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)