1. Incorporating the phonon distribution
Last time: heat capacity Cv = dU/dT where
Need to include the density of states D(E)
Got more questions than usual on DOS. Might have been because I said it would be on the test, but I should try to see if I can clear this up more. I actually felt like it was my best explanation so far, but maybe not. I moved to a similar explanation as in Kittel. Another discussion is on page 464 in Ashcroft and Mermin but it’s much less clear. Blackwell has an OK discussion but for electronic states in section 3.3. We didn’t get so far on other thermal stuff because of the questions, but that’s totally ok. DOS is more important that a detailed discussion of the other thermal properties. (Good question on why anharmonic effects can be treated as phonon collisions. I need to think about how to better explain that one.)
Pictures are reminders of analogies we used to help us understand density of states, density of available states, whether occupied or not.
Reviewing beginning of DOS conversation from last time
Plus for fermions, minus for bosons
Planck distribution

2. To make clear how important I think density of states is…
A density of levels and/or density of states question is guaranteed to be on a test, so ask questions if you don’t understand!
This is the first time I’m making this guarantee. I often asked a question about it before and people often did poorly and still did not learn this topic well enough. Since it’s an important topic that will come up in other classes, I decided I needed to emphasize it more. I figure if I tell you it will be on a test, you’re going to pay more attention. Am I right? 

3. Understanding DOS (confusing topic, see pages 109-112)
What is the density of points in the reciprocal lattice in one dimension?
1 point every 2/a, so density = #/length = a/2
Density of levels * dk = number of points in dk

4. Density of States (with degeneracy)
Density of Levels
Allowing  k, the # of states is
(sometimes only positive values of k, same result):
N = (L/2)3 (4 K3/3)
= V/83 (4 K3/3)
Dlevels = dN/dk
Times 2 if spin!
Note: sometimes you see the Volume term dropped, for example if finding energy per volume, it would be common to see DOS/V too.
If you are Einstein, you would take a degeneracy of 3 (one longitudinal and 2 transverse in 3D). Better yet, you can separate them into different frequencies, which I’ll show soon. For 2D, you’d only have degeneracy of 2.
For discussion of only positive k values, see extra slide at end.
Running waves allow the possibility of negative k values and allow energy transfer. Standing waves don’t (redundant). I mention this (even though I know it’s confusing) because you will see people using standing waves sometimes in CMP work and it’s useful to know why it’s slightly different and also that you should get the same result.
Density of States (with degeneracy)
What degeneracy for phonons?

5. Let’s Practice D(k): With a neighbor, figure out density of levels and density of states for a 2D system (like graphene).
How to get to D(E)?

6. Simpler: Finding the 3d density of states D(E) for Photons
(Same for spin since 2 polarization states)
Good analogy for low temp phonons. Why?
How to get energy in terms of k: Omega is 2pi times f, plug in for g with wave Velocity v= f times lambda, then finally k = 2pi over lambda

7. Density of States (for Photons)
Density of States (for Photons)
Phonon energy not just linear, so math gets a bit more challenging. You can approximate it as linear for low temperature acoustic phonons, though.
Why was this easier than for phonons?

8. Factor of 3 from one longitudinal and 2 transverse
Photons (before plug in c):
How about for Phonons?
Divided by two because light had 2 different polarizations
This density of states (which assumed Energy linear with k and 0 and k=0) would not work well for the optical (upper) branches. For that you have to take a more careful derivative of the phonon energies, assuming you know them. N
Factor of 3 from one longitudinal and 2 transverse

9. Another way to visualize
At low T’s only lattice modes having low frequencies can be excited from their ground states, since the energy distribution is heavily weighted toward low energies at low temperature
N() w
Low frequency
sound waves
at low T k
What’s another reason we might be limited to low temperature approximations?
We saw that our calculations of phonon dispersion curves were pretty accurate for low temperature. We could also only measure low q values by optical methods. Neutron measurements will give the whole curves but those are hard to come by.
Discussion of probabilities:
depends on the direction

10. How good is the Debye approximation at low T?
The lattice heat capacity of solids varies as at low temperatures; Debye law.
Excellent agreement for
non-magnetic insulators!
Motivation for Debye: The exact calculation of DOS is difficult for 3D unless you use a computer.
Debye obtained a good approximation to the heat capacity by neglecting the dispersion of the acoustic waves, i.e. assuming
for arbitrary wavenumber.
This approximation is still widely used today.
In a 1D crystal, Debye’s approximation gives the correct answer in both the high and low temperature limits.

