Talk:Curl (mathematics)/Archive 1

markup
Hooray for $$Math markup!$$ — Preceding unsigned comment added by 213.253.39.230 (talk • contribs) 17:35, 6 January 2003‎

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

tensor
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. — Preceding unsigned comment added by Paul Murray (talk • contribs) 05:45, 27 March 2004‎

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 — Preceding unsigned comment added by 131.155.116.96 (talk • contribs) 13:54, 29 November 2004‎


 * There is an invariant definition of curl at the end of the "usage" section. Holmansf (talk) 22:52, 27 February 2009 (UTC)

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)

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)

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)

Related to curvature?
Is this related to the curvature of a function? - SigmaEpsilon → &Sigma; &Epsilon; 11:51, 7 October 2006 (UTC)
 * I don't think it is. --BenWhitey 06:24, 14 January 2007 (UTC)

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


 * The curl operator is just a compact notation that ultimately creates a cross product - if you read about the cross product, you'll see that it can be defined for 3 and 7 dimensions. In 1948, Feynman evidently showed Freeman Dyson some kind of crazy proof of the 1st two of Maxwell's equations, starting with Newton's force law, which Dyson made public in 1990.  Anyhow, you can search on "Feynman maxwell's equations Silagadze" to find a paper where Silagadze extends Feynman's "proof" to 7 dimensions ... now we just have to figure out what good 7 dimensional electromagnetism is.Thomasfly (talk) 22:34, 27 December 2008 (UTC)


 * 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)


 * This is a very good point - curl as generally defined is only for vector fields in 3 dimensions; I've added this to the lede.
 * It can be defined in higher dimensions, but the most natural generalization (the infinitesimal rotation of a vector field) is no longer a vector field. I've explained the generalization in this revision.
 * —Nils von Barth (nbarth) (talk) 04:12, 12 February 2010 (UTC)

what is curl, really?
This article should explain what curl is in terms of geometry so that people can understand what it is. This is how my professor explained it. You take a pinwheel and put it into a river parallel to the bottom. The river is the flow (flux) of a vector field. The pinwheel is the device that you are using to calculate the curl. Depending on whether the river is flowing faster on one part of the pinwheel than the other then it will either spin or not spin. That is what curl is. Does that make sense? --BenWhitey 06:27, 14 January 2007 (UTC)


 * Curl is how much rotation there is at a designated point in a vector field. Its length gives the amount of rotation.  Its direction gives the direction of the rotation using the convention of the right-hand rule.

--Loodog 22:30, 22 May 2007 (UTC)


 * I think an example like this would be useful to readers, but it needs to be better explained. Whether or not the wheel will spin seems to be very directly a matter of torque (and is more indirectly related to curl). The thing is, curl has no use (that I know of) unless you're talking about vector fields. Then, all it tells you is how curvy the field lines are (generally or evaluated at a specific point) and which way they curve. In your example (or rather your professor's), the river, which can be regarded as a force field, doesn't curve at all. The torque doesn't either. What does curve is (among others), the acceleration field of the wheel. If we define a cylindrical coordinate system with the $$z$$-axis through the center of the wheel, and $$\rho$$- and $$\phi$$-axes in the plane of the wheel, then:
 * $$\vec{a} = \alpha \rho \hat{\phi}$$ and $$\nabla \times \vec a = \frac{1}{\rho} \frac{\partial}{\partial \rho} [\rho (\alpha \rho) ] \hat{z} = 2 \alpha \hat{z}$$
 * This tells you that the acceleration field curves constantly at $$2\alpha$$, which you can then use to relate to torque or something. In fact, showing that $$\tau = \frac{mr^2}{2} \frac{ \nabla \times \vec{a}(\rho) }{ 2 } = \frac{r^2}{4} \nabla \times \vec{f}(\rho)$$ (for a more or less uniform wheel) is straight-forward. I think that stating the example more like this would be better than simply stating that curl determines whether the wheel will or will not spin. Any other thoughts on this? --Misho88 (talk) 07:15, 20 December 2007 (UTC)


 * all this stuff about "rotation" is nonsense. That just replaces one word (curl) with another (rotation).  Plus it gives the misleading idea that the fluid particles actually rotate around some axis, which they don't (at least they don't rotate around the little paddle wheel).--345Kai (talk) 04:41, 21 November 2008 (UTC)


 * I agree that the example in the article is useful. I was discussing this with someone, and recalled something an old Russian geophysics professor of mine had said (Alex Kaufman, if you're interested): given that there are two generators of fields, sources and vortexes, the divergence of a field is a measure of the density of sources while the curl is a measure of the density of vortexes.  In his example, a vortex would be something like a conduction current.  Would anyone care to discuss the accuracy/usefulness of this description?  I have never seen or heard anyone else use the vortex nomenclature, and his EM books are WAAAY to expensive to buy. --Andykass (talk) 01:31, 28 September 2011 (UTC)
 * Your Russian geophysics professor was referring to Helmholtz decomposition. Many vector fields, including many in electromagnetism, may be described in terms of a scalar potential and a vector potential. Those things give the "sources" and the "vortexes" if you like. Such an approach is can be useful for deriving useful results in classical electromagnetism theory for example. Curl is a mathematical concept, not a physical one, so using fuzzy english terms like rotation can get things wrong unfortunately. It is possible to come up fields which in some sense are rotating, but, slightly counter-intuitively, are irrotational (that is curl is zero) everywhere, except perhaps at a finite set of points. JJ Harrison (talk) 09:48, 28 September 2011 (UTC)

