solid sate physics

8/12/2019 Solid sate physics

1/37


2/37

Prob. 1 Monatomic linear lattice

Consider a longitudinal wave:

us = u cos(t- sKa)which propagates in a monatomic linear lattice of atoms of

mass M, spacing a, and nearest- neighbor interaction C.

Victor Chikhani


3/37

(a) Show that the total energy of the wave is:E= M (dus/dt)

2 + C (us- us+1)2

The total energy is equal to the kinetic

energy ( Mv2) plus the potential energy( Cx2) for each atom, summed over allatoms.

M, and C are the same for all atoms v=(dus/dt).

E= M (dus/dt)2 + C (us-us+1)

2 (1)


4/37

(b) By substitution of u, in this expression, show that thetime- average total energy per atom is: M2u2 + C(1- cosKa)u2 = M2u2

Substitution of us=u cos(t - sKa) into (1)

E = M(2u2sin2(t-sKa)+ C[u2cos2(t-sKa)+u2cos2(t-(s+1)Ka)-2u2cos(t-sKa)cos(t-(s+1)Ka)]


5/37

(b) cont Integrate (from 0 to 2/) over time to find time- average

total energy:

= { M(2u2sin2(t- sKa) + C[u2cos2(t- sKa) +u2cos2(t- (s+1)Ka)- 2u2cos(t- sKa)cos(t- (s+1)Ka)]}dt

Knowing that sin2(t- sKa)dt = cos2(t- sKa)dt = cos2(t-(s+1)Ka)dt =

And using the trig. relation that :cos(t- sKa)cos(t- (s+1)Ka)dt=

[ cos[(t- sKa)- (t- (s+1)Ka)] +

cos[(t- sKa)- (t- (s+1)Ka)]dt = cos(Ka)


6/37

(b) cont

Term with t will cancel out and the remaining

terms become C (1 cos(Ka))u2

= M2u2 + C(1-cosKa)u2

From (9) 2 = (4C/M)sin2( Ka) and from therelation sin2() = (1-cos2) we get:

(1-cosKa) = 22M/4C

Therefore, M2u2+ C(22M/4C)u2 = M2u2


7/37

Show that for long wavelengthsthe equation of motion

reduces to the continuum elastic

equation:

Prob. 2 Continum wave equationJason Thorsen


8/37

Prove reduces to

Solution: The group velocity is given as:

Where the wavevectorFor large wavelengths K


9/37

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.

Prove reduces to


10/37

For 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 /max =

Prob. 3 Kohn AnomalyAdam Gray


11/37

Show Decoupling at K=/a

Starting with Equation 20:

CuiKaCvuM 2)]exp(1[12 +=

CviKaCuvM 2]1)[exp(2

2 +=

We then solve at .aK /max =

CuiCvuM 2)]exp(1[1

2 +=

CviCuvM 2]1)[exp(22 +=


12/37

This leaves:CuuM 2

1

2 =

CvvM 222 =

Which are decoupled with frequencies

1

2

1 M/C2=

2

2

2 M/C2=


13/37

At , the u lattice moves while the v

lattice is at rest.Likewise, at , the v lattice moves while

the u lattice is at rest.

1

2

1 M/C2=

2

2

2 M/C2=

CuiKaCvuM 2)]exp(1[12 +=Note: at 1

Requires v = 0 for any K. i.e., only u lattice moves.Likewise, at 2, only v lattice moves.


14/37

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 2 and also

2/K is infinite when K=k0. Thus plot 2 versus K or of versus K has avertical tangent at k0 (there is a kink in k0 in the phonon dispersion relation (K)).

Prob. 4 Kohn AnomalyDaniel Wolpert


15/37

Know: Cp = A (sin(pk0a)/pa)

A and k0 are constants

p is an integer

Eq. 16a) 2 = (2/M) p > 0(Cp (1-cos(pKa))

Substitute Cp into 16a : 2 = (2/M) p > 0((A (sin(pk0a)/pa))(1-cos(pKa))

Find d2/dK : = (2/M) * A * p > 0(sin(pk0a))(sin(pKa))

Apply the identity: sin(a) * sin(b) = cos(a-b) + cos(a+b)

d2/dK = p > 0 [cos(p(k0-K)) + cos(p(k0+K))]

When K = k0 = p > 0 [cos(p(k0-k0)) + cos(p(k0+k0))]

= p > 0 [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.

p > 0 [ + cos(p(2k0))]

d2/dK = p > 0 (diverge)

As this increments, it will cause the function d2

/dK to go to infinity


16/37

Plot 2 versus K to show there is a kink atk0

0 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

-0.2

0

0.2


17/37

Statement of the problem

5. Diatomic Chain. Consider the normal modes of a linearchain in which the force constants between nearest-neighboratoms are alternately C and 10C. Let the masses be equal, andlet the nearest-neighbor separation be a/2. Find (K) at K = 0and K = /2. Sketch in the dispersion relation by eye. Thisproblem simulates a crystal of diatomic molecules such as H2.

