Homework # 5 Chapter 5 Kittel Phys 175A Dr. Ray Kwok SJSU
Prob. 1 – Singularity in density of states Victor Chikhani
(a) From the dispersio dispersion n relation relation derive derived d in Chapter Chapter 4 for a monato monatomic mic linear lattice of N atoms with nearest-neighbor nearest-neighbor interactions, show that 1 the density of states is D(ω ) = 2 N π ω − ω ( ) where ωm is the maximum frequency. frequency. 2 ω Na dK Dispersion relation: ω (K ) = ω sin Ka K (ω ) = sin D(ω ) = 2
2
1
2
m
−1
2
m
Therefore,
dK
ω d ω
2 =
a
1
(ω
2
m
And,
( )=
D ω
2 N
π
2
− ω
1
(ω
2
m
2
− ω
1
)
2
1
)
2
a
ω
m
π d ω ω
(b) Suppose that an optical phonon branch has the form 2 ω(K)= ω -AK , near K=0 in three dimensions. Show that 0 L 2π for ω<ω0 and D(ω)=0 D(ω ) = (ω − ω ) 3
2
2π A 2 3
1
2
2
o
for ω>ω0 . Here the density of modes is discontinuous. From the dispersion relation: we get K (ω ) = 11 (ω 0 − ω )12 A
and
dK d ω
1
=
2 A
For 3-D we Therefore,
1
2
ω (K ) = ω o − AK
2
1
1 2 (ω − ω ) 2 o
L 3 know D(ω ) = (4π K 2 )dK 2π d ω 1 L 3 4 π (ω 0 − ω ) 1 L 3 2π 1 = D(ω ) = ω − ω ) 2 1 1 3 ( o 2π A A 2 2(ω o − ω ) 2 2π A 2
Prob. 2 – RMS thermal dilation of crystal cell Daniel Wolpert (a) Estimate for 300 K the RMS thermal dilation ∆V/V for a primitive cell of sodium. Take the bulk modulus as 7x1010 erg cm-3. Note that the Debye temperature 158 K is less than 300 K, so that the thermal energy is of the order kBT.
½ kBT = ½ B a3 (∆V/V)**
T = 300K
∆V/V = kBT/Ba3
B = 7x1010 erg cm-3
∆V/V = .125
a = 4.225 x10-10 m kB = 1.38062x10-16 erg/K
(b) Use this result to estimate the RMS thermal fluctuation ∆a/a of the lattice parameter
V = ½ a3
∆V/V = 3∆a/a ∆a/a = .125/3 = .04
** B = P/(∆V/V) U = ∫PdV = W (Q=0, adiabatic, no external heat) or U = PV average U = ½ kT (longitudinal only – bulk) V = ½ a3 (bcc, 2 atoms per cubic cell)
Prob. 3 – Zero point lattice displacement and strain Adam Gray
A) In the Debye approximation, show that the mean square displacement of an atom at absolute zero is = 3ħωd2 / 8πρv3 , where v is the velocity of sound. Start from the result (4.29) summed over the independent lattice modes: = (ħ / 2ρV) Σω-1 . We have included a factor of ½ to go from mean square amplitude to mean square displacement.
= (ħ / 2ρV) Σω-1
Σω-1 can be expressed as: ω d
∫
0
−1
d ω D(ω )ω
Where ωd is the cutoff frequency, and D( ω) is the density of states. Using the following two equations in Kittel: D(ω) = (VK2 /2π2)(dK/dω) (5.21) ω = vK We end up with D(ω) = (VK2 /2π2v) Plugging this into the integral above results in: (5.20)
Σω-1 = Vωd2 / 4π2v3 Plugging this into the above expression for : = (ħ / 2ρV) (Vωd2 / 4π2v3) = (ħ ωd2 / 8π2ρv3) The final missing factor of 3 comes from the phonon velocity being independent of the polarization according to Kittel.
B) Show that Σω-1 and diverge for a one-dimensional lattice, but the mean square strain is finite. Consider <(dR/dx)2)> = (1/ 4) ΣK2u02 as the mean square strain, and show that it is equal to ħωd2L/ 4πMNv3 for a line of N atoms each of mass M, counting longitudinal modes only. The divergence of R2 is not significant for any physical measurement. I think the equation for mean strain should be ¼ instead of ½ Consider R ~ uocos( ωt-kx), dR/dx would give you k·R. Averaging would give ½ for each time average and spatial average.
