Mathematical Physics Vol 1

4.6 Examples

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Following an analogous interpretation for j · φ i k · φ , we finally obtain A · φ = A 1 ( a 11 i + a 12 j + a 13 k )+ A 2 ( a 21 i + a 22 j + a 23 k )+ + A 3 ( a 31 i + a 32 j + a 33 k )= =( A 1 a 11 + A 2 a 21 + A 3 a 31 ) i +( A 1 a 12 + A 2 a 22 + A 3 a 32 ) j + +( A 1 a 13 + A 2 a 23 + A 3 a 33 ) k . Thus, the scalar product of vector and dyad A · φ is a vector. Exercise 74 Let the vector functions A and B , the scalar function φ , and the operator ∇ , be expressed with respect to the Cartesian coordinate system. a) Interpret the symbol A · ∇ , and then apply it to the scalar function φ . b) Give a possible definition of ( A · ∇ ) B . c) Is it possible to write this expression without brackets, as A · ∇ B , without chang ing its sense? Solution a) Let A be expressed as A = A 1 i + A 2 j + A 3 k . Then A · ∇ =( A 1 i + A 2 j + A 3 k ) · ∂ ∂ x i + ∂ ∂ y j + ∂ ∂ z k = A 1 ∂ ∂ x + A 2 ∂ ∂ y + A 3 ∂ ∂ z represents an operator. By applying this operator to the function φ , we obtain ( A · ∇ ) φ = A 1 ∂ ∂ x + A 2 ∂ ∂ y + A 3 ∂ ∂ z φ = A 1 ∂φ ∂ x + A 2 ∂φ ∂ y + A 3 ∂φ ∂ z

This is the same as A · ( ∇ φ ) . Thus, given that ( A · ∇ ) φ = A · ( ∇ φ ) , in this case the brackets can be omitted, that is ( A · ∇ ) φ = A · ( ∇ φ )= A · ∇ φ . b) If we replace φ by B = B 1 i + B 2 j + B 3 k in a) we obtain ( A · ∇ ) B = A 1 ∂ ∂ x + A 2 ∂ ∂ y + A 3 ∂ ∂ z B = = A 1 ∂ B ∂ x + A 2 ∂ B ∂ y + A 3 ∂ B ∂ z = = A 1 ∂ B 1 ∂ x + A 2 ∂ B 1 ∂ y + A 3 ∂ B 1 ∂ z i + + A 1 ∂ B 2 ∂ x + A 2 ∂ B 2 ∂ y + A 3 ∂ B 2 ∂ z j + + A 1 ∂ B 3 ∂ x + A 2 ∂ B 3 ∂ y + A 3 ∂ B 3 ∂ z k .