11. Visualizing the approximations we make
Einstein
Visualizing the approximations we make
Einstein approximation to the dispersion
Debye approximation to the dispersion

12. Rest of today: Anharmonic Effects (Thermal expansion and if time thermal conductivity)
Thermal expansion important for:
Buildings/bridges/roads
Computer chips/devices
Space applications
Thermal conductivity important for:
Heat flow (e.g. cooking, thermoelectrics)

13. Harmonic Theory Assumptions
The following are the assumptions made in the harmonic theory:
Two lattice waves do not interact with each other
There is no thermal expansion
The elastic constants are independent of pressure and temperature
None of the above assumptions is satisfied in real crystals!

14. Flashback: PROPERTIES FROM BONDING: Energy versus bond length
• Bond length, r
• Bond energy, Eo
For example, in the ionic bond, the ions attract each other because opposite charge. Repulsive force from the repulsion of the electron clouds, due to the pauli principle.
Consider Ionic

15. Harmonic vs. Anharmonic
For a harmonic oscillator with a displacement x from equilibrium <x> = 0, independent of amplitude. Therefore, no thermal expansion is possible.
Thermal expansion is only possible as a result of non-harmonicity in potential.

16. Anharmonic Effects
Due to the shape of the interatomic potential curve, real crystals resist compression to a smaller volume than its equilibrium value more strongly than expansion.
This is a departure from Hooke’s law.
Anharmonic effect due to the
higher order terms in potential.
Thermal expansion is an example.
(Calculate on board)
In harmonic approximation, phonons
do not interact with each other. In the absence of boundaries, lattice defects and impurities (which also scatter the phonons), the thermal conductivity is infinite.
Considering anharmonic effects, phonons collide with each other and these collisions limit thermal conductivity.

17. Analogy to Elastic Properties
• Elastic modulus, E (or Y)
E similar to spring constant
• E ~ curvature at ro
E is larger if curvature is larger.

18. PROPERTIES FROM BONDING: CTE or a
• Coefficient of thermal expansion, a
• a ~ symmetry at ro
is larger if modulus E is smaller
and curve very asymmetric.
a generally decreases with increasing bond energy.

19. THERMAL EXPANSION: COMPARISON
Thermal expansion mismatch is a major problem for design of everything from semiconductors to bridges.
Particularly an issue in applications where temperature changes greatly (esp. engines).
Selected values from Table 19.1, Callister 6e.

20. Thermal expansion example
An Al wire is 10 m long and is cooled from 38 to -1 degree Celsius. How much change in length will it experience?
Does it matter that I give temperature in Celsius? Change in temp in celcius is same as Kelvin. Have to change if formula needs absolute temperature (not just a change).
-9.2 mm
Note:  =2, = 3 (area and volume expansion, resp.)

21. A solid object has a hole in it
A solid object has a hole in it. Which of these illustrations more correctly shows how the size of the object and the hole change as the temperature increases? #1 #2
A. illustration #1
B. illustration #2
C. The answer depends on the material of which the object is made.
D. The answer depends on how much the temperature increases.
E. Both C. and D. are correct.
Answer: A

22. Thermal conduction by phonons
A flow of heat takes place from a hotter region to a cooler region when there is a temperature gradient in a solid (zeroth law of therm)
The most important contribution to thermal conduction comes from the flow of phonons in an electrically insulating solid.
Thermal conduction is an example of a transport property.
A transport property is the process in which the flow of some quantity occurs.
Thermal conductivity describes the rate of heat flow in a material.
The thermal conductivity of a phonon gas in a solid can be calculated by means of the elementary kinetic theory of the transport coefficients of gases.

23. Heat conduction in a phonon and real gas The essential differences between the processes of heat conduction in a phonon and real gas;
Phonon gas
Real gas
Speed is roughly constant.
Both the number density and energy density is greater at the hot end.
Heat flow is primarily due to phonon flow with phonons being created at the hot end and destroyed at the cold end.
Average velocity and kinetic energy per particle are greater at the hot end, but the number density is greater at the cold end, and the energy density is uniform due to uniform pressure.
Heat flow is solely by transfer of kinetic energy from one particle to another in collisions which is a minor effect in phonon case.
cold
hot
hot
cold

