Part II Materials Science and Metallurgy TENSOR PROPERTIES SYNOPSIS

Transcription

1 Part II Materials Science and Metallurgy TENSOR PROPERTIES Course C4 Dr P A Midgley 1 lectures + 1 examples class Introduction (1 1 / lectures) SYNOPSIS Reasons for using tensors. Tensor quantities and properties (field and matter tensors). Electrical conductivity as a simple example of a tensor property. General notation and Einstein summation convention. Transformation of axes. Tensor rank. Anisotropy and symmetry. Physical significance of tensor components. Stress and strain ( 1 / ) Introduction to elastic and plastic behaviour, and to failure modes of materials and structures. Definition of stress at a point. Tensor notation for stress. Examples for uniaxial tension and compression, hydrostatic tension and compression, pure shear. Stress tensor in cylindrical coordinates. Resolution of stresses: derivation of formula for normal and shear stresses on a plane. Example of stresses on plane in bar in uniaxial tension. Definition of strain at a point. Distinction between rigid body displacement and rotation, and shear. Symmetry of strain tensor. Properties of symmetrical second rank tensors: principal axes. Diagonalisation of general tensor to find principal stresses and strains. Diagonalisation in two dimensions: the Mohr circle construction. Examples, including representation of 3-D stress state. Hydrostatic and deviatoric components of the stress tensor. Dilatational and deviatoric components of the strain tensor. Elasticity () Isotropic medium: linear elasticity theory, Hooke's Law, principle of superposition. Poisson's ratio, shear modulus, bulk modulus, Lamé constants, interrelationships of elastic constants. General anisotropic medium: stiffness and compliance tensors. Simplification by symmetry: matrix notation. Strain energy density: symmetry of general stiffness and compliance matrices. Effects of crystal symmetry: specific example of cubic point group 3. Physical interpretation of three deformation modes in a cubic crystal. Anisotropy factor for several materials. Methods of experimental stress analysis (1 1 / ) Strain gauges: fundamentals underlying gauge factor. Practical details. Analysis of strain gauge results. Photoelasticity: birefringence, stress-optical co-efficients, isochromatic and isoclinic fringes, applications. Other methods of stress measurement: brittle coatings, X-ray diffraction, ultrasonics. Residual stresses: origin and measurement. Examples of stress and strain analysis (1) 1

2 Stresses and strains in thin films: origin, measurement, epitaxy. Stresses in thin-walled tubes. (cont.) Elastic stress distributions (1) General elasticity theory: stress equilibrium, strain compatibility, stress-strain relationship. St. Venant's principle. Examples of elastic stress distributions: beam bending, circular hole in plate, notch in plate, dislocations. Elastic waves ( 1 / ) Reasons for interest: dynamic fracture behaviour, ultrasonics. Wave equation for longitudinal wave in rod. Other waves: torsion in rod, dilatation and distortion in an infinite medium, Rayleigh waves. Comparison of wave velocities for steel, aluminium and rubber. Tensor properties other than elasticity () Piezoelectricity: tensor notation. Applications of piezoelectric effect, electrical transducers in a variety of devices. Ferroelectricity, polycrystalline piezoelectric transducers. Selection of piezoelectric materials. Optical properties: indicatrix, ray-velocity surfaces. Non-linear optics and introduction to higher order properties. Applications: introduction to optoelectronic devices. Brief mention of other tensor properties. Principal references: BOOK-LIST Nye: Physical Properties of Crystals, chapters 1,, 5, 6 & 8. (library code: Lj13b, ref) We use Nye's notation throughout. Fundamental basis for the description of material properties using tensors. Lovett: Tensor Properties of Crystals, all chapters. (NbA 73b) The poor man's Nye. The content of this book corresponds very closely to that of the course. Relatively cheap in paperback, purchase could be considered. Other references: Cottrell: Mechanical Properties of Matter, chapters 4, 5 & 6. (Lj18a, ref) Elasticity, stress distributions, elastic waves. Dieter: Mechanical Metallurgy, chapters 1 &. (Kz1a,c,e, ref) Analysis of mechanical failures, stress and strain, elasticity. Holister: Experimental Stress Analysis, chapters 1, & 4. (Ky) Strain gauges, photoelasticity. Kelly and Groves: Crystallography and Crystal Defects, chapters 4 & 5. (Ll30) Tensors, stress and strain, elasticity. Knott: Fundamentals of Fracture Mechanics, chapters 1 &. (Ke45a, ref) Modes of failure, stress concentrations. Le May: Principles of Physical Metallurgy, chapters 1, & 4. (Kz31b, ref) Stress and strain, criteria for failure, dislocations. Lovell, Avery and Vernon: Physical Properties of Materials, ch. 8 & 10. (AB59) Pyroelectricity, piezoelectricity, ferroelectricity, electro-optics, non-linear optics. Wilson and Hawkes: Optoelectronics: An Introduction, chapter 3. (LcD7) Electro-optic effect, non-linear optics. Wyatt and Dew-Hughes: Metals, Ceramics and Polymers, ch. 5 & 6. (Ab14b, ref)