Examples
I added some sorely needed mathematical exmples. Feel free to tweak with the format if you have a better aesthetic sense than I.--Loodog 03:28, 23 May 2007 (UTC)

Curl in 2D, a scalar ?
The article begins stating "In vector calculus, curl is a vector operator ...". In the article however, without saying that explicitely, everything is in 3D. I have the feeling that there should be at least a 2D equivalent: Imagine a vector valued function $$R^3 -> R^3$$ where z'=0 everywhere. If we now look into the z=0 plane and have a rotation in that plane, we will get a (3D) curl vector in z direction.
 * No, you're crazy.--Loodog 23:44, 5 November 2007 (UTC)

Example: F(x,y,z) = (-y, x, 0)

Now, obviously, the interesting things happen in the xy planes (e.g. z=0). If we forget about the z-dimension because the result does not depend on z (i.e. consider our function $$R^2 -> R^2$$), we might ask, how much in-plane rotation we have in 2D, which should be a scalar like dfy/dx-dfx/dy. This embeds into 3D and we always have a curl vector (0,0,x) where x is the 2D curl. By the way, this is the same as the simple example in the text. Such 2D vector fields appear on surfaces or in image processing. So what about 2D curl definition ?

Anyway, the article would benefit from a clear statement about the assumed 3D dimensionality.89.53.7.199 23:39, 5 November 2007 (UTC)
 * Hi. Points:
 * Curl for two-dimensional vector fields is already defined, and it's given an orientation that requires three dimensions, and with good reason.
 * If you want a scalar to represent the curl of a two-dimensional vector field, why commit original research and invent bizarre new math when you can just use:
 * $$(\vec{\nabla} \times \vec{F}) \cdot \boldsymbol{\hat{z}}$$

--Loodog 19:21, 6 November 2007 (UTC)
 * This is exactly what I was saying, although I am not used to thinking in $$\vec{\nabla}$$-notation. Sorry if I was unclear. I am not proposing anything or inventing bizarre math, but I am looking for a 2D curl definition... I dont have a function $$F: R^3 -> R^3$$ but a function $$F: R^2 -> R^2$$. The definition on the wikipedia curl page uses the cross product which doesnt exist in two dimensions, so it implies at least 3 dimensions, which, however, isnt stated anywhere. It's perhaps obvious to all those people who know what curl is. On the other hand people always tend to think "SOMETHING is in 3D", because they always work with SOMETHING in 3D and do not understand the math behind it. So basically you're saying that curl is an intrinsically 3D phenomenon and you need "artificial" 3D embedding (which results in a 3D curl vector = (0,0,dy/dx-dx/dy) ) to define it for 2D ?
 * What do you mean with "Curl for two-dimensional vector fields is already defined". Where ? This is what I am looking for... --134.245.253.140 17:44, 7 November 2007 (UTC)
 * Take a look at the examples on the page. All of those given are 2D vector fields.  To talk about the direction of their curl, you need another dimension perpendicular to those.  E.g.  Do time lapse photography on a second hand as it runs around the clock.  Take direction of the velocity at various points in time to be the vector field in question.  Obviously, the direction of the curl will be clockwise.  How are you going to draw the vector representing "clockwise" on the 2D face of a clock?  You don't.  You define, by convention, clockwise curl to be out of the clock coming towards you, requiring a third dimension.--Loodog 19:24, 7 November 2007 (UTC)
 * I have already seen the examples, all of these are handled in 3D. The basic difference between rotation in 2D and 3D is, that in 3D rotation is around an axis (with a direction), while in 2D rotation is around a point. Therefore the direction is a concept of 3D and makes no sense in 2D. Here the only relevant thing is the angle. In 4D curl might be an entity with even much higher dimension. Your example makes sense and so do the examples on the curl page, but they are biased towards 3D and do not give a 2d definition. The meaning of divergence and curl in the 2D plane explains what I was looking for:curl is a scalar (dfy/dx-dfx/dy) in 2D. Nevertheless, thanks for your feedback ! --89.53.33.139 22:20, 7 November 2007 (UTC)
 * Look. Curl is a mapping: R3->R3.  In the event that your domain is R2, you still get a mapping into R3 since R2 is a subset of R3.  Curl of a 2D field is a 3D vector confined to one dimension perpendicular to the plane, but can have dependence on x and y.  Curl of a 3D field is a vector in 3D that can have dependence on x, y, and z.  Curl of a 4D field has no meaning.  I don't know what else you want.--Loodog 23:57, 7 November 2007 (UTC)
 * If the curl of a 2D field is a 3D vector field confined to one direction perpendicular to the plane, without loss of generality I can define a scalar field whose magnitude at (x,y) is equal to the magnitude of the vector at (x,y), and whose sign is positive if the vector points up and negative if the vector points down. I think this was the point of the original poster. --JB Gnome (talk) 01:37, 2 December 2008 (UTC)
 * That's true. And that's also in agreement with the very first lines of this page where an anonymous user wrote
 * "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"
 * In 2D this would mean a 1D curl (interpretation: rotation angle), in 3D, this means a 3D vector (interpretation=rotation axis x angle) and there might be useful representations in higher dimensions.
 * So from the quote, there seems to be a more general definition involving the dimension n (although 3D is the most widely used case and should be the "example case" for the wikipedia page). This page is called "Curl_(mathematics)" so we shouldnt get stuck in 3D just because it's the case handled in all physics books. My French is not good enough too understand that article, but doesnt this look like a general definition: Curl in French wikipedia ?


