MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am a TA for a multivariable calculus class this semester. I have also TA'd this course a few times in the past. Every time I teach this course, I am never quite sure how I should present curl and divergence. This course follows Stewart's book and does not use differential forms; we only deal with vector fields (in $\mathbb{R}^3$ or $\mathbb{R}^2$). I know that div and curl and gradient are just the de Rham differential (of 2-forms, 1-forms, and 0-forms respectively) in disguise. I know that things like curl(gradient f) = 0 and div(curl F) = 0 are just rephrasings of $d^2 = 0$. However, these things are, understandably, quite mysterious to the students, especially the formula for curl, given by $\nabla \times \textbf{F}$, where $\nabla$ is the "vector field" $\langle \partial_x , \partial_y , \partial_z \rangle$. They always find the appearance of the determinant / cross product to be quite weird. And the determinant that you do is itself a bit weird, since its second row consists of differential operators. The students usually think of cross products as giving normal vectors, so they are lead to questions like: What does it mean for a vector field to be perpendicular to a "vector field" with differential operator components?! Incidentally, is the appearance of the "vector field" $\nabla = \langle \partial_x , \partial_y , \partial_z \rangle$ just some sort of coincidence, or is there some high-brow explanation for what it really is?

Is there a clear (it doesn't have to necessarily be 100% rigorous) way to "explain" the formula for curl to undergrad students, within the context of a multivariable calculus class that doesn't use differential forms?

I actually never quite worked out the curl formula myself in terms of fancier differential geometry language. I imagine it's: take a vector field (in $\mathbb{R}^3$), turn it into a 1-form using the standard Riemannian metric, take de Rham d of that to get a 2-form, take Hodge star of that using the standard orientation to get a 1-form, turn that into a vector field using the standard Riemannian metric. I imagine that the appearance of the determinant / cross product comes from the Hodge star. I imagine that one can work out divergence in the same way, and the reason why the formula for divergence is "simple" is because the Hodge star from 3-forms to 0-forms is simple. Is my thinking correct?

Stewart's book provides some comments about how to give curl and divergence a "physical" or "geometric" or "intuitive" interpretation; the former gives the axis about which the vector field is "rotating" at each point, the latter tells you how much the vector field is "flowing" in or out of each point. Is there some way to use these kinds of "physical" or "geometric" pictures to "prove" or explain curl(gradient f) = 0 and div(curl F) = 0? Is there some way to explain to undergrad students how the formulas for curl and div do in fact agree with the "physical" or "geometric" picture? Though such an explanation is perhaps less "mathematical", I would find an explanation of this sort satisfactory for my class.

Thanks in advance!

share|cite|improve this question
My advice ... at this level, strictly stick to the textbook. If a student comes to you outside of class and asks for more insight, go ahead. But any deviation from the text will likely cause far more confusion that it prevents! – Gerald Edgar Apr 19 '10 at 20:54
@Gerald: Thank you for your advice. I should say that I am certainly not going to try to teach my students about differential forms and the Hodge star. My main issue is just that every time curl comes up, there are inevitably some students who ask about where the "unnatural-looking" formula comes from. On the other hand, they usually don't ask such questions about gradient and divergence, because their formulas "look" natural to them. I don't want to tell them that the curl formula is just some magic formula that has these magical properties. – Kevin H. Lin Apr 19 '10 at 21:08
What I am really hoping for is some way to convey to them that curl is in fact as natural as gradient and divergence, despite initial appearances. – Kevin H. Lin Apr 19 '10 at 21:09
This has always been my intuitive picture of curl:… – Dan Piponi Apr 20 '10 at 1:52
I would like to add a comment which doesn't constitute an answer. I have always explained the definition of divergence and curl just as Qiaochu suggests, by starting a proof of Gauss's and Stoke's theorems, computing the flux or divergence integrals on small boxes and deriving the formulas for divergence and curl as a limit. This has the advantage that these two theorems, which are rarely explained or motivated in a calculus class, are essentially self-evident, if one is comfortable with some heuristics about the integral as summing up small contributions. – David Jordan Jul 14 '10 at 0:19

18 Answers 18

up vote 29 down vote accepted

To me, the explanation for the appearance of div, grad and curl in physical equations is in their invariance properties.

