1. Tensor of Rank 2 If Φ ij (x, y) satisfies: (a) having four components (9 for 3-D). (b) when the coordinate system is changed from x i to x i,

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1 1. Tensor of Rank 2 If Φ ij (x, y satisfies: (a having four components (9 for 3-D. Φ i j (x 1, x 2 = β i iβ j jφ ij (x 1, x 2. Example 1: ( Φ i j = ( To prove whether this is a tensor or not, the tensorial transformation rule needs to be examined for every index. For the indices (1, 1,,. Φ 1 1 = 1 β 1 iβ 1 j β 1 1β 1 1Φ 11 + β 1 1β 1 2Φ 12 + β 1 2β 1 1Φ 21 + β 1 2β 1 2Φ 22 = β 1 1β β 1 2β 1 2 = cos 2 θ + sin 2 θ = 1, so Φ 1 1 = β 1 iβ 1 jφ ij. This also holds for the rest of the indices so Φ ij is a second rank tensor. Example 2: A constant matrix: ( , Φ i j = ( This is not a second rank tensor. For the indices (1, 1, Φ 1 1 = 1 β 1 iβ 1 j β 1 1β 1 1Φ 11 + β 1 1β 1 2Φ 12 + β 1 2β 1 1Φ 21 + β 1 2β 1 2Φ 22 = cos 2 θ + 2 cos θ sin θ + sin 2 θ, so Example 3: Φ 1 1 β 1 iβ 1 jφ ij. ( x 2 xy xy y 2. 1

2 You can prove that this is a second rank tensor Tensor of Rank 3 If Φ ijk (x, y satisfies: (a having 8 components (27 for 3-D. 3. Tensor of Rank 4 If Φ ijkl (x, y satisfies: (a having 16 components (81 for 3-D. Φ i j k (x 1, x 2 = β i iβ j jβ k kφ ijk (x 1, x 2. Φ i j k l (x 1, x 2 = β i iβ j jβ k kβ l lφ ijkl (x 1, x 2. Fourth rank tensors are important in continuum mechanics as elastic constants are an example of fourth rank tensors. Differentiation In index notation, partial derivatives are expressed using a comma (, as Examples x i, i. (1 x 1,2 = x y = 0. x i,i = x 1,1 + x 2,2 + x 3,3 = x x + y y + z z = 3. (2 3. x i,j = x i x j = δ ij. (3 4. i.e. higher order derivatives can be specified by additional indices. 2 f f,12 = x 1 x 2 = 2 f x y, (4 1 This is known as the moment. 2

3 5. v,i = v x i = v, (5 i.e. this is the gradient operator (. 6. v i,i = v 1,1 + v 2,2 + v 3,3 = v x x + v y y + v z z = v, (6 i.e. this is the divergence operator (. Differentiation:theorem If v ij... is a tensor, its derivative, v ij...,kl... is also a tensor. 2 (Proof For simplicity, assume v i is a first rank tensor. Hence its transformation rule is Differentiating the both sides of eq.(7 with respect to x j yields v i = β i jv j. (7 v i,j = β i jv j,j 3 = β i jv j,k x k,j 4 = β i jv j,k ( β 1 kj = β i jv j,k ( β T kj = β i jβ j kv j,k. (8 Eq.(8 implies that v i,j is a second rank tensor. The rank of the resulting tensor depends on the operation performed. For example, if v is a scalar, v,i, is a first rank tensor, v,ij is a second rank tensor but v,ii is a scalar etc... Permutation Symbols The permutation symbol, ϵ ijk, is defined as 1 (ijk = (123, (231, (312 ϵ ijk = 1 (ijk = (213, (321, (132 0 otherwise Note that ϵ ijk is zero if two indices are the same. Most of the usage of ϵ ijk is either (1 vector products or (2 the curl operator (rotation. Example 1: (vector product Example 2: (curl, rotation (a b i = ϵ ijk a j b k. ( v i = ϵ ijk v j,k. 2 This is true for Cartesian tensors but not true for general tensors as will be revealed later. (9 3

4 Invariance of tensor equations Theorem 1 If A ij and B ij are both tensors, their linear combination, αa ij + βb ij, is also a tensor. (Proof Let C ij αa ij + βb ij, it follows C i j = αa i j + βb i j Theorem 2 If v ij = 0 in one coordinate system, v i j (Proof = αβ i iβ j ja ij + ββ i iβ j jb ij = β i iβ j j (αa ij + βb ij = β i iβ j jc ij. (10 v i j = β i iβ j jv ij = β i iβ j j0 vanishes in any other coordinate systems. = 0. (11 Theorem 3 If a ij = b ij holds in one coordinate system, it holds in any other coordinate system 5. (Proof Let C ij a ij b ij. It follows 1. C ij is a tensor (from Theorem Since C ij = 0, C i j = a i j b i j = 0 (from Theorem Hence, a i j = b i j. Quotient Rule If AB = C holds and A and C are both tensors, B must be a tensor. Examples: 1. The force, f, is a tensor as it is defined by fu i = W (12 where u i is the displacement and W is work done. As both the displacement, u i, (first rank tensor and the work, W, (0-th rank tensor are tensors, f, must be a tensor from the quotient rule. Moreover, f must have an index, i, for the both sides to be consistent. Thus, the force must be a first rank tensor, f i. 2. Elastic constants The elastic constant, C, is defined as the proportionality factor between the stress, σ ij and the strain, ϵ ij, as σ = Cϵ. (13 As both the stress and the strain are second rank tensors, the above becomes σ ij = C? ϵ kl. 6 (14 From the quotient rule, C must be a tensor as both σ and ϵ are tensors. Furthermore, C must have indices of ijkl for the both sides of the above to be consistent. Thus, we have 5 Because of this theorem, tensor equations are independent of the frame of reference. 6 The direction of σ is not necessarily colinear with the direction of ϵ, hence, different indices (ij and kl. 4

5 σ ij = C ijkl ϵ kl. The component, C ijkl, can be interpreted as the ij component of the stress for the kl component of the applied strain. 3. Thermal conductivity The thermal conductivity, k, is defined as the proportionality factor between the heat flux, h j, and the temperature gradient, T,i, as h j = kt,i. As both h j and T,i are first rank tensors, k must be a second rank tensor as h i = K ij T,j, where the minus sign is introduced so that heat flows from higher temperature to lower temperature. 5