 * You (and the anonymous user) are correct, curl can be defined in n dimensions, but is generally defined in 3 dimensions as this is most useful in physics and turns out most nicely. As you suggest, in 2 dimensions curl is a scalar, and the 2D examples given extend to 3 dimensions. I've written this up in this revision; trust this helps!
 * —Nils von Barth (nbarth) (talk) 04:15, 12 February 2010 (UTC)


 * I think it would also help to state that curl is actually defined as an operator acting on tensor fields of arbitrary order, not just vectors. The easiest definition that I can think of is that "curl(A)=0" is equivalent to "A=grad(B)" for a certain tensor field B (of one order less). Example of an application is in mechanics under the assumtion of small deformation, where A is the strain tensor field and B is the displacement. Lerichard (talk) 13:16, 26 January 2012 (UTC)


 * I'm not sure how that could serve as a definition. There is this: Tensor derivative (continuum mechanics) (see the curl section), which probably merits a mention here. Also, I'm not sure that your strain/displacement example is correct, although I'm not an expert on continuum mechanics. According to Infinitesimal strain theory, the curl of the strain tensor is not necessarily zero.Holmansf (talk) 21:44, 26 January 2012 (UTC)


 * The section Generalizations should fully cover this, but it is currently written very obscurely. The curl generalizes is a straightforward and natural manner to any number of dimensions in the form of the operator ∇∧ of geometric algebra.  The result is a bivector in a space of dimension n(n−1)/2.  In two dimensions this corresponds to a pseudoscalar, and in three dimensions to a pseudovector, and naturally manages the distinction of and properties of a pseudovector.  It finds direct application in the four-dimensional geometry of spacetime.  Though this should be made clearer in the article, anyone seriously intending to study or use a generalization will generally use something like geometric algebra or differential forms, and will end up not using the name "curl". So a brief answer to the OP is that the curl per se is defined only in three dimensions, but there is an equivalent in any number of dimensions (where the equivalent in three dimensions differs from the curl in that its result is in a distinct 3D space, the pseudovectors).  The article clearly needs to be updated to reflect this, as evidenced by this discussion.  — Quondum☏✎ 06:02, 27 January 2012 (UTC)


 * I agree that Generalizations is not very good at this point. It seems to be a brief discussion of de Rahm cohomology, which may be related to curl, but isn't the same thing. I would support rewriting that section to reflect what you said although I don't want to do it ;)Holmansf (talk) 15:03, 27 January 2012 (UTC)


 * A good title for a re-written section might be "bivector curl", explaining that if the curl is identified as a bivector quantity, that is something that can continue to be meaningful in arbitrary dimensional space. We have quite an extensive article on bivectors, even if it currently nowhere contains the words differential form -- which it probably should.  Jheald (talk) 15:57, 27 January 2012 (UTC)


 * While sounding reasonable, the name "bivector curl" does not do it. The generalization is probably best handled as a generalization of the cross product as applied to the nabla symbol.  One thing I overlooked in what I said above is that the curl (and the cross product generally) may be applied to pseudovectors.  The situation is complicated by this, and probably requires the Hodge dual to generalize cleanly.  Simply state that ∇×v=*(∇∧v) in 3D, but that it can serve as a useful generalization, and make the point of nature of the result in the cases of 1D (identically zero), 2D (a pseudoscalar), 3D, (a pseudovector), 4D (a bivector of 6 dimensions) and nD, but that using ∇∧v itself is generally worth using instead.  In all, it really makes most sense to use geometric algebra if one is using the generalization (except perhaps in the simple 2D case, which merits its own section). — Quondum☏✎ 05:53, 28 January 2012 (UTC)

smoothness conditions?
I presume that some smoothness (e.g. must be a differentiable, or perhaps continuously differentiable, field??) conditions are required so that the two limits used for the curl definition (one using the surface, and the other using the path, integral) can (a) exist; or (b) be equal to each other. I guess such conditions must be required e.g. on the vector field itself, whose curl is being defined, and (perhaps less important to know) on the set of surfaces of the volumes used to produce the limit as V -> 0. Can anyone who knows add brief inserts on these requirement? If necessary and sufficient conditions are too long, complex and distracting then maybe some well-known sufficient conditions would be nice instead. E.g. is it enough to guarantee existence and equivalence of curl defined in these ways that the vector field should be differentiable (in the sense of a function between Banach spaces - is this called Frechet - forgotten)? Thanks in advance.

