MSE 21A Thermodynamics and Phase Transformations Fall, 28 Problem Set No. 7 Problem 1: (a) Show that if the point group of a material contains 2 perpendicular 2-fold axes then a second-order tensor property of the material contains no more than 3 independent elements. Let the tensor be referred to axes e 1, e 2 and e 3, and let e 3 be a two-fold rotation axis. Rotation by π about e 3 replaces e 1 and e 2 by their negatives. If the tensor property is written in dyadic form, A = A ij e i e j 1.1 then it follows that any element, A ij that contains an odd number of indices 1 and 2 must vanish. Hence a two-fold rotational symmetry about e 3 has the consequence that A 13 = A 31 = A 32 = A 23 = 1.2 Now let there be a perpendicular two-fold axes, which we can take to be e 2 without loss of generality. Then we have, in addition to 1.2, A 12 = A 21 = 1.3 And only the diagonal elements remain. Since there is no symmetry constraint that makes these equal, the resulting tensor has three independent components. (b) Show that the same result obtains when the material has two perpendicular mirror planes. Suppose there is a mirror plane perpendicular to the coordinate axis e 3. Reflection through this plane takes e 3 into its negative without changing e 1 or e 2. It follows that all elements that contain the index 3 an odd number of times must vanish, generating conditions identical to those in eq. 1.1. If there is a second mirror plane perpendicular to e 2, then all elements that contain the index 2 an odd number of times must also vanish. This produces the additional conditions 1.3. As in part (a), the second-order tensor property must be diagonal. (c) Show that if the point group of a crystal has two perpendicular four-fold symmetry axes then its second-order tensor properties are spherical tensors. Since a four-fold axis is also a two-fold axis, the second order tensor property must be diagonal by the results of part (a). Now let e 3 be one of the four-old axes. A rotation by π/2 about this axis interchanges e 2 and e 1 with the consequence that
MSE 21A: Fall, 28 Problem Set No. 7 A 11 = A 22 1.4 Taking the perpendicular four-fold axis parallel to e 2 and applying the same argument we have It follows that A 11 = A 33 1.5 A ij = Aδ ij 1.6 where δ is the unit tensor. The tensor property A is spherical. Problem 2: (a) Identify the crystal symmetries that permit pyroelectricity in a solid. What characteristic is shared by these symmetries? Since pyroelectricity is the presence of an electric field in the relaxed state, and the electric field is a vector, pyroelectric materials must support a vector property. From the table on p. 415-16 of the notes, the symmetries that can do this are: Triclinic: C 1 Monoclinic: C 2, C 1h Orthorhombic: C 2V Tetragonal: C 4V Rhombohedral: C 3, C 3V Hexagonal: C 6, C 6V (b) Identify the crystal symmetries that permit piezoelectricity. characteristic do they share? What The symmetries that permit piezoelectricity are those that can be elastically strained into a symmetry that permits an electric field, i.e., can be strained to have a vector property. Since the property that governs the sequential application of strain and field is a third-order tensor (γ), these symmetries are those that permit third-order properties. All materials that allow 1 st order property also allow 3 rd order properties, so all pyroelectric materials are also piezoelectric. The additional symmetries that allow piezoelectricity, but not pyroelectricity, are: Orthorhombic: D 2 Tetragonal: S 4, D 4, D 2d Rhombohedral: D 3 Hexagonal: C 3h, D 6, D 3h Cubic: T, T d Page 2
