Begin2.DVI

Example 7-18. Show that ∇× (∇× A
) = ∇(∇·A
)−∇^2 A

Solution Calculate ∇× A
using determinants to obtain

∇× A
=

∣∣

∣∣

∣∣

ˆe 1 ˆe 2 ˆe 3

∂

∂x

∂

∂y

∂

∂z

A 1 A 2 A 3

∣∣

∣∣

∣∣

=

(

∂A 3

∂y −

∂A 2

∂z

)

ˆe 1 −

(

∂A 3

∂x −

∂A 1

∂z

)

ˆe 2 +

(

∂A 2

∂x −

∂A 1

∂y

)

eˆ 3

One can then calculate the curl of the curl as

∇× (∇× A
) =

∣∣

∣∣

∣∣

∣

ˆe 1 eˆ 2 eˆ 3

∂

∂x

∂

∂y

∂

( ∂z

∂A 3

∂y −

∂A 2

∂z

) (

∂A 1

∂z −

∂A 3

∂x

) (∂A 2

∂x −

∂A 1

∂y

)

∣∣

∣∣

∣∣

∣

(7 .66)

The eˆ 1 component of ∇× (∇× A
)is

ˆe 1

[

∂

∂y

(

∂A 2

∂x −

∂A 1

∂y

)

−

∂

∂z

(

∂A 1

∂z −

∂A 3

∂x

)]

=ˆe 1

[

∂^2 A 2

∂x ∂y −

∂^2 A 1

∂y^2 −

∂^2 A 1

∂z^2 +

∂^2 A 3

∂x ∂z

]

By adding and subtracting the term

∂^2 A 1

∂x^2

to the above result one finds the ˆe 1

component can be expressed in the form

ˆe 1

{[

−∂

(^2) A 1
∂x^2
−∂
(^2) A 1
∂y^2
−∂
(^2) A 1
∂z^2
]

• ∂
  ∂x
  (
  ∂A 1
  ∂x
  +∂A^2
  ∂y
  +∂A^3
  ∂z
  )}
  (7 .67)

In a similar fashion it can be verified that the ˆe 2 component of ∇× (∇× A
)is

ˆe 2

{[

−∂

(^2) A 2
∂x^2 −
∂^2 A 2
∂y^2 −
∂^2 A 2
∂z^2
]
+∂y∂
(
∂A 1
∂x +
∂A 2
∂y +
∂A 3
∂z
)}
(7 .68)

and the ˆe 3 component of ∇× (∇× A
)is

ˆe 3

{[

−

∂^2 A 3

∂x^2 −

∂^2 A 3

∂y^2 −

∂^2 A 3

∂z^2

]

+

∂

∂z

(

∂A 1

∂x +

∂A 2

∂y +

∂A 3

∂z

)}

(7 .69)

Adding the results from the equations (7.67), (7.68), (7.69) one obtains the result

∇× (∇× A
) = ∇(∇·A
)−∇^2 A
 (7 .70)

Directional Derivatives

Let
r =
r (s) denote an arbitrary space curve which passes through the point

P(x, y, z)of the region R, where the scalar function φ=φ(x, y, z)exists and has all first-

order partial derivatives which are continuous. Here the space curve is expressed