Notation Issue
In the section where the curl operator is represented by Einstein notation, shouldn't one of the l's and one of the m's be superscripts not subscripts? Isn't that how the summation part works?--SurrealWarrior 00:04, 1 December 2007 (UTC)
 * No. All Einstein Notation says is to not explicitly write summation signs when you have a repeated index in term since it's redundant.--Loodog 06:45, 1 December 2007 (UTC)

Deleted a Section
I stopped by this site to check out what info it had on the curl, and found that someone mistakenly offered the definition of the divergence as one definition of the curl, so I deleted it, leaving the "other" definition, which was correct. —Preceding unsigned comment added by 128.95.43.202 (talk) 05:32, 8 February 2008 (UTC)

CURL IS NOT DIVERGENCE
Perhaps, this section was here again. The article started from the "definition of curl", which is actually the Divergence Theorem, and the picture was wrong as well. Check it: // I deleted it The curl of a vector field $$\mathbf{F}$$ is defined as the limit of the ratio of the surface integral of the cross product of $$\mathbf{F}$$ with the normal $$\mathbf{n}$$ of closed surface $$S^{(2)}$$, over a closed surface $$S^{(2)}=\partial V^{(3)}$$, the boundary to the volume $$V^{(3)}$$ enclosed by the surface $$S^{(2)}$$, as the volume goes to zero:


 * $$ \operatorname{curl}(\mathbf{F}) = \lim_{V^{(3)} \to 0} \frac{1}{|V^{(3)}|} \,\iint_{\partial V^{(3)}}\mathbf{n}\times\mathbf{F}\,dS^{(2)} $$

More precisely, at each point $$p$$ in three-dimensional space, $$\operatorname{curl}(\mathbf{F})(p)$$ is given by the above limit, where the closed surfaces $$\partial V^{(3)}$$ all enclose $$p$$ and the diameter, not just the volume, of the enclosed region  tends to zero.

This definition is rather complex and not very useful, and the following alternative equivalent definition gives better measures to calculate components of $$\operatorname{curl}(\mathbf{F})$$. -- Commentor (talk) 03:15, 9 March 2008 (UTC)

Definition
I think that curl should be defined in the standard classical way in Cartesian coordinates (which is now in the second section). The current definition has a few issues. Foremost amongst these is that each component of curl(F) is only well defined modulo a factor of +/- 1. Aside from this however, it is also unclear what is meant by S^2 -> 0. I agree that the equivalence of this type of definition to the classical definition should be discussed because the current one does make the rotational nature of the curl apparent, but this should be reserved for a later section.

On another issue, I think that the relationship between curl and the exterior derivative should be given (which naturally shows that curl can be defined in a coordinate independent way).

Holmansf (talk) 05:09, 20 May 2008 (UTC)
 * User, please post new talk topics at the bottom.--Loodog (talk) 01:45, 21 May 2008 (UTC)

Proper definition
I don't know why people keep writing this as the definition:
 * $$\operatorname{curl}(\mathbf{F}) = \vec{\nabla} \times \vec{F}$$

This is NOT a definition, since the right side doesn't rigorously mean anything either. This is an assigment of notation which happens to function as a convenient mnemonic. A definition gives, in unambiguous terms, exactly what is meant by a concept, in all circumstances.

Besides: $$\vec{\nabla} \times \vec{F} $$ only even works as a good mnenomic in Cartesian coordinates and breaks down in every other coordinate system.--Loodog (talk) 14:29, 1 July 2008 (UTC)
 * I don't believe this to be true. In other coordinate systems, del is translated into that coordinate system and the identity holds. del in cylindrical and spherical coordinates. 92.39.204.109 (talk) 00:50, 11 November 2009 (UTC)

Integral definition and differentiability
Does anyone know if the existence of the curl as now defined implies that the vector field is smooth? I suspect that the current definition does not and is actually more general then the classical derivative definition (otherwise why bother with the integral?), but I don't know. It would be nice to have some mention of this either way. Also, someone mentioned this above, but I thought I'd reiterate that it would be nice to have some sufficient conditions on the vector field F so that the curl exists. Holmansf (talk) 22:25, 3 July 2008 (UTC)
 * . You don't need smoothness, you just need differentiability.  Then again, if you're willing to go hardcore and use the definition, you might even be able to find a curl in fields that have nondifferentiability.--Loodog (talk) 17:29, 11 July 2008 (UTC)

