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Part II Materials Science and Metallurgy

TENSOR PROPERTIES Course C4 Dr P A Midgley 12 lectures + 1 examples class

SYNOPSIS Introduction (11/2 lectures) 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 (21/2) 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 co-ordinates. 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 (2) 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 23. Physical interpretation of three deformation modes in a cubic crystal. Anisotropy factor for several materials. Methods of experimental stress analysis (11/2) 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)

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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 (1/2) 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 (2) 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.

BOOK-LIST Principal references: Nye: Physical Properties of Crystals, chapters 1, 2, 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 & 2. (Kz1a,c,e, ref) Analysis of mechanical failures, stress and strain, elasticity. Holister: Experimental Stress Analysis, chapters 1, 2 & 4. (Ky2) 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 & 2. (Ke45a, ref) Modes of failure, stress concentrations. Le May: Principles of Physical Metallurgy, chapters 1, 2 & 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)

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Tensile testing, yield point phenomena, elasticity.

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Simple example of a tensor property:

ELECTRICAL CONDUCTIVITY

Electrical conductivity relates two vectors: electric field E = (E1, E2, E3) [units V m–1] current density J = (J1, J2, J3) [units A m–2] Throughout, components are referred to a right-handed, orthogonal set of axes x1, x2, x3. Linearity: Throughout, we assume that J and E are linearly related. This permits us to apply the Principle of Superposition – if EA gives JA, and EB gives JB, then (EA + EB) will give (JA + JB). 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: J1 = σ11E1 + σ12E2 + σ13E3 J2 = σ21E1 + σ22E2 + σ23E3 J3 = σ31E1 + σ32E2 + σ33E3 The conductivity is described by a tensor of the second rank, written as:

σ11 σ12 σ13σ21 σ22 σ23σ31 σ32 σ33

The components of the tensor are σ11, σ12, 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:

Ji = σijEj.

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TRANSFORMATION OF AXES

Old axes: x1, x2, x3 New axes: x1', x2', x3' Axis x2', for example, makes angles θ21, θ22, θ23 with the x1, x2, x3 axes. We define direction cosines, a21 = cosθ21, a22 = cosθ22, a23 = cosθ23, etc., completely specifying the angular relationships between the axes. The nine direction cosines form a transformation matrix:

Old Axes x1 x2 x3

x1' a11 a12 a13 New Axes x2' a21 a22 a23 x3' a31 a32 a33

The nine direction cosines are not independent: aikajk = 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) pi' = aijpj pi = ajipj' 2 Tij' = aikajlTkl Tij = akialjTkl' 3 Tijk' = ailajmaknTlmn Tijk = aliamjankTlmn' 4 Tijkl' = aimajnakoalpTmnop Tijkl = amianjaokaplTmnop'

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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) 2 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 (2) temperature change (0) 3 27 direct piezoelectric polarisation (1) stress (2) effect electro-optic change in effect dielectric impermeability (2) field (1) 4 81 elastic compliance strain (2) stress (2) piezo-optical change in effect dielectric impermeability (2) stress (2) electrostriction strain (2) 2 electric field components (2 × 1)

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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".

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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 σij, for example, represents the component of force in the xi direction which is transmitted across the face of the cube perpendicular to the xj direction. Components with i = j are normal stresses. Components with i ≠ j are shear stresses. The nine components in the second rank tensor:

σ11 σ12 σ13σ21 σ22 σ23σ31 σ32 σ33

are not all independent since the cubic volume element must be in equilibrium, neither accelerating linearly nor rotating. This condition is achieved when σij = σji, 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 θ1, θ2, θ3 with the axes x1, x2, x3, and we define the direction cosines:

λ1 = cosθ1 λ2 = cosθ2 λ3 = cosθ3

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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 P1, P2, P3 and each of these should be balanced by the stress components acting on the other faces of the volume element OABC. In the x1 direction we have:

P1.ABC = σ11.OBC + σ12.OAC + σ13.OAB.

