PH606 GM Dynamics 2013 Just Red
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Transcript of PH606 GM Dynamics 2013 Just Red
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
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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