Ch4: THERMAL PROPERTIES OF SOLIDS Course instructor: Dr Indranil Bhaumik
Effects of Phonons The creation (or annihilation) of even a single phonon can lead to critical material characteristics as: –Thermal conduction –Specific Heats The Specific Heat of a Crystalline Solid Specific heat: The amount of heat energy required per mole to raise the temperature of a solid by a given amount ∂ Q C ′ = V ∂ T V
∂ Q C ′ = P ∂ T P
Cv′ = C ′p −
α TV K
where V = volume, α = coefficient of thermal expansion and K = compressibility More frequently, we use the heat capacity per unit mass for generality We will not talk in this course
Talking about
Contribution from free electron (imp. for metals) ………we’ll talk later
CV/3Nk
T3-law constant
1.0 exponential
0.5
0 0
0.5 T/ΘE
1. 0
T (K)
[1] 1819 experimentally observed the specific heat of all (what they measured!) solids was about the same (independent of temperature) Quantified in the law of Dulong & Petit:
cV ~ 6 cal/mol.K = 25 J/mol.K
[2] Classical theory: no idea about quantization of lattice vibration. assumed that atoms are like classical harmonic oscillator, oscillates independently E= 3NKT C = 3NK= 3R (same as D-P law) 1900 AD……… Planck gave the idea of quantization in black body radiation [3] 1906 AD…..Einstein – used the idea But approximated that the atoms vibrates INDEPENDENTLY with a frequency ωE Introduced temperature dependence in the Sp. heat expression:
CV ≈ e
− ω E kT
Still fails to explain the T dependence at lower temperature!
CV ≈ T 3
[4] 1912 AD.. Debye
Classical model of heat capacity Simplistic picture: an atom vibrates simple harmonically about its equilibrium lattice position because of its thermal energy Such an atom can be thought of as being like a sphere supported by springs
)
(HARMONIC OSCCILATOR
Energy of an oscillator Et= KE + PE = ½ mv2 + ½ fx2
Force constant
Expectation value of energy:
< E >=
∫∫ Et e ∫∫ e
−
−
Et kT
Et kT
= kT
dvdx
Total energy of all atoms (oscillators)
integrating over all v and x
dvdx
Theory of equi-partition of energy
No. of atom In a 3-dimensional solid, the oscillator has energy, E = 3kT This is the energy per atom. The total internal energy per mole is therefore E = 3N0kT
Cv =
∂ E = 3kN 0 = 3R = 24.9 J / mole − K ∂T
Since N0 = 6.02 x 1023 (g.mol)-1 and k = 1.38 x 10-23 J/K which agrees very well with Dulong-Petit law The only problem is that this predicts Cv to be independent of T, which we saw is not the case, in fact Cv is less than 3R at lower temperature
The Einstein model of a solid (independent quantum oscillator) Energies of the atomic simple harmonic oscillators comprising the solid are quantised.
1 En = (n + )ω0 ≈ nω0 2
U(x)
E E2 E1 E0 x
In order to oscillate at all, the oscillator must possess at least one quantum of energy (hω)…………….. we’ll again ignore zero point energy. How energy is distributed amongst the available energy states?
Energy distributions We have 3N independent simple harmonic oscillators (where N is the total number of atoms in the solid). Number of ways of distributing quanta of energy amongst these oscillators? Say we have 3 quanta of energy to distribute amongst 2 oscillators:
This is one possibility….. Sketch the remaining possibilities. How many possibilities in total are there?
3 ω 0
0ω 0
Oscillator 1
Total of 4 possibilities
Oscillator 2
Boltzmann distribution: The probability of finding a component of the system (e. g. an atom) in an energy state Εn at temp T is proportional to the Boltzmann factor:
exp(−
E3=
3ω
E2=
2ω
E n (NB: T is in Kelvin) ) kT
Number of oscillators, nn , in energy level En =
n0 exp(−
n(ω ) ) kT
where n0 is the number of oscillators in the ground state. E1= ω E0=0 (we’re ignoring zero point energy)(still occupied by n0 oscillators as this is actually non-zero) ω Let x = e x p − Therefore, n1 = n0x, n2 = n0x2, ……… nn = n0xn kT
Total energy of all the oscillators? If oscillator is in the first excited state (E1) its energy is ħω. The total number of oscillators in energy level 1 is n1. Total energy of oscillators in level 1: n1 . ω = n0x. ħω
E1 E0
If an oscillator is in the second level (E2) its energy is 2ħω . The total number of oscillators in energy level 2 is n2. Total energy of the oscillators in level 2 : n2 . 2 ħω = n0x2. 2ħω Add these all together to get:
Etotal = n1ω + n2 2ω + n3 3ω + ....... Using: n1 = n0x, n2 = n0x2, …… nn = n0xn
ETOTAL = n0. ω
(x + 2x2 + 3x3 + ….)
Similarly: NTOTAL = n0 (1 + x + x2 + x3 +….) The average energy, is: ∞
denominator =
∑ xn
ETOTAL n0 ω ( x + 2 x 2 + 3 x 3 + ....) < E >= = N TOTAL n0 (1 + x + x 2 + x 3 + ......)