This can be simplified by noting that:

volume of element OABC = OD.ABC/3 = OA.OBC/3 = OB.OAC/3 = OC.OAB/3

OD = OA.cosθ1 = OA.λ1 = OB.λ2 = OC.λ3 Hence:

OBC = ABC.λ1 OCA = ABC.λ2 OAB = ABC.λ3

P1 = σ11λ1 + σ12λ2 + σ13λ3

and similarly, P2 = σ21λ1 + σ22λ2 + σ23λ3 P3 = σ31λ1 + σ32λ2 + σ33λ3 These equations show that stress is a second rank tensor relating force and plane orientation according to: Pi = σijλj.

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NORMAL AND SHEAR STRESSES The total stress P acting on plane ABC can be resolved into components along the x1, x2, x3 axes:

P = (P1, P2, P3)

The magnitude of P is given by: P = (P12 + P22 + P32)1/2

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 Pn is given by:

Pn = P1λ1 + P2λ2 + P3λ3 = Piλi Pn = σijλjλi

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The shear stress Ps is obtained from P and Pn:

P2 = Pn2 + Ps2

Ps = (P2 – Pn2)1/2

STRAIN

As a result of strain in a material a point is moved from (x1, x2, x3) to (x1', x2', x3') by amounts ui such that xi' = xi + ui. For ui to indicate a strain (rather than a translation), ui must vary with position. We allow each component of u to depend linearly on each component of x:

u1 = e11x1 + e12x2 + e13x3 u2 = e21x1 + e22x2 + e23x3 u3 = e31x1 + e32x2 + e33x3

i.e., ui = eijxj

We have defined a second rank tensor with nine components eij. The component eij represents the movement of points on the xj axis in the direction of the xi axis.

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It is not useful to use eij directly as the tensor representing strain, because in this form the eij include both shear and rotation. We can separate these contributions by expressing eij as the sum of symmetrical (εij) and antisymmetrical (ωij) components:

eij = εij + ωij where εij = (eij + eji)/2 and ωij = (eij – eji)/2 Shape change (shear) is described by the symmetrical tensor εij. This is the strain tensor. Rotation is described by the antisymmetrical tensor ωij.

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. σij ≠ 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 λ1, λ2, λ3) for which the total force per unit area P is parallel to the normal. Let the magnitude of P be ξ. Then P1 = ξλ1, etc. But Pi = σijλj P1 = σ11λ1 + σ12λ2 + σ13λ3 = ξλ1 P2 = σ21λ1 + σ22λ2 + σ23λ3 = ξλ2 P3 = σ31λ1 + σ32λ2 + σ33λ3 = ξλ3

The three simultaneous equations have non-trivial solutions only if:

σ11 – ξ σ12 σ13

σ21 σ22 – ξ σ23

σ31 σ32 σ33 – ξ

= 0

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If one principal axis is already known, the tensor to be diagonalised is: σ11 σ12 0σ12 σ22 00 0 σ3

The diagonalisation required is in two dimensions, not three, and is comparatively straightforward. The determinant becomes:

σ11 – ξ σ12 0

σ12 σ22 – ξ 0

0 0 σ33 – ξ

= 0

Multilplying out to get the secular equation: (σ3 – ξ){(σ11 – ξ)(σ22 – ξ) – σ122} = 0

One solution is ξ = σ3 for the principal axis already known. The other principal stresses are obtained by solving the quadratic:

ξ2 - (σ11 + σ22)ξ + σ11σ22 – σ122 = 0

ξ = σ11 + σ22 ± σ11 + σ22

2 – 4(σ11σ22 – σ122)

2

= σ11 + σ222

± σ11 – σ222

2 + σ12

2

The angle θ between the principal axes and the original axes is given by:

tan 2θ = 2σ12σ11 –σ22

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 σ11, σ22, and σ12 is represented. The diagram is constructed as follows: On the horizontal line (representing normal stresses) mark points A and B such that OA = σ11 and OB = σ22. At A and B construct perpendiculars AC and BD of length σ12. The line CD is a diameter of the circle. The circle cuts the horizontal line at P and Q.