Physics undergrads are taught (aren't they?) Galileo's principle that physical laws should be invariant under inertial coordinate changes. So take a first-order differential operator $D$, mapping 3-vector fields to 3-vector fields. If it's to appear in any general physical equation, it must commute with with translations (and therefore have constant coefficients) and also with rotations. Just by considering rotations about the 3 coordinate axes, you can then check that $D$ is a multiple of curl.

If I want to devise a "physical" operator which has the same invariance property - and therefore equals curl, up to a factor - I'd try something like "the mean angular velocity of particles uniformly distributed on a very small sphere centred at $\mathbf{x}$, as they are carried along by the vector field." (This is manifestly invariant, but not manifestly a differential operator!)

[Here I should admit that, having occasionally tried, I've never convinced more than a fraction of a calculus class that it's possible to understand something in terms of the properties it satisfies rather than in terms of a formula. That's unsurprising, perhaps: it's not an obvious idea, and it's entirely absent from the standard textbooks.]

share|cite|improve this answer
Dear Tim: I'll have to check that argument for myself, but argh, that's beautiful!!! I feel frustrated that they don't seem teach this in undergrad multivariable calculus courses. This is certainly not in Stewart's book, and certainly this is the first time I've seen this myself. I hope that at least this is commonly taught in undergrad classical mechanics courses, but I never took a proper classical mechanics course myself. – Kevin H. Lin Apr 19 '10 at 23:39
Kevin, I'm glad if this is new to you. I admit that the argument itself may a bit fiddly for a mass-market calculus text (it reduces to showing that the cross product is characterized by bilinearity, rotation-invariance and scale), but the statement is clear enough. In its own terms, Stewart's text is solid enough, but he sometimes seems to me to present mathematics as a subject that fossilized some time in the Late Triassic...;) – Tim Perutz Apr 20 '10 at 1:56
Hmm, well, it's possible that someone showed this to me at some point, but I must have not been paying attention ;). Anyway, I feel that this explanation is so nice that it should be at least mentioned in these courses and books; it doesn't have to necessarily be worked out in explicit detail. And yeah, I agree with your feelings about Stewart, though to be fair I feel similarly about a lot of other calculus books as well. Some mention of more modern work would be helpful to counteract such perceptions. For instance, an introduction to the Galilean principle could be nicely followed ... – Kevin H. Lin Apr 20 '10 at 8:42
... up by, for instance, an introduction to some ideas from special relativity. (Ok, ok, ok, that's probably way too much to ask. But a mention would still be nice.) – Kevin H. Lin Apr 20 '10 at 8:45
I am sure you can work this out yourself, but just to note: div,grad, and curl are not diffeomorphism invariant, just invariant under isometries. This is of course due to the fact that $grad = \sharp\circ d$, $div = \delta \circ\flat$ and $curl = \sharp \circ d \circ * \circ \flat$, so the metric is involved heavily. This is also why it is somewhat more natural to work with d and forms instead of those operations on vectors in the context of differential topology. – Willie Wong Apr 20 '10 at 18:33

As far as explaining the formulas for div and curl, you should be able to do this starting with the definitions given in the Wikpedia articles by taking the corresponding integrals on rectangles and boxes. These definitions have fairly clear physical meanings, at least if your students are comfortable with line and surface integrals.

As far as explaining d^2 = 0: curl(grad f) = 0 because the line integral of a gradient about small circles is zero, so gradients can't curl. (In other words, if a vector field has nonzero curl at some p, you wouldn't be able to define a consistent potential about some small closed contour around p.) And div(curl F) = 0 because the surface integral of a curl about small spheres is zero, so curls can't diverge (that is, flow). These interpretations get used all the time in applications of Stokes' theorem to physics.

share|cite|improve this answer
Ah, awesome, thanks. I'll keep this in mind when we get to Stokes' theorem in the class. – Kevin H. Lin Apr 19 '10 at 20:29
When I was younger, I envisioned curl(grad f) = 0 as the mathematical version of the statement "you can't go uphill both ways to school". – Ryan Thorngren Apr 20 '10 at 0:38
2 – Asaf Karagila Oct 15 '12 at 11:53

I have taught multivariable calculus exactly once, to engineering students at Concordia University in Montreal. I found the course to be replete with expository challenges like the one you mention: namely, to explain what is going on with the various concepts of vector analysis in something like geometric terms, but of course without introducing anything like differential forms. [Conversely, it is possible to know Stokes' Theorem in the form $\int_{\partial M} \omega = \int_M d \omega$ and still not have any insight into flux, divergence and other such geometric and physical notions. I myself spent about 10 years in this position.]