Yes, the "hardcore" option was what I was talking about. Despite what it may say on the page Smooth function (actually that should probably be changed), there is no universally agreed upon convention as to whether smooth means C^k for some k, or C^infinity. See :, although I've actually heard smooth being used to refer to C^1 functions, which was what I was doing above. Holmansf (talk) 22:52, 27 February 2009 (UTC)

ball vs. paddle wheel
I should probably justify the use of a ball vs. the use of a paddle wheel in the intuitive definition of curl. A paddle wheels has an axis, and it can only rotate around its axis. A ball does not have a predetermined rotation axis. That's the advantage of the ball! Please don't call the ball stupid. Thanks.--24.85.68.231 (talk) 07:49, 22 November 2008 (UTC)
 * Far from calling it stupid, I actually quite liked your intuitive definition.Malatinszky (talk) 14:21, 5 February 2009 (UTC)
 * But the curl is the limit to a point of this motion, which is most surely circular? 92.39.204.109 (talk) 00:56, 11 November 2009 (UTC)

Definition of Curl
I really think that the normal derivative definition of curl should be given first, and then later the current definition should be shown to be equivalent. This would match the format of the Divergence page, which I think is much clearer. I can only imagine the number of vector calculus students who have gone away from this page in complete confusion. If no one objects in the next week or so, then I'm going to make the change.Holmansf (talk) 22:52, 27 February 2009 (UTC)
 * No. That is NOT the definition.  The definition is the limit of the line integral as it reduces to a point.  That definition is the only reason $$curl$$ has any physical meaning.  The derivative "definition" you keep bringing up is only a simpler formula that returns the same result in a very very specific situation, that of Cartesian coordinates.  Wikipedia is not Mathworld; our priority is an academic description, not a list of formulas for students.--Loodog (talk) 23:14, 25 March 2009 (UTC)
 * I have a few points.


 * 1) Don't call other editors tools.


 * 2) You're integral definition is coordinate dependent. The most apparent way to see this is that the definition involves the dot product, which is obviously dependent on the coordinate system. In fact, I'm fairly sure that the integral definition given in this article only works in orthonormal coordinate systems, but the cartesian derivative formula also works for curl in such coordinate systems, and so there is really no more generality using the integral definition (unless you can prove it applies to less than C^1 functions).


 * 3) The definition of curl in the single reference given is not the same as the definition given in this article. I'm fairly sure that the definition in the reference is just wrong. Most references I've seen give the cartesian definition first, unless they're about calculus on manifolds in which case they give the more general definition involving the exterior derivative, which is really the appropriate generalization, and a way to make the definition coordinate independent.


 * 4) One of the goals of wikipedia for technical articles is to make them as accessible to a general audience as possible, at least at the beginning of the article. I think the derivative definition is much simpler to understand for a general audience. I don't think academic description is the main goal for mathematics articles on wikipedia. See Make technical articles accessible.


 * 5) The key result that makes the curl useful in physics is Stokes' Theorem. The fact that the integral definition and the derivative definition are equivalent (for C^1 functions) is only one application of Stokes' Theorem.Holmansf (talk) 06:59, 29 March 2009 (UTC)


 * I have to say that I completely agree with Holmansf. The limit/integral definition of curl given here is simply inappropriate for a Wikipedia article.  If you look in math and physics textbooks that discuss the curl, virtually all of them introduce the curl using partial derivatives, though many do go on to interpret the curl as the limit of a line integral.  Wikipedia is not the place for an aesthetic or philosophical argument about what the definition should be &mdash; it should simply state what the standard definition is. Jim (talk) 15:23, 29 March 2009 (UTC)


 * Talk to mathworld or provide a source giving a more general, more rigorous definition and we'll use that instead. To Jim.belk, wikipedia is not a textbook: "The purpose of Wikipedia is to present facts, not to teach subject matter."  The Cartesian formula is included anyway.--Loodog (talk) 16:34, 29 March 2009 (UTC)


 * On a technical point, in Euclidean space, how is the dot product coordinate dependent?--Loodog (talk) 16:34, 29 March 2009 (UTC)


 * Mathworld was not the reference to which I was referring. Rather I was talking about the one entry in the reference section, although I've been doing some thinking about the definition given there and have concluded that it is correct. Interestingly the definition given there is as the limit of an integral over a surface rather than a line. It is the definition that whoever wrote the "Curl is not divergence" comment above deleted. The mathworld definition is essentially correct, although not entirely rigorous (in particular it doesn't specify the orientation of the line integral, which is an issue).


 * In reference to the dot product and coordinate independence- take any invertible linear map L:R^n -> R^n. This then provides a change of coordinates on R^n: y = Lx. The two coordinate systems are the y coordinates and the x coordinates. In the x coordinates the dot product is x^T x (this is x transpose times x). If you naively take the dot product in the y coordinates you get y^T y = x^T L^T L x. This is only equal to the dot product in the x coordinates for every x if L^T L is the identity matrix, or in other words that L is represented by an orthogonal matrix. From the point of view of differential geometry the dot product is an inner product defined on the tangent space of R^n at each point which is interpreted as a 2-tensor field- the Euclidean metric. In general, to define the curl you need a Riemannian metric.


