Mathematical Physics Vol 1

Chapter 4. Field theory

144

B 1 · i + B 2 · j + B 3 · k . It follows that grad B ∇ B = ∂ ∂ x i + ∂ ∂ x j + ∂ ∂ x

k ⊗ ( B 1 i + B 2 j + B 3 k )=

= + +

i ⊗ k + j ⊗ k +

∂ B 1 ∂ x ∂ B 1 ∂ x ∂ B 1 ∂ x

∂ B 2 ∂ x ∂ B 2 ∂ x

∂ B 3 ∂ x ∂ B 3 ∂ x

i ⊗ i + j ⊗ i + k ⊗ i +

i ⊗ j + j ⊗ j +

∂ B 2 ∂ x k ⊗ k , where i ⊗ i , i ⊗ j ,. . . , k ⊗ k represent the so called unit dyads 20 . The value represented in the form a 11 i ⊗ i + a 12 i ⊗ j + a 13 i ⊗ k + + a 21 j ⊗ i + a 22 j ⊗ j + a 23 j ⊗ k + + a 31 k ⊗ i + a 32 k ⊗ j + a 33 k ⊗ k is a dyad and the constants a 11 , a 12 , a 13 ... are its components. k ⊗ j + ∂ B 3 ∂ x

Exercise 73 Let the vector A be expressed in the following form A = A 1 i + A 2 j + A 3 k , anddyad φ as φ = a 11 i ⊗ i + a 12 i ⊗ j + a 13 i ⊗ k + a 21 j ⊗ i + a 22 j ⊗ j + a 23 j ⊗ k + a 31 k ⊗ i + a 32 k ⊗ j + a 33 k ⊗ k . Compute A · φ .

Solution Using the law of distribution we obtain

A · φ =( A 1 i + A 2 j + A 3 k ) · φ = A 1 i · φ + A 2 j · φ + A 3 k · φ . Let us first compute i · φ . This product is defined as the scalar product of the unit vector i with each component of the dyad φ . Typical elements are i · a 11 i ⊗ i , i · a 12 i ⊗ j , i · a 21 j ⊗ i , i · a 32 k ⊗ j , . . . Taking this into account, we obtain

i · a 11 i ⊗ i = a 11 ( i · i ) i = a 11 i , i · a 12 i ⊗ j = a 12 ( i · i ) j = a 12 j , i · a 21 j ⊗ i = a 21 ( i · j ) i = 0 , i · a 32 k ⊗ j = a 32 ( i · k ) j = 0 .

20 An ordered pair of vectors ( a , b ), denoted by a ⊗ b or ab (the older way of denoting), is called a dyad or tensor product or open product of vectors a and b . Adyad ab is defined as that associates each vector v with a vector ( b · v ) a , that is, ( a ⊗ b ) v =( b · v ) a , where the point denotes a scalar product.