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OP and OQ are the maximum and minimum principal stresses:

OE = σ11 + σ222

OP = OE + EP = OE + EC

OP = σ1 = σ11 + σ222

+ σ11 – σ222

2 + σ12

2

OQ = σ2 = σ11 + σ222

– σ11 – σ222

2 + σ12

2

The diagram also shows the angular relationship between the original and the principal axes. The axes are rotated by θ, where:

tan 2θ = 2σ12σ11 –σ22

HYDROSTATIC AND DEVIATORIC COMPONENTS OF STRESS

σ11 σ12 σ13σ21 σ22 σ23σ31 σ32 σ33

= σH 0 00 σH 00 0 σH

+ σ11 -σH σ12 σ13

σ21 σ22 - σH σ23σ31 σ32 σ33 - σH

hydrostatic deviatoric component component ⇒ volume change ⇒ shape change

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DILATATIONAL AND DEVIATORIC COMPONENTS OF STRAIN

∆VV

= ε11 + ε22 + ε33 = ∆ "dilatation"

ε11 ε12 ε13ε21 ε22 ε23ε31 ε32 ε33

=

13

∆ 0 0

0 13

∆ 0

0 0 13

∆

+

ε11 - 13

∆ ε12 ε13

ε21 ε22 - 13

∆ ε23

ε31 ε32 ε33 - 13

∆

dilatational deviatoric component component ⇒ volume change ⇒ 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:

ε1 = σ1 - ν(σ2 + σ3)E

ε2 = σ2 - ν(σ1 + σ3)

E

ε3 = σ3 - ν(σ2 + σ1)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 - 2ν) µ = E

2(1 + ν)

E =

9µK(µ + 3K)

ν = (3K - 2µ)(6K + 2µ)

Lamé's constants: Defining the quantities:

λ = νE

(1 + ν)(1 - 2ν) and ∆ = (ε11 + ε22 + ε33) we can write for normal stresses:

σ11 = 2µε11 + λ∆, etc

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and for shear stresses: σ12 = 2µε12, etc.

ELASTICITY IN GENERAL ANISOTROPIC MEDIA

Stress and strain can be related by either the stiffness tensor (Cijkl) or the compliance tensor (Sijkl), each of which is of fourth rank. For example a stress component is given in terms of the strain components by: σ11 = C1111ε11 + C1112ε12 + C1113ε13 + C1121ε21 + C1122ε22 + C1123ε23 + C1131ε31 + C1132ε32 + C1133ε33 Using the summation convention, this and the equations for the other eight stress components can be written:

σij = Cijklεkl

Or, using the compliance tensor: εij = Sijklσkl

Matrix notation: This notation allows the various equations to be expressed in a shorter form.

σ11 σ12 σ13σ21 σ22 σ23σ31 σ32 σ33

→ σ1 σ6 σ5σ6 σ2 σ4σ5 σ4 σ3

ε11 ε12 ε13ε21 ε22 ε23ε31 ε32 ε33

→ ε1 1/2 ε6 1/2 ε5

1/2 ε6 ε2 1/2 ε41/2 ε5 1/2 ε4 ε3

σij = Cijklεkl → σi = Cijεj

where Cijkl → Cmn for all m,n. For compliances:

εij = Sijklσkl → εi = Sijσj but Sijkl = Smn when m, n = 1, 2 or 3 2Sijkl = Smn when either m or n = 4, 5 or 6 4Sijkl = Smn when m, n = 4, 5 or 6

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EXAMPLES OF STRESS AND STRAIN ANALYSIS

Thin films: These are usually in a state of plane stress with biaxial tension or compression:

σ 0 00 σ 00 0 0

→ ε 0 00 ε 00 0 ε3

The biaxial stress σ and biaxial strain ε in the plane of the film (x1, x2 plane) are related by:

σ = E

(1 – ν) ε

ε = (1 – ν)

E σ

The normal strain ε3 in the thin film (determined, for example, by θ-2θ X-ray diffractometry) is given by:

ε3 = –2ν(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, σrr = 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:

σzz = F

2πrt A torque T on the tube gives a shear stress:

σθz = T

2πr2t An internal pressure P in the tube gives normal stresses:

σθθ = Pr

t (hoop)

σzz = Pr

2t (longitudinal)

