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Is it possible to write the curl in terms of the infinitesimal rotation tensor? Basically, we can write the curl as a matrix operator

$$ curl=\begin{bmatrix} 0 & -\partial z & \partial y\\\partial z & 0 & -\partial x\\-\partial y & \partial x & 0\end{bmatrix} $$ and we can write the infinitesimal rotation tensor as a matrix operator (modulo a 1/2 constant) $$ \nabla-\nabla^T = \begin{bmatrix} 0 & -(\partial x-\partial y) & \partial z-\partial x\\ \partial x-\partial y & 0 & -(\partial y-\partial z)\\ -(\partial z-\partial x) & \partial y-\partial z & 0 \end{bmatrix}. $$ The axial vector to the infinitesimal rotation tensor is $$ \begin{bmatrix} \partial y-\partial z\\ \partial z-\partial x\\ \partial x-\partial y \end{bmatrix}, $$ which looks kind of like the curl, except that this seems kind of sloppy since we have a vector with a bunch of differential operators inside of it with no clear way on how to apply it.

As such, again, is there a way to write the curl in terms of the infinitesimal rotation tensor?

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  • $\begingroup$ Probably - 'In the past physicists showed no hesitation in employing infinitesimal methods,' - The Continuous and the Infinitesimal, John L Bell. But I haven't seen this problem addressed in his books; you may have to work it out from scratch. $\endgroup$ – user117644 Jul 25 '14 at 16:35
  • $\begingroup$ I suspect that the right way to proceed is by using index notation rather than explicit matrices (e.g. $(\nabla\times \mathbf{A})_k=\epsilon_{ijk} \partial_i A_j.)$ That gets away from the sloppiness you're worried about. $\endgroup$ – Semiclassical Jul 25 '14 at 16:40
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My notation was screwed up and fixing it solves the problem. Basically, my definition above for the infinitesimal rotation tensor was incorrect because the rotation tensor applied to a vector is a matrix. Here's how it works.

The infinitesimal rotation tensor is really the skew symmetric part of the Fréchet derivative. Recall, $$ f^\prime(x)= \begin{bmatrix} \frac{\partial}{\partial x} f_1 & \frac{\partial}{\partial y} f_1 & \frac{\partial}{\partial z} f_1\\ \frac{\partial}{\partial x} f_2 & \frac{\partial}{\partial y} f_2 & \frac{\partial}{\partial z} f_2\\ \frac{\partial}{\partial x} f_3 & \frac{\partial}{\partial y} f_3 & \frac{\partial}{\partial z} f_3\\ \end{bmatrix}. $$ Hence, the skew-symmetric part (modulo a 1/2) of this operator is: $$ f^\prime(x)= \begin{bmatrix} 0 & -\left(\frac{\partial}{\partial x} f_2 - \frac{\partial}{\partial y} f_1\right) & \frac{\partial}{\partial z} f_1-\frac{\partial}{\partial x} f_3\\ \frac{\partial}{\partial x} f_2 - \frac{\partial}{\partial y} f_1 & 0 & -\left(\frac{\partial}{\partial y} f_3 -\frac{\partial}{\partial z} f_2\right)\\ -\left(\frac{\partial}{\partial z} f_1-\frac{\partial}{\partial x} f_3\right) & \frac{\partial}{\partial y} f_3 -\frac{\partial}{\partial z} f_2 & 0\\ \end{bmatrix}. $$ That means that the axial vector corresponding to this operator is: $$ \begin{bmatrix} \frac{\partial}{\partial y} f_3 -\frac{\partial}{\partial z} f_2\\ \frac{\partial}{\partial z} f_1-\frac{\partial}{\partial x} f_3\\ \frac{\partial}{\partial x} f_2 - \frac{\partial}{\partial y} f_1 \end{bmatrix}, $$ which is precisely the curl. Hence, $$ curl(f) = axial(f^\prime(x)-f^\prime(x)^*) = axial(infrot(f)). $$ Gurtin actually states this on page 32 of his 1981 book An Introduction to Continuum Mechanics.

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  • $\begingroup$ And there should probably be (x,y,z) everywhere since everything is evaluated at a point, but this should be close enough. $\endgroup$ – wyer33 Jul 25 '14 at 18:01
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I'm making this CW as it's not exactly an answer, but, this is how some physicists define curl and divergence and it does almost immediately connect curl to circulation and divergence to change in flux. Probably semiclassical's comment is most useful towards the path you begin to walk.

We usually see definitions for curl and divergence which were based in Cartesian coordinates. However, some authors actually use the identities below to define curl and divergence. Naturally, if you use these as definitions then the question of what div and curl mean are easily answered. However, on the other hand, in that approach you have no simple formula to calculate curl or div until you have mastered both surface and line integrals.

  1. Assume $E$ is a volume with piecewise smooth, outward oriented, boundary $\partial E$ where $E$ contains the point $P$. Then if we shrink the volume down to $P$ we obtain the divergence of a differentiable $\vec{F}$ as follows: $$ div( \, \vec{F}\, )(P) = \lim_{V \rightarrow 0^+}\frac{1}{V} \iint_{\partial E} \vec{F} \cdot d\vec{S}. $$ In invite the reader to show the formula above is true (in the sense that it matches the usual formula for divergence given in Cartesian coordinates) by an argument involving the divergence theorem.
  2. Assume $S$ is a surface with piecewise smooth, consistently oriented, boundary $\partial S$ where $E$ contains the point $P$. Then if we shrink the surface to $P$ we obtain the curl of the the vector field in the direction of the normal $\widehat{n}$ to $S$ at $P$ as follows: $$ \biggr[ curl ( \, \vec{F} \, )(P) \biggl] \cdot \widehat{n} = \lim_{A \rightarrow 0^+} \frac{1}{A} \oint_{\partial S} \vec{F} \cdot d\vec{r} $$ Once again, a theorem (Stokes') will show this formulation is simpatico with the usual definition of curl.
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