I thought hard and often found explanations that were much more satisfactory than the textbook, which was amazingly laconic. Or rather, I found explanations which were much more satisfactory to me. The students had a lot of trouble conceptualizing the material, to the extent that my lectures almost certainly would have been more successful if I hadn't tried to give geometric explanations and intuition but simply concentrated on the problems. Thus Gerald Edgar's comment rings true to me. But let me proceed on the happier premise that you want to give more motivation to the bright student who approaches you outside of class.

One thing which was useful for me was to read the "physical explanations" that the book sometimes gave and try to make some kind of mathematical sense of them. For context, I should say that I have never taken any physics classes at the university level and that I have rarely if ever met a mathematician who has less physics background than I. Moreover, when I took introductory multivariable calculus myself (around the age of 17), I found the physical explanations to be so vague and so far away from the mathematics as to be laughable. For instance, the geometric intuition for a curl involved some story about a paddlewheel.

So when I taught the class, I tried to make some mathematical sense out of the names "incompressible" (zero divergence) and "irrotational" (zero curl), and to my surprise and delight I found that this was actually rather straightforward once I stopped to think about it.

Let me also tell you about my one "innovation" in the course (I am sure it will be commonplace to many of the mathematicians here). It seems strange that there are two versions of Stokes' theorem in three-dimensional space (one of them is called Stokes' theorem and one of them is called Gauss' Theorem or -- better! -- the Divergence Theorem) whereas in the plane there is only Green's Theorem. Stokes' Theorem is about curl, whereas Gauss' theorem is about divergence. What about Green's Theorem?

The answer is that Green's theorem has a version for divergence -- i.e., a flux version involving normal line integrals -- and a version for curl -- a circulation version involving tangent line integrals -- but these two versions are formally equivalent. Indeed, one gets from one to the other by applying the "turning" operators L and R: L applied to a planar vector field rotates each vector 90 degrees counterclockwise, and R is the inverse operator. Then (with the convention that the curl of a planar vector field should always be pointing in the vertical direction, so we can make a scalar function out of it)

curl(L(F)) = div(F)

and by making this formal substitution one gets from one version of Green's Theorem to the other.


share|cite|improve this answer
A related problem (for math dept administrators)... How desirable/avoidable is it for physics & engineering students to be taught div,grad,curl and all that by an instructor who, himself, has never taken a physics course ?? – Gerald Edgar Apr 20 '10 at 14:20
Gerald, your remark seems rather harsh. I take my teaching responsibilities seriously, and I think that a PhD in mathematics is sufficient qualification for anyone to pick up concepts from freshman/sophomore physics as needed. I would certainly not be pleased to receive such an inquiry from my department head. – Pete L. Clark Apr 20 '10 at 15:26
Yes, in all of your examples it could only improve the course if the instructor knew those things. My point is that asking someone with a PhD in mathematics what courses they took as an undergraduate is both somewhat insulting and against the point of getting a PhD. I have never taken a course on Shimura varieties, but I taught one. If you wanted to teach an undergraduate course on algebraic number theory (or whatever is outside of your student training), I would trust you to prepare yourself properly to do so. – Pete L. Clark Apr 20 '10 at 18:01
When did taking a course in a subject become necessary OR sufficient for understanding a subject? – B. Bischof Apr 21 '10 at 13:27
@Gerald: Somewhere between your first and second comments you switched from formal qualifications (taking a course) to competence. IMO, it's reasonable to demand some degree of competence in physics of an instructor in "calculus for physicists and engineers" but it is unreasonable to demand that the instructor have taken a physics course. The latter kind of bean-counting is precisely the kind of thing that gives administrators a bad name. – Timothy Chow Oct 22 '10 at 23:12

This is perhaps a crude (and certainly non-rigorous) explanation, but it's always how I've thought of motivating it.

