Physical Properties. Reading Assignment: 1. J. F. Nye, Physical Properties of Crystals -chapter 1

Transcription

1 Physical Properties Reading Assignment: 1. J. F. Nye, Physical Properties of Crystals -chapter 1

2 Contents 1 Physical Properties 2 Scalar, Vector 3 Second Rank Tensor 4 Transformation 5 Representation Quadric 6 Neumann s Principle

3 Physical Properties of Crystals - crystalline- translational symmetry, long range order - amorphous- no long range order ex) glass - physical properties amorphous- isotropic crystalline- anisotropic magnitude of physical properties depends on direction

4 Crystalline vs. Non-crystalline Crystalline materials... atoms pack in periodic, 3D arrays typical of: -metals -many ceramics -some polymers Non-crystalline materials... atoms have no periodic packing occurs for: -complex structures -rapid cooling amorphous" = non-crystalline crystalline SiO2 Si Oxygen noncrystalline SiO2

5 Thermal Conductivity ex) gypsum (CaSO 4 2H 2 O) Monoclinic heated metal tip wax gypsum *ellipsoidal rather than circular Simon & Schuster s Guide to Rocks and Minerals

6

7 Physical Properties - scalar (zero rank tensor)- non-directional physical quantities, a single number ex) density, temperature - vector (first rank tensor)- magnitude and direction an arrow of definite length and direction ex) mechanical force, electric field, temperature gradient three mutually perpendicular axes Ox 1,Ox 2,Ox 3 components E = [ E1, E2, E3]

8 J. F. Nye, Physical Properties of Crystals

9 - second rank tensor - mechanical analogy Physical Properties central ring-2 pairs of springs at right angle springs on opposite sides are identical but have a different spring constant to perpendicular pair force (cause vector) displacement (effect vector) if a force is applied in a general direction, the displacement will not be in the same direction as the applied force (depends on relative stiffness) A. Putnis, Introduction to Mineral Science

10 Second Rank Tensor - problem solving 1. find components of the force F in the direction of each of the two springs 2. work out the displacement which each force component would produce parallel to each spring 3. combine two orthogonal displacement to find the resultant displacement

11 Second Rank Tensor F sinθ y kf 2 sinθ F cosθ θ kf 1 cosθ φ resultant displacement F x 1. force F = [ Fcos θ, Fsin θ] 2. spring constant along x and y are k and k,respectively 3. displacement [ kfcos θ, kfsin θ] 1 2 resultant displacement tanφ = k k tanθ

12 Second Rank Tensor - consequences 1. in an anisotropic system, the effect vector is not, in general, parallel to the applied cause vector. 2. in two-dimensional example, there are two orthogonal directions along which the effect is parallel to the cause. 3. an anisotropic system can be analyzed in terms of components along these orthogonal principal directions, termed principal axes. along these principal axes, the values of the physical property are termed the principal values.

13 Second Rank Tensor -in 3-D, general direction- direction cosines, l,m,n -a force F is applied in a general direction resulting in a displacement D at some angle ϕ to F -component of in the direction of D F = Dcosϕ D Dcosϕ DF K = = F F K = K( k1, k2, k3) -component of F along principal axes Fx = lf, Fy = mf, Fz = nf - component of D along principal axes D = klf, D = k mf, D = k nf x 1 y 2 z 3 F A. Putnis, Introduction to Mineral Science

14 D = D l+ D m+ D n K F x y Second Rank Tensor = kfl + kfm + kfn kl 1 + k2m + k3n F = ( ) DF = = kl F + k m + k n z - variation of a property K with direction - representation surface direction cosine lmn,, x y z l =, m=, n= r r r x y z K = kl 1 + k2m + k3n = k1( ) + k2( ) + k3( ) r r r A. Putnis, Introduction to Mineral Science

15 2 let 1, 1/ kx + k y + kz = Second Rank Tensor rk= r= K if k, k, k are positive, k x + k y + k z = 1 (ellipsoid) normal form of the equation of an ellipsoid x y z + + = 1 ( abc,. : semiaxes) a b c representation surface semiaxes: 1 k 1 1 1,, k k 2 3 In any general direction, the radius is equal to the value of 1/ K in that direction. A. Putnis, Introduction to Mineral Science

16 Second Rank Tensor - electric field E current density i) if conductor is isotropic and obeys Ohm's law j = σ E E = [ E, E, E ] j = [ j, j, j ] j = σe, j = σe, j = σe ii) if conductor is anisotropic j = σ E + σ E + σ E j = σ E + σ E + σ E j = σ E + σ E + σ E j isotropic anisotropic J. F. Nye, Physical Properties of Crystals

17 Second Rank Tensor - physical meaning of σ ij if field is applied along x, E = [ E,0,0] 1 1 j = σ E j = σ E j = σ E conductivity - nine components speified in a square array σ11 σ12 σ13 σ21 σ22 σ 23 σ31 σ32 σ 33 second rank tensor, components, leading diagonal * the number of subscripts equals the rank of tensor J. F. Nye, Physical Properties of Crystals

