PH606 GM 3

Kinetic theory of gases

• for ideal monatomic gas of N molecules

• energy U = N 1/2mvrms2

• pressure P = d(mv)/dt � A = N 1/3mvrms2 � V

• observe P V = N kB T– Boltzmann constant kB relates T to energy

• equipartition of energy: 1/2kBT for each d.o.f– 1/2mvrms

2 = 3/2kBT where vx,rms2 = 1/3vrms

2

• molar specific heat Cv = ∆U/∆T = 3/2R– gas constant R = kBNA

• diatomic gas has 3 trans., 2 rot. and 2 vib. d.o.f– where 2 vib. d.o.f are kinetic and potential

– molar specific heat Cv = ∆U/∆T = 7/2R

PH606 GM 4

Kinetic theory of crystals

• for ideal monatomic cubic crystal of N atoms

• energy U = N (1/2mvrms2 + 1/2κ(x-a)rms

2)

• equipartition of energy: 1/2kBT for each d.o.f

• atom in crystal vib. d.o.f. =6 (3xKE+3xPE)– molar specific heat Cv = ∆U/∆T = 3R

• this model is too simple:

• quantum description of vibrations– CV → 0 as T → 0

• anharmonic potential– U = 1/2k(x-a)2 - c(x-a)3 ...

PH606 GM 6

Vibrations of monatomic lattice in 1D

• positions xj=ja and displacements xj→xj+uj

• assume force due to nearest neighbours– Fj = κ(uj+1 - uj) - κ(uj - uj-1) = κ(uj+1 + uj-1 - 2uj)

• assume wave solution uj=Aexp(i(kxj-ωt))– wavevector k=2π/λ and angular frequency ω=2πf

• Fj = maj hence d2uj/dt2 = -ω2uj

– satisfied if ω2=(κ/m) [2- uj+1/uj - uj-1/uj]

– use uj+1/uj=exp(ika) and uj-1/uj=exp(-ika)

– ω2 = (κ/m) [2 - exp(ika) - exp(-ika)] = 2(κ/m) [1 - cos(ka)] = 4(κ/m) sin2(ka/2)

PH606 GM 7

• ω = ωmaxsin(±ka/2) with ωmax=2(κ/m)1/2

– k has continuous values and ±k denotes direction

• for large λ, have k→0 and sin(ka/2)→ka/2– ω=2(k/m)1/2ka/2

– wave velocity v0=ω/k=a(κ/m)1/2

– group velocity v=dω/dk=v0

– (atoms in phase)

– elastic wave: v=(B/ρ)1/2, B=κa, ρ=m/a

• for small λ, have ω→ωmax as k→π/a– group velocity v=dω/dk=v0cos(ka/2)

– group velocity→0 as k→π/a

– (atoms in antiphase)

1st Brillioun zone

k<<π/a, in phase

at rest

PH606 GM 8

• the first Brillioun zone: -π/a<k<π/a

• is a region in the reciprocal lattice– halfway to reciprocal lattice points K=±2π/a

• note: uj(k+K)=exp(iKxj)uj(k)=uj(k)

• if K is a reciprocal lattice vector– because Kxj=nπ and exp(iKxj)=1

– k>π/a is equivalent to k'=k-K and -π/a<k'<0

– k<-π/a is equivalent to k'=k+K and 0<k'<π/a

• the Brillioun zone describes all unique waves– wave with k=π/a is standing wave due to Bragg reflection

– 2dsinθ=λ with d=a and 2θ=180° hence k=2π/λ=π/aat rest

k=π/a, out of phase

PH606 GM 10

• wave solutions uj=εsexp(i(k.rj-ωst))– where ωs and εs depend on wavevector k

• angular frequency ωs(k) is function in 3D

• 3 polarisation vectors εs(k)– simplified for special geometries

– 1 longitudinal (L), with u parallel to k

– 2 transverse (T), with u perpendicular to k

– referred to as T/L "modes" or "branches"

• in general εs have mixture of T/L

• ωs(k) are different for L and T modes– L involves stretching of bonds, higher ω– T involves bending of bonds, lower ω

PH606 GM 11

• 3D dispersion ωs(k)– consider ωs(|k|) for a particular direction k^

specified by reciprocal lattice vector K, e.g. (1,0,0)

• ωs(|k|) has similar behaviour to 1D crystal– for large λ, have |k|→0 and find ω=v0|k|

– wave velocity = group velocity = v0

– (atoms in phase)

– for small λ, have ω→ωmax as |k|→|K|/2– standing wave due to Bragg reflection

– (atoms in antiphase)

PH606 GM 12

Vibrations of diatomic lattice in 1D

• positions xj=ja/2 and displacements xj→xj+uj

– even j for atom type M

– odd j for atom type m

• assume force due to nearest neighbours– Fj = κ(uj+1 - uj) - κ(uj - uj-1) = κ(uj+1 + uj-1 - 2uj)

• for even j, assume wave solution uj=Aexp(i(kxj-ωt))

• for odd j, assume wave solution uj=Bexp(i(kxj-ωt))

PH606 GM 13

• Fj=Maj satisfied if -Mω2uj = κ [uj+1/uj + uj-1/uj - 2]– hence -Mω2 = κ [Bexp(ika/2)/A + Bexp(-ika/2)/A - 2] = κ [2Bcos(ka/2)/A - 2]

• Fj=maj satisfied if -mω2uj = κ [uj+1/uj + uj-1/uj - 2]– hence -mω2 = κ [Aexp(ika/2)/B + Aexp(-ika/2)/B - 2] = κ [2Acos(ka/2)/B - 2]

