A Curl in cylindrical coordinates -- seeking a deeper understanding

[LagrangeEuler]

I calculate that \mbox{curl}(\vec{e}_{\varphi})=\frac{1}{\rho}\vec{e}_z, where $\vec{e}_{\rho}$, $\vec{e}_{\varphi}$, $\vec{e}_z$ are unit vectors of cylindrical coordinate system. Is there any method to spot immediately that $\mbox{curl}(\vec{e}_{\varphi}) \neq 0$ without employing any calculations.

 

[hutchphd]

Not sure what you are asking. Does the fact that our cartesian coordinates are based upon $$\hat {\mathbf x} \times \hat {\mathbf y}=\hat {\mathbf z}$$ make it easier?

 

[ergospherical]

Recall some facts about how to interpret the curl, specifically as\begin{align*}
[\nabla \times \mathbf{e}_{\varphi}]_z =\lim_{A \rightarrow 0} \frac{1}{A} \oint_C \mathbf{e}_{\varphi} \cdot d\mathbf{r}
\end{align*}Let the curve $C$ be a "curvy rectangle" around the point at which you want the curl (by which I mean start at $(\rho,\varphi) \rightarrow (\rho, \varphi + \delta\varphi) \rightarrow (\rho + \delta\rho, \varphi + \delta\varphi) \rightarrow (\rho + \delta\rho, \varphi)$ and back to the start). Then\begin{align*}
[\nabla \times \mathbf{e}_{\varphi}]_z = \frac{1}{\rho \delta \rho \delta \varphi} \left\{ (\rho + \delta \rho)\delta \varphi - \rho \delta \varphi \right\} = \frac{1}{\rho}
\end{align*}Although this is an explicit calculation, you can see that the line integral will be non-zero since the outer side is longer than the inner side.

 

[Delta2]

The simplest intuitive rule (but not always correct, see counter example below) is that when the vector field rotates around something, then its curl is not zero. $\hat \phi$ rotates around the origin.

However $$\nabla\times \frac{1}{\rho}\hat \phi=0$$ !

 

[Delta2]

Another counter example, in cartesian coordinates this time, the field $$y\hat x$$ doesn't seem to rotate or form closed loops, yet its curl equals $$-\hat z$$.

 

[wrobel]

Angular velocity of a rigid body is given by the formula $\boldsymbol \omega=\frac{1}{2}\mathrm{curl}\,\boldsymbol v,$ where $\boldsymbol v$ is a velocity field of rigid body's points;
thus my guess is $$\boldsymbol e_z=\frac{1}{2}\mathrm{curl}\,(\rho\boldsymbol e_{\varphi})=\frac{1}{2}(\nabla\rho\times \boldsymbol e_\varphi+\rho\, \mathrm{curl}\,\boldsymbol e_{\varphi}),\quad \nabla\rho=\boldsymbol e_\rho$$

 

[Orodruin]

The simplest intuitive rule (but not always correct, see counter example below) is that when the vector field rotates around something, then its curl is not zero. $\hat \phi$ rotates around the origin.

However $$\nabla\times \frac{1}{\rho}\hat \phi=0$$ !

That sounds like a more global perspective whereas curl is local. The typical intuitive rule is to imagine an infinitesimal paddle wheel in the flow of the vector field. This also works on your later example because the direction of the flow is the same everywhere but larger for larger y - thus leading to a net ”torque” on the paddle wheel.

 

[wrobel]

omething, then its curl is not zero. ϕ^ rotates around the origin.

However ∇×1ρϕ^=0 !

such a vector field has singularity at $\rho=0$

 

[Delta2]

such a vector field has singularity at $\rho=0$

Ah yes correct the field and hence its curl and divergence are not defined for $\rho=0$. The limits $$\lim_{\rho \to 0} |\vec{F}|=\infty, \lim_{\rho \to 0}|\nabla\times\vec{F}|=0$$.

 

[Delta2]

That sounds like a more global perspective whereas curl is local. The typical intuitive rule is to imagine an infinitesimal paddle wheel in the flow of the vector field. This also works on your later example because the direction of the flow is the same everywhere but larger for larger y - thus leading to a net ”torque” on the paddle wheel.

Yes, this is the typical, more accurate and more local rule, but not as simple to apply if you ask me , you have to take a look at how components vary spatially, not just to take a look of how the field lines look.

 

[Orodruin]

Another counter example, in cartesian coordinates this time, the field $$y\hat x$$ doesn't seem to rotate or form closed loops, yet its curl equals $$-\hat z$$.

Regarding this field, we can subtract the gradient of $xy/2$, which is $\nabla(xy/2) = (x\hat y + y\hat x)/2$. The gradient being curl free, the fields $y\hat x$ and $(y\hat x - x\hat y)/2 = -\rho \vec e_\phi/2$ have the same curl.

 

[Orodruin]

Ah yes correct the field and hence its curl and divergence are not defined for $\rho=0$. The limits $$\lim_{\rho \to 0} |\vec{F}|=\infty, \lim_{\rho \to 0}|\nabla\times\vec{F}|=0$$.

In the distributional sense those quantities are defined. The curl has a delta distribution along the z-axis that points in the z-direction.