Lattice Dynamics
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Transcript of Lattice Dynamics
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21 February, 2007 U. Milan Short Course
Lattice Dynamics
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Thermal Expansion• As temperature changes,
density changes• Thermodynamics
– Relates this change to changes in other properties
– Cannot tell the magnitude or even the sign!
• Why positive alpha?• What value vs. P,T,X?• Macroscopic to Microscopic• Thermodynamics to
Statistical Mechanics
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Interatomic Forces• Ambient Structure
– Minimum• Bulk Modulus
– Curvature• Thermal
Expansivity– Beyond harmonic– Molecules– Solids
Potential EnergyDistance
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One Dimensional Lattice
€
Vn = V0 + 12
∂ 2Vn,n + p
∂un2 (un + p − un )2 + H.O.T.
p=−N
N
∑
Fn = −∂Vn
∂un
= K p (un + p − un )p=−N
N
∑
K p =∂ 2Vn,n + p
∂un2
m ∂2un
∂t 2 = K p (un + p − un )p=−N
N
∑
un = u0 exp i nka −ωt( )[ ]
ω2 = 4m
K p sin2 pka2
⎛ ⎝ ⎜
⎞ ⎠ ⎟
p=1
N
∑
ω = 2 Km
sin ka2 ⎛ ⎝ ⎜
⎞ ⎠ ⎟
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One Dimensional Lattice• Periodicity reflects
that of the lattice• Brillouin zone
center, k=0: =0• Brillouin zone
edge, k=/a: =maximum
• All information in first Brillouin zone
1st BrillouinZone
0
Wavevector, k
Frequency,
2(K/m)1/2
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One Dimensional Latticek=/a=2/; =2a
k0;
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Acoustic Velocities
– k0– w=2(K/m)1/2ka/2– w/k=dw/dk=a(K/m)1/2
• Acoustic Velocity– v=a(K/m)1/2
• Three dimensions– Wavevector, ki
– Polarization vector, pi
– For each ki, 3 acoustic branches
– ki pi longitudinal (P) wave– kipi transverse (S) waves (2)
€
=2 Km
sin ka2 ⎛ ⎝ ⎜
⎞ ⎠ ⎟
Frequency
1.00.80.60.40.20.0
Wavevector (/a)
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Polyatomic Lattice
• Unit cell doubled• Brillouin Zone Halved• Acoustic Branches
folded• New, finite frequency
mode at k=0• Optic Branch
Frequency
1.00.80.60.40.20.0
Wavevector (/a)
a
New BrillouinZone
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General Lattice• Number of modes = 3N
organized into 3Z branches– Z= number of atoms in unit
cell• 3 Acoustic branches• 3Z-3 optic branches• Experimental Probes
– Optic zone center• Raman• Infrared
– Acoustic near zone center• Brillouin
– Full phonon spectrum• Inelastic neutron scattering Quartz
Fumagalli et al. (2001) EPSL
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MgSiO3 Perovskite Movie
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Internal Energy
• Sum over all vibrational modes
• Energy of each mode depends on– Frequency– Population
• Frequency• Temperature
Energy
Displacement
n=0
n=1
n=2
n=3
€
En = hω 12
+ n ⎛ ⎝ ⎜
⎞ ⎠ ⎟
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Heat Capacity
• or• CV=3R per mol of
atoms (Dulong-Petit)
€
Uvib = 12
hωi + n ihωii=1
3N
∑i=1
3N
∑
n i = exp hωi
kT ⎛ ⎝ ⎜
⎞ ⎠ ⎟−1
⎡ ⎣ ⎢
⎤ ⎦ ⎥
−1
Uvib = 12
hωi + hωi
exp hωi
kT ⎛ ⎝ ⎜
⎞ ⎠ ⎟−1i=1
3N
∑i=1
3N
∑
High Temperature
Uvib ≈ 12
hωi + kTi=1
3N
∑i=1
3N
∑
CV = ∂U∂T ⎛ ⎝ ⎜
⎞ ⎠ ⎟V
= 3Nk
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Thermal Pressure 1
• Compression Increases– Vibrational
frequencies– Vibrational energy
Thermal pressure
€
FTH = 12
hωii=1
3N
∑ + kT ln 1− exp − hωi
kT ⎛ ⎝ ⎜
⎞ ⎠ ⎟
⎡ ⎣ ⎢
⎤ ⎦ ⎥
i=1
3N
∑
PTH = − ∂FTH
∂V ⎛ ⎝ ⎜
⎞ ⎠ ⎟T
PTH = γV
UTH
γ = γ iωii=1
3N
∑ ωii=1
3N
∑
γ i = −∂ lnωi
∂ lnV
Energy
Displacement
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Thermal Pressure 2Thermal Pressure
Bulk Modulus
Volume dependence of
€
PTH ≈ γρ 3RT
KTH ≈ γρ γ +1− q( )
q = − ∂ lnγ∂ lnρ
Thermal pressure
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Interatomic Forces• Ambient Structure
– Minimum• Bulk Modulus
– Curvature• Thermal
Expansivity– Beyond harmonic– Molecules– Solids
Potential EnergyDistance
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Fundamental Thermodynamic Relation
€
F(V ,T)
Helmholtz free energy as a function of volume and temperatureComplete information of equilibrium states/properties
Divide into purely volume dependent “cold” part and a thermal part
€
F(V ,T) = F0 + F(V ,T0) + FTH (V ,T)Recall we already have an expression for the “cold” part
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140
120
100
80
60
40
20
0
Pressure (GPa)
1.000.950.900.850.800.750.70
Volume, V/V0
MgSiO3Perovskite
300 K
F=af 2
Cold part• Start from fundamental relation• Helmholtz free energy
– F=F(V,T,Ni)
• Isotherm, fixed composition– F=F(V)
• Taylor series expansion• Expansion variable must be V or
function of V– F = af2 + bf3 + …
• f = f(V) Eulerian finite strain• a = 9K0V0
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Thermal part
€
FTH = 12
hωii=1
3N
∑ + kT ln 1− exp − hωi
kT ⎛ ⎝ ⎜
⎞ ⎠ ⎟
⎡ ⎣ ⎢
⎤ ⎦ ⎥
i=1
3N
∑
Not easily evaluated, need to know all vibrational frequencies at all pressures
This information not available for ANY mantle mineral!
What to do?
Frequencies only appear in sums
Thermodynamics insensitive to details of distribution of frequencies
Assume i=E, all i, where E is a characteristic frequency of the material
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Comparison to experiment
70
60
50
40
30
20
10
02000150010005000
Temperature (K)
Anorthite
S
CP
(H-H0)/T
-10
70
60
50
40
30
20
10
02000150010005000
Temperature (K)
Forsterite
S
CP
(H-H0)/T
-10
70
60
50
40
30
20
10
02000150010005000
Temperature (K)
CP
(H-H0)/T
Corundum
S
-10
Characteristic (Debye) frequencies
Anorthite 522 cm-1
Forsterite 562 cm-1
Corundum 647 cm-1