Mathematical Methods for Physics and Engineering - Matematica.NET

26.8 THE TENSORS δ ij AND ɛ ijk◮Write the following as contracted Cartesian tensors: a·b, ∇ 2 φ, ∇×v, ∇(∇·v), ∇×(∇×v),(a × b) · c.The corresponding (contracted) tensor expressions are readily seen to be as follows:a · b = a i b i = δ ij a i b j ,∇ 2 φ =∂2 φ ∂ 2 φ= δ ij ,∂x i ∂x i ∂x i ∂x j∂v k(∇×v) i = ɛ ijk ,∂x j[∇(∇ · v)] i =∂ ( ) ∂vj ∂ 2 v j= δ jk ,∂x i ∂x j ∂x i ∂x k( )∂ ∂v m∂ 2 v m[∇×(∇×v)] i = ɛ ijk ɛ klm = ɛ ijk ɛ klm ,∂x j ∂x l ∂x j ∂x l(a × b) · c = δ ij c i ɛ jkl a k b l = ɛ ikl c i a k b l . ◭An important relationship between the ɛ- andδ- tensors is expressed by theidentityɛ ijk ɛ klm = δ il δ jm − δ im δ jl . (26.30)To establish the validity of this identity between two fourth-order tensors (theLHS is a once-contracted sixth-order tensor) we consider the various possiblecases.The RHS of (26.30) has the values+1 if i = l and j = m ≠ i, (26.31)−1 if i = m and j = l ≠ i, (26.32)0 for any other set of subscript values i, j, l, m. (26.33)In each product on the LHS k has the same value in both factors and for anon-zero contribution none of i, l, j, m can have the same value as k. Since thereare only three values, 1, 2 and 3, that any of the subscripts may take, the onlynon-zero possibilities are i = l and j = m or vice versa but not all four subscriptsequal (since then each ɛ factor is zero, as it would be if i = j or l = m). Thisreproduces (26.33) for the LHS of (26.30) and also the conditions (26.31) and(26.32). The values in (26.31) and (26.32) are also reproduced in the LHS of(26.30) since(i) if i = l and j = m, ɛ ijk = ɛ lmk = ɛ klm and, whether ɛ ijk is +1 or −1, theproduct of the two factors is +1; and(ii) if i = m and j = l, ɛ ijk = ɛ mlk = −ɛ klm and thus the product ɛ ijk ɛ klm (nosummation) has the value −1.This concludes the establishment of identity (26.30).943