Isotropic Tensors

APPENDIX

Isotropic Tensors

C

Isotropic tensors have the same components in any frame of reference. In this section, we present some useful results for isotropic tensors. Proofs of these results are left as an exercise for the reader. Let v be a vector (or first-order tensor). Then v is isotropic, i.e., v = Qv

(C.1a)

for all proper orthogonal tensors Q, if and only if v = 0.

(C.1b)

Hence, the only isotropic first-order tensor is the zero vector. Let T be a second-order tensor. Then T is isotropic, i.e., T = QTQT

(C.2a)

for all proper orthogonal tensors Q, if and only if T = λI,

(C.2b)

where λ is a scalar and I is the identity tensor. Let S(3) be a third-order tensor with Cartesian components Sijk . Then S(3) is isotropic, i.e., Sijk = Qil Qjm Qkn Slmn

(C.3a)

for all proper orthogonal Qij , if and only if Sijk = λijk ,

(C.3b)

where λ is a scalar and ijk is the permutation symbol. Let C(4) be a fourth-order tensor with Cartesian components Cijkl. Then C(4) is isotropic, i.e., Cijkl = Qim Qjn Qkp Qlq Cmnpq

(C.4a)

for all proper orthogonal Qij , if and only if Cijkl = λδij δkl + μδik δjl + γ δil δjk ,

(C.4b)

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APPENDIX C Isotropic Tensors

where λ, μ, and γ are scalars and δij is the Kronecker delta. If we further restrict C(4) such that Cijkl = Cjikl, then μ = γ in (C.4b), which implies that Cijkl = λδij δkl + μ(δik δjl + δil δjk ).

Note that (C.5) implies the additional symmetry Cijkl = Cijlk = Cklij .

(C.5)