1. Defining sign of stress tensor Kittel’s Fig. 15 may be confusing about sign of T xx (which he calls X x ) Stress tensor component T xx is defined as the x-component of force, transmitted in –x direction. The compressive force shown (labeled X x ) is in the –x direction (note that the x axis points left!) and is transmitted in the –x direction, so the sign of T xx is negative. Tension would be positive.

2. Elasticity (Chapter 3) – only 6 of 9 elements are independent. Replace 3x3 matrix by a 6-component column vector: Lecture 21 Oct. 8, 2010 PH 481/581 Recall Young’s modulus Y: Stress = elasticity constant x strain In general, stress and strain are 3x3 matrices, so the most general relation would be Where M has 4 indices (3x3x3x3 =81 elements) Fortunately, not all elements of T and  are independent: these matrices are symmetric: ←L+  L → ← L → F F

3. Elasticity (continued) T is symmetric matrix – only 6 of 9 elements are independent. Replace 3x3 matrix By 6-element column vector: How about strain tensor? Again replace the 3x3 symmetric matrix By a 6-element column vector: Note factor of 2, only in off- diagonal elements. (Later, makes energy formula simpler.) We will use  and to represent the composite indices 1..6, and  and  to represent x,y,z.

4. Elasticity (continued) Now we can write the most general linear relationship between stress and strain as a matrix equation e = S T: Recalling that the 1..6 indices are short for (xx),...(xy), this is

5. Elasticity (continued) In the case of cubic symmetry, the axes are equivalent, so that S (xx)(xx) = S (yy)(yy), i.e. S 11 = S 22. Similarly, S (xy)(xy) = S (yz)(yz), so S 66 = S 44. Also, S 13 = S 12 -- all the off-diagonal elements in the upper left 3x3 matrix are equal. Also, anything like S (xx)(xy) that has a single y index must vanish due to the y↔ -y symmetry. This leads to We call S (in e = S T) the compliance matrix. We denote the inverse matrix (“stiffness matrix”) by C = S -1, so that T = C e. In a system with no symmetry, all 36 components of C and S are independent. Cubic symmetry only The stiffness matrix C has the same form -- a cubic material has 3 independent stiffness coefficients C 11, C 12, and C 44.

6. Elasticity (continued) Important special case: uniaxial stress, T xx >0, other T  = 0. Then This experiment defines Young’s modulus Y = (F/A)/(  L/L) = T xx /  xx = 1/S 11, as well as Poisson’s ratio P = -  yy /  xx = -S 12 /S 11. Note that  yy < 0 when stress is applied along x only, so S 12 < 0. if cubic symmetry ←L+  L → ← L → F F Elastic Energy: energy/volume U is (C  can be chosen to be symmetric)

7. Elastic Waves (Chapter 3) Lecture 23 Oct. 13, 2010 PH 481/581 Write equation of motion for a volume element from position r, now at displaced position r + R(t): becomes F F But displacement R is related to the strain tensor, which is related to the stress tensor T. Expressing everything in terms of R, Origin r R So F=ma becomes Like a wave equation.

8. Elastic Waves (continued) Derived a wave equation for displacement R: Solutions: try a plane wave Writing part of the  =x component explicitly, This is Eq. (57a) in Kittel. Try k = (K, 0, 0) & R along x: Soand the wave speed is so

9. Elastic Waves (Chapter 3) Lecture 24 Oct. 13, 2010 PH 481/581 Review: becomes in terms of displacement R – wave equation. Insert a plane wave Try k along x, i.e. (100) & R along 100:  v 2 = C 11 & k along (100) & R along 010:  v 2 = C 44 & k along (110) & R along 110:  v 2 = ½(C 11 + C 12 + 2C 44 ) & k along (110) & R along 001:  v 2 = C 44 & k along (110) & R along (1-10):  v 2 = ½(C 11 - C 12 ) Condition for isotropy: C 11 - C 12 = 2C 44 Kittel gives table – W is close to isotropic Lamé constants: T = 2  + tr  1; derive C 11 =2  C 12 =  C 44 = 