4.4 Conservative Fields/Exact Forms

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Theorem 4.4.1. If V is a conservative vector field on Rn, ie V = r' for

' : Rn ! R differentiable, and if c : [a, b] ! R is any smooth curve, then

Z

c

V =

Z b

a

V(c(t)) q c0(t)dt

= '(c(b)) − '(c(a))

Proof

Z

c

V =

Z b

r=a

V(c(t)) q c0(t)dt

write

c(t) =

2

6664

x1(t)

x2(t)

...

xn(t)

3

7775

Then:

40 CHAPTER 4. FIELDS AND FORMS

Z

c

V =

Z

2

6664

@'

@x1

(c(t))

@'

@x2

(c(t))

...

@'

@x (c(t))

3

7775

q

2

6664

dx1

dt

dx2

dt

...

dxn

dt

3

7775

t

dt

=

Z _X@'i

@xi

_

c(t)

_

dxi

dt

_

t

dt

=

Z t=b

t=a

d

dt

(' _ c) dt (chain rule)

=

Z t=b

t=c

d(' _ c)

= 'c(a) − ' _ c(b)

_

In other words, it’s the chain rule.

Corollary 4.4.1.1. For a conservative vector field, V, the integral over a

curve c

R

cV depends only on the end points of the curve and not on the

path. _

Remark 4.4.1. It is possible to ask an innocent young student to tackle a

thoroughly appalling path integral question, which the student struggles for

days with. If the result in fact doesn’t depend on the path, there could be

an easier way.

Example 4.4.1.

V

_

x

y

_

,

_

2x cos(x2 + y2)

2y cos(x2 + y2)

_

Let c be the path from

_

1

0

_

to

_

0

1

_

that follows the curve shown in figure

4.4, a quarter of a circle with centre at [1, 1]T .

Find

R

cV

The innocent student finds c(t) =

_

1 − sin t

1 − cos t

_

t 2 [0, _/2] and tries to evaluate

Z _/2

t=0

_

2(1 − sin t) cos((1 − sin t)2 + (1 − cos t)2)

2(1 − cos t) cos((1 − sin t)2 + (1 − cos t)2)

_

q

_

−cos t

sin t

_

dt

4.4. CONSERVATIVE FIELDS/EXACT FORMS 41

Figure 4.4: quarter-circular path

The lazy but thoughtful student notes that V = r' for ' = sin(x2 + y2)

and writes down '

_

0

1

_

− '

_

1

0

_

= 0 since 12 + 02 = 02 + 12 = 1.

Which saves a lot of ink.

Remark 4.4.2. It is obviously a good idea to be able to tell if a vector field

is convervative:

Remark 4.4.3. If V = rf for f : Rn −! R we have

V

_

x

y

_

, P

_

x

y

_

dx + Q

_

x

y

_

dy

is the corresponding 1-form where

P =

@f

@x

Q =

@f

@y

.

In which case

@P

@y

=

@2f

@y@x

=

@2f

@x@y

=

@Q

@x

So if

@P

@y

=

@Q

@x

then there is hope.

Definition 4.15. A 1-form on R2 is said to be closed iff

@P

@y

−

@Q

@x

= 0

42 CHAPTER 4. FIELDS AND FORMS

Remark 4.4.4. Then the above argument shows that every exact 1-form is

closed. We want the converse, but at least it is easy to check if there is hope.

Example 4.4.2.

V

_

x

y

_

=

_

2x cos(x2 + y2)

2y cos(x2 + y2)

_

has

P = 2x cos(x2 + y2), Q = 2y cos(x2 + y2)

and

@P

@y

= (−4xy sin(x2 + y2)) =

@Q

@x

So there is hope, and indeed the field is conservative: integrate P with respect

to x to get, say, f and check that

@f

@y

= Q

Example 4.4.3. NOT!

! = −ydx + xdy

has

@P

@y

= −1

but

@Q

@x

= +1

So there is no hope that the field is conservative, something our physical

intuitions should have told us.

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