Let $F = (F_1, F_2, F_3)$ denote a vector field in $\mathbb{R}^3$, and write $\text{curl}\ F = (G_1, G_2, G_3)$. We would like a situation where $G_1$ describes the "instantaneous" rotation of $F$ about the $x$-axis, $G_2$ the rotation about the $y$-axis, and $G_3$ the rotation about the $z$-axis.

So let's think of vector fields which do just that. Three simple (linear!) ones which come to mind are $$H_1(x,y,z) = (0, -z, y)$$ $$H_2(x,y,z) = (z, 0, -x)$$ $$H_3(x,y,z) = (-y, x, 0)$$ So in order to measure how much $F$ rotates about, say, the $z$-axis, it makes sense to look at something that compares how similar $F$ is to $H_3$. The dot product $F(x,y,z) \cdot H_3(x,y,z)$ seems reasonable, which is precisely $-yF_1(x,y,z) + xF_2(x,y,z).$

This suggests that defining $$G_1(x,y,z) \approx -zF_2(x,y,z) + yF_3(x,y,z)$$ $$G_2(x,y,z) \approx zF_1(x,y,z) - xF_3(x,y,z)$$ $$G_3(x,y,z) \approx -yF_1(x,y,z) + xF_2(x,y,z)$$ might give something close to what we want. But this is a very crude way to measure "instantaneous" rotation -- in fact, one might say it's a sort of linear approximation. Thus, we are led to replacing the linear terms with their corresponding derivations: $$G_1(x,y,z) = -\frac{\partial}{\partial z}F_2 + \frac{\partial}{\partial y}F_3$$ $$G_2(x,y,z) = \frac{\partial}{\partial z}F_1 - \frac{\partial}{\partial x}F_3$$ $$G_3(x,y,z) = -\frac{\partial}{\partial y}F_1 + \frac{\partial}{\partial x}F_2,$$ which is precisely the curl.

This heuristic also works with divergence, but instead consider $(H_1, H_2, H_3) = (x,y,z)$.

share|cite|improve this answer
Oh, this is really nice! – Kevin H. Lin Jul 14 '10 at 0:36
Thanks! I like it also because it reiterates the idea that derivatives, differential operators, etc. are good linear approximations (which wasn't emphazised when I took Calc 3). – Jesse Madnick Jul 14 '10 at 0:45

Depending (*) on the underlying degree of analyticity (**) in your calculus course, it might be just as well to start with the Stokes theorem, stating it as an existence and uniqueness theorem:

Theorem (Stokes) Given a differentiable vector field $X$ on a region $U$ of $\mathbb{R}^3$ there is a unique continuous vector field $\operatorname{curl} X$ such that for any regularly parametrized surface $(u,v):D^2\rightarrow U$ with normal field $\hat{\mathbf{n}}$ and boundary tangent field $\mathbf{s}$, the integrals $$\iint (\operatorname{curl} X)\cdot \hat{\mathbf{n}} dA $$ and $$ \oint X \cdot \mathbf{s}\ dt $$ are equal

From there you can proceed by deducing properties of $\operatorname{curl} X$ sufficient to give its formula in coordinates and at the same time prove the theorem.

Note for instance that even stated only as an existence theorem, it already says there's a sufficient local criterion for local integrability of a vector field; the actual formula for curl then tells you what the criterion is.

This style of approach also gives you a quick proof that $\operatorname{div}\operatorname{curl}(X)=0$, because a sphere can be regularly parametrized by a disk such that $\mathbf{s} \equiv 0$.

(*) Here I mean roughly that if you show them sufficient variations of the mean value theorem to prove that itterated derivatives commute when they're continuous, then it should be feasible to give this construction with comparable rigor.

(**) none of these words used here with any standard mathematical sense.

share|cite|improve this answer
Thanks. Yeah, I think your the idea of your proposal is essentially contained in the wikipedia links in Qiaochu's answer. (Sigh -- another example of me not first looking things up on wikipedia before asking on MO -- sorry.) – Kevin H. Lin Apr 19 '10 at 21:14
oh! me too, for not thoroughly investigating what other people already said... – some guy on the street Apr 19 '10 at 21:17
This is exactly how I learned about div and curl in my physics course on electromagnetic theory, where they talk about the integral form of Maxwell's equations (which we call Stokes' theorem) and take a limit to derive the differential form of Maxwell's equations (containing the div and curl). Although I generally do not like the way physicists present mathematics, this is one case, where I think the physicists do it better. – Deane Yang Apr 20 '10 at 0:05
An old officemate of mine tried to teach his multivariate calc course this way (integrals first, then the derivative operators). The students who were taking simultaneously a physics course which "counted on div,grad,curl being taught first" (of course the physics department never bothered to communicate that fact to the maths department) were understandably confused and upset. So just as a protocol I suggest checking with departmental administrators before changing the order of the course too much! – Willie Wong Apr 20 '10 at 18:37

This could be a comment on Qiaochu Yuan's answer but I don't have enough reputation.

