Vector Algebra Using Tensorial Notation

Vector algebra using tensorial notation ⎪⎧1 if i = j Definition of the Kronecker Delta ⇒ δij = ⎨⎪ ⎩0 if i ≠ j and the Levi-Civita tensor ⇒ εijk εijkεlmn δil δim δin = δ jl δ jm δ jn δkl δkm δkn ⎧0 for i = j or i = k or j = k ⎪ = ⎨-1 for an odd index permutation ⎪1 for an even index permutation ⎩ in particular, note that if i=l then εijkεimn = δ jmδkn − δ jnδ km RULE: Whenever an index is repeated, a summation is assumed over that index 3 AiBi = ∑ AiBi in 3D i=1   Dot product: A ⋅ B = AiBi   Cross product: A ×B = ε ijk A j Bk ( ) Gradient: ∇ i = i ∂ ∂x i The k index was permuted to the left 2 times, this gives a (+1) That’s all we need to know, here are a few examples:          1) Prove that A × B × C = B(A ⋅ C) − C(A ⋅ B)      A ×B × C = εijk A j B × C = ε ijkA jεklm BlC m = ε ijkεklm A jBl Cm = εkijε klm A j BlC m k i     = δ ilδ jm − δ imδ jl A j Bl C m = A j Bi C j − A j B j Ci ≡ Bi A ⋅ C − Ci A ⋅ B ( ) ( ( ) ) ( ) ( ) The i index was permuted once with j, thus giving a (–1)  2) Prove that ∇ ⋅ ∇ × A = 0   ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∇⋅∇×A= ∇×A = εijk Ak = ε ijk Ak = −ε jik A i ∂x i ∂xi ∂x j ∂ x i ∂x j ∂x i ∂x j k ∂ ∂ ∂ ∂ = −ε jik Ak = −ε ijk A =0 ∂x j ∂x i ∂xi ∂x j k ( The derivation order was inverted, no harm done since xj and xi are independent variables. ) No index permutation, we just decided to change notation such that all i’s transform into j’s and vice versa. These two expressions are identical except for their signs. This contradiction can only be avoided if the expression is equal to zero.       3) Prove that ∇ ⋅ A × B = ∇ × A ⋅ B− ∇ × B ⋅ A ⎛ ⎞   ∂   ∂ ∂ ∂ ∂ ∇ ⋅ A× B = εijk A jBk = εijk A jBk = εijk ⎜⎜ A j Bk + Bk A j ⎟⎟ A× B = i ∂x i ∂x i ∂x i ∂x i ⎠ ⎝ ∂x i ∂ ∂ ∂ ∂ = εijk A j Bk + ε ijkBk A j = ε ikj Ak B j + ε kijBk A ∂x i ∂x i ∂x i ∂x i j ( ( ) ) ( ) ( ) ( ) Again, no index permutation, we changed all j’s for k’s and vice versa. = −εkij Ak Two permutations of k to the left yield a (+1) ∂ ∂ ∂ ∂ B j + εkij Bk A j = −A kεkij B j + Bkεkij A ∂x i ∂x i ∂x i ∂x i j A single permutation of k to the left gives a (-1)     = −A ⋅ ∇ × B + B⋅ ∇ × A ( ) ( ) ⎛ B2 ⎞     4) Prove that B × ∇ × B = ∇ ⎜⎜ ⎟⎟ − B⋅ ∇ B ⎝2⎠   ∂ ∂ ∂ Bm = εkijε klm B j Bm = (δilδ jm − δimδ jl )B j B B× ∇ × B = ε ijkB jεklm i ∂x l ∂x l ∂x l m 1 ∂ ∂ ∂ ∂ ∂ ⎛⎜ B2 ⎞⎟  =Bj B j − Bj Bi = BjBj − B j Bi = − B⋅ ∇ Bi 2 ∂x i ∂x i ∂x j ∂x j ∂xi ⎝⎜ 2 ⎠⎟ ( [ ( ) ( ) )] ( ) ∂ ∂ B B = 2B j B ∂x i j j ∂x i j In this way, we can deal with arbitrarily complicated vector representations by following simple rules. As an exercise, try other identities on your own.