Whys does 2 dimensional curl always measure twice the angular velocity of rotational component of velocity field?

Question

I thought I understood curl in 2 D until I saw the above statement in the manual. Why is this necessarily true?. I used the vector field F= <-y,x> and yes I can see when curl is calculated by using the second derivatives and subtracting them in the correct order it does in fact = 2. But this is one example, how is it that the twist will always "double up". Intuitively the force field is twisting in both x and y but they may not be twisting in the same direction and when the stronger direction wins out why would it be double ??? Should it not be a fraction of the stronger twist since the directions are different unless it is the case they always twist in the same direction. I am convinced there is an intuition failure on my part here.

Answer 1

It's because the curl fundamentally measures the shear gradient of a velocity field rather than rotation. Suppose you have a velocity field with a uniform shear gradient: $$ \vec{v} = -y\,\hat{x} $$ Then the curl of this is \begin{align} \vec\nabla\times\vec{v} &= \partial_x v_y - \partial_y v_x\\ &= 1\, . \end{align} But a rotating velocity field, $$ \vec{v} = -y\,\hat{x} + x\,\hat{y}\, , $$ is made of two shear gradients...

Answer 2

I think you should be thinking about Green's/Stokes' theorem here. In particular, consider the circulation around the circle of radius $R$ centered at the origin. You have

$$C=\int_0^{2\pi} |-R \sin(\theta),R\cos(\theta)|^2 d \theta = 2\pi R^2.$$

Green's theorem also tells you that this is going to be

$$\int_0^R \int_0^{2\pi} |\nabla \times F| r d \theta dr.$$

If $|\nabla \times F|$ is a constant $c$ then this is $\pi R^2 c$. So in this respect the $2$ is the same $2$ in the discrepancy between the area and circumference of a circle. It is geometric, not analytic, if that means anything to you.

Answer 3

There is no connection between the the angular velocity of the fluid described by the field ${\bf v}$ and the curl of ${\bf v}$, hence this factor $2$ is purely coincidental, and could as well be $0$ or $\pi$.

Note that ${\rm curl}({\bf v})$ is a quantity that is determined at each point ${\bf z}\in\dot{\mathbb R}^2$ by doing some subtle local measurements, whereas the global angular velocity can be seen only "from high above", and is not felt by the individual particles.

As an example consider the solenoidal field $${\bf a}(x,y):=\left({-y\over x^2+y^2}, {x\over x^2+y^2}\right)\qquad(=\nabla{\rm arg})\ .$$ From high above we can clearly see that something is flowing around the origin (albeit with diminishing angular speed, as $x^2+y^2$ increases). Nevertheless one easily computes ${\rm curl}({\bf a})\equiv0$.