A book that takes the approach mentioned on those wikipedia pages (defining divergence as flux density and curl as the vector in the direction of greatest circulation density ) is Multivariable Calculus by McCallum, Hughes-Hallett et. al.

Another resource worth looking at is Bridging the Vector Calculus Gap.

share|cite|improve this answer

When I did this course myself I was deeply confused and distressed by it all. I know that everyone learns differently but when we are shown all these mnemonics for remembering formula that treat partials as if there were really numbers - it's not so bad, until they start getting used in proofs.. then it drags up all these memories about people saying "well it's not really a fraction but lets treat it that way anyway". You can pass this class (with a very high mark) by memorizing these formula and I suppose that is all that really matters for beginner classes like that (being able to solve lots of problems, without necessarily knowing why or what) but it can be a bit stomach churning and unpleasant to sit through a semester of not having a clue what any of this means while still getting all the right answers. There is a terrible sense of being lied to, when people try to dumb down things in the hope that it makes it simpler and easier to learn. It's only when I found a book on differential forms (which unified all these different concepts in one of the "applications" chapters) that I started to get the impression this was real mathematics and not just a strange act of going through the motions of writing lots of integral signs and so on. Although I do appreciate the remark in the preface of one of the many text books I read cover to cover in a worried haste to try and make sense of all this notational juggling, which said roughly that having illustrative notations like this fuels the intuition (and gave an example of that curl formula coming from the visual that the determinant had two equal rows in it), I didn't personally find this reassuring. Maybe there is value in teaching it this way to scare people into studying hard for it but I don't think I personally got anything out those months of work myself except for the vague geometric intuitions about div, grad and curl (which you could explain in a day by showing a video). If people learned about differential forms earlier on (rather than being told "it's what Leibniz did so we do to") maybe this course could be taught by developing the abstract setting a bit more and then specializing it down for the 3 dimensional case - which you could then get a got grip on by applying it to physics problems. I've not taught any class myself so I just wanted to give an account of what it can feel like sitting (or being dragged through backwards) this type of class.

share|cite|improve this answer
I almost feel like the exterior derivative definition of curl is kind of like a mneumonic - all you remember is that differential forms are anti-commutative, and then you basic just work out the curl formula from that! – David Corwin Jul 14 '10 at 0:23
I agree wholeheartedly -- despite having the highest score in my vector calculus course, I had absolutely no idea what any of these things meant. I even ended up failing a physics class after finding myself completely unable to make sense of the divs, grads, and curls. Several years later, while trying to get a feeling for homological algebra, I picked up a book talking about de Rham cohomology since I'd heard that it was a good source of practical examples. By the time I was six or seven pages in, I suddenly understood everything I'd spent years struggling with in completely futility. – Daniel McLaury Apr 20 '12 at 4:25
Soon, I was able to learn what had been a full year's worth of material in a few weeks. Whenever I bring this up in mathematical company, I'll always hear the same thing -- "Yeah, me too. I didn't understand any of that stuff until I learned differential geometry." – Daniel McLaury Apr 20 '12 at 4:28

I don't know that I have anything new to add. I've taught vector calculus a few times, and I have to admit I really enjoy it. The last few times I actually used a combination of traditional vector notation and differential forms, and it seemed to work OK (i.e. it wasn't a complete disaster). I'm linking my recently revised notes. Please keep in mind they were written for undergraduate science majors, so the standards probably fall short of what most people here might like.

share|cite|improve this answer
Your notes look very nice to me. I would say they are towards the high end of the spectrum of what could be done in an American sophomore level multivariable calculus course. – Pete L. Clark Apr 21 '10 at 19:15
Thanks. I threw in a lot of things that could or should not teach in such a class. But I was hoping that some students would keep reading. – Donu Arapura Apr 21 '10 at 19:47

This was originally a comment that got too big. Since I wont address the real heart of your question, it is CW. I hope this is ok :/

I am also teaching Multivar out of Stewart this semester. As it has been suggested, I stick fairly close to the book, even working through some of the same examples he does. I focus very hard on motivating the ideas from the ground up.