3 Tensile testing, yield point phenomena, elasticity. 3

4 Simple example of a tensor property: ELECTRICAL CONDUCTIVITY Electrical conductivity relates two vectors: electric field E = (E 1, E, E 3 ) [units V m 1 ] current density J = (J 1, J, J 3 ) [units A m ] Throughout, components are referred to a right-handed, orthogonal set of axes x 1, x, x 3. Linearity: Throughout, we assume that J and E are linearly related. This permits us to apply the Principle of Superposition if E A gives J A, and E B gives J B, then (E A + E B ) will give (J A + J B ). Isotropic material: The resultant vector is always parallel to the applied vector, i.e., J = σe, where σ is a constant. General anisotropic material: The relationship between J and E can be expressed as a series of equations relating their components: J 1 = σ 11 E 1 + σ 1 E + σ 13 E 3 J = σ 1 E 1 + σ E + σ 3 E 3 J 3 = σ 31 E 1 + σ 3 E + σ 33 E 3 The conductivity is described by a tensor of the second rank, written as: σ 11 σ 1 σ 13 σ 1 σ σ 3 σ 31 σ 3 σ 33 The components of the tensor are σ 11, σ 1, etc. Using the dummy suffix notation and the Einstein summation convention (implying summation over a repeated suffix), the equations relating J and E can be written in a shortened form: J i = σ ij E j. 4

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6 TRANSFORMATION OF AXES Old axes: x 1, x, x 3 New axes: x 1 ', x ', x 3 ' Axis x ', for example, makes angles θ 1, θ, θ 3 with the x 1, x, x 3 axes. We define direction cosines, a 1 = cosθ 1, a = cosθ, a 3 = cosθ 3, etc., completely specifying the angular relationships between the axes. The nine direction cosines form a transformation matrix: Old Axes x 1 x x 3 x 1 ' a 11 a 1 a 13 New Axes x ' a 1 a a 3 x 3 ' a 31 a 3 a 33 The nine direction cosines are not independent: a ik a jk = 1 when i = j = 0 when i j. TRANSFORMATION LAWS FOR TENSORS These laws are the basis for defining tensors of a given rank. Transformation law rank of tensor new from old old from new 0 (scalar) φ' = φ φ = φ' 1 (vector) p i ' = a ij p j p i = a ji p j ' T ij ' = a ik a jl T kl T ij = a ki a lj T kl ' 3 T ijk ' = a il a jm a kn T lmn T ijk = a li a mj a nk T lmn ' 4 T ijkl ' = a im a jn a ko a lp T mnop T ijkl = a mi a nj a ok a pl T mnop ' 6

7 MATERIAL PROPERTIES REPRESENTED BY TENSORS Rank No. of Property Relating the quantities (with rank) components 0 1 density mass (0) volume (0) heat capacity temperature (0) energy (0) 1 3 pyroelectricity polarisation (1) temperature change (0) electrocaloric effect entropy change (0) electric field (1) 9 electrical conductivity current density (1) electric field (1) thermal conductivity heat flow (1) temperature gradient (1) permittivity dielectric displacement (1) electric field (1) permeability magnetic induction (1) magnetic field (1) thermal expansion strain () temperature change (0) 3 7 direct piezoelectric polarisation (1) stress () effect electro-optic change in effect dielectric impermeability () field (1) 4 81 elastic compliance strain () stress () piezo-optical change in effect dielectric impermeability () stress () electrostriction strain () electric field components ( 1) 7

8 These material properties are matter tensors and must obey Neumann's Principle: "The symmetry elements of a material property must include the symmetry elements of the point group of the material". 8