24. Phonon-phonon collisions
The coupling of normal modes by the unharmonic terms in the interatomic forces can be pictured as collisions between the phonons associated with the modes. A typical collision process of
phonon1
After collision another phonon is produced
phonon2
and
conservation of energy
conservation of momentum

25. Phonons are represented by wavenumbers with
If lies outside this range add a suitable multible of to bring
it back within the range of Then, becomes
This phonon is indistinguishable from a phonon with wavevector
where , , and are all in the above range.
Longitudinal
Transverse
Umklapp process
(reverses the direction of energy transport)
Normal process
Phonon3 has
; Phonon3 has and Phonon3=Phonon3’

26. Normal and Umklapp Processes
N processes do not offer resistance because there is no change in direction or energy
U processes offer resistance to phonons because they turn phonons around k1 k2 k3 G
k’3 k1 k2 k3

27. Thermal Conductivity 
Measures the rate at which heat can be transported through a medium per unit area per unit temperature gradient.
Thermal conductivity due to phonons
(derivation in extra slides)
Mean free path
Phonon velocity
Heat capacity per unit volume

28. Temperature dependence of thermal conductivity K
Approximately equal to velocity of sound and so ~temperature independent.
Tends to classical value at high T’s
Mean free path = average velocity times scattering time ?
Temperature dependence of phonon mean free length is determined by phonon-phonon collisions at low temperatures
Since heat flow is associated with phonon flow, the most effective collisions for limiting the flow are those in which the phonon group velocity is reversed. Thus, Umklapp processes best limit the thermal conductivity.

29. What if scatter off more than phonons?
how often scatter from impurities
Mathiesen’s Rule
how often scatter total
how often scattered from phonons
Applies to electron and phonon interactions
Independent scattering processes means the RATES can be added.
5 phonons per sec. + 7 impurities per sec.
= 12 scattering events per second

30. Size Effect
When the mean free path becomes comparable to the size of the sample, the thermal conductivity also depends on the size of the sample. This is known as the size effect.
Imperfections such as dislocations, grain boundaries and impurities will scatter phonons. But at low temperatures the dominant phonon wavelength is so long, that these imperfections do not affect it.

31. Phonon Thermal Conductivity
Matthiessen Rule:
Kinetic Theory:
Phonon Scattering Mechanisms
Decreasing Boundary
Separation
Boundary Scattering
Defect & Dislocation Scattering
Phonon-Phonon Scattering l
Increasing
Defect
Concentration
Boundaries change the spring stiffness  crystal waves scatter when encountering a change of elasticity (similar to scattering of EM waves in the presence of a change of an optical refraction index)
Phonon Scattering
Defect
Boundary
0.01
0.1
1.0
Temperature, T/qD

32. Thermal conductivity optimization
To maximize thermal conductivity, there are several options:
Provide as many free electrons (in the conduction band) as possible
free electrons conduct heat more efficiently than phonons.
Make crystalline instead of amorphous
irregular atomic positions in amorphous materials scatter phonons and diminish thermal conductivity
Remove grain boundaries
gb’s scatter electrons and phonons that carry heat
Remove pores (air is a terrible conductor of heat)

33. Thermal conduction at low temperatures
for phonon-phonon collisions becomes very long at low T’s and eventually exceeds the size of the solid, because
number of high energy phonons necessary for Umklapp processes decay exponentially as
is then limited by collisions with the sample surface, i.e.
Sample width/length/diameter
T dependence of K comes from which obeys law in this region
Temperature dependence of dominates.

34. Low Temperature Cryostats
Solid State Physics has greatly benefited by the ability to study materials at low temperatures where the physics can be simplified.
The cryogens (e.g. liquid nitrogen or liquid helium) can however be quite expensive, particularly in West Virginia.
Closed-cycle cryostats consist of a chamber through which cold helium vapor is pumped. A mechanical refrigerator extracts the warmer helium exhaust vapor, which is cooled and recycled. Closed-cycle cryostats consume a larger amount of power, but need not be refilled with helium and can run continuously for an indefinite period. Objects may be cooled by attaching them to a metallic coldplate inside a vacuum chamber which is in thermal contact with the helium vapor chamber.

