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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
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(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)
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(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)]
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(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)
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(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
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Show that for long wavelengthsthe equation of motion
reduces to the continuum elastic
equation:
Prob. 2 Continum wave equationJason Thorsen
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Prove reduces to
Solution: The group velocity is given as:
Where the wavevectorFor large wavelengths K
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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
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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
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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 +=
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This leaves:CuuM 2
1
2 =
CvvM 222 =
Which are decoupled with frequencies
1
2
1 M/C2=
2
2
2 M/C2=
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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.
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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
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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
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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
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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
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us vs us+1 vs+1
m m m m mm
C C C10C 10C
|------- a/2-------| K
vs-1us-1
Equations of motion
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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
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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
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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
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us vs us+1 vs+1
m m m m mm
C C C10C 10C
|------- a/2-------| K
vs-1us-1
0
K
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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
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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
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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-
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The frequency is given by
By plug in the value of C in this equationwe will get
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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
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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
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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
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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
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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 ]
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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
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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+ + + =
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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.
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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
+ + =
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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 + + =( ...)
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Thank god its over
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