18 in general p= [ p, p, p ] q = [ q, q, q ] p = T q + T q + T q p = T q + T q + T q p = T q + T q + T q Second Rank Tensor T T T T T T T T T J. F. Nye, Physical Properties of Crystals

19 Second Rank Tensor p = T q + T q + T q = T q j j= p = T q + T q + T q = T q p = T q ( i= 1, 2,3) j j i ij j j= 1 j= j j i ij j j= 1 3 p = T q + T q + T q = T q p = T q ( i= 1, 2,3) -Einstein summation convention: when a letter suffix occurs twice in the same term, summation with respect to that suffix is to be automatically understood. j dummy suffix, i free suffix p = T q = T q i ij j ik k j

20 Second Rank Tensor -in an equation written in this notation, the free suffixs must be the same in all the terms on both sides of the equation: while the dummy suffixs must occur as pairs in each term. ex) A + B C D = E F ij ik kl lj ik kj i, j free suffixs k, l dummy suffixs ( C B D = B C D ) kl ik lj ik kl lj -in this book, the range of values of all letter suffixs is 1,2,3 unless some other things is specified.

21 p = T q + T q + T q p = T q + T q + T q p = T q + T q + T q Transformation - q p ( T determine), arbitrary axes chosen j i ij - different set of axes different set of coefficients - both sets of coefficents equally well represent the same physical quantity - there must be some relation between them - when we change the axes of reference, it is only our method of representing the property that changes; the property itself remains the same. T ij

22 - transformation of axes a change from one set of mutually perpendicular axes to another set with same origin first set: x, x, x, second set: x', x', x' angular relationship old x' a a a new x' a a a x' a a a a : cosine of the angle between x ' and x ( a ):matrix ij Transformation x x x i j ij J. F. Nye, Physical Properties of Crystals

23 Direction Cosines, a ij - ( )-nine component- not independent a ij - only three independent quantities are needed to define the transformation. - six independent relation between nine coefficients a + a + a = a a a a ik jk ij + a a + = = δ a13a23 0 ( orthogonality relation) Kronecker delta δ = 1 ( i = j) ij 0 ( i j)

24 Transformation - transformation of vector components p p, p, p with respect to x, x, x p', p', p' with respect to x', x', x' p' = p cos x x' + p cos x x' + p cos x x' = a p + a p + a p p' = a p + a p + a p p' = a p + a p in dummy suffix notation new in terms of old: p' old in terms of new: a p p 3 a p i ij j = = a p ' i ji j J. F. Nye, Physical Properties of Crystals

25 - transformation of components of second rank tensor p = T q with respect to x, x, x i ij j p' = T' q' with respect to x', x', x' i ij j p' p q q' ( : in terms of) p' = a p p = T q q = a q' i ik k k kl l l jl p' = a p = a T q = a T a q' p' = T' q' T T ' i ik k ik kl l ik kl jl j i ij j a a T ij ik jl kl = = a a T ij ki lj kl ' Transformation j

26 Transformation T aat aat aat aat ' ij = ik jl kl = ik j 1 k 1+ ik j 2 k 2 + ik j 3 k 3 = aat + aa T + aa T i1 j1 11 i1 j2 12 i1 j a a T + a a T + a a T i2 j1 21 i2 j2 22 i2 j a a T + a a T + a a T i3 j1 31 i3 j2 32 i3 j3 33 J. F. Nye, Physical Properties of Crystals

27 Definition of a Tensor -a physical quantity which, with respect to a set of axes xi, has nine components Tij that transform according to equations ' ij = a ik a jl T kl -a second rank tensor- physical quantity existing in its own right, and quite independent of the particular choice of axes -when we change the axes, the physical quantity does not change, but only our method of representing it. ( ) a ij - : array of coefficient relating two set of axes T T = -symmetric ij ji anti-symmetric (skew-symmetric) T T ij = T ji

28 Representation Quadric - geometrical representation of a second rank tensor - consider the equation -if Sxx = 1 S:coefficients ij i j ij S x + S x x + S x x S x x + S x + S x x S x x + S x x + S x = S ij = S ji S x + S x + S x + 2S x x + 2S x x + S x x = general equation of a second-degree surface (quadric) referred to its center as origin

29 - transformed to new axes Ox ' x = a x' x = a x' i ki k j lj l Saax' x' = 1 ij ki lj k l Representation Quadric S' x' x' = 1 where S' = a a S kl k l kl ki lj ij - compared with second rank tensor transformation law T' ij = aikajltkl (identical) if S = S ij ji coefficient S of the quadric transform like the components ij of a symmetrical tensor of the second tank. i

30 Representation Quadric - a representation quadric can be used to describe any symmetrical second-rank tensor, and in particular, it can be used to describe any crystal property which is given by such a tensor. - principal axes principal axes- three directions at right angles such that = 1 takes the simpler form Sxx ij i j Sx + Sx + Sx =

31 Representation Quadric S11 S21 S31 S1 0 0 S = S S S 0 S 0 ij S31 S32 S S 3 S, S, S : principal components x y z + + = a b c representation quadric- semi axes,, S S S 1 2 3

32 Representation Quadric J. F. Nye, Physical Properties of Crystals

33 Representation Quadric - in a symmetric tensor refered to arbitrary axes, the number of independent components is six. - if the tensor is refered to its principal axes, the number of independent components is reduced to three. - the number of degree of freedom is nevertheless still six, for three independent quantities are needed to specify the directions of the axes, and three to fix the magnitudes of the principal components.

