Mathematical Physics Vol 1

4.6 Examples

137

Solution

∇ × ( r f ( r ))= ∇ × ( xf ( r ) i + yf ( r ) j + zf ( r ) k )=

i

j

k

∂ ∂ x ∂ ∂ z xf ( r ) yf ( r ) zf ( r ) ∂ ∂ y

=

=

= z

∂ f ∂ z

i + x

∂ f ∂ x

j + y

∂ f ∂ y

∂ f ∂ y −

∂ f ∂ z −

∂ f ∂ x −

k .

y

z

x

Let us now compute the partial derivatives of the function f by x ∂ f ∂ x = d f d r ∂ r ∂ x = d f d r ∂ ∂ x ( p x 2 + y 2 + z 2 )= f ′ ( r )

x p x 2 + y 2 + z 2

x r

= f ′

,

d f d r ≡ f ′ . In a similar way we can compute the remaining two partial derivatives

where

∂ f ∂ y

∂ f ∂ z

y r

z r

= f ′

= f ′

i

,

and thus finally obtain

∇ × ( r f ( r ))= z y r f ′ − y z r

f ′ i + x

f ′ j + y

f ′ k = 0 .

z r

x r

x r

y r

f ′ − z

f ′ − x

Exercise 66 Prove that ∇ × ( ∇ × V )= − ∇ 2 V + ∇ ( ∇ · V ) .

Solution

i

j

k

∂ ∂ x V 1

∂ ∂ y V 2

∂ ∂ z V 3

∇ × ( ∇ × V )= ∇ ×

=

= ∇ ×

∂ z

i +

∂ x

j +

∂ y

k =

∂ V 3

∂ V 2

∂ V 1

∂ V 3

∂ V 2

∂ V 1

∂ y −

∂ z −

∂ x −

i

j

k

∂ ∂ x

∂ ∂ y

∂ ∂ z

=

=

∂ V 3

∂ V 2 ∂ z

∂ V 1

∂ V 3 ∂ x

∂ V 2

∂ V 1 ∂ y

∂ y −

∂ z −

∂ x −