 * Another way to see that this integral definition is coordinate dependent is to simply think about how you would actually use it to calculate the curl. The first thing you have to do is find the plane perpendicular to a given vector. How do you do this in a general coordinate system? You can't unless you know the Euclidean metric in that coordinate system, and in order to find this you must find the change of coordinates maps from your coordinates to cartesian coordinates, etc. The point is of course that in order to use this integral definition you still have to return to cartesian coordinates.


 * Here are two references that give a derivative definition of the curl. For physicists-


 * This one gives the cartesian derivative definition first, but later derives the line integral formula and acknowledges that it could be used as a definition.


 * For mathematicians-


 * This one gives a definition using the exterior derivative.Holmansf (talk) 14:37, 30 March 2009 (UTC)


 * Here's where we're differing. The dot product I had in mind was only of transformations to orthonormal systems from Cartesian that preserved length.  That dot product is coordinate independent, as is the integral definition when restricting oneself to those transformations.  Short of GR, these are the only ones you'll need in physics and engineering.--Loodog (talk) 16:14, 30 March 2009 (UTC)


 * Also, haven't we specified the orientation of the line integral in the article?--Loodog (talk) 21:45, 30 March 2009 (UTC)

The curl is the differential d : &Omega;1(R3)→&Omega;2(R3) in the de Rham complex with &Omega;1 and &Omega;2 identified (under a choice of orientation) and each of them identified with the module of vector fields. This is the natural (and generalizable) definition of the curl. This says the curl should be defined as a derivative. In addition, the definition most often seen in calculus classes and textbooks is as a derivative. I'm pretty sure the curl was originally defined as a derivative, too. I thus see no reason to define the curl by using line integrals. That the line integral interpretation as a circulation density is important is certainly true, however it is not the only reason curl has physical meaning. Its definition as the differential above relates it to the topological structure of the underlying space and non-trivial de Rham cohomology thus leads to physical effects such as the Aharonov–Bohm effect. The existence of vector potentials of divergence-free vector fields on R3 is also a simple consequence of the above definition of the curl (and the de Rham cohomology of Euclidean 3-space). RobHar (talk) 14:16, 1 April 2009 (UTC)


 * I didn't follow all that because I do physics, not math, but here's my thinking in terms of a good encyclopedia article structure: do it epistemologically i.e. define the concept in terms of the problem or context it was originally developed for. Then, explain what it is today.  Now, I know very little of history of math, but it seemed more likely to me someone sat down thinking, "I need a way to quantify circulation/rotation" rather than "Let me define a mess of differential operators and hope it has physical significance".


 * Since the prevailing physical significance in physics: (1) closed line integrals around an area are essentially equivalent to sums of the curl at every point inside and consequently (2) beautifully simple relationships between (a) current, magnetic fields, and electric fields, and (b) electric fields and magnetic fields, I assumed the line integral idea was used first.--Loodog (talk) 20:33, 1 April 2009 (UTC)


 * Ah I see. Well, the nabla operator was introduced by Hamilton (as a quaternion) $$\nabla=i\frac{d}{dx}+j\frac{d}{dy}+k\frac{d}{dz}$$. The curl showed up as the "imaginary part" of the product of nabla with an imaginary quaternion:
 * $$\nabla q=-\operatorname{div}\ q+\operatorname{curl}\ q$$
 * where q = ai+bj+ck can be identified with a vector (a,b,c). This equation appears on page 610 of Hamilton's "Lectures on Quaternions" (1952) (available here) (in different notation). It appears that the quantity curl was first given a name by Maxwell in the early 1870s, and it was given this name because of what it computes (i.e. the circulation density); it was however defined as a quaternion differential operator. Take a look at and  for some more history. RobHar (talk) 23:34, 1 April 2009 (UTC)
 * Even if it were not true that the curl was originally defined as a certain differential operator, it would still be appropriate to put that definition first. It is direct, simple, and by far the most common; and it's not our job on WP to innovate (that's what papers are for). As Loodog points out, the definition isn't immediately clear. But fixing that is a matter of good exposition.  Good exposition would explain why this definition is interesting and useful, and as part of that, it would explain that the curl computes circulation density. You might even say something like, "This apparently silly definition reveals itself as follows" (David Mumford, as quoted in Eisenbud's Commutative algebra). Because, after all, the physical significance of the curl is that it computes circulation density, but that doesn't make circulation density the definition. Ozob (talk) 18:53, 3 April 2009 (UTC)

In any case, arguing that there is only one definition of curl and that it is the "right" one because it is historically the first (although apparently not as Rob points out above), is all beside the point. Holman, above, is correct when pointing out that according to Make technical articles accessible we should put more accessible parts first. The definition of curl using differentiation in Cartesian coordinates is basic and often the first definition seen. There is also WP:Manual_of_Style_(mathematics), which recommends putting more accessible portions first. These guidelines reflect consensus from many math editors. --C S (talk) 00:48, 3 April 2009 (UTC)


