# Curl in 2D is a vector or a scalar?

1. Mar 26, 2014

### Jhenrique

Vector, by definition, have 2 or 3 scalar components (generally), but the curl of a vector field f(x,y) in 2D have only one scalar component: $$\left ( \frac{\partial f_y}{\partial x} -\frac{\partial f_x}{\partial y} \right )dxdy$$
So, the Curl of a vector field in 2D is a vector or a scalar?

2. Mar 26, 2014

### PeroK

You can think of 2D curl as a special case of 3D where the only component is in the k direction.

In Green's theorem, for example, the normal to the area is also k, so the integral reduces to the scalar value.

I'd say that technically it's a vector, but it's always in the same direction, so it's never going to change direction like curl in 3D.

3. Mar 26, 2014

### chogg

A curl is always the same type of beast in any number of dimensions. It's neither a vector nor a scalar; it's a bivector.

(Or a two-form; I'm not sure which. The point is that it's an intrinsically two-dimensional object.)

In 2D, the dual to a bivector is a scalar. In 3D, the dual to a bivector is a vector. Typically, students learn only about the vector, because bivectors are not typically taught.

So to answer your question: the curl in 2D is definitely not a vector. If you think of the 3D curl as a vector, you should think of the 2D curl as a scalar.

4. Mar 28, 2014

### Jhenrique

So is wrong to say that $$\oint_{s}\vec{f}\cdot d\vec{s} = \iint_{A}\vec{\nabla}\times \vec{f} d^2A$$ ? Because in the left side we have an scalar and in the right side we have a vector and a scalar is not equal to a vector... BUT, if I use the full definition, the equation will be: $$\oint_{s}\vec{f}\cdot d\vec{s} = \iint_{A}\vec{\nabla}\times \vec{f} \cdot \hat{n} d^2A$$ However, we can't define a normal vector to xy plane because is assumed that we are in the operating in 2D... So, is necessary use the modulus in the equation: $$\oint_{s}\vec{f}\cdot d\vec{s} = \iint_{A}|\vec{\nabla}\times \vec{f}| d^2A$$ But wich is the difference between $|\vec{\nabla}\times \vec{f}|$ and $\vec{\nabla}\times \vec{f}$, in the pratice, none...

5. Mar 29, 2014

### AlephZero

$|-1| \ne -1$.

6. Mar 29, 2014

### chogg

$d^2A$ should also be a bivector, and you should take the dot product between the two bivectors. This works equally well in 2D or 3D.

Notice how much more elegant this is: in particular, you never need to make a completely arbitrary choice between a "right hand rule" and a "left hand rule" (which would give you the same answer in the end).