9 NOTATION FOR STRESSES Stress is a second rank tensor. Referred to right-handed Cartesian axes, its components may be visualised acting on a cubic volume element: The component σ ij, for example, represents the component of force in the x i direction which is transmitted across the face of the cube perpendicular to the x j direction. Components with i = j are normal stresses. Components with i j are shear stresses. The nine components in the second rank tensor: σ 11 σ 1 σ 13 σ 1 σ σ 3 σ 31 σ 3 σ 33 are not all independent since the cubic volume element must be in equilibrium, neither accelerating linearly nor rotating. This condition is achieved when σ ij = σ ji, i.e., when the tensor is symmetrical and has six independent components. RESOLUTION OF STRESSES We consider the action of a vector force P per unit area (or "stress") on a plane ABC whose orientation is specified by its normal vector n. The vector n makes angles θ 1, θ, θ 3 with the axes x 1, x, x 3, and we define the direction cosines: λ 1 = cosθ 1 λ = cosθ λ 3 = cosθ 3 9

10 Triangle ABC, for example, will be taken to have area ABC. The force per unit area P acting on face ABC can be resolved into its components P 1, P, P 3 and each of these should be balanced by the stress components acting on the other faces of the volume element OABC. In the x 1 direction we have: P 1.ABC = σ 11.OBC + σ 1.OAC + σ 13.OAB. This can be simplified by noting that: volume of element OABC = OD.ABC/3 = OA.OBC/3 = OB.OAC/3 = OC.OAB/3 Hence: OD = OA.cosθ 1 = OA.λ 1 = OB.λ = OC.λ 3 OBC = ABC.λ 1 OCA = ABC.λ OAB = ABC.λ 3 P 1 = σ 11 λ 1 + σ 1 λ + σ 13 λ 3 and similarly, P = σ 1 λ 1 + σ λ + σ 3 λ 3 P 3 = σ 31 λ 1 + σ 3 λ + σ 33 λ 3 These equations show that stress is a second rank tensor relating force and plane orientation according to: P i = σ ij λ j. 10

11 NORMAL AND SHEAR STRESSES The total stress P acting on plane ABC can be resolved into components along the x 1, x, x 3 axes: P = (P 1, P, P 3 ) The magnitude of P is given by: P = (P 1 + P + P 3 ) 1/ In many cases it is of interest to resolve P into normal and shear components. To obtain the normal stress, we resolve the components along the axes in the direction of the plane normal n. The normal stress P n is given by: P n = P 1 λ 1 + P λ + P 3 λ 3 = P i λ i P n = σ ij λ j λ i 11

12 The shear stress P s is obtained from P and P n : P = P n + P s P s = (P P n ) 1/ STRAIN As a result of strain in a material a point is moved from (x 1, x, x 3 ) to (x 1 ', x ', x 3 ') by amounts u i such that x i ' = x i + u i. For u i to indicate a strain (rather than a translation), u i must vary with position. We allow each component of u to depend linearly on each component of x: u 1 = e 11 x 1 + e 1 x + e 13 x 3 u = e 1 x 1 + e x + e 3 x 3 u 3 = e 31 x 1 + e 3 x + e 33 x 3 i.e., u i = e ij x j We have defined a second rank tensor with nine components e ij. The component e ij represents the movement of points on the x j axis in the direction of the x i axis. 1

13 It is not useful to use e ij directly as the tensor representing strain, because in this form the e ij include both shear and rotation. We can separate these contributions by expressing e ij as the sum of symmetrical (ε ij ) and antisymmetrical (ω ij ) components: e ij = ε ij + ω ij where ε ij = (e ij + e ji )/ and ω ij = (e ij e ji )/ Shape change (shear) is described by the symmetrical tensor ε ij. This is the strain tensor. Rotation is described by the antisymmetrical tensor ω ij. DIAGONALISATION OF STRESS TENSOR Stress is a symmetrical second rank tensor. As such there are three principal axes for stress, i.e., directions along which the applied and resultant vectors (for stress these are the force vector and the plane normal vector) are parallel. When the principal axes are taken as reference axes, only normal stresses appear in the tensor, i.e. σ ij 0 only when i = j. The process of referring a symmetrical second rank tensor to its principal axes is called diagonalisation. We need to find the orientation of a plane normal n (specified by direction cosines λ 1, λ, λ 3 ) for which the total force per unit area P is parallel to the normal. Let the magnitude of P be ξ. Then P 1 = ξλ 1, etc. But P i = σ ij λ j P 1 = σ 11 λ 1 + σ 1 λ + σ 13 λ 3 = ξλ 1 P = σ 1 λ 1 + σ λ + σ 3 λ 3 = ξλ P 3 = σ 31 λ 1 + σ 3 λ + σ 33 λ 3 = ξλ 3 The three simultaneous equations have non-trivial solutions only if: σ 11 ξ σ 1 σ 13 σ 1 σ ξ σ 3 σ 31 σ 3 σ 33 ξ = 0 13

