8Some applications of integral theorems
IA Vector Calculus
8.2 Conservative fields and scalar products
Definition (Conservative field). A vector field F is conservative if
(i) F = ∇f for some scalar field f; or
(ii)
R
C
F · dr is independent of C, for fixed end points and orientation; or
(iii) ∇ × F = 0.
In R
3
, all three formulations are equivalent.
We have previously shown (i) ⇒ (ii) since
Z
C
F · dr = f(b) − f(a).
We have also shown that (i) ⇒ (iii) since
∇ × (∇f) = 0.
So we want to show that (iii) ⇒ (ii) and (ii) ⇒ (i)
Proposition. If (iii) ∇ × F = 0, then (ii)
R
C
F · dr is independent of C.
Proof.
Given
F
(
r
) satisfying
∇ × F
= 0, let
C
and
˜
C
be any two curves from
a
to b.
a
b
˜
C
C
If S is any surface with boundary ∂S = C −
˜
C, By Stokes’ theorem,
Z
S
∇ × F · dS =
Z
∂S
F · dr =
Z
C
F · dr −
Z
˜
C
F · dr.
But ∇ × F = 0. So
Z
C
F · dr −
Z
˜
C
F · dr = 0,
or
Z
C
F · dr =
Z
˜
C
F · dr.
Proposition.
If (ii)
R
C
F ·
d
r
is independent of
C
for fixed end points and
orientation, then (i) F = ∇f for some scalar field f.
Proof.
We fix
a
and define
f
(
r
) =
R
C
F
(
r
0
)
·
d
r
0
for any curve from
a
to
r
.
Assuming (ii),
f
is well-defined. For small changes
r
to
r
+
δr
, there is a small
extension of C by δC. Then
f(r + δr) =
Z
C+δC
F(r
0
) · dr
0
=
Z
C
F · dr
0
+
Z
δC
F · dr
0
= f(r) + F(r) · δr + o(δr).
So
δf = f(r + δr) − f(r) = F(r) · δr + o(δr).
But the definition of grad is exactly
δf = ∇f · δr + o(δr).
So we have F = ∇f .
Note that these results assume
F
is defined on the whole of
R
3
. It also
works of
F
is defined on a simply connected domain
D
, ie a subspace of
R
3
without holes. By definition, this means that any two curves
C,
˜
C
with fixed
end points can be smoothly deformed into one another (alternatively, any loop
can be shrunk into a point).
If we have a smooth transformation from
C
to
˜
C
, the process sweeps out a
surface bounded by C and
˜
C. This is required by the proof that (iii) ⇒ (ii).
If
D
is not simply connected, then we obtain a multi-valued
f
(
r
) on
D
in
general (for the proof (ii)
⇒
(i)). However, we can choose to restrict to a subset
D
0
⊆ D such that f(r) is single-valued on D
0
.
Example. Take
F =
−y
x
2
+ y
2
,
x
x
2
+ y
2
, 0
.
This obeys
∇ × F
= 0, and is defined on
D
=
R
3
\ {z-axis}
, which is not
simply-connected. We can also write
F = ∇f,
where
f = tan
−1
y
x
.
which is multi-valued. If we integrate it about the closed loop
x
2
+
y
2
= 1
, z
= 0,
i.e. a circle about the
z
axis, the integral gives 2
π
, as opposed to the expected 0
for a conservative force. This shows that the simply-connected-domain criterion
is important!
However f can be single-valued if we restrict it to
D
0
= R
3
− {half-plane x ≥ 0, y = 0},
which is simply-connected. (Draw and check!) Any closed curve we can draw in
this area will have an integral of 0 (the circle mentioned above will no longer be
closed!).