For instance when talking about curvature, I made the students try to define curvature for themselves. Then proceeded to find little issues with their definitions, until we arrived at something that was pretty close to the standard defn(with lots of urging).

Just yesterday Terry Tao posted a link to this video;

which talks about this exact style of teaching. I think that it works very well for Calc 3, where many of the students are pretty decent at math to get to that point. In so far as specific examples and motivation, I really enjoy the notes by Oliver Knill;

There are some nice diagrams, examples, and explanations of pretty much everything in Calc 3. In particular, he gives real physical applications of the ideas, and at the end he gives a "calc beyond calc" intro.

share|cite|improve this answer

There's one simple answer, which I'm very surprised hasn't been mentioned. Use the book Div, Grad, Curl, and All That.

share|cite|improve this answer

This is a good question and there are already a lot of good answers. Why add one more answer? Because there is a pedagogic option that nicely synthesizes these good questions and answers.

(1) The question recognizes undergraduate students exposed to div, grad, curl, etc. commonly ask questions whose answers are associated to higher-level concepts like one-forms, Hodge duals, etc.

(2) Numerous comments and answers recognize that wading into these advanced topics will slow, confuse, discourage, and distress many undergraduate students. How then to proceed?

(3) A reasonable response is to refer that subset of students (generally a minority) who are willing to work (for no academic credit) toward a broader understanding, to a (free) on-line book by William L. Burke titled Div, Grad, Curl are Dead.

Note: do not confuse Div, Grad, Curl are Dead with the above-recommended book Div, Grad, Curl and All That: the former is modern and polemic, while the latter is traditional and tutorial. Students who like either one generally will not like the other one; it is useful therefore to point students toward both books.

The on-line text of Div, Grad, Curl are Dead is an uncorrected publisher's proof because sadly, Prof. Burke was killed in an accident before the final corrections were done. So the Burke's proof pages have to be read carefully and critically, imperfections and all, this in itself is a good practice for thoughtful students.

Some of Burke's lively prose:

I am going to include some basic facts on linear algebra, multilinear algebra, affine algebra, and multi-affine algebra. Actually I would rather call these linear geometra, etc., but I follow the historical use here. You may have taken a course on linear algebra. This to repair the omissions of such a course, which now is typically only a course on matrix manipulation.

Another example is:

Mathematician: When do you guys [scientists and engineers] treat dual spaces in linear algebra?
Scientist / Engineer: We don't.
Mathematician: What! How can that be?

Burke's lively exposition supplies in abundance what many undergraduates crave: a dramatic narrative about why the geometric aspects of differential calculus are useful and important ... and a dawning realization that undergraduate mathematics is just the preliminary chapter of a wonderful story.

Bill Thurston's Foreword to Mircea Pitici's recent book The Best Writing on Mathematics: 2010 makes this same point, and is recommended to undergraduates who ask "How can two mathematics texts on the same topic be so very different?"

For many undergraduate students, Div, Grad, Curl are Dead will be the wrong textbook. But for those students who ask tough questions and refuse to accept glib answers, it's an excellent textook that gives undergraduate students explicit permission—indeed, seduces them—into reading more deeply.

share|cite|improve this answer
That book looks nice. Thanks for your vivid advertisement :-) – Kevin H. Lin Feb 16 '11 at 16:48
Thanks ... I added another juicy quote (Burke's text has many such), also a link to a recent essay by William Thurston. That mathematical writing can be lively is a welcome revelation to many undergraduates. – John Sidles Feb 16 '11 at 17:45
Well, I for one have just discovered that I like both books! – Toby Bartels Apr 4 '11 at 1:04