• require (2κ - Mω2)(2κ - mω2) = 4κ2cos2(ka/2)– this is quadratic in ω2 with 2 solutions

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PH606 GM 14

• for large λ, have k→0 and sin(ka/2)→ka/2 with ka/2<<1– can rearrange to get two solutions

• low frequency solution ω=v0k– wave velocity = group velocity = v0 = (κa2/2(M+m))1/2

– same as for monatomic lattice in 1D

– called acoustic (like sound waves)

• high frequency solution ω=(2κ(M+m)/Mm)1/2

– group velocity=0

– called optical (because interact with photons)

• for small, have k→π/a with sin(ka/2)→1– can rearrange to get two solutions, ω2=2κ/M or 2κ/m

– group velocity=0

( )2/1

222 2/sin4)()(

−

+

±+

=��

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��

�� κκϖ

PH606 GM 15

PH606 GM 16

Vibrations of diatomic lattice in 3D

• as for the monatomic lattice in 3D– wave solutions uj=εsexp(i(k.rj-ωst))

• dispersion ωs(k) is function in 3D– consider ωs(|k|) for a particular direction k^

• 3 polarisation vectors εs(k)– 1 longitudinal (L) and 2 transverse (T)

• as for diatomic lattice in 1D– M and m atoms can vary in phase

• low frequency solution– called acoustic (M and m atoms in phase)

• high frequency solution– called optical (M and m atoms out of phase)

PH606 GM 26

Classical approach

• energy of atom in 1-dimension E = 1/2mvx2 + 1/2kx2

– need 3 dimensions in total

• average over Boltzmann distribution

– <E> = ∫∫Eexp(-E/kBT)dvdx / ∫∫exp(-E/kBT)dvdx

• gives <E> = kBT– i.e. equipartition of energy: 1/2kBT for each d.o.f.

• hence U = 3NkBT or CV = 3NKB = 3R– gas constant R = kBNA

• agrees with experiment at high T but not at low T

• wrong because doesn't include quantisation of E– due to quantum mechanics

PH606 GM 27

Quantum mechanics approach

• Planck proposed electromagnetic wave E=n�ω*

• Einstein proposed n photons with E=�ω• vibrational waves are quantised as phonons**

– energy of vibrational wave is n�ω– i.e. n phonons each with energy �ω

• average over quantised energies

– <E> = Σn En exp(-En/kBT) / Σn exp(-En/kBT)

• gives <E> = �ω / [exp(�ω/kBT) - 1] = <n>�ω– <n> = 1 / [exp(�ω/kBT) - 1]

– corresponds to Bose-Einstein distribution

– phonons are a type of boson (integer spin)

** Note: phonons used in spectroscopy lectures

* Note: QM of simple harmonic motionsolutions are quantised with E=(n+1/2)�ω

Maxwell-BoltzmanBose-EinsteinFermi-Dirac

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PH606 GM 28

Einstein model• ������������������������������������ω�ω�

– (like an optical vibration)

– total of 3N different vibrational modes

• hence U = 3N�ωE / [exp(�ωE/kBT) - 1]– CV = 3NKB FE(ωE,T), with function FE

• temperature dependence of CV due to FE

– FE(ωE,T) =

(�ωE/kBT)2 exp(�ωE/kBT) / [exp(�ωE/kBT) - 1]2

• FE(ωE,T) ≈ 1 at high T– because exp(�ωE/kBT)≈1+�ωE/kBT

• FE(ωE,T) ≈ (�ωE/kBT)2exp(-�ωE/kBT) for T<<TE

– where Einstein temperature TE=�ωE/kB

• does not agree with experiment – because experimental CV proportional to T3 at low T

PH606 GM 29

Debye model

• ��������������ω����– (like an acoustic vibration)

– total of 3N different vibrational modes

• U = ∫ �ω g(ω) / [exp(�ωD/kBT) - 1] dω• density of states g(ω)*

– gives number of modes at each ω

– require ∫ g(ω) dω = 3Ν

• find g(ω) = 3Vω2 / 2π2v03 for 0<ω<ωD

– 3N = VωD3 / 2π2v0

3

• U =

[3V� / 2π2v03] ∫ ω3 / [exp(�ωD/kBT) - 1] dω

* g(ωE) = 3N in Einstein model

PH606 GM 30

• U = [3V� / 2π2v03] ∫ ω3 / [exp(�ωD/kBT) - 1] dω

– change to dimensionless variable x=(�ω/kBT)

• U = [9NKBT4 / θD3] ∫ x3 / [exp(x) - 1] dx

– where Debye temperature θD = TD = �ωD/kB

• CV = [9NKBT3 / θD3] ∫ x4 exp(x) / [exp(x) - 1]2 dx

– integral ≈ 1/3(θD/T)3 at high T

– integral ≈ π4/15 at low T

• CV = [12π4NKB/15] (T/θD)3 at low T– agrees with experiment

– (metals have CV= AT + BT3 due to electrons)

PH606 GM 33

Phonon density of states in 3D

• assume 3D modes are equally spaced in k– maxiumum crystal dimension Lx=N1/3a

– continuous boundary conditions: kx=±2π/Lx, ±4π/Lx, ...

– g(k)=(∆k)-1=V/(2π)3

– if k taken to be |k| then g(ω)=4πk2g(k)dk/dω– g(k) = (Vk2/2π2)

– need extra factor of 3 because L/T modes in 3D

• density of modes g(ω)=g(k)dk/dω• assume ω = v0k

– this is Debye approximation

– g(ω) = 3Vk2 / 2π2v0 = 3Vω2 / 2π2v03