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Curl in 2D is a vector or a scalar?

  1. Mar 26, 2014 #1
    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: [tex]\left ( \frac{\partial f_y}{\partial x} -\frac{\partial f_x}{\partial y} \right )dxdy[/tex]
    So, the Curl of a vector field in 2D is a vector or a scalar?
  2. jcsd
  3. Mar 26, 2014 #2


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    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.
  4. Mar 26, 2014 #3
    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.
  5. Mar 28, 2014 #4
    So is wrong to say that [tex]\oint_{s}\vec{f}\cdot d\vec{s} = \iint_{A}\vec{\nabla}\times \vec{f} d^2A[/tex] ? 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: [tex]\oint_{s}\vec{f}\cdot d\vec{s} = \iint_{A}\vec{\nabla}\times \vec{f} \cdot \hat{n} d^2A[/tex] 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: [tex]\oint_{s}\vec{f}\cdot d\vec{s} = \iint_{A}|\vec{\nabla}\times \vec{f}| d^2A[/tex] But wich is the difference between ##|\vec{\nabla}\times \vec{f}|## and ##\vec{\nabla}\times \vec{f}##, in the pratice, none...
  6. Mar 29, 2014 #5


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    ##|-1| \ne -1##.
  7. Mar 29, 2014 #6
    [itex]d^2A[/itex] 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).
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