One way of motivating the derivative definition of curl (or rather scalar curl in 2D) is to prove a "Mean Value Theorem" for Rectangles. Suppose that $\mathbf{F} = (M, 0)$ is a vector field defined on the rectangle $R = [a,b] \times [c,d]$. Write the integral $\frac{1}{\operatorname{area}(R)}\int_{\partial R} \mathbf{F} \cdot d\mathbf{s}$ as $\frac{-1}{b-a}\int_a^b \frac{M(t,d) - M(t,c)}{d-c}$. Apply the MVT for integrals and then the MVT for derivatives to conclude that there is a point $\mathbf{x} \in R$ such that $$\frac{1}{\operatorname{area}(R)}\int_{\partial R} \mathbf{F} \cdot d\mathbf{s}= -\frac{\partial M}{\partial y}(\mathbf{x}).$$ Do the same for vector fields of the form $(0, N)$. Then if $\mathbf{F}$ is C$^1$ and if $R_n$ is a sequence of rectangles with horizontal and vertical sides converging to a point $\mathbf{a}$, then we obtain: $$\lim_{n \to \infty} \frac{1}{\operatorname{area}(R_n)}\int_{\partial R_n} \mathbf{F} \cdot d\mathbf{s} = \frac{\partial N}{\partial y}(\mathbf{a}) - \frac{\partial M}{\partial x}(\mathbf{a}).$$ This is enough to explain what curl is measuring, without using Green's theorem.

Some students and I in a recent paper use this result to give a rather more intuitive proof of Green's theorem than those usually found in vector calc texts. See

share|cite|improve this answer

Note: I wrote this when the title was still "What is the divergence? What is the curl?"

A nice geometric interpretation of the divergence is that it measures the rate of expansion of a fixed volume in the flow defined by the vector field. There is a very concrete way to see this by comparing the volume of an small cube to the volume of a parallelepiped given by considering where the corners of the cube are dragged by the flow in an infinitesimal length of time (it is easiest to work out the analogous case of a square with corners (x,y), (x+dx,y), (x,y+dy), (x+dx,y+dy) in the plane). I imagine the fact that the determinant measures volume can be used to explain the presence of the cross product. A more sophisticated (and conceptual) way to prove this fact is to note that the divergence can be defined on any manifold with a volume form as the Lie derivative of the volume form (contract the volume form with the vector field and then take the exterior derivative).

I have also seen a result generalizing the curl to arbitrary dimension by noting that we can identify 2-forms with skew-symmetric matrices (elements of the Lie algebra of SO(n)) if we have an inner product. I'd be curious to hear how exponentiating this infinitesimal rotation relates to integrating the vector field.

share|cite|improve this answer

I agree that it's very confusing that curl is again a vector field, and that there's this wacky determinant. I recommend doing the 2-d version, and "discovering" that curl is more naturally thought of as scalar-valued there, but of course you can think of it as that scalar field times the constant vector orthogonal to the plane.

Basically, I think it's reasonable to tell people that curl is measuring curl, and that it should really be fed two vectors. In the plane, you don't have to tell it the two vectors, and in 3-space you can cheat differently by feeding it the third, orthogonal vector.

share|cite|improve this answer

Here as another way to define curl:

1) Curl is a vector describing "instantaneous rotation". The line integral over a gradient vector field is zero on any closed curve, so whatever instantaneous rotation is, it should have the property $curl(\nabla f) = 0$.

2) Every symmetric $3\times 3$ matrix is $D^2f$ for some homogeneous polynomial $f$ in three variables. Thus a vector field $F:\mathbb{R}^3 \to \mathbb{R}^3$ is locally well-approximated by an irrotational gradient vector field when $DF$ is symmetric.

3) The function $g(\vec v) = \vec v \times \vec p$, where $\vec p$ is fixed but arbitrary, is both the general rotational velocity field about an axis $\vec p$ through the origin, with angular velocity $||\vec p ||$, as well as the linear function described by an arbitrary antisymmetric matrix. When $DF$ is antisymmetric, $F$ is locally well-approximated by a rotational velocity field.

