Phys 446: Solid State Physics / Optical Properties Lattice Vibrations: Thermal, Acoustic, and Optical Properties
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Phys 446: Solid State Physics / Optical Properties Lattice vibrations: Thermal, acoustic, and optical properties Fall 2015 Lecture 4 Andrei Sirenko, NJIT 1 Solid State Physics Lecture 4 Last weeks: (Ch. 3) • Diffraction from crystals • Scattering factors and selection rules for diffraction Today: • Lattice vibrations: Thermal, acoustic, and optical properties This Week: • Start with crystal lattice vibrations. • Elastic constants. Elastic waves. • Simple model of lattice vibrations – linear atomic chain 2 • HW1 and HW2 discussion 1 Material to be included in the 1st QZ • Crystalline structures. Diamond structure. Packing ratio 7 crystal systems and 14 Bravais lattices • Crystallographic directions n and Miller indices dhkl 1 2 h2 k 2 l 2 2 2 2 a b c • Definition of reciprocal lattice vectors: • What is Brillouin zone • Bragg formula: 2d·sinθ = mλ ; k = G 3 •Factors affecting the diffraction amplitude: Atomic scattering factor (form factor): f n(r)eikrl d 3r reflects distribution of electronic cloud. a r0 sinΔk r In case of spherical distribution f 4r 2n(r) dr a 0 Δk r •Structure factor 2i(hu j kv j lw j ) F faje j where the summation is over all atoms in unit cell •Be able to obtain scattering wave vector or frequency from geometry and data for incident beam (x-rays, neutrons or light) 4 2 Material to be included in the 2nd QZ TBD Elastic stiffness and compliance. Strain and stress: definitions and relation between them in a linear regime (Hooke's law): ij Cijkl kl ij Sijkl kl kl kl 2 2 C u C u eff •Elastic wave equation: eff x sound velocity v t 2 x2 5 • Lattice vibrations: acoustic and optical branches In three-dimensional lattice with s atoms per unit cell there are 3s phonon branches: 3 acoustic, 3s - 3 optical • Phonon - the quantum of lattice vibration. Energy ħω; momentum ħq • Concept of the phonon density of states • Einstein and Debye models for lattice heat capacity. Debye temperature Low and high temperatures limits of Debye and Einstein models Formula for thermal conductivity 1 • K Cvl 3 • Be able to obtain scattering wave vector or frequency from geometry and data for incident beam (x-rays, neutrons or light) 6 3 7 Elastic properties Elastic properties are determined by forces acting on atoms when they are displaced from the equilibrium positions Taylor series expansion of the energy near the minimum (equilibrium position): U 1 2U U (R) U 0 (R R0 ) 2 (R R0 ) ... R R 2 R 0 R0 U For small displacements, neglect higher terms. At equilibrium, 0 R R0 2 So, ku 2 U U (R) U where k 0 2 R2 R0 8 u = R - R0 - displacement of an atom from equilibrium position 4 U force F acting on an atom: F ku R k - interatomic force constant. This is Hooke's law in simplest form. Valid only for small displacements. Characterizes a linear region in which the restoring force is linear with respect to the displacement of atoms. Elastic properties are described by considering a crystal as a homogeneous continuum medium rather than a periodic array of atoms In a general case the problem is formulated as follows: • Applied forces are described in terms of stress , • Displacements of atoms are described in terms of strain . • Elastic constants C relate stress and strain , so that = C. In a general case of a 3D crystal the stress and the strain are tensors 9 Stress has the meaning of local applied “pressure”. Applied force F(Fx, Fy, Fz) Stress components ij (i,j = 1, 2, 3) x 1, y 2, z 3 General case for stress: i.e ij ij Fj x j FdVij dV dS iijj VVx j S Shear forces must come in pairs: ij = ji (no angular acceleration) stress tensor is diagonal, generally has 6 components 10 5 Stress has the meaning of local applied “pressure”. Applied force F(Fx, Fy, Fz) Stress components ij (i,j = 1, 2, 3) x 1, y 2, z 3 General case for stress: i.e ij ij Fj x j Hydrostatic pressure – stress tensor is equivalent to a scalar: i.e xx =yy =zz p 0 0 ˆ [ ] 0 p 0 ij 0 0 p Stress tensor is a “field tensor” that can have any symmetry not related to the crystal symmetry. Stress tensor can change the crystal symmetry 11 Stress has the meaning of local applied “pressure”. Applied force F(Fx, Fy, Fz) Stress components ij (i,j = 1, 2, 3) x 1, y 2, z 3 Compression stress: i= j, i.e xx , yy , zz Fx xx Ax Shear stress: i ≠ j, i.e xy , yx , xz zx , yz , zy Fy yx Ax Shear forces must come in pairs: ij = ji (no angular acceleration) stress tensor is diagonal, generally has 6 components 12 6 Strain tensor ui 3x3 ij x j In 3D case, introduce the displacement vector as u = uxx + uyy + uzz Strain tensor components are defined as ux xx 0 0 xx ˆ x [ij ] 0 yy 0 ux 0 0 xy zz y VdV' Share deformations: xx yy zz Tr() ˆ dV xx yy zz Tr() ˆ 0 Can be diagonalized in x-y-z coordinates at a ˆ certain point of space In other points the tensor is not necessarily 13 diagonal Strain tensor components are defined as ui ij x j u x xx x u x xy y Since ij and ji always applied together, we can define shear strains symmetrically: 1 u u i j ij ji 2 x j xi So, the strain tensor is also diagonal and has 6 components14 7 Elastic stiffness (C) and compliance (S) constants relate the strain and the stress in a linear fashion: ij Cijkl kl kl This is a general form of the Hooke’s law. ij Sijkl kl 6 components ij, 6 ij 36 elastic constants kl Notations: Cmn where 1 = xx, 2 = yy, 3 = zz, 4 = yz, 5 = zx, 6 = xy For example, C11 Cxxxx , C12 Cxxyy , C44 Cyzyz Therefore, the general form of the Hooke’s law is given by 15 Elastic constants in cubic crystals Due to the symmetry (x, y, and z axes are equivalent) C11 = C22 = C33 ; C12 = C21 = C13 = C31 = C23 = C32 ; C44 = C55 = C66 Also, the off diagonal shear components are zero: C45 = C54 = C46 = C64 = C56 = C65 and mixed compression/shear coupling does not occur: C45 = C54 = C46 = C64 = C56 = C65 the cubic elastic stiffness tensor has the form: only 3 independent constants 16 8 Elastic constants in cubic crystals Longitudinal compression xx F A C11 (Young’s modulus): xx u L L Transverse expansion: xx C12 yy Shear modulus: xy F A C44 xy 17 Uniaxial pressure setup for optical characterization of correlated oxides Pressure control •Variables: Uniaxial pressure Temperature External magnetic field Measured sample properties: Far-IR Transmission / Reflection Optical Raman scattering cryostat sample Low T 18 9 Elastic waves Considering lattice vibrations three major approximations are made: • atomic displacements are small: u << a , where a is a lattice parameter • forces acting on atoms are assumed to be harmonic, i.e. proportional to the displacements: F = - Cu (same approximation used to describe a harmonic oscillator) • adiabatic approximation is valid – electrons follow atoms, so that the nature of bond is not affected by vibrations The discreteness of the lattice must be taken into account For long waves >> a, one may disregard the atomic nature – solid is treated as a continuous medium. Such vibrations are referred to as elastic (or acoustic) waves. 19 Elastic waves First, consider a longitudinal wave of compression/expansion mass density segment of width dx at the point x; elastic displacement u 2u 1 F xx where F/A = t 2 A x x xx Assuming that the wave propagates along the C [100] direction, can write the Hooke’s law in the form xx 11 xx 2 2 u u C11 u x Since x get wave equation: xx x t 2 x 2 20 10 A solution of the wave equation - longitudinal plane wave i(qxt) u(x,t) Ae where q - wave vector; frequency ω = vLq C v 11 - longitudinal sound velocity L Now consider a transverse wave which is controlled by shear stress and strain: In this case 2 u xy u where xy C44 xy and xy t 2 x x - transverse wave equation is 2 2 C u C44 ux 44 sound velocity vT t 2 x2 21 Two independent transverse modes: displacements along y and z For q in the [100] direction in cubic crystals, by symmetry the velocities of these modes are the same - modes are degenerate Normally C11 > C44 vL > vT We considered wave along [100]. C In other directions, the sound velocity depends v eff on combinations of elastic constants: Ceff - an effective elastic constant. For cubic crystals: Relation between ω and q - dispersion relation. For sound ω = vq 22 11 Model of lattice vibrations one-dimensional lattice: linear chain of atoms harmonic approximation: force acting on the nth atom is equation of motion (nearest neighbors interaction only): 2u M F C(u u ) C(u u ) C(2u u u ) t 2 n n1 n n1 n n n1 n1 M is the atomic mass, C – force constant Now look for a solution of the form u(x,t) Aei(qxn t) where xn is the equilibrium position of the n-th atom xn = na obtain 23 the dispersion relation is Note: we change q q + 2/a the atomic displacements and frequency ω do not change these solutions are physically identical can consider only i.e.