Curl of $u=(u_1,u_2)$

Question

If I have $u=(u_1,u_2)$ is a vector of two components, then what is $\text{curl}\,u:=\nabla\times u$? where $\nabla=(\partial_x,\partial_y)$.

I think it should give a scalar and not a vector.

Answer 1

The curl of a vector field is only really defined on vector fields $\mathbf{F}: \mathbb{R}^3 \to \mathbb{R}^3$. For a two-dimensional field $u = (u_1(x,y),u_2(x,y))$, the $z$ component is taken to be zero. The resulting curl has non-zero components only in the $z$-direction:

$$\nabla \times (u_1, u_2, 0) = \left(0, 0, \frac{\partial u_2}{\partial x} - \frac{\partial u_1}{\partial y}\right)$$

so people sometimes call the following scalar quantity the "curl":

$$\frac{\partial u_2}{\partial x} - \frac{\partial u_1}{\partial y}$$

Answer 2

The curl of a 2D vector can be defined with the following "intuition": The 2D vector $u=(u_1, u_2)$ can be "extended" to $\mathbb{R}^3$ as $\hat{u}=(u_1,u_2,0)$. Now we can take the "usual" curl of this $\hat{u}$ vector: $\hat{v}:=\nabla \times \hat{u}=(0, 0, v_3)$ and we identify $v_3$ with the curl of $u$, i.e. $$\text{curl}(u_1, u_2)=\varepsilon_{321} \partial_2 u_1 + \varepsilon_{312}\partial_1 u_2 = \frac{\partial u_2}{\partial x} - \frac{\partial u_1}{\partial y}$$