# 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?

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 $$\displaystyle \frac{\partial c_r}{\partial \theta} = c_\theta$$.