We’d like to know more about what makes a vector field independent of path
(IOP). Is there a visual difference between a vector field that is IOP and one that
is not IOP? From lecture 19 we considered the two vector fields
F(x, y) =
−y
x
,
x2 + y2 x2 + y2
G(x, y) = h−y, xi .
We discovered that the former is conservative within the first quadrant while the
latter is conservative nowhere. Consider the effect that these two vector fields have
on the circular paddle pictured below.
y
y
x
x
The vector field F is pictured in blue on the left, and the vector field G is pictured
in red on the right. We notice that the trajectory of the paddle will be the same
whether it lie in F and G, that is, in a circular arc counterclockwise about the
origin. On the other hand, it would appear that the paddle may have a tendency
to spin in G whereas perhaps not in F. Perhaps this is the visual difference we’d
like to have in vector fields that are IOP versus those that are not? Perhaps vector
fields that are IOP simply displace paddles like the one above without rotating or
spinning them versus non-IOP vector fields that do spin them? This is, in fact, the
case. Indeed, click here to run a java applet to see this phenomenon in real-time.
How do we put these intuitive ideas on solid ground? If we want to know whether
or not F is “rotational”, then we should compute the average tangential component
of F on the oriented boundary of a paddle. If C is the oriented boundary of a
paddle, we already know that the average tangential component of F on C is
1
`(C)
Z
F · dr.
C
1
2
The paddle will not rotate if this integral is zero. In fact, if all paddles of all shapes1
in a region D don’t rotate in the presence of F, then
Z
1
F · dr = 0
`(C) C
for all smooth closed curves C, i.e., smooth loops contained in D. From the previous
lecture, we know this is equivalent to F being IOP which is equivalent to F being
conservative on D!
Even better: suppose we wish to know the average angular velocity of F around
a circle of radius r centered at the origin, say C, oriented counterclockwise. This is
computed by the same formula as above, but divided by r:
Z
1
F · dr.
2πr2 C
Letting r → 0+ should give us the angular velocity of F about the origin. Let
F = hP, Qi.
Claim 1. lim+
r→0
1
2πr2
Z
F · dr =
C
Qx (0, 0) − Py (0, 0)
.
2
Proof. We parameterize C via
x(t) = r cos t
Then
1
2πr2
Z
F · dr
=
C
=
y(t) = r sin t.
Z 2π
1
hP(r cos t, r sin t), Q(r cos t, r sin t)i · h−r sin t, r cos ti dt
2πr2 0
Z
1 2π
Q(r cos t, r sin t) cos t − P(r cos t, r sin t) sin t dt
2πr 0
Consider the Q part of the above integral. We use integration by parts with
u = Q(r cos t, r sin t)
dv = cos t dt.
Then the chain rule gives us
du = Qx (r cos t, r sin t)(−r sin t) + Qy (r cos t, r sin t)r cos t dt
v = sin t dt.
So
1
2πr
Z 2π
Q(r cos t, r sin t) cos t
=
0
=
1...with a smooth boundary.
2π
Z
1 2πr
1
Q(r cos t, r sin t) sin t −
v du
2πr
2π 0
0
Z
−1 2π
v du.
2πr 0
3
Now let r → 0+ on both sides of the above, so
Z 2π
Z
1
Qx (0, 0) 2π 2
lim+
Q(r
cos
t,
r
sin
t)
cos
t
dt
=
sin t dt −
r→0 2πr2 0
2π
0
Z
Qy (0, 0) 2π
−
cos t sin t dt
2π
0
2π
Qx (0, 0) t sin 2t
=
−
2π
2
4
0
=
Qx (0, 0)
.
2
An identical argument shows that
Z 2π
−Py (0, 0)
1
−P(r cos t, r sin t) sin t dt =
.
lim+
2
r→0 2πr
2
0
Hence
lim+
r→0
1
2πr2
Z
F · dr =
C
Qx (0, 0) − Py (0, 0)
,
2
as claimed.
Repeating the above arguments where C is centered about (x, y), we can show
that
Qx (x, y) − Py (x, y)
2
represents the angular velocity of F about the point (x, y). So on the one hand
Z
1
F · dr
2πr2 C
represents the average angular velocity of F about the oriented circle C. On the
other hand,
Qx (x, y) − Py (x, y)
2
represents the angular velocity of F about the point (x, y). Does the fundamental
theorem of calculus have meaning here? Let D be the interior of the circle C. Should
it be that the average angular velocity of F around C be equal to the average value
of the angular velocities of F about the points interior to C, i.e., is it true that
Z
ZZ
Qx − Py
1
1
F
·
dr
=
dA ?
2πr2 C
πr2 D
2
Yes! It is true. In fact, in a greater generality:
Theorem 1. (Green) Let C be the smooth boundary, oriented counterclockwise, of
a simply connected2 region D. Suppose F = hP, Qi with P and Q having continuous
first partial derivatives on D. Then
Z
ZZ
F · dr =
Qx − Py dA.
C
2A region essentially without holes.
D
4
We now see that if Qx − Py = 0, i.e., if Qx = Py on a simply connected region
D, then F is conservative or IOP on D since
Z
F · dr = 0
C
for all loops C contained in D. We now have the following very interesting theorem.
Theorem 2. Let D be a simply connected region, and suppose F = hP, Qi where
P and Q have continuous first partial derivatives on D. Then the following are
equivalent.
(i) F is conservative.
(ii) F is independent of path.
(iii) F is irrotational in D, that is, Qx (x, y) = Py (x, y) for all (x, y) ∈ D.