Why is the magnitude of the curl of a vectorfield twice the angular velocity?

Question

(if V is a vectorfield describing the velocity of a fluid or body, and $x\in R^3$)

I agree that it should be when you look at the calculation, but intuitively speeking... If $\nabla \times V(x)= curl(V(x))$, couldn't one interpret the curl to be the change of velocity orthogonally to the flow line at the given point, x, and thus the length of the curl to be the angular velocity, $\omega$?

Or is this a flawed argument due to $\nabla$ having no direction in itself$\dots$?

To clarify, why the 2 in $\| curl(V(x)) \| = 2\omega(x)$?

Answer 1

Denote by ${\bf v}$ the velocity field of a planar fluid rotating with angular velocity $\omega$ around the origin ${\bf 0}$ of the ${\bf z}$-plane. At distance $r$ from ${\bf 0}$ we then have $|{\bf v}|=r\omega$. Two rays from ${\bf 0}$ at arguments $\phi$ and $\phi+d\phi$, together with two concentric circles of radii $r$ and $r+dr$ determine an infinitesimal quadrangle $Q$ with center ${\bf z}_0$. Let us compute the circulation of ${\bf v}$ around $\partial Q$, i.e., the line integral $\int_{\partial Q}{\bf v}\cdot d{\bf z}$.

Along the (counterclockwise) outer arc we obtain the contribution $\omega(r+dr)\cdot(r+dr)\>d\phi$, along the (clockwise) inner arc the contribution $-\omega r\cdot r\>d\phi$, and the two radial segments give no contribution since they are orthogonal to ${\bf v}$. It follows that, up to first order in $dr$ we obtain $$\int_{\partial Q}{\bf v}\cdot d{\bf z}\doteq2\omega r\>dr \>d\phi\ .\tag{1}$$ On the other hand, by Green's (or Stokes') theorem we have $$\int_{\partial Q}{\bf v}\cdot d{\bf z}=\int_Q{\rm curl}({\bf v}({\bf z}))\>{\rm d}({\bf z})\doteq {\rm curl}({\bf z}_0)\>{\rm area}(Q)\doteq{\rm curl}({\bf z}_0)\>dr\>r\>d\phi\ .\tag{2}$$ Comparing $(1)$ and $(2)$ we can see that necessarily ${\rm curl}({\bf z}_0)=2\omega$, independently of ${\bf z}_0$. Looking back we recognize the reason for this factor of $2$: Going outwards not only the speed $|{\bf v}|$ increases, but also the length of the arcs corresponding to a given $d\phi$.

A student confronted with the above field ${\bf v}$ or ${\rm curl}$ the first time is inclined to think that ${\rm curl}({\bf v})$ is somehow concentrated at the origin ${\bf 0}$. But this is not the case: ${\rm curl}({\bf v})=2\omega$ can be felt everywhere!

Answer 2

Hello I would say for a problem with the radius of a circle denoted by $\mathbf{r}$, the tangential velocity $\mathbf{v}$ and the angular velocity $\boldsymbol{\omega}$ we have \begin{equation} \nabla \times \mathbf{v} = \nabla \times (\boldsymbol{\omega} \times \mathbf{r}) \end{equation} Using the Graßmann identity \begin{equation} \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c} \end{equation} gives \begin{equation} \nabla \times (\boldsymbol{\omega} \times \mathbf{r}) = (\nabla \cdot \mathbf{r})\boldsymbol{\omega} - (\nabla \cdot \boldsymbol{\omega})\mathbf{r} \end{equation} The second term vanishes since $\boldsymbol{\omega}$ has only a $z$-component and therefore also $\nabla \cdot \boldsymbol{\omega}$ while the corresponding $z$-component of $\mathbf{r}$ is $0$. In the first term the divergence of \begin{equation} \mathbf{r} = x\mathbf{e}_1 + y\mathbf{e}_2 \end{equation} is calculated yielding \begin{equation} \nabla \cdot \mathbf{r} = \partial_x x + \partial_y y = 2 \end{equation} Finally the result is \begin{equation} \nabla \times \mathbf{v} = \nabla \times (\boldsymbol{\omega} \times \mathbf{r}) = 2\boldsymbol{\omega} \end{equation}