Solid sate physics

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 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