Lecture 7 - Elasticity Elasticity Elastic Behavior is the fundamental Stress and Strain in Crystals distinction between solids and liquids •Similartity: both are “condensed matter” Kittel – Ch 3 •A solid or liquid in equilibrium has a definite density (mass per unit volume measured at a given temperature) •The energy increases if the density (volume) is changed from the equilibrium value - e.g. by applying pressure Change of Pressure applied volume to all sides Physics 460 F 2006 Lect 7 1 Physics 460 F 2006 Lect 7 2 Elastic Behavior is the fundamental Strain and Stress distinction between solids and liquids Strain is a change of relative positions of the parts of the material •Difference: •A solid maintains its shape Stress is a force /area applied to the material to cause the strain •The energy increases if the shape is changed – “shear” •A liquid has no preferred shape •It has no resistance to forces that do not change the volume Volume dilation Two types of shear Two types of shear Physics 460 F 2006 Lect 7 3 Physics 460 F 2006 Lect 7 4 Pressure and Bulk Modulus Total Energy of Crystal • Consider first changes in the volume – applies to liquids and any crystal The general shape applies for any type of binding • General approach: E(V) where V is volume Can use ether Ecrystal(Vcrystal) or Ecell(Vcell) since Ecrystal= N Ecell and Vcrystal = N Vcell P = -dE/dV • Pressure = P = - dE/dV (units of Force/Area) = 0 at the minimum Energies of Crystal of Energies • Bulk modulus B = - V dP/dV = V d2E/dV2 (same B = - V dP/dV = V d2E/dV2 units as pressure ) proportional to curvature at the minimum • Compressibility K = 1/B Volume Physics 460 F 2006 Lect 7 5 Physics 460 F 2006 Lect 7 6 1 Lecture 7 - Elasticity Elasticity Elastic Equations • Up to now in the course we considered only • The elastic equations describe the relation of perfect crystals with no external forces stress and strain • Elasticity describes: • Linear relations for small stress/strain • Change in the volume and shape of the crystal when Stress = (elastic constants) x Strain external stresses (force / area) are applied • Sound waves • Large elastic constants fi the material is stiff - a given strain requires a large applied stress • Some aspects of the elastic properties are determined by the symmetry of the crystal • We will give the general relations - but we will consider only cubic crystals • Quantitative values are determined by strength • The same relations apply for isotropic materials like a glass and type of binding of the crystal? • More discussion of general case in Kittel Physics 460 F 2006 Lect 7 7 Physics 460 F 2006 Lect 7 8 Elastic relations in general crystals Elastic Properties of Crystals • Strain and stress are tensors • Definition of strain • Stress eij is force per unit area on a surface Normal n Six independent variables: • Force is a vector F , F , F x y z Force F Using the relation e1 ≡ exx , e2 ≡ eyy , e3 ≡ ezz , • A surface is defined by the exy = eyx etc. e4 ≡ eyz , e5 ≡ exz , e6 ≡ exy normal vector nx, ny, nz • 3 x 3 = 9 quantities • Stress Here Xy denotes force in x direction applied σ1 ≡ σxx = Xx , σ2 ≡ Yy , σ3 ≡ Zz • Strain σij is displacement per unit distance in a to surface normal to y. σ4 ≡ Yz , σ5 ≡ Xz , σ6 ≡ Xy particular direction Displacement u σxy = σyx etc. • Displacement is a vector u , u , u x y z • Linear relation of stress and strain • A position is a vector R , R , R Position R x y z Elastic Constants C • 3 x 3 = 9 quantities ij σi = Σj Cij ej , (i,j = 1,6) -1 ( Also compliances Sij = (C ) ij) Physics 460 F 2006 Lect 7 9 Physics 460 F 2006 Lect 7 10 Strain energy Symmetry Requirements Cubic Crystals • For linear elastic behavior, the energy is quadratic in the strain (or stress) • Simplification in cubic crystals due to symmetry Like Hooke’s law for a spring since x, y, and z are equivalent in cubic crystals • Therefore, the energy is given by: • For cubic crystals all the possible linear elastic information is in 3 quantities: C11 = C11 = C22 = C33 E = (1/2) Σi ei σi = (1/2) Σij ei Cij ej , (i,j = 1,6) C12 = C13 = C23 C = C = C • Valid for all crystals 44 55 66 • Note 21 independent values in general • Note that by symmetry C14 = 0, etc (since Cij = Cji ) • Why is this true for cubic crystals? Physics 460 F 2006 Lect 7 11 Physics 460 F 2006 Lect 7 12 2 Lecture 7 - Elasticity Elasticity in Cubic Crystals Elasticity in Cubic Crystals • Elastic Constants Cij are completely specified by • Elastic Constants Cij are completely specified by 3 values C11 , C12 , C44 3 values C11 , C12 , C44 σ1 = C11 e1 + C12 (e2 + e3) , etc. σ1 = C11 e1 + C12 (e2 + e3) , etc. σ4 = C44 e4 , etc. σ4 = C44 e4 , etc. Pure change in volume – compress equally in x, y, z Two types of shear – no change in volume •For pure dilation δ = ∆V / V e1 = e2 = e3 = δ / 3 C11 -C12 C44 •Define ∆E / V = 1/2 B δ2 No change in volume •Bulk modulus B = (1/3) (C + 2 C ) 11 12 if e2 = e3 = -½ e1 Physics 460 F 2006 Lect 7 13 Physics 460 F 2006 Lect 7 14 Elasticity in Cubic Crystals Elastic Waves • The general form of a displacement pattern is • Pure uniaxial stress and strain ∆r (r ) = u(r ) x + v(r ) y + w(r ) z • σ1 = C11 e1 with e2 = e3 = 0 • A traveling wave is described by 2 ∆r (r ,t) = ∆r exp(ik . r -iωt) • ∆E = (1/2) C11 (δx/x) • Occurs for waves where there is • For simplicity consider waves along the x direction in a no motion in the y or z directions cubic crystal Also for a crystal under Longitudinal waves (motion in x direction) are given by σ1 ≡ Xx stress u(x) = u exp(ikx -iωt) if there are also stresses Transverse waves (motion in y direction) are given by σ2 ≡ Yy , σ3 ≡ Zz of just the right magnitude so that e2 = e3 = 0 v(x) = v exp(ikx -iωt) Physics 460 F 2006 Lect 7 15 Physics 460 F 2006 Lect 7 16 Waves in Cubic Crystals Elastic Waves • Propagation follows from Newton’s Eq. on each volume element ∆V= ∆x ∆y ∆z • Longitudinal waves: ∆y ρ∆V d2 u / dt2 = ∆xdX/ dx = ∆xC d2 u / dx2 x 11 ∆z ∆x y x (note that strain is e1 = d u / dx) •Since ∆V / ∆x = area and ρ area = mass/length = ρL, this leads to z 2 • Variations in x direction ρL u / dt = C11 du/ dx or • Newton’s Eq: ma = F Net force in x direction 2 2 ω = (C11 / ρL ) k • Longitudinal: displacement u along x, 2 2 2 2 • Transverse waves (motion in the y direction) are given ρ∆V d u / dt = ∆xdXx/ dx = ∆xC11 d u / dx by • Transverse: displacement v along y, ω2 = (C / ρ ) k2 2 2 2 2 44 L ρ∆V d v / dt = ∆xdYx/ dx = ∆xC44 d v / dx Physics 460 F 2006 Lect 7 17 Net force in y direction Physics 460 F 2006 Lect 7 18 3 Lecture 7 - Elasticity Sound velocities Young’s Modulus & Poisson Ratio • The relations before give (valid for any elastic wave): • Consider crystal under tension (or compression) in x direction 2 2 ω = (C / ρL ) k or ω = s k • If there are no stresses σ2 ≡ Yy , σ3 ≡ Zz then the crystal will also strain in the y and z directions • where s = sound velocity • Poisson ratio defined by (dy/y) / (dx/x) • Different for longitudinal and transverse waves • Young’s modulus defined by • Longitudinal sound waves can happen in a liquid, gas, Y = tension/ (dx/x) or solid Homework problem • Transverse sound waves exist only in solids to work this out for a cubic crystal y x • More in next chapter on waves Physics 460 F 2006 Lect 7 19 Physics 460 F 2006 Lect 7 20 When does a crystal break? Next Time • Vibrations of atoms in crystals • Consider crystal under tension (or compression) in x direction • Normal modes of harmonic crystal • For large strains, when does it break? • Role of Brillouin Zone • Crystal planes break apart – or slip relative to one another • Quantization and Phonons • Governed by “dislocations” • Read Kittel Ch 4 • See Kittel – Chapter 20 Physics 460 F 2006 Lect 7 21 Physics 460 F 2006 Lect 7 22 4.