= 1 + x + x 2 + ..... =
n=0
1 for x < 1 1− x
1 d numerator = ω ∑ nx n = ω x ∑ x n = ω x (1 − x )2 dx n=0 ∞
ω E =
x ω − 2 (1 − x ) = ω x = ω e kT ω 1 − 1− x 1 − e kT 1− x
E =
ω exp(ω / k T ) − 1
[
]
Expectation value of energy
Expand model to include N atoms and 3 degrees of freedom within the crystal
U=
3 Nω E [exp(ω E / kT ) − 1]
(ω E / kT ) 2 exp(ω E / kT ) ∂U Cv = = 3 Nk . ∂T V [exp(ω E / kT ) − 1]2 = 3 Nk ⋅ FE (ω E , T ) Where FE is the Einstein Function
FE (ωE , T ) =
Where Einstein temp.
( ωE / kT ) 2 exp(ωE / kT ) 2
[exp(ωE / kT ) − 1] 2
TE exp (T / T ) T E = 2 [exp(TE / T ) − 1]
T E = ( ω E / k ) At high temperatures, (T>>TE ).
2
2
TE exp 2(T / 2T ) TE exp(T / 2T ) E E T T = f2 = FE (ω E , T ) = 2 [exp(TE / T ) − 1] [exp(TE / T ) − 1] TE 1 + + ...... T T 2T f T →∞ = E ≅ 1+ E ≅1 T 2T 1 + TE + ...... − 1 T
Cv= 3Nk…..classical value Below TE, the energy deviates exponentially from the classical expectation T>>TE
Cv=3NR
T<
The Debye Model (Coupled quantum oscillator) Can we do better? We begin again with the same postulate that suggests 3N vibrational modes in a solid with N atoms. Debye also postulated that solid is a system of coupled oscillator. They oscillate with a range of frequency dependent on wave vector k. Debye
Einstein
ω are equal for all oscillators
ω1≠ω 2; no. of oscillator having distributed ω
Frequency of oscillation were assumed to lie in a range from 0 to a maximum ωD such that
ωD
3N =
∫ g (ω ) ⋅ dω 0
g(ω) g(ω) is density of state, number of oscillators with freq. ω
Debye spectrum
Total mode = 3N
Experimentally get g(ω) ωD
Calculate ωD
ω
U Debye =
ω
D
∫
E (ω ) g (ω )dω
0
ω
=
D
∫ 0
(ω ) ⋅ g (ω ) ⋅ dω [exp(ω / k T ) − 1]
In the limit T 0 CV ∝ T3. This model explains the low temperature behaviour.
Derivation excluded Can be seen in Kittel
Thermal expansion of solid
E
E ro
atomic separation
Harmonic Oscillator x
increasing thermal energy
Eb
Harmonic potential U(x)= cx2
average inter-atomic distance increases Anharmonic oscillator U(x)= cx2+gx3-fx4+…
•Because the curve is not symmetric, the increased energy of the atoms
with the increase in temp. leads to a change of average atomic spacing •If the curve is more symmetric, the effect is reduced (becomes 0 if it is symmetric!) –and the coefficient of thermal expansion is lower. (See the proof at the end of this handout) Thermal conductivity Q (heat flow) Hot Th
T −T dT Q = − kA h c = −kA Cold L dx Tc
Thermal conductivity L Heat is carried by: electrons and phonons Depending on the material involved, one or other species tends to dominate Examples: Insulator vs. metal -1 -1 In electrical insulators, no free electrons, Diamond 2000(W.m .K ) BN (cubic) 1300 so the heat conducted by SiC 490 lattice vibrations or phonons Silver 406.0 Copper 385.0
Looking for an expression for k contributed by phonon: To do this, we consider a bar of material with a thermal gradient dT/dx We calculate the flow of energy through a volume due to the temperature gradient by calculating the energy carried by the “hot” phonons Phonons will move randomly and collide with each other and transfer energy. The average distance a phonon travels between collisions is called the mean free path l (MFP) If c is the specific heat of a phonon moving from a local temp. T+δT to a region of local temperature T gives energy c δT, travels a path l (MFP) δT= (dT/dx). l = dT/dx. vxτ τ= l /vx is mean time between two collision vx is the velocity along x direction Transfer of energy per unit area per unit time Q = energy carried by no of phonon in pipe of length vx of cross section unit area = -No. cδT = -(n.vx). c (dT/dx. vxτ) n=concentration of phonon = -n.vx2. c (dT/dx) τ 2 = -C. 1/3 v . (dT/dx) τ [nc=C, v2 = vx2+vy2+vz2 ~ 3vx2 = - 1/3 C v l dT/dx vτ=l = -K dT/dx So, What restricts mean free path ‘l’? Phonon Scattering Mechanisms: • Boundary Scattering • Defect & Dislocation Scattering • Phonon-Phonon Scattering
Effect of temperature (i.e., no. of phonon), boundary separation and defect concentration on mean free path Decreasing Boundary Separation
1 1 1 1 = + + l defect boundary phonon
l
Increasing Defect Concentration
Boundary 0.01
Defect
phonon ~ exp(θD/bT)
~ 1/T
Phonon Scattering
0.1 Temperature, T/θD
1.0
At high temperatures, the number of phonons ∝ T, As the temperature decreases further: No. of defects and phonon decreases and the mean free path, hence the conductivity goes up:
Still dominated by p-p scattering At lower temperatures K exhibits a peak. No. of phonon is low scattering from defect restricts l The height of the peak depends on the purity of the sample. For very pure materials it may even be limited by the size of the sample. The limiting process is scattering of phonons by lattice imperfections, grain boundaries, impurities... At still lower temp., scattering being very low, thermal conductivity becomes dominated by the Specific heat term (rem: Debye relation).
Appendix: Thermal expansion (derivation)
also see Kittel