Talk:Curl

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I was hoping to find the formula for curl in polar coordinates... cylindrical coordinates to be precise. <sigh> moink 02:10, 16 Jan 2004 (UTC)

Hooray for <math>Math markup!</math>

My understanding is that cul is actually a tensor, with n * (n-1) / 2 free components. It only comes out nice in 3 dimensions because 3*2/2 = 3.

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[edit] invariance

not only does curl give a pseudovector depending if the (orthogonal) coordinate system is left or right-handed but it is clearly invariant for an arbitrary coordinate system. It would be nice if this fact was linked to the field of differential geometry and tensor calculus in which the curl operation has a far more natural counterpart. m3n0

[edit] Gilbert Strang and Curl

Gilbert Strang, in Introduction to Applied Mathematics introduces the Curl as

\begin{pmatrix} 0 & {-\frac{\partial}{\partial z}} & {\frac{\partial}{\partial y}} \\ \\ {\frac{\partial}{\partial z}} & 0 & {-\frac{\partial}{\partial x}} \\ \\ {-\frac{\partial}{\partial y}} & {\frac{\partial}{\partial x}} & 0 \end{pmatrix} \begin{pmatrix} F_x \\ \\ F_y \\ \\ F_z \end{pmatrix}

A few problems are worked out using this notation for the curl before Strang introduces the more convenient method of the cross product of del and F. I'm not sure where this notation comes from or how (if at all) it's better, so I didn't want to include it on the page. --- Trevie 17:07, 23 September 2005 (UTC)

It looks like this is just another longer way of representing

\begin{pmatrix} {\frac{\partial F_z}{\partial y}} - {\frac{\partial F_y}{\partial z}} \\ \\ {\frac{\partial F_x}{\partial z}} - {\frac{\partial F_z}{\partial x}}\\ \\ {\frac{\partial F_y}{\partial x}} - {\frac{\partial F_x}{\partial y}} \end{pmatrix}

--- Trevie 18:56, 23 September 2005 (UTC)

I've seen that notation for general cross products. A dynamics book I have uses it a lot. Basically, it turns the cross from a binary operator to a unary operator on the first vector. That is

\vec{v}^\times = \begin{pmatrix} 0 & -v_3 & v_2 \\ v_3 & 0 & -v_1 \\ -v_2 & v_1 & 0 \end{pmatrix}

so that you get

\vec{v} \times \vec{u}=\vec{v}^\times \vec{u} which is a simple matrix-vector multiplication. The author liked it because he could use expressions like \vec{w}^\times \vec{w}^\times in things like the expressions for transforming between frames of reference, and they became meaningful. moink 08:18, 26 September 2005 (UTC)

[edit] Relation to Divergence

I think there are a few things that could be added to this article, they may be trivial but:

 -The curl of a gradient of a scalar function is always zero (equivalently
the curl of a conservative function is zero)
 -The divergence of the curl is zero
 -Green's Theorem can be expressed as a double integral of curl dot-producted with kdA

--205.188.116.195 00:31, 14 December 2005 (UTC) Duke of Worcestershire

I think some of that is already at del. There is also Lagrange's formula. I would encourage you to indeed add some of the stuff you suggested, and also where appropriate refer to the two articles above. (It is OK if what you add to this article is already covered in the two, a bit of duplication does not hurt as long as it is connected to the matter at hand). Oleg Alexandrov (talk) 01:42, 14 December 2005 (UTC)


[edit] Rot

I think there should be reference to the fact "rotation" is also used as a synonym for curl, with operator rot(·). —DIV

I agree. --Vladimír Fuka 22:32, 29 September 2006 (UTC)

[edit] Related to curvature?

Is this related to the curvature of a function? - SigmaEpsilon → ΣΕ 11:51, 7 October 2006 (UTC)

[edit] n dimensions

Is there a way to define curl in n dimensions? unsigned

I would think not. And even if one could define it, it would be some kind of mathematical artifact without any useful physical interpretations or properties. But I am not sure. Oleg Alexandrov (talk) 03:55, 31 October 2006 (UTC)
I think this page should point out clearly that curl only applies in three dimensions, in contrast to gradient and divergence, which apply in any dimension. It is possible to define analogous operations to curl in higher dimensions. In two dimensions we have two operations: gradient and divergence, and applying divergence to gradient gives you zero. In three dimensions, gradient, curl and divergence - and applying two in sequence gives you zero. In four dimensions, there are four operations, e.g. gradient, "twist", "spin", and divergence - and applying two in sequence gives you zero, In five dimensions, five operations and so on. While gradient turns a scalar field into a vector field, curl turns a vector field into a vector field, and divergence turns a vector field into a scalar field, "twist" turns a vector field into a skew-symmetric matrix field, and spin turns a skew-symmetric matrix field into a vector field. The number of degrees of freedom in the field outputs follow Pascal's triangle: in 2 dimensions, the gradient takes a scalar field (1 degree of freedom) to a vector field (2 degrees) which is what divergence takes to a scalar field (1). In three dimensions, gradient goes from (1) to (3) which curl takes to (3) which divergence takes to (1). In four dimensions, you get 1->4->6->4->1 because a 4x4 skew-symmetric matrix has 6 degrees of freedom. Thus is Pascal's triangle spelled out. If I knew tensor calculus, I bet there would be a neater way to say this (and you need them to go above 5 dimensions here). Qseep 06:17, 7 December 2006 (UTC)
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