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F · dS ωF
S S
1
(∇ — F) · dS dωF
S S

3
f dV ωf
„¦ „¦
2
(∇ · F)dV dωF
„¦ „¦

Let us now apply the Generalized Stokes™ Theorem to various situations. First,
we start with a parameterization, φ : [a, b] ’ σ ‚ R3 , of a curve in R3 , and a
function, f : R3 ’ R. Then we have


∇f · ds ≡ f = f (φ(b)) ’ f (φ(a))
df =
σ σ ‚σ
This shows the independence of path of line integrals of gradient ¬elds. We can
use this to prove that a line integral of a gradient ¬eld over any simple closed curve is
0, but for us there is an easier, direct proof, which again uses the Generalized Stokes™
Theorem. Suppose σ is a simple closed loop in R3 (i.e. ‚σ = …). Then σ = ‚D, for
some 2-chain, D. We now have
4. VECTOR CALCULUS AND THE MANY FACES OF STOKES™ THEOREM 79




∇f · ds ≡ df = ddf = 0
σ σ D

Now, suppose we have a vector ¬eld, F, and a parameterized surface, S. Yet
another application of the Generalized Stokes™ Theorem yields

1 1
F · ds ≡ dωF ≡ (∇ — F) · dS
ωF =
‚S ‚S S S

In vector calculus we call this equality “Stokes™ Theorem”. In some sense, ∇ — F
measures the “twisting” of F at points of S. So Stokes™ Theorem says that the net
twisting of F over all of S is the same as the amount F circulates around ‚S.



Example 5.12. Suppose we are faced with a problem phrased thusly: “Use
F · ds, where C is the curve of intersection of
Stokes™ Theorem to calculate
C
2 2
the cylinder x + y = 1 and the plane z = x + 1, and F is the vector ¬eld
’x2 y, xy 2, z 3 .”
We will solve this problem by translating to the language of di¬erential forms, and
F · ds =
using the Generalized Stokes™ Theorem instead. To begin, note that
C
1 1 2 2 3
= ’x y dx + xy dy + z dz.
ωF , and ωF
C
Now, to use the Generalized Stokes™ Theorem we will need to calculate

dωF = (x2 + y 2 ) dx § dy.
1


Let D denote the subset of the plane z = x + 1 bounded by C. Then ‚D = C.
Hence, by the Generalized Stokes™ Theorem we have
1 1
(x2 + y 2 ) dx § dy
ωF = dωF =
C D D

The region D is parameterized by Ψ(r, θ) = (r cos θ, r sin θ, r cos θ + 1), where
0 ¤ r ¤ 1 and 0 ¤ θ ¤ 2π. Using this one can (and should!) show that
(x2 + y 2 ) dx § dy = 8π.
D
80 5. STOKES™ THEOREM

Exercise 5.17. Let C be the square with sides (x, ±1, 1), where ’1 ¤ x ¤ 1 and
(±1, y, 1), where ’1 ¤ y ¤ 1, with the indicated orientation (see Figure 3). Let F be
the vector ¬eld xy, x2 , y 2 z . Compute F · ds. Answer: 0
C

z




C 1


-1


y
-1 1

1

x

Figure 3.

Suppose now that „¦ is some volume in R3 . Then we have

2 2
F · dS ≡ dωF ≡ (∇ · F)dV
ωF =
„¦ „¦
‚„¦ ‚„¦
This last equality is called “Gauss™ Divergence Theorem”. ∇ · F is a measure
of how much F “spreads out” at a point. So Gauss™ Theorem says that the total
spreading out of F inside „¦ is the same as the net amount of F “escaping” through
‚„¦.

Exercise 5.18. Let „¦ be the cube {(x, y, z)|0 ¤ x, y, z ¤ 1}. Let F be the vector ¬eld
xy 2 , y 3, x2 y 2 . Compute F · dS. Answer: 4
3
‚„¦
CHAPTER 6


Applications

1. Maxwell™s Equations

As a brief application we show how the language of di¬erential forms can greatly
simplify the classical vector equations of Maxwell. These equations describe the
relationship between electric and magnetic ¬elds. Classically both electricity and
magnetism are described as a 3-dimensional vector ¬eld which varies with time:

E = Ex , Ey , Ez


B = Bx , By , Bz

Where Ex , Ez , Ez , Bx , By , and Bz are all functions of x, y, z and t.
Maxwell™s equations are then:

∇·B = 0
‚B
+∇—E = 0
‚t
∇ · E = 4πρ
‚E
’ ∇ — B = ’4πJ
‚t
The quantity ρ is called the charge density and the vector J = Jx , Jy , Jz is called
the current density.
We can make all of this look much simpler by making the following de¬nitions.
First we de¬ne a 2-form called the Faraday, which simultaneously describes both the
electric and magnetic ¬elds:

F = Ex dx § dt + Ey dy § dt + Ez dz § dt
+Bx dy § dz + By dz § dx + Bz dx § dy
81
82 6. APPLICATIONS

Next we de¬ne the “dual” 2-form, called the Maxwell:

F = Ex dy § dz + Ey dz § dx + Ez dx § dy
+Bx dt § dx + By dt § dy + Bz dt § dz

We also de¬ne the 4-current, J, and it™s “dual”, — J:

J= ρ, Jx , Jy , Jz

J = ρ dx § dy § dz
’Jx dt § dy § dz
’Jy dt § dz § dx
’Jz dt § dx § dy

Maxwell™s four vector equations now reduce to:

dF = 0
d— F = 4π — J

Exercise 6.1. Show that the equation dF = 0 implies the ¬rst two of Maxwell™s
equations.