35. Cryogens Can be Dangerous: Some Examples
If you are going to deal with liquid cryogens, you need to take a safety course (e.g., Shared Facilities, Harley Hart)!
Eye protection: safety glasses with sides or face shield
Avoid eye or skin contact by wearing longsleeve shirt and long pants that cover the tops of closed shoes, and insulated gloves big enough to shed easily. Remove jewelry. Why?
Cryogens can condense oxygen out of the atmosphere.
Avoid transporting a container of liquid cryogen (>10 liters) on an elevator with any person in the elevator (asphyxiation).
Transfer cryogens slowly to prevent pressure buildup.
Do not interchange adapters between different cryogens
But there is one good thing about liquid nitrogen (besides the science). See next slide

36. Liquid Nitrogen Ice Cream
The basic recipe for liquid nitrogen ice cream is 2 parts heavy cream to 1 part milk with sugar and vanilla added to taste. The basic science idea is that by quickly cooling the liquids you form very small crystals that make the ice cream smooth.
Ingredients:
Milk - 1 pint (16 oz.)
Vanilla - 3 tablespoons
Sugar - 1/2 cup
Liquid nitrogen - 5 liters
Half and Half (or heavy cream) - 1 quart (32 oz)
Cooking Tools:
Large stainless steel mixing bowl
Wooden mixing spoons
Small deli cups
Gloves & safety glasses
Demonstration?

37. Typical high temperature trend
For an impure or polycrystalline specimen the maximum can be broad and low (see a), whereas for a carefully prepared single crystal (see b), the maximum is quite sharp and conductivity reaches a very high value, of the order that of the metallic copper in which the conductivity is predominantly due to conduction electrons. 10
10-1 10
10-1
Typical high temperature trend
(b)Thermal conductivity of artificial sapphire rods of different diameters
(a)Thermal conductivity of a quartz crystal

38. Wiedemann-Franz Law
Good heat conductors are usually good electrical conductors.
Metals & Alloys: free e- pick up energy due to thermal vibrations of atoms as T increases and lose it when it decreases.
Insulators: no free e-. Phonons are created as T increases, eliminated as it decreases.
Major Assumption: thermal = electronic
very high T & very low T (not intermediate)
The ratio of the thermal conductivity and electrical conductivity at a constant temperature is a constant for metals
Link to importance of Seebeck effect/thermoelectrics

39. THERMOELECTRIC COOLING & HEATING
Two different materials are connected at the their ends and form a loop. One junction is heated up.
There exists a potential difference that is proportional to the temperature difference between the ends.

40. THERMOELECTRIC COOLING & HEATING
Reversion of the Seebeck effect is the Peltier Effect.
A direct current flowing through heterojunctions causes one junction to be cooled and one junction to be heated up.
Lead telluride and or bismuth telluride are typical materials in thermoelectric devices that are used for heating and refrigeration.

41. THERMAL PROTECTION SYSTEM
Fig. 19.2W, Callister 6e. (Fig. 19.2W adapted from L.J. Korb, C.A. Morant, R.M. Calland, and C.S. Thatcher, "The Shuttle Orbiter Thermal Protection System", Ceramic Bulletin, No. 11, Nov. 1981, p )
• Application:
Space Shuttle Orbiter
Fig. 23.0, Callister 5e. (Fig courtesy the National Aeronautics and Space Administration.
• Silica tiles ( C):
--large scale application
Fig. 19.3W, Callister 5e. (Fig. 19.3W courtesy the National Aeronautics and Space Administration.
--microstructure:
~90% porosity!
Si fibers
bonded to one
another during
heat treatment.
Fig. 19.4W, Callister 5e. (Fig W courtesy Lockheed Aerospace Ceramics
Systems, Sunnyvale, CA.)

42. THERMOELECTRIC COOLING & HEATING
Why does this happen?
When two different electrical conductors are brought together, e- are transferred from the material with higher EF to the one with the lower EF until EF (material 1)= EF (material 2).
Material with smaller EF will be (-) charged. This results in a contact potential which depends on T.
e- at higher EF are caused by the current to transfer their energy to the material with lower EF, which in turn heats up. Material with higher EF loses energy and cools down.
Honestly, the Seebeck effect makes more sense after electrical conductivity, but unclear if there will be enough time to sneak into those lectures.

43. THERMOELECTRIC COOLING & HEATING
Peltier–Seebeck effect, or the thermoelectric effect, is the direct conversion of thermal differentials to electric voltage and vice versa.
The effect for metals and alloys is small, microvolts/K. For Bi2Te3 or PbTe (semiconductors), it can reach up to millivolts/K.
Applications: Temperature measurement via thermocouples (copper/constantan, Cu-45%Ni, chromel, 90%Ni-10%Cr,…); thermoelectric power generators (used in Siberia and Alaska); thermoelectric refrigerators; thermal diode in microprocessors to monitor T in the microprocessors die or in other thermal sensor or actuators.

