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As a trivial example in our vector analysis class, we did the following computation.

Let $\overrightarrow{\omega} = (\omega_1, \omega_2, \omega_3)$ be the angular velocity and $\overrightarrow{r} =(x,y,z)$ be the position. Then we have a vector field $\overrightarrow{R} = \overrightarrow{v} = \overrightarrow{\omega} \times \overrightarrow{r}$.

We quickly calculated the rotor and got:

$\text{rot} \overrightarrow{v} = 2 \overrightarrow{\omega}$.

The calculation is trivial of course, but I can't see any physical meaning behind this. But the equation is so simple that there must be some neat way to interpret this! Does anyone know of a nice intuitive explanation of why this equality holds?

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  • $\begingroup$ $\vec{\nabla}\times\vec{v},\,\vec{\omega}$ are both axial vectors of dimension $\mathsf{T}^{-1}$, so by symmetries must be equal up to a dimensionless factor. Explaining why one is twice the other is the hard part. $\endgroup$
    – J.G.
    Commented Aug 1, 2022 at 17:55
  • $\begingroup$ On math.stackexchange there is a question titled: Why does 2 dimensional curl always measure twice the angular velocity of rotational component of velocity field? Incidentally, I googled 'rotor', I assume you are talking about 'curl'. I surmise 'rotor' is the name of it in your language. $\endgroup$
    – Cleonis
    Commented Aug 1, 2022 at 18:15
  • $\begingroup$ I don't think that $~\rm rot\, \vec v=2\,\vec\omega~$? I get $~\rm {rot} \vec v=6\,(\vec\omega-\vec r)~$? $\endgroup$
    – Eli
    Commented Oct 10, 2022 at 16:58
  • $\begingroup$ \begin{align*} \rm rot \vec v=\vec\nabla\times\vec{\omega}\times\vec{r}= \vec{\omega}\,\left(\vec\nabla\cdot\vec{r}\right)- \vec{r}\left(\vec\nabla\cdot\vec\omega\right)+ \left(\vec{r}\cdot\vec{\nabla}\right)\vec{\omega}- \left(\vec\omega\cdot\vec{\nabla}\right)\vec{r} \end{align*} $\endgroup$
    – Eli
    Commented Oct 10, 2022 at 17:13

1 Answer 1

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I think the comments of @J.G. and @Cleonis add up to a complete answer. In particular see the answer of Ian to that maths stack exchange question: the factor of 2 is geometric in origin. The geometric definition of the curl, or rot, at a point P is the limit of the loop integral around an arbitrary loop enclosing P divided by the area. If we subdivide the area into many triangles apex at P and base on the loop, then the contribution of each triangle to the loop integral is $(\omega h) b$ where $h$ is the height of the triangle and $b$ is the base, and the contribution to the area integral is $bh/2$. Thus the factor of 2 comes directly from the geometry.

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