4.2 Integrating 1-forms (vector fields) over curves.

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Definition 4.12. I = {x 2 R : 0 _ x _ 1}

Definition 4.13. A smooth curve in Rn is the image of a map

c : I −! Rn

that is infinitely differentiable everywhere.

It is piecewise smooth if it is continuous and fails to be smooth at only a

finite set of points.

Definition 4.14. A smooth curve is oriented by giving the direction in which

t is increasing for t 2 I _ R

Remark 4.2.1. If you decided to use some interval other than the unit

interval, I, it would not make a whole lot of difference, so feel free to use,

for example, the interval of points between 0 and 2_ if you wish. After all, I

can always map I into your interval if I feel obsessive about it.

Remark 4.2.2. [Motivation] Suppose the wind is blowing in a rather erratic

manner, over the great gromboolian plain (R2). In figure 4.2 you can

see the path taken by me on my bike together with some vectors showing the

wind force.

4.2. INTEGRATING 1-FORMS (VECTOR FIELDS) OVER CURVES. 33

Figure 4.2: Bicycling over the Great Gromboolian Plain

Figure 4.3: An infinitesimal part of the ride

We represent the wind as a vector field F : R2 −! R2. I cycle along a curved

path, and at time 0 I start out at c(0) and at time t I am at c(t). I stop at

time t = 1.

I am interested in the effect of the wind on me as I cycle. At t = 0 the wind

is pushing me along and helping me. At t = 1 it is against me. In between it

is partly with me, partly at right angles to me and sometimes partly against

me.

I am not interested in what the wind is doing anywhere else.

I suppose that if the wind is at right angles to my path it has no effect

(although it might blow me off the bike in real life).

The question is, how much net help or hindrance is the wind on my journey?

I solve this by chopping my path up into little bits which are (almost) straight

line segments.

F is the wind at where I am time t.

My path is, nearly, the straight line obtained by differentiating my path at

time t, c0(t) This gives the “infinitesimal length” as well as its direction. Note

34 CHAPTER 4. FIELDS AND FORMS

that this could be defined without reference to a parametrisation.

The component in my direction, muliplied by the length of the path is

F _ c(t) q c0(t) for time4t

(approximately.) q denotes the inner or dot product.

(The component in my direction is the projection on my direction, which is

the inner product of the Force with the unit vector in my direction. Using c0

includes the “speed” and hence gives me the distance covered as a term.)

I add up all these values to get the net ‘assist’ given by F.

Taking limits, the net assist is

Z t=1

t=0

F(c(t)) q c0(t)dt

Example 4.2.1. The vector field F is given by

F

_

x

y

_

,

_

−y

x

_

(I am in a hurricane or cyclone)

I choose to cycle in the unit (quarter) circle from

_

1

0

_

to

_

0

1

_

my path

(draw it) c is

c(t) ,

_

cos _

2 t

sin _

2 t

_

for t 2 [0, 1]

Differentiating c we get:

c0(t) =

_

−_

2 sin _

2 t

_

2 cos _

2 t

_

My ’assist’ is therefore

Z t=1

t=0

_

−sin _

2 t

cos _

2 t

_

q

_

−_

2 sin _

2 t

_

2 cos _

2 t

_

dt

=

_

2

Z t=1

t=0

h

sin2 _

2

t + cos2 _

2

t

i

dt

=

_

2

4.2. INTEGRATING 1-FORMS (VECTOR FIELDS) OVER CURVES. 35

This is positive which is sensible since the wind is pushing me all the way.

Now I consider a different path from

_

1

0

_

to

_

0

1

_

My path is first to go

along the X _ axis to the origin, then to proceed along the Y -axis finishing at

0

1

_

. I am going to do this path in two separate stages and then add up

the answers. This has to give the right answer from the definition of what a

path integral is. So my first stage has

c(t) ,

_

−t

0

_

and

c0(t) =

_

−1

0

_

(which means that I am travelling at uniform speed in the negative x direction.)

The ‘assist’ for this stage is

Z 1

0

_

0

−t

_

q

_

−1

0

_

which is zero. This is not to surprising if we look at the path and the vector

field.

The next stage has a new c(t):

c(t) ,

_

0

t

_

which goes from the origin up a distance of one unit.

Z 1

0

F(c(t)) q c0dt =

Z _

−t

0

_

q

_

0

1

_

dt

=

Z

0dt

so I get no net assist.

This makes sense because the wind is always orthogonal to my path and has

no net effect.

Note that the path integral between two points depends on the path, which

is not surprising.

36 CHAPTER 4. FIELDS AND FORMS

Remark 4.2.3. The only difficulty in doing these sums is that I might describe

the path and fail to give it parametrically - this can be your job. Oh,

and the integrals could be truly awful. But that’s why Mathematica was

invented.

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