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I want to try making some calculation-less questions about vector calculus identities that are solely based upon picture diagrams of vector fields, or fields that could be sketched out by hand. The purpose of this is to introduce the ideas so that students aren't just calculating and using formulas without thinking about what they mean.

It seems that the most common analogies to these vector calculus identities are the following:

  • Gradient is the steepest path up a mountain at a location.
  • Divergence is the way dust would disperse (or contract) at a specific location. Usually questions involve trying to find out if the field acts like a source, a sink, or "zero" (or maybe there's a better description for zero divergence?)
  • Curl is how a paddle wheel (at a location) would rotate. Usually questions involve comparing vectors contributing to clockwise and counterclockwise motions of the wheel, and seeing which is net positive.

I can't remember an analogy for the Laplacian but it seems that it could be understood as the "average changing".

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    $\begingroup$ I don’t remember hearing of one for the Laplacian, but it’s something like the tension if the surface were a rubber sheet: If it’s negative the function value would decrease if the surface were relaxed and the opposite if it were positive. (Think of the diffusion/heat equation.) $\endgroup$ – user1027 Jul 5 at 3:46
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    $\begingroup$ I do this kind of thing in this book: lightandmatter.com/mod . You may also want to look at Purcell, Electricity and Magnetism. $\endgroup$ – Ben Crowell Jul 5 at 18:08
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    $\begingroup$ The Laplacian is the combination of gradient and divergence so it seems whatever pictures you promote for those ought to combine to form the picture for the Laplacian; $\nabla^2 f = \nabla \cdot \nabla f$ (ignoring the sign-convention issue, some folks throw in a minus for the Laplacian...) $\endgroup$ – James S. Cook Jul 6 at 2:20
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    $\begingroup$ This is a physics education question! $\endgroup$ – Alexander Woo Jul 7 at 1:17
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    $\begingroup$ @AlexanderWoo vector calculus is a Math topic $\endgroup$ – David Jul 13 at 12:01
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I motivate these grad, curl, and div for myself as being the things which would make the respective version of Stokes' theorem true infinitesimally. You can read about this interpretation in the context of differential forms here:

https://math.stackexchange.com/a/614473/34287

Given that interpretation, we can see

  1. The gradient at a point $p$ is the vector which represents the map $df|_p$. In other words $df|_p(\vec{v}) = \nabla f|_p \cdot \vec{v}$. You can see that the dot product is maximized when $\vec{v} = \nabla f|_p$, which justifies the "steepest slope" interpretation.
  2. Given a vector field $F$, convert it into a covector field $\omega$ via $\omega(\vec{v}) = F \cdot \vec{v}$. Then the integral of the differential form $\omega$ along a curve is the same as the "line integral" of the vector field $F$. Following my answer above, you can define $d\omega$ in terms of the line integral around little parallelograms. $d\omega$ is a 2-form, but it can be converted into a vector field by a similar representation trick (this one only works in $\mathbb{R}^3$): define $\textrm{curl}(F)$ as the vector for which $\textrm{curl}(F) \cdot (\vec{v} \times \vec{w}) = d\omega(\vec{v},\vec{w})$. The RHS is maximized when the line integral around a small parallelogram defined by $\vec{v}$ and $\vec{w}$ is maximized and then the LHS would require that $\textrm{curl}(F)$ be perpendicular to that parallelogram, and pointing in the direction given by the right hand rule so as to make the integral positive.
  3. Given a vector field $F$, convert it into a $2$-form $\omega$ via $\omega(\vec{v},\vec{w}) = \textrm{Det}(\vec{v},\vec{w},\vec{F}) = \vec{F} \cdot ( \vec{v} \times \vec{w})$. The integral of $\omega$ over a surface is the same as the surface integral of $F$. We can define $d\omega(\vec{v},\vec{w},\vec{b})$ in terms of the integral of $\omega$ around the boundary of the small oriented parallelepiped defined by the three vectors. This is the same as the surface integral of $F$ over that same surface. This is a top level form in $\mathbb{R}^3$, so it is a multiple of the volume form, i.e. $\textrm{div}(F)\textrm{Det}(\vec{v},\vec{w},\vec{b}) = d\omega(\vec{v},\vec{w},\vec{b})$. So the sign of the divergence can be "seen" according to whether most of the vectors $F$ are pointing "into" or "out of" a small parallelepiped based at $p$.

You can replace the parallelogram with a circle and the parallelepiped with a sphere to obtain a more "symmetric" representation, but you loose the easy connection with the differential forms. Also, it is much easier to decompose surfaces into small parallelograms (and solids into small parallelepipeds), which leads to the global Stokes' theorems being just "telescoping" sums, with cancellation of all interior terms.

I think the real difficulty with motivating these things comes from the "unnatural" conversion of differential forms into scalar and vector fields, in ways which only work in $\mathbb{R}^3$. They are somewhat arbitrary. They are also unnecessary, as any work you want to do can be done with the differential forms themselves.

These will let you draw pictures though! In each case, you have to imagine many different "test" domains of integration, and try to figure out which domain will maximize the integral. The grad, curl, and div connect to those maximizing domains.

EDIT: I forgot to mention the laplacian, which doesn't really fit into this framework. The interpretation you link to seems pretty much optimal though.

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    $\begingroup$ The Laplacian is formed by a combination of the exterior derivative and the coderivative which takes $p$ to $p-1$ forms. When I translated the tensor equations in Griffith's Electromagnetics text ( very popular text last couple decades) two of Maxwell's equations are formed from $dF=0$ and the other two from $\delta F = \star J$ where $J$ is the 4-current. It seems like the equations which Physicists like to put in books involve more than just the exterior derivative. Great answer. $\endgroup$ – James S. Cook Jul 28 at 3:26

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