Section 5_6 early version

This is the Title PhL 11.20.15 1.15.16. This version seems older than what is now installed, so file this into old versions folder. 5.6 The Tensor Product of two or more tensors in T(V) The tensor algebra T(V) shown in (5.4.1) is closed under both + and . It seems evident how one would add two tensors of T(V) of the form (5.4.2), but how would one multiply two tensors? During this set of steps, we try to gracefully transition into multiindex notation. Consider two tensors of rank k and k' expanded as in (5.2.1), T = Σii....i Tii....i (ei ei ..... ei) . rank k, T ϵ Vk (5.6.1) S = Σjj....j Sjj....j (ej ej ..... ej) rank k', S ϵ Vk' . (5.6.2) Multiplying these together with one gets, using the rules (5.3.1), TS = [Σii....iTii....i(ei ei ..... ei)][Σjj....jSjj....j(ej ej ..... ej)] (a) = Σii....i Σjj....jTii....i Sjj....j(ei ei ..... ei) (ej ej ..... ej) (b) = Σii....ijj....jTii....i Sjj....j(ei ei ..... ei ej ej ..... ej) (c) = Σii....iii....i Tii....i Sii....i (ei ei ...... ei) . (d) = ΣI,I'TI SI' (eI eI') I ≡ i1, i2...ik I' ≡ ik+1, ik+2, ....ik+k' eI ≡ (ei ei .... ei) eI' ≡ (ei...ei) (e) = ΣI (TS)I eI where eI ≡ (ei ei .... ei) I ≡ I, I' = i1,i2...ik+k' (5.6.3) Notice that the (2.8.22) associativity of is used going from (a) to (b). In step (c) we renamed the dummy jr summation indices so that j1 = ik+1 , j2 = ik+2 and so on. Step (d) rewrites step (c) in multiindex notation, then step (e) uses the outer product form (3.1.14) to replace TI SI' = (TS)I,I' = (TS)I . The object eI is a basis vector in Vk+k', so the object TS is therefore a tensor in Vk+k'. So we have just shown that if T ϵ Vk and S ϵ Vk', then TS ϵ Vk+k. Thus we have strengthened the claim made in (5.4.5) that T(V) is closed under the operation : the tensor product of an Vk tensor with an Vk' tensor lies in Vk+k' which is in T(V). It is easy then to show that this is true for the tensor product of any two multivectors as defined below (5.4.6). We shall now undertake the tensor product of three tensors T,S,R of ranks k,k',k" by mimicking the above step of steps, but leaning more heavily now on multiindex notation. TSR = [ΣITIeI][ΣJ SJeJ][ΣK SKeK] (a) = ΣI,J,K TISJRK (eI) (eJ) (eK) (b) = ΣI,J,K TISJRK (eI eJ eK ) // associative of used here (d) = ΣI,I',I" TISI'RI" eI eI' eI" // rename multiindices J→I',K→I" I ≡ i1, i2...ik I' ≡ ik+1, ik+2, ....ik+k' I" ≡ ik+k'+1, ik+k'+2, ....ik+k'+k" eI ≡ (ei.... ei) eI' ≡ (ei...ei) eI" ≡ (ei...ei) (e) = ΣI (TSR)I eI eI ≡ (ei.... ei) I ≡ I, I',I" = i1,i2...ik+k'+k" (5.6.4) Now the outer product form is TISI'RI" = (TSR)I,I',I" = (TSR)I . Since the eI are now basis vectors in Vk+k'+k", we conclude that TSR ϵ Vk+k'+k" . Our conclusions so far are: TS = ΣI (TS)I eI I ≡ I, I' = i1,i2...ik+k' TSR = ΣI (TSR)I eI I ≡ I, I',I" = i1,i2...ik+k'+k" (5.6.5) Repeating the development leads to this conclusion for the tensor product of N tensors Ti of rank ki: T1T2...TN = ΣI (T1IT2I .... TNI) eI = ΣI (T1T2....TN)I eI (5.6.6) Ti = tensor of rank ki Ii = multiindex range of ir values for tensor Ti I1 = i1, i2.....ik, I2 = ik+1, ik+2.....ik+k, etc. I = I1 , I2 .... , IN = i1, i2......ik+k+...k eI = ei ei ..... ei The rank of this product tensor is then K = Σi=1N ki and the tensor is an element of VK T(V). Example 1: The tensor product of two rank-1 tensors. TS = Σii[Ti Si] (ei ei) = Σij TiSj (eiej) = Σij (TS)ij (eiej) (TS)ab = Σij TiSj (eiej)ab = Σij TiSj δiaδjb = TaSb (5.6.7) Example 2: The tensor product of two rank-2 tensors. TS = Σiiii Tii Sii (ei ei ei ei ) (TS )abcd = TabScd (5.6.8) In both examples the evaluation of components produces the expected outer product forms. Special cases of the tensor product TS. Assume T and S have rank k and k'. If S = κ' ϵ K = a scalar, then rank(S) = k' = 0 and (5.6.4) reads, TS = Σii....ijj....jTii....i Sjj....j(ei ei ..... ei ej ej ..... ej) = Σii....iTii....i (κ') (ei ei ..... ei) = κ'T and ST = Σjj....jii....i Sjj....jTii....i (ej ej ..... ej ei ei ..... ei) = Σii....i (κ') Tii....i ( ei ei ..... ei) = κ'T so we find that TS = ST = κ'T . If T = κ and S = κ', the result above would be TS = κκ' and ST = κ'κ and so TS = ST = κκ'. Thus, TS = κS = ST = Sκ = κS if T = κ ϵ V0 TS = Tκ' = ST = κ'T = κ'T if S = κ' ϵ V0 TS = κκ' = ST = κ'κ = κκ' if T,S = κ,κ' ϵ V0 (5.6.9)