# Is it possible to write the curl in terms of the infinitesimal rotation tensor?

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?

• 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. – user117644 Jul 25 '14 at 16:35
• 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. – Semiclassical Jul 25 '14 at 16:40

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.
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.