owsicag7

Answered

2022-07-18

If I have $u=({u}_{1},{u}_{2})$ is a vector of two components, then what is $\text{curl}\phantom{\rule{thinmathspace}{0ex}}u:=\mathrm{\nabla}\times u$? where $\mathrm{\nabla}=({\mathrm{\partial}}_{x},{\mathrm{\partial}}_{y})$

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

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

Answer & Explanation

abortargy

Expert

2022-07-19Added 19 answers

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:

$\mathrm{\nabla}\times ({u}_{1},{u}_{2},0)=(0,0,\frac{\mathrm{\partial}{u}_{2}}{\mathrm{\partial}x}-\frac{\mathrm{\partial}{u}_{1}}{\mathrm{\partial}y})$

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

$\frac{\mathrm{\partial}{u}_{2}}{\mathrm{\partial}x}-\frac{\mathrm{\partial}{u}_{1}}{\mathrm{\partial}y}$

$\mathrm{\nabla}\times ({u}_{1},{u}_{2},0)=(0,0,\frac{\mathrm{\partial}{u}_{2}}{\mathrm{\partial}x}-\frac{\mathrm{\partial}{u}_{1}}{\mathrm{\partial}y})$

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

$\frac{\mathrm{\partial}{u}_{2}}{\mathrm{\partial}x}-\frac{\mathrm{\partial}{u}_{1}}{\mathrm{\partial}y}$

PoentWeptgj

Expert

2022-07-20Added 6 answers

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: $\hat{v}:=\mathrm{\nabla}\times \hat{u}=(0,0,{v}_{3})$ and we identify ${v}_{3}$ with the curl of u, i.e.

$\text{curl}({u}_{1},{u}_{2})={\epsilon}_{321}{\mathrm{\partial}}_{2}{u}_{1}+{\epsilon}_{312}{\mathrm{\partial}}_{1}{u}_{2}=\frac{\mathrm{\partial}{u}_{2}}{\mathrm{\partial}x}-\frac{\mathrm{\partial}{u}_{1}}{\mathrm{\partial}y}$

$\text{curl}({u}_{1},{u}_{2})={\epsilon}_{321}{\mathrm{\partial}}_{2}{u}_{1}+{\epsilon}_{312}{\mathrm{\partial}}_{1}{u}_{2}=\frac{\mathrm{\partial}{u}_{2}}{\mathrm{\partial}x}-\frac{\mathrm{\partial}{u}_{1}}{\mathrm{\partial}y}$

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