Mathematical Physics Vol 1

Chapter 4. Field theory

146

c) Let us now use the previously computed expression for ∇ B (see Example 72) A · ∇ B =( A 1 i + A 2 j + A 3 k ) · ∇ B = = A 1 i · ∇ B + A 2 j · ∇ B + A 3 k · ∇ B = A 1 ∂ B 1 ∂ x i + ∂ B 2 ∂ x j + ∂ B 3 ∂ x k + A 2 ∂ B 1 ∂ y i + ∂ B 2 ∂ y j + ∂ B 3 ∂ y k + + A 3 ∂ B 1 ∂ z i + ∂ B 2 ∂ z j + ∂ B 3 ∂ z k . It is obvious from here that the result is the same as the one obtained under b), namely ( A · ∇ ) B = A · ( ∇ B )= A · ∇ B . We have used here the results from Example 73 as well.

Exercise 75

If

A = 2 yz i − x 2 y j + xz 2 k B = x 2 i + yz j − xy k φ = 2 x 2 yz 3

find

a) ( A · ∇ ) φ , b) A · ∇ φ , c) ( B · ∇ ) A , d) ( A × ∇ ) φ , e) A × ∇ φ .

Solution a)

( A · ∇ ) φ = 2 yz i − x 2 y j + xz 2 k ∂ ∂ x i + ∂ ∂ y j + ∂ ∂ z

k φ =

∂ ∂ z

= 2 yz

∂ ∂ y

∂ ∂ x −

2 x 2 yz 3 = 2 x 2 yz 3 + xz 2

x 2 y

+ xz 2

= 2 yz 2 x 2 yz 3 = =( 2 yz ) 4 xyz 3 − x 2 y 2 x 2 z 3 + xz 2 6 x 2 yz 2 = = 8 xy 2 z 4 − 2 x 4 yz 3 + 6 x 3 yz 4 . ∂ ∂ x 2 x 2 yz 3 − x 2 y ∂ ∂ y ∂ ∂ z