We explore the relationship between the gradient, the curl, and the divergence of a vector field.

At this point in our study, we have many fundamental theorems. Let’s try to use them together, and see what we can discover.

1 A two-dimensional dream

So far we have two fundamental theorems of calculus for functions of two variables.

  • The fundamental theorem of line integrals:

    \[ \int _C \grad F \dotp \d \vec {p} = F(\vec {b}) -F(\vec {a}) \]

  • Green’s theorem:

    \[ \iint _R \curl \vec {F} \d A = \oint _{\partial R} \vec {F} \dotp \d \vec {p} \]

Wouldn’t it be cool if we could combine these theorems to make something like:

\[ \iint _R \curl \grad F \d A = \oint _{\partial R} \grad F \dotp \d \vec {p} = F(\vec {b}) - F(\vec {a}) \]

A double integral, equaling a single integral, equaling a difference! It’s an amazing idea, and while it is true, we shouldn’t get too excited. In fact each expression above is equal to zero. Let’s see why.

First note that both sides of the equation:

\[ \oint _{\partial R} \grad F \dotp \d \vec {p} = F(\vec {b}) - F(\vec {a}) \]

are zero. This is because, with a closed curve, \(\vec {a}=\vec {b}\), and so \(F(\vec {a}) = F(\vec {b})\). But this means that

\[ \iint _R \curl \grad F \d A = 0 \]

Note, this is totally independent of the region of the region \(R\) and the surface \(F\). The upshot: \(\curl \grad F = 0\) for all functions \(\vec {F}:\R ^2\to \R ^2\) with continuous second derivatives!

Let \(F:\R ^2\to \R \) be a function with continuous second derivatives. Which of the following make sense?
\(\grad F\) \(\curl F\) \(\divergence F\) \(\curl \grad F\) \(\curl \divergence F\) \(\divergence \grad F\) \(\divergence \curl F\)
Of the choices above that make sense, which must be equal to \(0\)?
\(\grad F\) \(\curl \grad F\) \(\divergence \grad F\)

Of course, you could poo-poo the work above and say “I already knew that! This is just the Clairaut gradient test!” Well, sure, but what we just presented is another reason that \(\curl \grad F = 0\). Moreover, this line of reasoning will lead us to new ideas too. Read-on young mathematician, there is not much further to go in this course.

2 A three-dimensional dream

Working in a similar way to how we worked above, let us recall the fundamental theorems of calculus for functions of three variables.

  • The fundamental theorem of line integrals:

    \[ \int _C \grad F \dotp \d \vec {p} = F(\vec {b}) - F(\vec {a}) \]

  • Stokes’ Theorem:

    \[ \iint _R \curl \vec {F}\dotp \uvec {n} \d S = \oint _{\partial R} \vec {F} \dotp \d \vec {p} \]

  • The Divergence Theorem:

    \[ \iiint _R \divergence \vec {F} \d V = \oiint _{\partial R} \vec {F}\dotp \uvec {n} \d S \]

Putting Stokes’ Theorem and the Divergence Theorem together, we find the beautiful expression:

\[ \iiint _R \divergence \left (\curl \vec {F}\right )\d V = \oiint _{\partial R} \left (\curl \vec {F}\right )\dotp \uvec {n} \d S = \oint _{\partial \partial R} \vec {F}\dotp \d \vec {p} \]

Again, what a fantastic idea, a triple integral, equaling a double integral, equaling a single integral! However, there is again a rub, the boundary of a closed curve is empty. A closed curve has no boundary! Hence:

\[ \oint _{\partial \partial R} \vec {F}\dotp \d \vec {p} = 0 \]

This fact, in turn means that all of the integrals above, including:

\[ \iiint _R \divergence \left (\curl \vec {F}\right )\d V = \oiint _{\partial R} \left (\curl \vec {F}\right )\dotp \uvec {n} \d S \]

Are all equal to to zero! We thus conclude:

  • \(\oiint _{\partial R} \left (\curl \vec {F}\right )\dotp \uvec {n} \d S=0\) tells us that the circulation along any closed surface is zero, and
  • \(\iiint _R \divergence \left (\curl \vec {F}\right )\d V = 0\) tells us that for any vector field \(\vec {F}:\R ^3\to \R ^3\), \(\divergence \left (\curl \vec {F}\right )=0\).

Let’s see what you can do now.

Let \(F:\R ^3\to \R \) be a function with continuous second derivatives and let \(\vec {G}:\R ^3\to \R ^3\) be a vector field. Which of the following make sense?
\(\curl F\) \(\curl \vec {G}\) \(\divergence F\) \(\curl \grad F\) \(\curl \divergence F\) \(\curl \divergence \vec {G}\) \(\divergence \grad F\) \(\divergence \left (\curl \vec {G}\right )\)
Of the choices above that make sense, which must be equal to \(0\) or \(\vec {0}\)?
\(\curl \vec {G}\) \(\curl \grad F\) \(\divergence \grad F\) \(\divergence \left (\curl \vec {G}\right )\)

3 The shape of things to come

Recalling that \(C^\infty (A,B)\) is the set of differentiable functions from \(A\) to \(B\) where all of the derivatives are continuous, we can make the following “chain” of derivatives:

From our work above (and from previous parts of this course) we will always have that \(\curl \grad F = 0\).

Since the Fundamental Theorem of Line Integrals (FTLI) is in some sense about “undoing” the gradient and Green’s Theorem is in some sense about “undoing” the scalar curl, we can also place these in the diagram:

Working in a similar way, we can make a chain of derivatives in three-dimensions as well:

From our work above, we always will have that \(\curl \grad F = \vec {0}\) and \(\divergence \left (\curl \vec {G}\right ) = 0\).

Since Stokes’ Theorem is in some sense about “undoing” the curl and the Divergence Theorem is in some sense about “undoing” the divergence, we can place these theorems into the diagram as well:

Now let’s get crazy. Let’s put all of our diagrams together into one big diagram:

Checkout the values for \(n\) in each of the \(C^\infty (\R ^m,\R ^n)\):

Here they are by themselves:

\[ \begin{array}{ccccccc} & & 1 & & 1 & & \\ & 1 & & 2 & & 1 & \\ 1 & & 3 & & 3 & & 1 \end{array} \]

These are rows of Pascal’s Triangle! For each row we find more fundamental theorems of calculus. Moreover, the triangle can be extended, for example:

In the fourth row, there are four(!) fundamental theorems of calculus. Moreover, this triangle can be extended infinitely! The details of this are beyond the scope of this course. Math folks who know this eventually start calling all of these fundamental theorems “ Stokes’ Theorem.”

Alas, our course has come to an end. However, it is really much more of a beginning, as there is so much more to know.