14 If one principal axis is already known, the tensor to be diagonalised is: σ 11 σ 1 0 σ 1 σ σ 3 The diagonalisation required is in two dimensions, not three, and is comparatively straightforward. The determinant becomes: σ 11 ξ σ 1 0 σ 1 σ ξ σ 33 ξ = 0 Multilplying out to get the secular equation: (σ 3 ξ){(σ 11 ξ)(σ ξ) σ 1 } = 0 One solution is ξ = σ 3 for the principal axis already known. The other principal stresses are obtained by solving the quadratic: ξ - (σ 11 + σ )ξ + σ 11 σ σ 1 = 0 ξ = σ 11 + σ ± σ 11 + σ 4(σ 11 σ σ 1 ) = σ 11 + σ ± σ 11 σ + σ1 The angle θ between the principal axes and the original axes is given by: tan θ = σ 1 σ 11 σ MOHR'S CIRCLE Mohr's circle is a geometrical method for solving the above equations, which helps in the visualisation of diagonalisation. A stress state in two dimensions with σ 11, σ, and σ 1 is represented. The diagram is constructed as follows: On the horizontal line (representing normal stresses) mark points A and B such that OA = σ 11 and OB = σ. At A and B construct perpendiculars AC and BD of length σ 1. The line CD is a diameter of the circle. The circle cuts the horizontal line at P and Q. 14

15 OP and OQ are the maximum and minimum principal stresses: OE = σ 11 + σ OP = OE + EP = OE + EC OP = σ 1 = σ 11 + σ + σ 11 σ + σ1 OQ = σ = σ 11 + σ σ 11 σ + σ1 The diagram also shows the angular relationship between the original and the principal axes. The axes are rotated by θ, where: tan θ = σ 1 σ 11 σ HYDROSTATIC AND DEVIATORIC COMPONENTS OF STRESS σ 11 σ 1 σ 13 σ H 0 0 σ 1 σ σ 3 = 0 σ H 0 + σ 31 σ 3 σ σ H hydrostatic component volume change σ 11 -σ H σ 1 σ 13 σ 1 σ - σ H σ 3 σ 31 σ 3 σ 33 - σ H deviatoric component shape change 15

16 DILATATIONAL AND DEVIATORIC COMPONENTS OF STRAIN V V = ε 11 + ε + ε 33 = "dilatation" ε 11 ε 1 ε 13 ε 1 ε ε 3 = ε 31 ε 3 ε dilatational component volume change + ε ε 1 ε 13 ε 1 ε ε 3 ε 31 ε 3 ε deviatoric component shape change ELASTICITY IN ISOTROPIC MEDIA In isotropic media there are just two independent elastic moduli. These can be taken to be Young's modulus E and Poisson's ratio ν, in which case the principal stresses and strains are related by: ε 1 = σ 1 - ν(σ + σ 3 ) E ε = σ - ν(σ 1 + σ 3 ) E ε 3 = σ 3 - ν(σ + σ 1 ) E Other moduli are the bulk modulus K and the shear modulus µ. Since only two moduli are independent, there are relationships between them: K = E 3(1 - ν) µ = E (1 + ν) E = 9µK (µ + 3K) ν = (3K -µ) (6K + µ) Lamé's constants: Defining the quantities: λ = we can write for normal stresses: σ 11 = µε 11 + λ, etc νe (1 + ν)(1 - ν) and = (ε 11 + ε + ε 33 ) 16

