Proving $(\nabla \times \mathbf{v}) \cdot \mathbf{c} = \nabla \cdot (\mathbf{v} \times \mathbf{c})$ using cylindrical coordinates Assuming the form of divergence in polar coordinates is known, I am attempting to use the following definition of the curl of a vector field to determine the form of the curl in cylindrical coordinates. The definition (see section 4 of this tensor derivative Wikipedia page) states that the curl of vector field $\bf{v}$ and an arbitrary vector $\bf{c}$ satisfies:$$(\nabla\times\mathbf{v})\cdot \mathbf{c}=\nabla \cdot (\mathbf{v}\times\mathbf{c}).$$
My attempted derivation follows:
It is known that the divergence of a vector field $\mathbf{u}$ in cylindrical coordiantes is
$$\nabla\cdot\mathbf{u}=\frac{\partial u_{r}}{\partial r}+\frac{u_r}{r}+\frac{1}{r}\frac{\partial u_{\theta}}{\partial \theta}+\frac{\partial u_{z}}{\partial z}.$$
In polar coordinates, if $$\mathbf{v}=v_{r}\hat{r}+v_{\theta}\hat{\theta}+v_{z}\hat{z}\\
\mathbf{c}=c_{r}\hat{r}+c_{\theta}\hat{\theta}+c_{z}\hat{z}$$ then
$$\mathbf{v}\times\mathbf{c}=(v_{\theta}c_{z}-v_{z}c_{\theta})\hat{r}+(v_{z}c_{r}-v_{r}c_{z})\hat{\theta}+(v_{r}c_{\theta}-v_{\theta}c_{r})\hat{z}.$$
Using the equation for divergence above gives
\begin{align}
\nabla \cdot (\mathbf{v}\times\mathbf{c}) &= \frac{\partial (v_{\theta}c_{z}-v_{z}c_{\theta})}{\partial r} + \frac{(v_{\theta}c_{z}-v_{z}c_{\theta})}{r} + \frac{1}{r}\frac{\partial (v_{z}c_{r}-v_{r}c_{z})}{\partial \theta} + \frac{\partial (v_{r}c_{\theta}-v_{\theta}c_{r})}{\partial z}\\
&= c_{r}\left(\frac{1}{r}\frac{\partial v_{z}}{\partial \theta}-\frac{\partial v_{\theta}}{\partial z}\right) + c_{\theta}\left(\frac{\partial v_{r}}{\partial z}-\frac{\partial v_z}{\partial r}-\frac{v_z}{r}\right) + c_z\left(\frac{\partial v_{\theta}}{\partial r} +\frac{v_{\theta}}{r}-\frac{1}{r}\frac{\partial v_{r}}{\partial\theta}\right).
\end{align}
Referring back to the given definition of the curl, we see that
$$(\nabla\times\mathbf{v})_r=\frac{1}{r}\frac{\partial v_{z}}{\partial \theta}-\frac{\partial v_{\theta}}{\partial z}
 \\(\nabla\times\mathbf{v})_{\theta}=\frac{\partial v_{r}}{\partial z}-\frac{\partial v_z}{\partial r}-\frac{v_z}{r} \\
(\nabla\times\mathbf{v})_z=\frac{\partial v_{\theta}}{\partial r} +\frac{v_{\theta}}{r}-\frac{1}{r}\frac{\partial v_{r}}{\partial\theta}.$$
I know that the $v_z/r$ term in the $\theta$-component should not be there, but everything else seems correct. Can anyone spot the flaw in my derivation?
 A: The issue here is that even if $\mathbf{c}$ is an arbitrary fixed vector, $c_r$ and $c_\theta$ are not constant with respect to $r$ and $\theta$. So, you're missing some other terms in your computation.
Adding the missing terms, we get:
\begin{align}
\nabla \cdot (\mathbf{v}\times\mathbf{c}) &= \frac{\partial (v_{\theta}c_{z}-v_{z}c_{\theta})}{\partial r} + \frac{(v_{\theta}c_{z}-v_{z}c_{\theta})}{r} + \frac{1}{r}\frac{\partial (v_{z}c_{r}-v_{r}c_{z})}{\partial \theta} + \frac{\partial (v_{r}c_{\theta}-v_{\theta}c_{r})}{\partial z}\\
&= c_{r}\left(\frac{1}{r}\frac{\partial v_{z}}{\partial \theta}-\frac{\partial v_{\theta}}{\partial z}\right) + c_{\theta}\left(\frac{\partial v_{r}}{\partial z}-\frac{\partial v_z}{\partial r}-\frac{v_z}{r}\right) + c_z\left(\frac{\partial v_{\theta}}{\partial r} +\frac{v_{\theta}}{r}-\frac{1}{r}\frac{\partial v_{r}}{\partial\theta}\right) + v_z \left(\frac{1}{r}\frac{\partial c_r}{\partial \theta} - \frac{\partial c_\theta}{\partial r} \right).
\end{align}
Now, $\displaystyle \frac{\partial c_\theta}{\partial r} = 0$ and $\displaystyle \frac{\partial c_r}{\partial \theta} = c_\theta$. Thus, this cancels with the extra term that you found, and proves your formula.

Why is $\displaystyle \frac{\partial c_\theta}{\partial r} = 0$?
For each point $P$ in $\mathbb{R}^3$, we have an orthonormal basis $\{ \hat{r}(P), \hat{\theta}(P), \hat{z}(P) \}$ at that point. Now, consider two points $P$ and $Q$ in $\mathbb{R}^3$ such that $Q - P = (\Delta r) \hat{r}(P)$ for some small $\Delta r > 0$. That is, $Q$ is obtained from $P$ by a small shift along the $\hat{r}(P)$ direction. Notice that this does not change the directions of the corresponding unit vectors at $Q$! Hence, the projection of $\mathbf{c}$ along each of the basis vectors $\hat{r}(P), \hat{\theta}(P), \hat{z}(P)$ at $P$ is identical to the projection along the corresponding basis vectors $\hat{r}(Q), \hat{\theta}(Q), \hat{z}(Q)$ at $Q$. Thus, taking the limit as $\Delta r$ tends to zero, we have $\displaystyle \frac{\partial c_\theta}{\partial r} = 0$.
Why is $\displaystyle \frac{\partial c_r}{\partial \theta} = c_\theta$?
Consider two points $P$ and $Q$ in $\mathbb{R}^3$ such that $\angle(\hat{r}(Q),\hat{r}(P)) = \Delta \theta$ for some small $\Delta \theta > 0$. That is, $Q$ is obtained from $P$ by a small rotation in the anti-clockwise direction about the origin. Since the projection of $\mathbf{c}$ along the $\hat{z}$ direction never changes, there is no loss of generality in assuming that $\mathbf{c}$ is an arbitrary fixed vector lying in the plane. Now, a little bit of trigonometry shows that if $\mathbf{c} = c_r \hat{r}(P) + c_\theta \hat{\theta}(P) = \tilde{c_r} \hat{r}(Q) + \tilde{c_\theta} \hat{\theta}(Q)$, then
$$
\frac{\tilde{c_r} - c_r}{\Delta \theta} = c_r \left( \frac{\cos(\Delta \theta) - 1}{\Delta \theta} \right) + c_\theta \frac{\sin (\Delta \theta)}{\Delta \theta}.
$$
Thus, in the limit as $\Delta \theta$ tends to zero, we have $\displaystyle \frac{\partial c_r}{\partial \theta} = c_\theta$.