Prob. 5 Diatomic ChainBrian Jennings


18/37

us vs us+1 vs+1

m m m m mm

C C C10C 10C

|------- a/2-------| K

vs-1us-1

Equations of motion


19/37

us vs us+1 vs+1

m m m m mm

C C C10C 10C

|------- a/2-------| K

vs-1us-1

Substitute the travelling wave equations

and

to get


20/37

us vs us+1 vs+1

m m m m mm

C C C10C 10C

|------- a/2-------| K

vs-1us-1

Solve as a quadratic equation

Which is

Set the determinant to zero


21/37

us vs us+1 vs+1

m m m m mm

C C C10C 10C

|------- a/2-------| K

vs-1us-1

And the final equation is

At K=0, the radical becomes and

At K= , the radical becomes and


22/37

us vs us+1 vs+1

m m m m mm

C C C10C 10C

|------- a/2-------| K

vs-1us-1

0

K


23/37

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

Prob. 6 Atomic Vibrations in a metalNabel Alkhawani


24/37

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


25/37

1- the electric force by the electron sea on the displaced

ion is where n( r ) is the number of electrons

2- n(r)=

3- Plug- in the value of n(r) will yield

4-

5-


26/37

The frequency is given by

By plug in the value of C in this equationwe will get


27/37

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


28/37

Third objective

The maximum wave vector K should be in

the order of 1010 m-1

Assume the oscillation frequency isassociated with the maximum wave vector

v= /k will yield 3*103 m


29/37

Line of ions of equal mass but alternatingin charge ep = e(-1)

p as the charge on thepth ion. Inter-atomic potential is the sum of

two contributions: (1) a short-rangeinteraction of force constant C = , and (2)a coulomb interaction between all ions.

Prob. 7 Soft Phonon modesGregory Kaminsky

S f


30/37

Show that the contribution of the

coulomb interaction to the atomicforce constants is

Well ion feels a force due to all other ions.

I expanded the force using the Taylor expansionand 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 roleso only the restoring force, second term matters. F= kx. The second term is the k (the force

constant).

C

e

p apC

p

=

2 1 2

3 3

( )

F x F xdF

dx( ) * ( )= +0 0


31/37

Taking the derivative of the

force at x = 0. Now the minuscan be ignored since we knowit is the restoring force.

What happened to the 4 Idont know it seems to have

mysteriously vanished fromthe answer in the book.

F epa x

p

=

+( ) *

* ( )1

4

2

2

Fe

pa

p

=

2 1

4

2

3

* ( ) *

* ( )

[=F(0)]

[ note: The Coulomb constant = 1 in cgs units ]


32/37

From 16a show that the dispersion

relation may be written as

Well considering that

Dividing w2 by we get this equation aftercanceling out all the terms. You can try ityourself.

w w Ka pKa pp

p

2

0

2 2 3

1

1

2 1 1/ sin ( ) ( cos )= +

=

with w M =02 4 /

ande

a =

2

3*

w C M Ka M C pKapCp

2 2

1

41

2

2 1= + =

( / ) sin ( / ) ( cos )

w02


33/37

Show that w2 is negative (unstablemode) at the zone boundary Ka =

if > 0.475 or 4/7(3)

Using a calculator I summed this up approximately

to the seventh term (I got lazy afterwards) andyeah if = 0.475 all works out and if > 0.475then w2 is negative.

w w pp

p

2

0

2 3

1

1 1 1 0/ ( ( ) )= =

=

* * ( ...)2 11

3

1

5

1

713 3 3+ + + =


34/37

The sum of (3) even terms =1.0505307

2*0.475* 1.0505307 = 1

The sum of (3) = 1.202, if > 4/7 (3)it also works out since 4/7 (3) = 0.475.


35/37

Show that the speed of sound at small Ka is

imaginary if > (2ln2)-1

Lets do several

approximations.We got:

w w Ka pKa p

p

p

2

0

2 2 3

1

1

2 1 1/ sin ( ) ( cos )= +

=

sin ( cos )2 2 2

1

2

1

2 1

1

4Ka Ka K a= =

Then (1 cos pKa)/p3 =K a

p

2 2

2

To get imaginary speed, w2 < 0.

1

4 2 2 3 4 5 0

2 2

2 2 2 2

( ) (( )( ) ( ) ( ) ( )

...)Ka KaKa Ka Ka Ka

+ + =


36/37

Canceling all out, going waco on the

formula

This series is the alternating harmonic, itconverges to ln2.

- *(ln2)/2=0. If > (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, ) if 0.475 < < 0.721

1

4 2 1

1

2

1

3

1

4

1

5 0 + + =( ...)


37/37

Thank god its over

solid sate physics

Documents