17 and for shear stresses: σ 1 = µε 1, etc. ELASTICITY IN GENERAL ANISOTROPIC MEDIA Stress and strain can be related by either the stiffness tensor (C ijkl ) or the compliance tensor (S ijkl ), each of which is of fourth rank. For example a stress component is given in terms of the strain components by: σ 11 = C 1111 ε 11 + C 111 ε 1 + C 1113 ε 13 + C 111 ε 1 + C 11 ε + C 113 ε 3 + C 1131 ε 31 + C 113 ε 3 + C 1133 ε 33 Using the summation convention, this and the equations for the other eight stress components can be written: σ ij = C ijkl ε kl Or, using the compliance tensor: ε ij = S ijkl σ kl Matrix notation: This notation allows the various equations to be expressed in a shorter form. σ 11 σ 1 σ 13 σ 1 σ σ 3 σ 31 σ 3 σ 33 σ 1 σ 6 σ 5 σ 6 σ σ 4 σ 5 σ 4 σ 3 ε 11 ε 1 ε 13 ε 1 ε ε 3 ε 31 ε 3 ε 33 ε 1 1/ ε 6 1/ ε 5 1/ ε 6 ε 1/ ε 4 1/ ε 5 1/ ε 4 ε 3 σ ij = C ijkl ε kl σ i = C ij ε j where C ijkl C mn for all m,n. For compliances: ε ij = S ijkl σ kl ε i = S ij σ j but S ijkl = S mn when m, n = 1, or 3 S ijkl = S mn when either m or n = 4, 5 or 6 4S ijkl = S mn when m, n = 4, 5 or 6 17

18 EXAMPLES OF STRESS AND STRAIN ANALYSIS Thin films: These are usually in a state of plane stress with biaxial tension or compression: σ 0 0 ε σ 0 0 ε ε 3 The biaxial stress σ and biaxial strain ε in the plane of the film (x 1, x plane) are related by: σ = E (1 ν) ε ε = (1 ν) E σ The normal strain ε 3 in the thin film (determined, for example, by θ-θ X-ray diffractometry) is given by: ν ε 3 = (1 ν) ε Thin-walled tubes: The tube has its axis parallel to the z-axis, with radial (r) and tangential (θ) co-ordinates about this. For ideally thin walls, σ rr = 0, i.e., the walls are always in a state of plane stress. The tube has radius r and wall thickness t. A tensile load F on the tube gives a normal stress: σ zz = F πrt A torque T on the tube gives a shear stress: σ θz = T πr t An internal pressure P in the tube gives normal stresses: σ θθ = Pr t σ zz = Pr t (hoop) (longitudinal) 18

19 ELASTIC STRESS DISTRIBUTIONS St. Venant's principle: If the forces acting on a small part of the surface of a body are replaced by statically equivalent forces (i.e., same resultant and couple), the stress state is negligibly changed at large distance. Bending of beams: Consider a volume element in the beam, of thickness δy, and a distance y from the neutral surface. The longitudinal strain in this element is y/ρ, where ρ is the radius of curvature of the beam. The longitudinal stress in element is then Ey/ρ, where E is the Young's modulus of the material (assumed isotropic). Total tensile force exerted by the element: = force cross-sectional area = Ey ρ bδy The bending moment exerted by the element is the force y, so that the total bending moment over the beam cross-section M is given by: M = E ρ + h h by dy = E ρ bh 3 1 = EI ρ where I = bh 3 /1 is the second moment of area of the rectangular cross-section. For small deflections, displacements of the beam due to its curvature can be calculated using: d y dx = 1 ρ 19

20 Elastic stress concentrations: Hole in a plate: σ rr = ( 1 / )σ{1 r 0 /r + (1 + 3r 0 4/r 4 4r 0 /r )cosθ} σ rθ = ( 1 / )σ(1 3r 0 4/r 4 + r 0 /r )sinθ σ θθ = ( 1 / )σ{1 + r 0 /r (1 + 3r 0 4/r 4 )cosθ} Screw dislocation in an infinite medium: σ θz = σ zθ = µb/πr (all other components zero) Edge dislocation: µb sinθ σ rr = σ θθ = π(1 ν)r σ rθ = µb cosθ π(1 ν)r µbν sinθ σ zz = ν(σ rr + σ θθ ) = π(1 ν)r σ rz = σ θz = 0 ELASTIC WAVES In a rod: v 0 = E ρ longitudinal waves ("rod waves") µ v t = ρ shear waves (torsion) 0

21 In an infinite medium: λ + µ v 1 = ρ longitudinal waves (compression) µ v = ρ tranverse waves (shear, distortional) Elastic wave velocities (m s 1 ) steel aluminium rubber rod waves v bulk, longitudinal v bulk,shear v surface v s PIEZOELECTRICITY Direct piezoelectric effect: In this effect an electric polarisation P arises as a result of an applied stress σ. P i = d ijk σ jk For example: P 1 = d 111 σ 11 + d 11 σ 1 + d 113 σ 13 + d 11 σ 1 + d 1 σ + d 13 σ 3 + d 131 σ 31 + d 13 σ 3 + d 133 σ 33 The 7 d ijk are the piezoelectric moduli and form a third rank tensor. As d ijk = d ikj, there are up to 18 independent moduli. Converse piezoelectric effect: In this effect a strain arises as a result of an applied electric field. The moduli are the same as for the direct effect. ε jk = d ijk E i FERROELECTRICITY A ferroelectric material can show a spontaneous electric polarization, reversible by applying a sufficiently large electric field. The polarization is along a unique direction and is possible in 1

