 owsicag7

2022-07-18

If I have $u=\left({u}_{1},{u}_{2}\right)$ is a vector of two components, then what is $\text{curl}\phantom{\rule{thinmathspace}{0ex}}u:=\mathrm{\nabla }×u$? where $\mathrm{\nabla }=\left({\mathrm{\partial }}_{x},{\mathrm{\partial }}_{y}\right)$
I think it should give a scalar and not a vector. abortargy

Expert

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 $u=\left({u}_{1}\left(x,y\right),{u}_{2}\left(x,y\right)\right)$the z component is taken to be zero. The resulting curl has non-zero components only in the z-direction:
$\mathrm{\nabla }×\left({u}_{1},{u}_{2},0\right)=\left(0,0,\frac{\mathrm{\partial }{u}_{2}}{\mathrm{\partial }x}-\frac{\mathrm{\partial }{u}_{1}}{\mathrm{\partial }y}\right)$
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

The curl of a 2D vector can be defined with the following "intuition": The 2D vector $u=\left({u}_{1},{u}_{2}\right)$ can be "extended" to ${\mathbb{R}}^{3}$ as $\stackrel{^}{u}=\left({u}_{1},{u}_{2},0\right)$. Now we can take the "usual" curl of this $\stackrel{^}{u}$ vector: $\stackrel{^}{v}:=\mathrm{\nabla }×\stackrel{^}{u}=\left(0,0,{v}_{3}\right)$ and we identify ${v}_{3}$ with the curl of u, i.e.
$\text{curl}\left({u}_{1},{u}_{2}\right)={\epsilon }_{321}{\mathrm{\partial }}_{2}{u}_{1}+{\epsilon }_{312}{\mathrm{\partial }}_{1}{u}_{2}=\frac{\mathrm{\partial }{u}_{2}}{\mathrm{\partial }x}-\frac{\mathrm{\partial }{u}_{1}}{\mathrm{\partial }y}$

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