 * As I've said before, my sense was the best article would include history in terms of the motivation for developing the concept. Any good encyclopedia article would have that.  On top of that, we should obvious also have the formula used in Cartesian because it's useful.  I just got annoyed that people would keep defining the curl as $$\nabla \times \boldsymbol{F}$$ as if this abuse of notation rigorously meant anything, as if the curl were arbitrarily invented simply because someone got bored with differential operators.--Loodog (talk) 20:09, 3 April 2009 (UTC)


 * I'd like to start off by saying that I think C S has a point about where this discussion should be, i.e. what is the most accessible definition. My opinion is that the definition in terms of differentiation in Cartesian coordinates is the most accessible. Firstly, it involves the simplest amount of mathematics; secondly, it doesn't require physical intuition. People who are physicists tend to think that most things are better described intuitively in a physical manner, without regard to the fact that a large portion of the population lacks sufficient physics intuition. It is true that an unmotivated mathematical definition can be a drag as well, however, thirdly, I would say that most people that come here will have seen the definition in terms of differentiation in Cartesian coordinates and familiarity leads to accessibility.


 * To address your comments, Loodog: the "abuse of notation" &nabla;&times;F does mean something rigourously: in terms of quaternions, it is the vector-component of the product of two quaternions, the first quaternion being a differential operator, the second quaternion being a function; even in terms of vectors it is rigorous since it is a defintion. It is a completely common thing in math (and physics) to use a strongly suggestive notation to mean something, and there's nothing wrong with this. We do this because our brains aren't perfect. It would be much harder (for most people, at least) to remember how to compute the curl if it wasn't written as it is. Though I don't know about why Hamilton invented the curl, it does certainly seem possible it could've been related to being bored with differential operators and complex numbers. I feel like the facts are getting in the way of the narrative you would like to be true. Anyway, we should concentrate on the discussion at hand (i.e. C S's point). If you'd like to talk more about this paragraph, we should probably do that at one of our talk pages. Cheers, RobHar (talk) 14:14, 5 April 2009 (UTC)
 * Read abuse of notation. I never said it was a BAD thing. Abuse of notation has saved me countless hours with Gamma matrices. You just can't use it for a definition.  Sure, given adequate additional specifications, &nabla;&times;F can mean something rigorously, but as these symbols are commonly used it's only a suggestive expression, not a definition.--Loodog (talk) 20:10, 5 April 2009 (UTC)

Definition in Cartesian Coordinates
I corrected the expression of Curl(F) in Cartesian Coordinates to the following:

$$\left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right) \mathbf{i} + \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right) \mathbf{j} + \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right) \mathbf{k}$$

CDiPoce (talk) 08:05, 21 March 2009 (UTC)

Problem with assertion and reason
At the end of the introductory paragraph we see the statement,

''Unlike the gradient and divergence, curl does not generalize as simply to other dimensions; some generalizations are possible, but only in three dimensions is the geometrically defined curl of a vector field again a vector field, because in 3 dimensions, bivectors (2-vectors) can be identified with vectors (1-vectors). This is the same phenomenon as in the 3 dimensional cross product, and the connection is reflected in the notation $$\nabla \times$$ for the curl.''

While this statement contains true assertions, I'm not convinced that the reasons given here for the true assertions are correct. There is also a seven-dimensional cross product which appears to have been overlooked by the author. This statement requires the attention of somebody who is an expert on bivectors. David Tombe (talk) 19:04, 17 June 2010 (UTC)
 * It seems fine to me. The key part is 'geometrically defined', as this can be done in three dimensions: "the vector perpendicular to x and y with magnitude xy sin θ", together with some convention such as the right hand rule. It also relates to the bivector-pseudovector relation, as the above paragraph suggests. But this can't be done in 7D as there's not a unique result: you have to make an arbitrary choice, of e.g. a product rule or a trivector with particular properties. So it can't just be defined geometrically in 7D, you need an additional algebraic part to the definition.
 * In theory you could define curl in 7D but you'd have the same problem: you'd have to make an arbitrary choice and there would be an infinity of possible curl operators depending on your choice, not a single one geometrically defined. The justification for curl is physical, i.e. it models certain physical phenomena, which also limits it to 3D where our physical world is based. So both theoretically and practically it's hard to think of a useful curl in 7D.-- JohnBlackburne wordsdeeds 00:37, 18 June 2010 (UTC)
 * This is a deep subject, perhaps. I'd hope that John Blackburne might explain how the 3-D curl might be generalized to some other form that degenerates into the 3D curl as dimensions n→3. It isn't clear to me that ‘an additional algebraic part to the definition’ is a deal breaker. Does such an extension to higher n exist in mathematical form that becomes (in part, anyway; maybe some extra parts remain?) the usual 3D curl when n =3? What sources would provide some assistance on this topic? Brews ohare (talk) 20:32, 18 June 2010 (UTC)
 * the vector form doesn't generalise but it generalises if you relax the condition that the result is a vector and replace it with a bivector - the dual of the vector - in 3D. This is already covered in the article, using the language of differential forms.-- JohnBlackburne wordsdeeds 21:42, 18 June 2010 (UTC)

John, I was thinking specifically about the relationship to the determinant of a 3x3 matrix, which lies at the core of both the 3D curl and the 3D cross product. I can't envisage a 7D curl at all, but that's not for the reasons stated in this article. This article gives reasons which suggest that the author may never have heard of the seven-dimensional cross product. I still think that the article is stating a correct assertion, but for the wrong reasons. I don't see why 7D should be excluded, based on the reasons given in the article. In fact, ignoring the fact that a 5D cross product wouldn't satisfy the Lagrange identity, I can't see any reason why 5D should be excluded based on the reasons given in the article. Does this table of a 5D cross product not satisfy the condition that two vectors can be identified as one vector?