44. Mathiesen’s Rule Resistivity
If the rates add, then resistivities also add:
Resistivities Add (Mathiesen’s Rule)

45. Derivation of thermal conductivity A material's ability to conduct heat.
Electric current density
Heat current density
Fourier's Law for heat conduction.
Heat current density
 = Energy per particle
v = velocity
n = N/V

46. About half the particles are moving right, and about half to the left.
Thermal conductivity
Heat current density
Heat Current Density jtot through the plane: jtot = jright - jleft
Heat energy per particle passing through the plane started an average of “l” away.
About half the particles are moving right, and about half to the left. x

47. Thermal conductivity
Heat current density x
Limit as l gets small:

48. Thermal conductivity
Heat current density x

49. Thermal conductivity
Heat current density x

50. Thermal conduction at high temperatures
At temperatures much greater then the Debye temperature D the heat capacity is given by temperature-independent classical result of
The rate of collisions of two phonons  phonon density.
At high temperatures the average phonon density is constant and the total lattice energy  T ; phonon number  T , so
Scattering rate T and mean free length
Then the thermal conductivity of
Need to review why v turns out to be T independent.

51. Conduction at intermediate temperatures
Referring to figure a
At T< D , the conductivity rises more steeply with falling temperature, although the heat capacity is falling in this region. Why?
This is due to the fact that Umklapp processes which will only occur if there are phonons of sufficient energy to create a phonon with So
Energy of phonon must be ~ the Debye energy ( k D )
The energy of relevant phonons is thus not sharply defined but their number is expected to vary roughly as exp(-D/2T)
when T~D,
Then Scattering length l  exp(-D/2T)
This exponential factor dominates any low power of T in thermal conductivity, such as a factor of T3 from the heat capacity.

52. Thermal conductivity of Metals
Scales linearly with temperature for a metal

53. Shorter Derivation: Thermal Conductivity
The thermal conductivity coefficient is defined with respect to the steady-state flow of heat down a temperature gradient, i.e.
To arrive at the expression for the thermal conductivity, we start from
where
Substituting back the above results leads to

54. Lindemann melting criterion[1]
A lattice will melt when the objects (atoms, electrons, molecules,…) residing on the lattice sites travel, on average, more than a critical distance out of their lattice site.
For electrons on a flat surface, Grimes and Adams observed classical melting when the electrons travel more than 13% of the distance between the the lattice points[2].
Motion out of the lattice site can be increased through:
Decreasing lattice parameter
(pressurizing)
Increasing temperature:
classical melting
Analogy
[1] F. A. Lindemann, Phys. Z. 11, 609 (1910).
[2] C. C. Grimes and A. Adams, Phys. Rev. Lett. 42, 795 (1979).

55. Values of c range from 0.1–0.3 for most materials
Lindemann criterion
Assumes all atoms in a crystal vibrate with the same frequency ν.
Average thermal energy can be estimated using the equipartition theorem as
                                
where m is atomic mass, u is average vibration amplitude, and T is temp.
If the threshold value of u2 is c2a2 where c (=f) is the Lindemann constant and a is the atomic spacing, then the melting point is estimated as
                          
                     
From the expression for the Debye frequency for ν, we have
                            
where θD is the Debye temp.
Values of c range from 0.1–0.3 for most materials
½ k x2, omega is square root of k/m

56. Phonons generated in the hot region travel toward the cold region and thereby transport heat energy. Phonon-phonon unharmonic interaction generates a new phonon whose momentum is toward the hot region.

57. The Old Version of Density of Levels
Allowing  k (or running waves):
Result same for standing waves (only positive k)!
4k2 dk
L/2
per dimension
N=V g(k)
I said I would put this in the extra slides. I was trying to make it more clear. If any of you like this approach better, let me know and ideally why you like it better. Note that the D we get here is a little different than what we discussed in class. Normally, I don’t see D dependent on V, but that’s what Kittel does, so I adjusted.
If used running waves, would lose the 1/8, but would gain 2^3 so same thing
Running waves allow the possibility of negative k values. Standing waves don’t (redundant). I mention this (even though I know it’s confusing) because you will see people using standing waves sometimes in CMP work and it’s useful to know why it’s slightly different and also that you should get the same result.
Times 2 if spin!
N=V g(k)
k Space