# Intuition behind $\nabla \times \mathbf{F}$

Is there a simple explanation why this form for the curl of a vector field $\mathbf{F}$, $$\nabla \times \mathbf{F}=\begin{vmatrix} \hat{x} & \hat{y} &\hat{z} \\ \frac{\partial}{\partial x}& \frac{\partial}{\partial y} &\frac{\partial}{\partial z} \\ F_x& F_y &F_z \end{vmatrix}$$ Corresponds to the amount of 'twiting' of $\mathbf{F}$ (and any other qulities of $\nabla \times \mathbf{F}$)?

When I first saw the equation, it seemed, very roughly, to be a measure of how much a component of $\mathbf{F}$ is affected by the other two components. However, this only really differentiates between $0$ and ' not $0$' curl, and anyway there are thousands of possible equations that would give the same first impression. What's so unique about this one?

• A really good intuitive explanation is the one Feynman gives in the first chapters of his second volume of "Lectures in Physics". The first two chapters are about vector calculus, and he explains this stuff really good. – MyUserIsThis Mar 10 '13 at 19:39
• @MyUserIsThis Looking through it, it seems that, uncharacteristically, he Just stated the mathematics without a physical underpinning. – Meow Mar 10 '13 at 19:50
• I remember him doing just the opposite, something about drawing crystal tubes in a fluid and seeing what it happened to the fluid int he tube and how it's related to the curl. – MyUserIsThis Mar 10 '13 at 19:51
• Not in the first couple of chapters. I'll keep searching. – Meow Mar 10 '13 at 19:52

## 1 Answer

This is completely explained by Stokes' theorem: $$\oint_{\partial S} \vec{F} \cdot d\vec{r} = \iint_S (\nabla\times\vec{F}) \cdot d\vec{n},$$ where $d\vec{n}$ is the infinitesimal normal to the surface $S$ and $d\vec{r}$ is the infinitesimal tangent to its boundary $\partial S$, oriented "positively" (according to the right-hand rule). If you take $S$ to be an infinitesimal disk of radius $\epsilon$ oriented perpendicular to a fixed vector $\vec{u}$ and centered at some $\vec{v}$, then you get $$\iint_S (\nabla \times \vec{F}) \cdot d\vec{n} \approx \pi\epsilon^2 (\nabla \times \vec{F})(\vec{v}) \cdot \vec{u}$$ while the line integral around $\partial S$ is interpreted, physics-ly, as the work done by $\vec{F}$ around the circular boundary. Therefore, dividing and taking a limit, you get $$(\nabla \times \vec{F})(\vec{v}) \cdot \vec{u} = \lim_{\epsilon \to 0} \frac{1}{\pi \epsilon^2} \int_{\partial S} \vec{F} \cdot d\vec{r},$$ meaning that the component of the curl of $\vec{F}$ along a particular direction given by $\vec{u}$ is (basically) the work done by $\vec{F}$ while moving in a small circle around the $\vec{u}$ axis.

This is likely explained in most multivariable calculus textbooks; I know it's in Stewart, from which I lifted this almost verbatim from my memory of the class I taught from it last semester.

• I don't quite see what the purpose of (multiplying by) $\mathbf{v}$ is- I assume it's 'normalising' the direction of $(\nabla \times \mathbf{F})$, but don't get how. – Meow Mar 10 '13 at 20:05
• Excellent answer (as far as I can tell), by the way. – Meow Mar 10 '13 at 20:05
• @Alyosha You mean $\vec{u}$ or $\vec{v}$ (seems like you wrote the latter)? The reason for $\vec{u}$ is that for this choice of $S$, we have $d\vec{n} = \vec{u}\, dA \approx \pi \epsilon^2 \vec{u}$, where $dA$ is the infinitesimal surface area. For $\vec{v}$, it's just because we are approximating the function $\nabla \times \vec{F}$ by its value at the center of the tiny disk. – Ryan Reich Mar 10 '13 at 20:07
• Also, you should take note of the word "component" in the interpretation. Since the curl is a vector (field), you should expect to compute its projections along various axes, not the whole thing at once. – Ryan Reich Mar 10 '13 at 20:10
• @Alyosha: $\vec{v} = (x_0, y_0, z_0)$ in coordinates. – Ryan Reich Mar 10 '13 at 20:11