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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 = Eyρ

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 = Eρ

by2dy– h

2

+ h2

= Eρ

bh3

12 = EI

ρ

where I = bh3/12 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:

d2ydx2

= – 1ρ

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Elastic stress concentrations: Hole in a plate:

σrr = (1/2)σ{1 – r02/r2 + (1 + 3r04/r4 – 4r02/r2)cos2θ} σrθ = – (1/2)σ(1 – 3r04/r4 + 2r02/r2)sin2θ σθθ = (1/2)σ{1 + r02/r2 – (1 + 3r04/r4)cos2θ} Screw dislocation in an infinite medium: σθz = σzθ = µb/2πr (all other components zero) Edge dislocation:

σrr = σθθ = –

µb sinθ2π(1 – ν)r

σrθ =

µb cosθ2π(1 – ν)r

σzz = ν(σrr + σθθ) = –

µbν sinθπ(1 – ν)r

σrz = σθz = 0

ELASTIC WAVES

In a rod:

v0 = E

ρ vt =

µρ

longitudinal waves shear waves ("rod waves") (torsion)

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In an infinite medium:

v1 =

λ + 2µρ

v2 = µρ

longitudinal waves tranverse waves (compression) (shear, distortional)

Elastic wave velocities (m s–1) steel aluminium rubber rod waves v0 5190 5090 46 bulk, longitudinal v1 5940 6320 1040 bulk,shear v2 3220 3100 27 surface vs 2980 2920 26

PIEZOELECTRICITY Direct piezoelectric effect: In this effect an electric polarisation P arises as a result of an applied stress σ.

Pi = dijkσjk

For example: P1 = d111σ11 + d112σ12 + d113σ13 + d121σ21 + d122σ22 + d123σ23 + d131σ31 + d132σ32 + d133σ33 The 27 dijk are the piezoelectric moduli and form a third rank tensor. As dijk = dikj, 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.

εjk = dijkEi

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

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any of the ten polar classes (1, 2, m, mm2, 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 ∆Pi is linked to the temperature change ∆T by:

∆Pi = pi ∆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:

εjk = d˚ijk Ei + γiljk Ei El The term with d˚ijk represents (converse) piezoelectricity. At zero field the coefficients are equivalent to the dijk 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 γiljk 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 γiljkEl is a third rank tensor and can be regarded as a correction term to the piezoelectric coefficients:

εjk = (d˚ijk + γiljk El) Ei.

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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:

σ1 0 00 σ2 00 0 σ3

The applied field E has components (λ1E, λ2E, λ3E). The resulting current density J has components (σ1λ1E, σ2λ2E, σ3λ3E). The component of J resolved parallel to E , J(parallel) is:

J(parallel) = (λ1)2σ1E + (λ2)2σ2E + (λ3)2σ3E, so that the conductivity σ parallel to E is:

σ = (λ1)2σ1 + (λ2)2σ2 + (λ3)2σ3 = σij λi λj. This conductivity can be represented by the surface

σij xi xj = 1

which will be an ellipsoid if all the coefficients are positive. A radius in a given direction (λ1, λ2, λ3) will have length r and intersect the surface at a point (x1, x2, x3) given by:

x1 = r λ1, x2 = r λ2, x3 = r λ3. With these coordinates,

r2 σij λi λj = 1

r2 σ = 1

r = (σ)–1/2. In general for a second-rank tensor property, the radius length in any direction is the (property)–1/2 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

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surface is parallel to the resultant vector (e.g., current density). (For discussion of this radius-normal property, see Nye p. 28.)

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 = κ0 K E, where κ0 is the permittivity of vacuum and K the dielectric constant. The real refractive index n is (K)1/2. Anisotropic material: In this case the electric displacement and field strength are not necessarily parallel. Their vector components are related by:

Di = κ0 Kij Ej or Ei = (1/κ0) Bij Dj,

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where Kij is the tensorial dielectric constant, and Bij is the relative dielectric impermeability. On principal axes, the real refractive index n = (K)1/2 = (B)–1/2. The representation surface for Bij is therefore a plot of refractive index — this is the optical indicatrix.