converges to ln2. ¼- σ*(ln2)/2=0. If σ > (2ln2)-1 then w2 is imaginary. So w2 goes to zero and the lattice is unstable for some value of Ka in the interval (0, π) if 0.475 < σ < 0.721
Thank god it’s over
