Is curl actually a cross product?

Question

I tried deriving curl for some vector $\vec{v}$, everything in a cylindrical coordinate system. I looked at cross products in this coordinates: https://bridge.math.oregonstate.edu/workshops/mini04/13.pdf

So when I calculate the Curl, I get $$\vec{\nabla} \times \vec{V} = \begin{vmatrix} \hat{r}\ & \hat{\theta} & \hat{z} \\ \partial/\partial r & (1/r)\partial/\partial \theta & \partial/\partial z \\ 0 & V_{\theta} & 0 \end{vmatrix}$$ Which are the Del-operator components, and the vector components. However, upon looking at curl formulas, that determinant is incorrect. If curl was a cross product between two vectors that exist in an orthonormal basis, then it should have this form. How come is the curl different than the way we would go about doing a cross product.

Answer 1

As has been stated, the curl is not defined to be the cross product. In what follows, I'll give an operational definition of what the curl is. It won't be pretty and it won't be convenient, but hopefully it will motivate the equations with which you are already familiar.

Consider a 2D vector field $\mathbf V$. The circulation of $\mathbf V$ around a simple closed curve $C$ is defined to be $\oint_C \mathbf V \cdot d\mathbf r$ - that is, the line integral of $\mathbf V$ around $C$.

Example: Let $\mathbf V = x^2y\hat x + x\hat y$, and let $C$ be a square of side length $a$. The circulation of $\mathbf V$ around $C$ is not too hard to calculate, as we can break the integral up into four parts: $$\oint_C\mathbf V \cdot d\mathbf r = \int_{-a/2}^{a/2} \frac{a}{2} dy + \int_{a/2}^{-a/2} x^2\left(\frac{a}{2}\right) dx + \int_{a/2}^{-a/2} \left(-\frac{a}{2}\right) dy+\int_{-a/2}^{a/2}x^2\left(-\frac{a}{2}\right) dx$$ $$=\frac{a^2}{2} -\frac{2}{3}\left(\frac{a}{2}\right)^4 + \frac{a^2}{2} - \frac{2}{3}\left(\frac{a}{2}\right)^4$$ $$ = a^2 - \frac{a^4}{12}$$

Next, we consider the circulation of $\mathbf V$ around $C$ divided by the area enclosed by $C$. To my knowledge, this quantity doesn't have a good name, but it's essentially a measure of the average circulation. In particular, we are interested in the limit of this average circulation as the loop shrinks down to a point.

Example: In the previous example, the circulation of $\mathbf V = x^2y\hat x + x \hat y$ around a square of side length $a$ centered at the origin was found to be $a^2 - \frac{a^4}{12}$. The area of this square is simply $a^2$, and so the "average circulation" is immediately seen to be $1-\frac{a^2}{12}$. Taking the limit as the square shrinks down to a point (the origin), we simply get $1$.

---

This is the definition of the curl. In order to compute the curl of a vector field $\mathbf V$ at a point $p$, we choose a curve $C$ which encloses $p$ and evaluate the circulation of $\mathbf V$ around $C$, divided by the area enclosed. We then take the limit of this quantity as $C$ shrinks down to $p$.

One might immediately ask if there is a more efficient means to calculate this quantity, and the answer is yes. If we use Cartesian coordinates where $p=(x,y)$, then the most efficient thing is to encircle our point of interest with a square of side length $a$. If $a$ is very small, we find

$$\oint_C \mathbf V \cdot d\mathbf r \approx a\bigg(V_y(x+a/2,y)-V_x(x,y+a/2)-V_y(x-a/2,y)+V_x(x,y-a/2)\bigg)$$ Dividing by $a^2$ and taking the limit as $a\rightarrow 0$ yields the quantity $$\frac{\partial V_y}{\partial x}-\frac{\partial V_x}{\partial y}$$.

If our coordinates are not cartesian, then a square is not a good shape. For polar coordinates, we should use a wedge as shown e.g. here:

Evaluating this circulation yields $$\oint \mathbf V \cdot d\mathbf r \approx V_r\left(r,\theta-\frac{\Delta\theta}{2}\right)\Delta r +V_\theta\left(r+\frac{\Delta r}{2},\theta\right)\cdot \left(r+\frac{\Delta r}{2}\right) \Delta \theta$$ $$ - V_r\left(r,\theta+\frac{\Delta\theta}{2}\right)\Delta r - V_\theta\left(r-\frac{\Delta r}{2},\theta\right)\left(r-\frac{\Delta r}{2}\right)\Delta \theta$$

The area of this wedge is approximately $r(\Delta r )(\Delta \theta)$. Dividing and taking the limit as $\Delta r$ and $\Delta \theta$ go to zero yields the quantity

$$\frac{\partial V_\theta}{\partial r} + \frac{V_\theta}{r} - \frac{1}{r} \frac{\partial V_r}{\partial \theta} = \frac{1}{r}\frac{\partial (rV_\theta)}{\partial r} - \frac{1}{r}\frac{\partial V_r}{\partial \theta}$$

---

Lastly, of course curl is a 3D operation while we've been working in 2D. In 3D, each unit vector defines a plane perpendicular to it, with an orientation given by the right hand rule; the corresponding component of the curl is this average infinitesimal circulation as computed in that plane. For instance, what I calculated above is the $\hat z$ component of the curl in cartesian and cylindrical coordinates, because it was computed in the plane perpendicular to $\hat z$.

Answer 2

Even in cartesian coordinates, the curl isn't really a cross product.

A cross product is a map with the following properties:

1. It takes two vectors from $\mathbb{R}^3$ and outputs a third vector in $\mathbb{R}^3$;
2. It's anticommutative;
3. It's rotationally invariant.

The curl has only property 3, not 1 or 2. It doesn't even make sense to discuss property 2, since it doesn't make sense to write the partial derivative operator on the right.

The cross-product notation for the curl is just a helpful mnemonic for certain properties of the curl, such as the fact that its output is a vector, it has a handedness, and that the divergence of a curl is zero.