4) $DF = \frac{(DF + DF^T) + (DF - DF^T)}{2}$. The first summand is irrotational, the second, a rotational velocity field. Define the instantaneous rotation field for F to be the linear function given by $(DF - DF^T)$ (ignoring the factor of $2$). The vector $\vec p$ which gives us the corresponding function $g$ is easily determined from the coefficients of $(DF - DF^T)$, and is precisely $curl(F)$.

share|cite|improve this answer
"derivative of the gradient vector field for [a function]": why not just call it the Hessian? – Willie Wong Oct 22 '10 at 19:06
Edited to $D^2f$. – Tobias Hagge Oct 26 '10 at 5:07

I find that a reference to something physical really helps. In physics classes, references to calculus are inherent to the process. Similarly, calculus teachers can accelerate understanding by incorporating references to physics during lectures. To explain divergence, make reference to light from a bulb. Flux and divergence are simple to picture here, and it really helps the student. Similarly, for curl, hook up twp soda bottles and make a tornado. Emphasize the concept of source and sink. The calculus becomes intuitive after that.

share|cite|improve this answer

Well, $\nabla\times\nabla f=0$ if $f \in C^2$ can be proved by contradiction. Suppose the curl wasn't 0: then, if you chose a contour integral around the area with non-zero curl, it will be different than if you just stayed at that point. This doesn't make sense, because, given that the gradient describes change in the function, line integrals on it should be path invariant. You can make this visual with some arrows on a chalk board.

(You could explain that $\oint_a^b \nabla F\cdot dx = F(a)-F(b)$. Also, if you've explained differential forms, you can explain this with exact forms...)

And I'm rather pleased with the line ``Divergence Theorem is a grandiose way of saying that the amount of water than comes out of the tap is the amount that overflows the bath tub''...

As for $\nabla$, isn't it just a vector of unbound operators? ($(\mathbb{R}\to\mathbb{R})^3$?) And the multiplication/differentiation analogy seems not only to go back quite a ways, but useful given that it forms some sort of nice algebraic structure relative to addition... In fact, given continuity, it very nearly forms a field.

In any case, I'm a high school student and have explained this to one of my freinds in this manner. It seems to work fairly well in my one point data set...

share|cite|improve this answer

Many great answers have already been provided to your earlier questions; this one is to speculate on the final one about the history of the operators. I am almost certain the operators (in various guises) were first used in fluid/continuum/elastic mechanics, rather than in electromagnetism, based purely on knowledge that the (mathematical) development of the former predates the latter by almost a century. Consider that Euler wrote down his eponymous equation in fluid dynamics in 1757, some half century before Oersted's discovery that electricity and magnetism are connected. So at the very least the Div and Grad operators predates electromagnetism.

The Div operator probably first appeared in consideration of the continuity equations, which roughly says $$ \partial_t \mbox{density} + \nabla\cdot \mbox{flux} = \mbox{source} $$ but I am not certain about the history here. (I've made this answer a CW so people can add references.)

As to the Curl, I am much less certain. The name (also its alternate name the Rotor) suggests to me that, like the divergence, it arises from considering fluid flow, and in fact, the vorticity ($\omega = \nabla\times v$) is immensely useful for the study of fluids. Since in the (modern re-)derivative of Stoke's law it is used, and since the existence of the Kelvin-Stokes theorem, it is likely that George Stokes, at least, made use of the expression which now we call the Curl.

Of course, the name itself is unreliable as an arbiter, since it is most likely a modern invention. See, for example, this article in the Monthly. You should also take all of the above with a grain of salt: the operators were almost certainly not so named, nor their modern symbols used, in the 18th and early 19th centuries. So the most you can say is that some equivalent mathematical expression was found useful way back when.

share|cite|improve this answer
And... great. Just as I typed up an answer to your final question, you deleted it! – Willie Wong Apr 21 '10 at 18:44
Well, I felt that the history part was kind of distracting to the rest of the question, so I removed it... – Kevin H. Lin Apr 21 '10 at 19:11
Sorry... Maybe someone (you?) can ask about this in a new question? :) – Kevin H. Lin Apr 21 '10 at 19:12
Nah, if this weren't community wiki, I would've just deleted it. Now I am just not sure what to do about the response. – Willie Wong Apr 21 '10 at 22:35
I don't know how it was originally, but in Maxwell's day, the divergence was called the ‘convergence’ and measured with the opposite sign. When Gibbs and Heaviside pushed using 3-vectors with the dot and cross products in place of Hamilton's quaternions, they also pushed for using divergence in place of convergence. (Divergence comes directly from a dot product, as we all know from their notation which is still used today, while the quaternionic product has this term with the opposite sign.) – Toby Bartels Apr 4 '11 at 0:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.