22 any of the ten polar classes (1,, m, mm, 3, 3m, 4, 4mm, 6, 6mm). These are a subset of the twenty piezoelectric classes. Ferroelectricity is used in poling polycrystalline piezoelectrics. PYROELECTRICITY If the spontaneous polarization in a ferroelectric material is temperature-dependent, the material exhibits pyroelectricity. Pyroelectric materials come from the same crystal classes (the ten polar classes) as ferroelectrics. The change in polarization P i is linked to the temperature change T by: P i = p i T. Primary pyroelectricity is observed if the volume of the sample is held constant. The additional change in polarization arising from the volume change associated with the temperature change is secondary pyroelectricity. Normal measurements (without holding volume constant) give the sum of the primary and secondary effects. ELECTROSTRICTION Electrostriction arises if the piezoelectric effect is not linear. Whereas centrosymmetric crystals cannot show piezoelectricity, they can show electrostriction. For example, in the converse piezoelectric effect there may be an additional term: ε jk = d ijk E i + γ iljk E i E l The term with d ijk represents (converse) piezoelectricity. At zero field the coefficients are equivalent to the d ijk defined above. The term is linear in applied field (if the field is reversed, the strain is reversed), and there can be no effect if the crystal is centrosymmetric. The term with γ iljk represents electrostriction. The effect is quadratic in applied field (if the field is reversed, the strain is the same) and can appear for centrosymmetric crystals. The quantity γ iljk E l is a third rank tensor and can be regarded as a correction term to the piezoelectric coefficients: ε jk = (d ijk + γ iljk E l ) E i.

23 REPRESENTATION SURFACE FOR SECOND-RANK TENSORS For second-rank tensors there exists a simple geometrical representation which is useful, for example, in deriving the magnitude of a material property in a given direction. Taking electrical conductivity as an example, the property referred to principal axes is: σ σ σ 3 The applied field E has components (λ 1 E, λ E, λ 3 E). The resulting current density J has components (σ 1 λ 1 E, σ λ E, σ 3 λ 3 E). The component of J resolved parallel to E, J(parallel) is: J(parallel) = (λ 1 ) σ 1 E + (λ ) σ E + (λ 3 ) σ 3 E, so that the conductivity σ parallel to E is: σ = (λ 1 ) σ 1 + (λ ) σ + (λ 3 ) σ 3 = σ ij λ i λ j. This conductivity can be represented by the surface σ ij x i x j = 1 which will be an ellipsoid if all the coefficients are positive. A radius in a given direction (λ 1, λ, λ 3 ) will have length r and intersect the surface at a point (x 1, x, x 3 ) given by: x 1 = r λ 1, x = r λ, x 3 = r λ 3. With these coordinates, r σ ij λ i λ j = 1 r σ = 1 r = (σ) 1/. In general for a second-rank tensor property, the radius length in any direction is the (property) 1/ in that direction. For an applied vector quantity parallel to the radius (e.g., field), the normal to the representation surface at the point where the radius intersects the 3

24 surface is parallel to the resultant vector (e.g., current density). (For discussion of this radiusnormal property, see Nye p. 8.) The representation surface (or quadric) for a second-rank tensor property, shown in this case for conductivity. OPTICAL PROPERTIES Isotropic material: For an isotropic material the electric displacement D is related to the electric field strength E by D = κ 0 K E, where κ 0 is the permittivity of vacuum and K the dielectric constant. The real refractive index n is (K) 1/. Anisotropic material: In this case the electric displacement and field strength are not necessarily parallel. Their vector components are related by: D i = κ 0 K ij E j or E i = (1/κ 0 ) B ij D j, 4

25 where K ij is the tensorial dielectric constant, and B ij is the relative dielectric impermeability. On principal axes, the real refractive index n = (K) 1/ = (B) 1/. The representation surface for B ij is therefore a plot of refractive index this is the optical indicatrix. 5