To show the divergence, Equation (5.15) gives that D(ω) = L/ πv If we use this to evaluate the same integral for Σω-1 from earlier, the result is
Σω-1 = (L/ πv)(Ln(ωd)-Ln(0)) Which diverges at low values, since Ln(0) goes to ∞.
Starting with <(dR/dx)2)> = (1/ 4) ΣK2u02 Similar to part a) we can replace the sum by an integral over the frequencies with K2u02 multiplied by D(ω):
∫
ω d
dωD(ω)K2u02(1/ 4) From Kittel we have (4.29) u02 = 4(n+1/2)ħ / ρVω (5.15) D(ω) = L/ πv (5.21) K = ω /v Plugging these in to the above integral yields ω dω(L/ πv)(ω /v)2(2ħω / ρVω)(1/4) 0 (ħL/ πρVv3) 0 dω ω 0
∫
d
∫
d
Which results in: (ħLωd2 / 4πρVv3) But, ρV=NM for the given problem. So: <(dR/dx)2)> = (1/ 4) ΣK2u02 = (ħLωd2 / 4πΝΜv3)
Prob. 4 – Heat capacity of layer lattice Gregory Kaminsky
Heat capacity of layer lattice. (a) Consider a dielectric crystal made up of layers of atoms, with rigid coupling between layers, so that the motion of the atoms is restricted to the plane of the layer. Show that the phonon heat capacity in the Debye approximation in the low temperature limit is proportional to T2. (b) Adjacent layers are now weakly bound, what form would the phonon heat capacity approach at extremely low temperatures.
Energy is:
U =
hw K , p
∑ ∑ exp(hw K
U
=
p
∑ ∫ dwD
p
p
( w)
K,p
/ τ ) − 1
hw
exp( hw / τ ) − 1
D(w) is the number of modes per unit frequency range , also called the Density of states.
Apply periodic boundary conditions over N2 primitive cells within a square of side L
K
is determined by
exp[i ( K x x + K y y )] = exp[i ( K x ( x + L) + K y ( y + L)] Whence
K x , K y
=
0;
±
2π L
;
±
4π L
;
±
Therefore there is one allowed value of K per area (2π /L)^2 in K space or area per K is (L/ 2π)2 = A/4π2
N π L
;
The total number of modes with wavevector less then K is (L/2π)2 times the area of the sphere of radius K. N = (L/ 2π)2K2π
Density of states = number of modes per unit frequency range The density of states for each polarization is D(w) = dN/dw = 2*(L/ 2π)2K(dK/dw) π = (KA/2π)*dK/dw
w = vK D(w) = wA/(2πv2) If there are N primitive cells in the specimen, the total number of acoustic phonons is N. A cutoff frequency wd is determined by using equation for N. N = (L/ 2π)2K2π = (L/ 2πv)2w2π wd = (2πv/L)*(N/ π)1/2 For each polarization type, energy is given by:
U
U
=
=
hw
∫ dwD( w) exp(hw / τ ) − 1
∫
w D
0
dw
wA 2
hw
2π v exp( hw / τ ) − 1
Assuming that phonon velocity is independent of the polarization type, we multiply by the factor of 2.
U =
With
U =
Ah π v
2
x
∫
w D
0
=
dw
w
2
C lat
exp( hw / τ ) − 1
hw
3 B 2
3
Ak T π v h
2
x D
And
k B T
∫
x D
0
dw
x
=
2
exp( x ) − 1
hw D
k B T
=
∂ U ∂ T
= θ / T
C lat
=
∂ U ∂ T
=
3 Ak B3 T 2 2
π v h
At low temperature limit x D
∫
∞
dw
0
→
2
∫
x D
0
dw
x
exp( x ) − 1 x D
=
∞
x
2
hw D
k B T
= θ / T
2
This is some finite constant.
exp( x ) − 1
Thus phonon heat capacity is proportional to T2.
Adjacent layers are now weakly bound, what form would the phonon heat capacity approach at extremely low temperatures.
If adjacent layers are the only thing that can vibrate (at low T), then the problem now is a 1D and phonon heat capacity will approach T at low temperature, extrapolating from 2D and 3D models. [in-plane coupling is relatively too stiff so the only motion would be between planes.]
Prob. 5 – Gruneisen Constant Michael Tuffley
