Question: How to measure heat capacity?

1. Question: How to measure heat capacity?

2. Where we left off (almost done) At low T’s only lattice modes having low frequencies can be excited from their ground states, since the Bose-Einstein function falls to zero for small values of  w Low frequency sound waves at low T k Only a good approximation at low temperature depends on the direction

3. Approx. dispersion relation of any branch by a linear extrapolation Ensure correct number of modes by imposing a cut-off frequency , above which there are no modes. The cut-off freqency is chosen to make the total number of lattice modes correct. Since there are 3N lattice vibration modes in a crystal having N atoms, we choose so that: Debye approximation has two main steps Einstein approximation to the dispersion Debye approximation to the dispersion

4. Einstein Debye Accordingly, the Debye spectrum may be written as Actual Notes: The Debye spectrum is only an idealization of the actual situation obtaining in a solid; it may be compared with a typical spectrum. While for low-frequency modes (the so called acoustical modes) the Debye approximation is reasonably valid, there are serious discrepancies in the case of high-frequency modes ( the optical modes). For “averaged” quantities, such as the specific heat, the finer details of the spectrum are not very important. Though not as common, the longitudinal and the transverse modes of the solid should have their own cut-off frequencies, D,L and D,T say, rather than having a common cut-off at D. Accordingly, we should have:

5. The lattice vibration energy of becomes and, First term is the estimate of the zero point energy, and all T dependence is in the second term. The heat capacity is obtained by differentiating above eqn wrt temperature. Debye Lattice Energy and Heat Capacity define the Debye temperature

6. How does limit at high and low temperatures? High temperature >> ~ is small >> Low temperature << For low temperature the upper limit of the integral is ~infinite; the integral is then a known integral of . << We obtain the Debye law in the form

7. 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-magneticinsulators! Motivation for Debye: The exact calculation of DOS is difficult for 3D. 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.

8. 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!

9. Flashback: PROPERTIES FROM BONDING:Energy versus bond length • Bond length, r • Bond energy, Eo Consider Ionic

10. 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.

11. 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.

12. 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.

13. 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.

14. 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.

15. Thermal expansion example 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? -9.2 mm Note:  =2, = 3 (area and volume expansion, resp.)

16. 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.

17. 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.

18. Heat conduction in a phonon and real gasThe essential differences between the processes of heat conduction in a phonon and real gas; Phonongas 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

19. 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 and phonon2 conservation of energy conservation of momentum

20. 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’

21. k2 k1 k3 k3 k’3 k1 k2 G 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

22. 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 Kittel and extra slides) Mean free path Phonon velocity Heat capacity per unit volume

23. Temperature dependence of thermal conductivity K Approximately equal to velocity of sound and so ~temperature independent. Tends to classical valueat 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.

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

25. 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.

26. 1.0 0.01 0.1 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) PhononScattering Defect Boundary Temperature, T/qD

27. 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)

28. 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. Samplewidth/length/diameter T dependence of K comes from which obeys law in this region Temperature dependence of dominates. Impurities have little affect at low temp

29. We are about to move into conductivity 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-. Phononsare created as T increases, eliminated as it decreases. Major Assumption: thermal = electronic Good @ 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

30. 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.

31. 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.

32. 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. 1189.) • Silica tiles (400-1260C): --large scale application --microstructure: ~90% porosity! Si fibers bonded to one another during heat treatment. Fig. 19.3W, Callister 5e. (Fig. 19.3W courtesy the National Aeronautics and Space Administration. Fig. 19.4W, Callister 5e. (Fig. 219.4W courtesy Lockheed Aerospace Ceramics Systems, Sunnyvale, CA.) • Application: Space Shuttle Orbiter Fig. 23.0, Callister 5e. (Fig. 23.0 courtesy the National Aeronautics and Space Administration.

33. 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.

34. 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.

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

36. 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

37. 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

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

39. Thermal conductivity Heat current density x

40. Thermal conductivity Heat current density x

41. 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 .

42. 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.

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

44. 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

45. Motion out of the lattice site can be increased through: Increasing temperature: classical melting Decreasing lattice parameter (pressurizing) 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]. Analogy [1] F. A. Lindemann, Phys. Z. 11, 609 (1910). [2] C. C. Grimes and A. Adams, Phys. Rev. Lett. 42, 795 (1979).

46. 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

47. 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.

48. 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.

49. 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

50. 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