# The meaning of curl operator

mathematics physics vector analysis### Contents

### Introduction

Quoting the wikipedia definition of the curl vector operator:

In vector calculus, the curl is a vector operator that describes the infinitesimal rotation of a vector field in three-dimensional Euclidean space. At every point in the field, the curl of that point is represented by a vector. The attributes of this vector (length and direction) characterize the rotation at that point.

The devil in this definiton lies in the word *infinitestimal*. I was under the impression that curl was related to the *macroscopic* rotation, but I couldn’t be more wrong! Let me show what I mean. Consider the vector field defined by \(\mathbf{F}(x,y,z) = \left(-y/(x^2+y^2), x/(x^2+y^2), 0\right)\).

By looking at these images my first reaction was that this field is most certainly a rotational one. I mean look at how “swirly” it is! Imagine my surprise when I actually did the math and my intuition proved to be completely wrong:

\[\begin{align} \nabla \times \mathbf{F} &= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[5pt] {\dfrac{\partial}{\partial x}} & {\dfrac{\partial}{\partial y}} & {\dfrac{\partial}{\partial z}} \\[10pt] F_x & F_y & F_z \end{vmatrix}\\ &= \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right) \mathbf{i} + \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) \mathbf{j} + \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \mathbf{k}\\ &= (0 - 0)\mathbf{i} + (0 - 0) \mathbf{j} + \left[\frac{-x^2+y^2}{(x^2+y^2)^2} - \frac{-x^2+y^2}{(x^2+y^2)^2}\right] \mathbf{k} = \vec{0} \end{align}\]How can it be that this plot corresponds to an irrotational field? Well, it depends on which rotation you are referring to (macroscopic vs. microscopic or global vs. local). **Curl measures the local rotation!** Imagine that this vector field describes the flow of water in a pool with a sink at the bottom that sucks the water out it. If we put a small ball on the surface of the water, then the ball may move in two distinct ways:

- The
**general rotation of the flow around the z-axis**(z-axis is perpendicular to your monitor) in the*counterclockwise*direction, along the direction of the stream lines (blue arrows). - Since the arrows of the field are longer the closer we are to the z-axis, the field tends to push the ball more strongly on the side closest to the z-axis, rather than the opposite side. The “differential” push on the two sides of the ball would tend to make it
**rotate around itself**in the*clockwise*direction.

These two opposite effects may cancel out (as in our case) and then the curl is zero. **The ball still moves inside the pool around the z-axis, but it doesn’t rotate around itself, which is what the curl operator measures.**

### Another way to view curl

Please mind that the image above is drawn in a large scale. In reality the green circle is *infinitestimal*. Another way to look at curl is as *the average circulation of a field in a region that shrinks around a point*, i.e.:

Where \(A\) is the green area in the image above, as it shrinks into a point. Recall though that the curl is a vector, so the correct way to connect the above formula with the curl is:

\[(\nabla \times \mathbf{F}) \cdot \hat{\mathbf{n}} = \lim_{A\to 0} \left( \frac{1}{A} \oint_C \mathbf{F} \cdot \operatorname{d}\!\mathbf{r} \right)\]where \(\hat{\mathbf{n}}\) is the normal vector to the point \(O\) where we measure the curl.

### Relation of curl with the angular velocity at some point

By now, it shouldn’t come as a surprise that the curl of a vector field calculated at some point \(O\), is related to the angular velocity of a rotating object with its center fixed at \(O\). Let’s do the math!

#### First method

Let us calculate the curl of \(\mathbf{v}\):

\[\nabla \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[5pt] {\dfrac{\partial}{\partial x}} & {\dfrac{\partial}{\partial y}} & {\dfrac{\partial}{\partial z}} \\[10pt] \nu_x & \nu_y & \nu_z \end{vmatrix}\]The \(x\) component of \(\nabla \times \mathbf{v}\) is (for brevity we write \(\partial_x\) instead of \({\partial}/{\partial_x}\)):

\[\left( \nabla \times \mathbf{v} \right)_x = {\partial_y \nu_z} - {\partial_z \nu_y}\]Recall though that \(\mathbf{v} = \boldsymbol{\omega} \times \mathbf{r} \Rightarrow \nu_z = \omega_x y-\omega_y x\) and similarly \(\nu_y = \omega_x z - \omega_z x\). Therefore:

\[\begin{align} \left( \nabla \times \mathbf{v} \right)_x &= {\partial_y \nu_z} - {\partial_z \nu_y}\\ &= \partial_y (\omega_x y - \omega_y x)- \partial_z (\omega_x z - \omega_z x)\\ &= \omega_x + \omega_x = 2\omega_x \end{align}\]Similarly it is \(\left( \nabla \times \mathbf{v} \right)_y = 2 \omega_y\) and \(\left( \nabla \times \mathbf{v} \right)_z = 2\omega_z\). Therefore the curl is twice the angular velocity:

\[\nabla \times \mathbf{v} = 2 \boldsymbol{\omega}\]#### Second method

Another way to attack the problem is by calculating the average circulation of the vector field \(\mathbf{v}\) around the point \(O\). For simplicity let us assume that we are working on a vector field in two dimensions (\(xy\) plane):

\[\begin{align*} \mathbf{v} &= \boldsymbol{\omega} \times \mathbf{r} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[5pt] {\omega_x} & {\omega_y} & {\omega_z} \\[10pt] x & y & z \end{vmatrix} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[5pt] {0} & {0} & {\omega_z} \\[10pt] x & y & z \end{vmatrix}\\ &= (-\omega_z y)\mathbf{i} - (- \omega_z x) \mathbf{j} = -\omega_z y\mathbf{i} + \omega_z x \mathbf{j} \end{align*}\] \[I = \oint_C \mathbf{v} \cdot \operatorname{d}\!\mathbf{r} = \oint_C (\boldsymbol{\omega} \times \mathbf{r}) \cdot \operatorname{d}\!\mathbf{r} = \oint_C(-\omega_z y\mathbf{i} + \omega_z x \mathbf{j}) \cdot \operatorname{d}\!\mathbf{r}\]We use the parameterization \(\mathbf{r}(t) = \rho \cos t \mathbf{i} + \rho \sin t \mathbf{j} \Rightarrow \mathbf{r}'(t) = -\rho \sin t \mathbf{i} +\rho \cos t \mathbf{j}\), with \(t = [0, 2\pi]\).

Therefore:

\[\begin{align*} I &= \int_0^{2\pi} (-\omega_z \rho \sin t\mathbf{i} + \omega_z \rho \cos t \mathbf{j}) \cdot (-\rho \sin t \mathbf{i} + \rho \cos t \mathbf{j}) \operatorname{d}\!t\\ &= \int_0^{2\pi} \omega_z \rho^2 \sin^2 t + \omega_z \rho^2 \cos^2 t \operatorname{d}\!t \\ &= \int_0^{2\pi} \omega_z \rho^2 \operatorname{d}\!t = 2\pi\rho^2 \omega_z \end{align*}\]Therefore:

\[(\nabla \times \mathbf{v}) \cdot \hat{\mathbf{n}} = \lim_{A\to 0} \left( \frac{1}{A} \oint_C \mathbf{v} \operatorname{d}\!\mathbf{r} \right) = \lim_{\rho \to 0} \left( \frac{1}{\pi \rho^2} 2\pi \rho^2 \omega_z\right) = 2\omega_z\]