MSE 21A: Fall, 28 Problem Set No. 7 (c) A ferroelectric solid undergoes a structural phase transformation (mutation) that permits it to become pyroelectric below the transition temperature. Assuming that the starting structure is cubic (perovskites like BaTiO 3 are good examples) describe how the transformation can change the symmetry to permit pyroelectrictity. The mutation that introduces pyroelectricity (the ferroelectric transition) must create a vector property. That is, it must alter the symmetry of the unit cell so that a distinguishable vector appears. The classic example is the ferroelectric transition in the perovskite structure, e.g., barium titanate. In this transition the central ion, Ti +4, is displaced slightly in a <1> direction. The result is an ion desplacement in a particular direction (i.e., a vector displacement) that identifies a vector in the lattice and creates a permanent dipole moment. After the transformation, the symmetry includes a vector. Problem 3: If a symmetric tensor is referred to a coordinate system whose unit vectors parallel its eigenvectors, only its diagonal elements are non-zero. (a) Since we can diagonalize any symmetric second-order tensor in a material of arbitrary (triclinic) symmetry, is it correct to say that a second-order tensor property of a triclinic material has 6 independent elements (as in the text), or should we say it only has 3? The eigenvectors, a i (i = 1,2,3) of the symmetric matrix, A, are solutions to the three equations A a i = λ i a i 3.1 This equation is automatically satisfied if A is written in the form A = λ i a i a i (I = 1,2,3) 3.2 Which is the dyadic form of a symmetric tensor that has only the diagonal elements λ I when it is referred to the orthogonal set of axes, a i. There are six independent elements, since the three principal axes (eigenvectors) are not known a priori. If A is referred to arbitrary axes it is not diagonal, and all six independent elements of A are required to find the axes that diagonalize it (via eq. 3.1). (c) According to the text, a second-order tensor in an orthorhombic crystal has three independent elements. However, if the crystal is written for an arbitrary set of axes, all of its nine elements may be non-zero. In what sense is the statement in the text correct? A symmetric tensor is only assured to be diagonal when it is referred to its principal axes (eigenvectors). However, when the symmetry is orthorhombic, the principal axes are known they are the orthogonal axes of the orthorhombic unit cell. Page 3
MSE 21A: Fall, 28 Problem Set No. 7 This information can be used to diagonalize the tensor and reduce the number of independent elements to 3. Problem 4: (a) Show that a material with the hcp crystal structure does not have a six-fold axis, but does have a three-fold rotation axis and a perpendicular mirror plane. The hcp unit cell is shown below Fig. 7.1: The unit cell of the hcp crystal structure. While the basal planes on top and bottom have six-fold symmetry, the central plane of atoms has only three-fold symmetry. However, the cell is clearly symmetry under reflection through a plane parallel to the base at half its height. Hence the vertical (zaxis) of the hcp cell is a three-fold rotation axis with a perpendicular mirror plane. (b) Find the most general form of the 1 st through 4 th order tensor properties for this case. [Hint: both the restrictions for the mirror plane and those for a 3-fold rotation axis derived in the text must apply. You can assume the results for these symmetry elements that are given in the text.] Vector property (α): The 3-fold rotation axis on e 3 has the consequence that any vector property must parallel e 3. The mirror plane perpendicular to e 3 has the consequence that there can be no vector property parallel to e 3. Hence α =. 2 nd order tensor property (β): Given the 3-fold rotation axis on e 3 the most general form for β is, in dyadic form, β = β 11 (e 1 e 1 +e 2 e 2 ) + β 33 e 3 e 3 4.1 Page 4
MSE 21A: Fall, 28 Problem Set No. 7 This tensor is compatible with the mirror plane since it is unchanged by reflection across the e 1,e 2 plane. 3 rd order tensor (γ): In the Voit notation, the results derived in the notes show that = 11 21 31 61 12 22 32 62 43 53 = 11-11 41 51-2 22-22 22 51-41 -2 11 where the first form is set by the mirror plane and the second by the 3-fold rotation axis. Since the terms in the two matrices must be equal, the most general form for the thirdorder property is = 11-11 -2 22 which has only two non-zero values. - 22 22-2 11 13 13 33 4.2 (2 elements) 4.3 4 th order tensor (λ): As derived in the notes, in Voit notation the mirror plane requires = 61 22 23 26 while the 3-fold rotation requires 23 33 36 45 45 55 16 26 36 66 (13 elements) Page 5
