ψis any scalar function, is also a vector satisfying F = curl V∗.This result is verified
by using the distributive property of the curl since
F= curl V∗= curl (V +∇ψ) = curl V + curl (∇ψ) = curl V. (9 .51)
together with the fact that curl (∇ψ) = curl grad ψ= 0.
Since the vector potential V is not uniquely determined, it is only necessary to
exhibit one vector potential of F . Toward this end make note of the fact that F can
be expressed in the form
F=F 1 ˆe 1 +F 2 ˆe 2 +F 3 ˆe 3 =∇× V
=
(
∂V 3
∂y
−∂V^2
∂z
)
ˆe 1 +
(
∂V 1
∂z
−∂V^3
∂x
)
ˆe 2 +
(
∂V 2
∂x
−∂V^1
∂y
)
ˆe 3.
(9 .52)
Show that if the component V 3 = 0 , then the components of F must satisfy the
equations
F 1 =−
∂V 2
∂z , F^2 =
∂V 1
∂z , F^3 =
∂V 2
∂x −
∂V 1
∂y. (9 .53)
An integration of the first two equations in (9.53) produces
V 1 =
∫z
z 0
F 2 dz +f 1 (x, y )
V 2 =−
∫z
z 0
F 1 dz +f 2 (x, y ),
(9 .54)
where f 1 , f 2 are arbitrary functions which are held constant during the partial differ-
entiation processes used to calculate ∂V^1
∂z
and ∂V^2
∂z
. The functions f 1 and f 2 must be
selected in such a way that the last equation in (9.53) is also satisfied for all values
of x, y, and z. Substitution of equations (9.54) into the last equation of (9.53) informs
us that
∂V 2
∂x −
∂V 1
∂y =−
∫z
z 0
∂F 1
∂x dz −
∫z
z 0
∂F 2
∂y dz +
∂f 2
∂x −
∂f 1
∂y
=−
∫z
z 0
(
∂F 1
∂x +
∂F 2
∂y
)
dz +
∂f 2
∂x −
∂f 1
∂y.
(9 .55)
Now by assumption, F is a solenoidal vector field and consequently
div F=
∂F 1
∂x +
∂F 2
∂y +
∂F 3
∂z = 0.
We therefore can write
∂F 1
∂x +
∂F 2
∂y =−
∂F 3
∂z