Exercise 6.2. Show that the equation d— F = 4π — J implies the second two of Maxwell™s
equations.

The di¬erential form version of Maxwell™s equation has a huge advantage over the
vector formulation: it is coordinate free! A 2-form such as F is an operator that “eats”
pairs of vectors and “spits out” numbers. The way it acts is completely geometric...
that is, it can be de¬ned without any reference to the coordinate system (t, x, y, z).
This is especially poignant when one realizes that Maxwell™s equations are laws of
nature that should not depend on a man-made construction such as coordinates.

2. Foliations and Contact Structures

Everyone has seen tree rings and layers in sedimentary rock. These are examples
of foliations. Intuitively, a foliation is when some region of space has been “¬lled up”
with lower dimensional surfaces. A full treatment of foliations is a topic for a much
larger textbook than this one. Here we will only be discussing foliations of R3 .
2. FOLIATIONS AND CONTACT STRUCTURES 83

Let U be an open subset of R3 . We say U has been foliated if there is a family
φt : Rt ’ U of parameterizations (where for each t the domain Rt ‚ R2 ) such that
every point of U is in the image of exactly one such parameterization. In other words,
the images of the parameterizations φt are surfaces that ¬ll up U, and no two overlap.
Suppose p is a point of U and U has been foliated as above. Then there is a
t
unique value of t such that p is a point in φt (Rt ). The partial derivatives, ‚φ (p) and
‚x
‚φt
are then two vectors that span a plane in Tp R3 . Let™s call this plane Πp . In
(p)
‚y
other words, if U is foliated then at every point p of U we get a plane Πp in Tp R3 .
The family {Πp } is an example of a plane ¬eld. In general a plane ¬eld is just a
choice of a plane in each tangent space which varies smoothly from point to point in
R3 . We say a plane ¬eld is integrable if it consists of the tangent planes to a foliation.
This should remind you a little of ¬rst-term calculus. If f : R1 ’ R1 is a
di¬erentiable function then at every point p on its graph we get a line in Tp R2 (see
Figure 2). If we just know the lines and want the original function then we are
integrating.
There is a theorem that says that every line ¬eld on R2 is integrable. The question
we would like to answer in this section is whether or not this is true of plane ¬elds
on R3 . The ¬rst step is to ¬gure out how to specify a plane ¬eld in some reasonably
nice way. This is where di¬erential forms come in. Suppose {Πp } is a plane ¬eld. At
each point p we can de¬ne a line in Tp R3 (i.e. a line ¬eld) by looking at the set of all
vectors that are perpendicular to Πp . We can then de¬ne a 1-form ω by projecting
vectors onto these lines. So, in particular, if Vp is a vector in Πp then ω(Vp ) = 0.
Another way to say this is that the plane Πp is the set of all vectors which yield zero
when plugged into ω. As shorthand we write this set as Ker ω (“Ker” comes from
the word “Kernel”, a term from linear algebra). So all we are saying is that ω is a
1-form such that Πp = Ker ω. This is very convenient. To specify a plane ¬eld all
we have to do now is write down a 1-form!




Example 6.1. Suppose ω = dx. Then at each point p of R3 the vectors of
Tp R3 that yield zero when plugged into ω are all those in the dy-dz plane. Hence,
Ker ω is the plane ¬eld consisting of all of the dy-dz planes (one for every point
84 6. APPLICATIONS

of R3 ). It is obvious that this plane ¬eld is integrable; at each point p we just
have the tangent plane to the plane parallel to the y-z plane through p.

In the above example note that any 1-form that looks like f (x, y, z)dx de¬nes the
same plane ¬eld, as long as f is non-zero everywhere. So, knowing something about
a plane ¬eld (like the assumption that it is integrable) seems like it might not say
much about the 1-form ω, since so many di¬erent 1-forms give the same plane ¬eld.
Let™s investigate this further.
First, let™s see if there™s anything special about the derivative of a 1-form that
looks like ω = f (x, y, z)dx. This is easy: dω = ‚f dy § dx + ‚f dz § dx. Nothing too
‚y ‚z
special so far. How about combining this with ω? Let™s compute:
‚f ‚f
ω § dω = f (x, y, z)dx § dy § dx + dz § dx = 0
‚y ‚z
Now that™s special! In fact, recall our emphasis earlier that forms are coordinate
free. In other words, any computation one can perform with forms will give the same
answer regardless of what coordinates are chosen. The wonderful thing about folia-
tions is that near every point you can always choose coordinates so that your foliation
looks like planes parallel to the y-z plane. In other words, the above computation is
not as special as you might think:

Theorem 6.1. If Ker ω is an integrable plane ¬eld then ω § dω = 0 at every
point of R3 .

It should be noted that we have only chosen to work in R3 for ease of visualization.
There are higher dimensional de¬nitions of foliations and plane ¬elds. In general, if

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