MSE 21A: Fall, 28 Problem Set No. 7 = 14-25 - 14 25 33 14-14 25-25 25 14 25 14 1 2 (- ) (7 elements) Setting these two matrices equal gives the result = 33 1 2 (- ) (5 elements) So there are five independent fourth-order tensor properties. For example, a cph material has 5 independent elastic moduli. Problem 5: (a) Consider a material with a cubic symmetry that is equivalent to that of a tetrahedron embedded in a cube. Show that this material cannot be pyroelectric, but can be piezoelectric. Fig. 5.1: Tetrahedral symmetry in a cube. Page 6
MSE 21A: Fall, 28 Problem Set No. 7 The symmetry of a tetrahedron in a cube is shown in Fig. 5.1. Equivalent directions point toward the corners of the tetrahedron (alternate corners of the cube) and are described by the vectors: e 1 +e 2 +e 3, -e 1 -e 2+ e 3, e 1 -e 2 -e 3, -e 1 +e 2 -e 3 5.1 where the origin or coordinates is in the centre of the cube. The symmetry includes three perpendicular 4-fold improper rotations, one along each of the coordinate axes. As described in section 15.5.5 of the notes, a four-fold improper rotation interchanges the directions in its perpendicular plane and changes the sign of two of the three directions. It follows that the three perpendicular 4-fold improper rotations include operations that interchange all three coordinate axes, and transform each into its negative, but in such a way that negative directions occur in pairs, as in eq. 5.1. Since there exist symmetry operations that replace each element by its negative this material cannot have a vector property, hence cannot be pyroelectric. A general element of the third order tensor, γ, is, in dyadic form, γ ijk e i e j e k. Since there exists an operation in the symmetry group that replaces e 1 by e 1, we must have γ 111 = -γ 111 = 5.2 γ 222 and γ 333 must vanish by the same argument. Since there exists an operation that replaces e 1 by e 1 without inverting e 3, we must have γ 133 = -γ 331 = 5.3 In fact, relations like 5.3 hold for all elements γ ijk for which two indices are the same. It follows that γ ijk = (i=j or j=k or i=k) 5.4 The only elements of the tensor that can be non-zero are those for which (ijk) is some permutation of (123). In this case the sign is unchanged by any of the symmetry operations, but the indices are interchanged, so all of these non-zero terms must be equal. It follows that the only non-zero elements of the 3 rd order tensor are γ 123 = γ 132 = γ 213 = γ 231 = γ 312 = γ 321 5.5 In Voit notation the matrix of 3 rd order properties is Page 7
MSE 21A: Fall, 28 Problem Set No. 7 γ = 41 41 41 (1 element) 5.6 It follows that material with this cubic/tetrahedral symmetry (the groups with Schoenflies symbols T and T d ) cannot be pyroelectric, but can be (and ordinarily are) piezoelectric. (b) Show that the only non-zero piezoelectric coefficient produces an electric field that is perpendicular to the plane of mechanical shear. Apologies for the poor wording of this question, which is ambiguous on the meaning of the term plane of mechanical shear. If we define the symmetry of the material (as we should) as the symmetry in the relaxed state, the piezoelectric coefficient is the cross-derivative between strain and electric displacement: E k = F V D k Tε = 2 F v ε ij D ε ij = γ ijk ε ij 5.7 k As discussed above, the only non-zero elements of the tensor, γ, are those in which the indices ijk are all different. All examples are symmetrically identical to that in which ijk = 123. In that case, E 3 = γ 123 ε 12 5.8 The strain ε 12 is the shear caused by the differential displacement in the e 1 direction over the face perpendicular to e 2. The piezoelectric field, E = E 3 e 3, is directed normal to the plane defined by e 1 and e 2, that is (hence the ambiguity) it lies in the plane perpendicular to e 2, in the direction perpendicular to e 1. Page 8