34 Representation Quadric - simplification of equations when referred to pricipal axes p = S q ( T replaced by symmetric S ) i ij j ij ij p = S q, p = S q, p = S q for example, consider electrical conductivity j1 = σ1e1, j2 = σ2e2, j3 = σ 3E3 ( σ1, σ2, σ3: principal conductivities) - if E is parallel to Ox1, so E2 = E3 = 0 j2 = j3 = 0 j is parallel to Ox1 - if E = [ E, E,0], 1 2 j1 = σ1e1, j2 = σ2e2, j3 = 0 E and j not parallel J. F. Nye, Physical Properties of Crystals

35 Effect of Crystal Symmetry on Crystal Properties - Neumann's Principle the symmetry elements of any physical properties of a crystal must include the symmetry elements of the point group of the crystal - physical properties may, and often do, possess more symmetry than the point group. - ex1) cubic crystals - optically isotropic physical property (isotropic) possesses the symmetry elements of all the cubic point groups.

36 Effect of Crystal Symmetry on Crystal Properties - ex2) trigonal system (tourmaline, 3 m) - optical properties (variation of refractive index with direction - indicatrix) indicatrix for 3m- ellipsoid of revolution about triad axis (optic axis) ellipsoid of revolution- vertical triad axis three vertical planes of symmetry (extra- center of symmetry, other symmetry elements) - the symmetry of a physical property a relation between certain measurable quantities associated with the crystal

37 Effect of Crystal Symmetry on Crystal Properties - all second-rank tensor properties are centrosymmetric. p = T q i ij j - p = T ( q ) T : unchanged i ij j ij - symmetric second-rank tensor- 6 independent components - symmetry of crystal reduces the number of independent components - consider representation quadric for symmetric second rank tensor

38 J. F. Nye, Physical Properties of Crystals

39 Fick's first law Ji = Dij x j o at 1150 C Anisotropic Diffusion of Ni in Olivine c ex) Ni diffusion in olivine((mg,fe) SiO, orthorhombic) 2 4 D =4.40x10 cm /s, D =3.35x10 cm /s, D =124x10 cm /s x y z a:b:c= : : = 0.48 : 0.55 : 0.09 D D D x y z A. Putnis, Introduction to Mineral Science

40 Magnitude of a Property in a Given Direction - defintion in general, if pi = Sq ij j, the magnitude Sof the property [ Sij] in a certain direction is obtained by applying q in that direction and measuring p / q, where p is the componet of p parallel to q - ex) electrical conductivity the conductivity σ in the direction of E is defined to be the component of j parallel to E divided by E, that is, j / E J. F. Nye, Physical Properties of Crystals

41 Magnitude of a Property in a Given Direction - analytical expression (i) referred to principal axes direction cosine: l1, l2, l3 E = [ le 1, le 2, le 3 ] j = [ σ1le 1, σ2le 2, σ3le 3 ] component of j parallel to E j = l 2 1 σ E+ l σ E+ l σ E magnitude of conductivity in the direction l i σ = l σ + l σ + l σ

42 Magnitude of a Property in a Given Direction - analytical expression (ii) referred to general axes l : direction cosine of E referred to general axes i Ei = Eli component of j parallel to E je i / E in suffix notation je / E conductivity in the direction je E σ= σ l l i i ij j σ = = 2 2 ij i j σ EE E i l i i i

43 Geometrical Properties of Representation Quadric - length of the radius vector let P be a general point on the ellipsoid: σ xx = 1 direction cosines of OP: l x = rl where OP = r σ 1 ( σ σ ) 2 r ijll i j = = ijll i j 2 1/ 1/ σ = r r = σ i i i ij i j special cases- radius vectors in the directions of semi-axes of lengths 1/ σ, 1/ σ,1/ σ J. F. Nye, Physical Properties of Crystals

44 Geometrical Properties of Representation Quadric - in general, any symmetric second-rank tensor property S ij 2 S = 1/ r r = 1/ S - the length r of any radius vector of representation quadric is equal to the reciprocal of square root of magnitude S of the property in that direction

45 Geometrical Properties of Representation Quadric - radius-normal property Ox i principal axes of σ ij E = [ le 1, le 2, le 3 ] j= [ σ1le 1, σ2le 2, σ3le 3 ] direction cosines of j are proportional to σ l, σ l, σ l if P is a point on σ1x1+ σ x such that OP is parallel to E P= ( rl, rl, rl ) where OP= r tangent plane at P rl σ x + rl σ x + rl σ x = σ x = J. F. Nye, Physical Properties of Crystals

46 Geometrical Properties of Representation Quadric - radius-normal property normal at P has direction cosines proportional to lσ, l σ, l σ hence normal at P is parallel to if pi = Sijqj, the direction of p for a given q may be found by first drawing, parallel to q a radius vector OP of the representation quadric, and then taking the normal to the quadric at P. j