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?



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> 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?
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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!



How should one present curl and divergence in an undergraduate multivariable calculus class?
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.]