! |z = x × y ! |x × y !i !j !k !l !m |} David Tombe (talk) 21:57, 18 June 2010 (UTC)
 * j×m, l×k
 * i×l, k×m
 * i×j, l×m
 * i×m, j×k
 * i×k, j×l


 * That's not a cross product as you know - it doesn't exist in 5D. The (3D) cross product and curl can be defined geometrically and so independently of a choice of basis - they do not depend on the determinant or matrix, those are just useful techniques for calculating them. Other than that I don't know what you're asking. The article is clear and makes sense and there's no curl in 7D other than the generalisations given.-- JohnBlackburne wordsdeeds 00:27, 19 June 2010 (UTC)

John, I know fine well that there is no cross product in 5D that satisfies the Lagrange identity, and hence the distributive law. But let's not go into that here. I could have used the 7D cross product example to make my point,


 * {| class="wikitable" style="text-align: center;"

! |z = x × y ! |x × y !i !j !k !l !m !n !o |}
 * j×l, k×o, and n×m
 * i×l, k×m, and n×o
 * - i×o, j×m, and l×n
 * i×j, k×n, and m×o
 * i×n, j×k, and l×o
 * -i×m, k×l, and j×o
 * i×k, j×n, and l×m

but I deliberately chose the 5D cross product thinking that it would also be adequate for the purpose.

Anyway, you seem to be saying the same as myself as far as the assertion is concerned regarding that curl only exists in 3D. I was pleasantly surprised to see that the article was pushing the idea that curl only exists in 3D, because that is exactly what I believe. You weren't involved in recent discussions over at Pythagorean theorem in which I was trying to point out that Pythagoras's theorem, cross product, dot product, rotation, curl, and angle are all one single package in geometry, and necessarily in 3D. Curl and cross product in particular are closely related to rotation in 3D. And yes, the matrix determinant is indeed a useful technique, but it is a technique that seems to be restricted to 3D cases.

The point that I am making is that the assertions in the article are correct, but that the reasons given are wrong, and indicate a lack of full understanding on the part of the author, hence leading to an overly complicated article which is very hard to follow. I didn't follow all that stuff about bivectors, but it seems to me to be wrong when one considers the 7D cross product. David Tombe (talk) 09:57, 19 June 2010 (UTC)
 * It is an advanced topic - a degree level one - while the generalisations including into higher dimensions are more advanced, but I can't see anything wrong with it. Perhaps you could highlight the incorrect reasoning and why it is wrong.-- JohnBlackburne wordsdeeds 11:29, 19 June 2010 (UTC)

John, It was the statement,

because in 3 dimensions, bivectors (2-vectors) can be identified with vectors (1-vectors)

We know that even in 7 dimensions 2 vectors can be identified with 1 vector. So although I agree with the assertion that we can only have curl in 3D, I don't agree with the reasons that have been given for this assertion. David Tombe (talk) 16:24, 19 June 2010 (UTC)
 * They bivectors and vectors are only identified, i.e. are isomorphic as vector spaces, in three dimensions as only there do they have the same dimension, 3. In seven dimensions the space of bivectors has 21 dimensions, so is not isomorphic to vectors. So that statement is correct. But it's not what I write here that's important, but what's in the article. What in the article do you take issue with ?

John, we are talking about this sentence here,

because in 3 dimensions, bivectors (2-vectors) can be identified with vectors (1-vectors)

I am not doubting that this sentence is correct. And I am not doubting the main assertion in the associated paragraph in the inroduction, regarding that curl can only exist in 3D. What I am doubting is that the above sentence is the reason for the assertion. I would say that the sentence above would equally be true in 7 dimensions. And even if I am wrong, and it is only true in 3 dimensions, the question still holds regarding whether this sentence is the reason why curl can only exist in 3D. David Tombe (talk) 19:09, 19 June 2010 (UTC)
 * I'm sorry if you don't understand it but it makes sense to me. Another approach is to not try and justify it: most people think of things like the cross product and curl operator as just being defined in 3D. When told that e.g. the dot product, grad and div operators generalise to higher dimensions but the cross product and and curl operator don't, or at least not straightforwardly, they just accept it. To understand why you perhaps need a good understanding of higher dimensional algebra and geometry, some of which you've already said you "don't follow". You could consult other articles or some of the sources for more information on this, as it's only touched on here.-- JohnBlackburne wordsdeeds 19:58, 19 June 2010 (UTC)

John, Actually your second sentence above states exactly what my point of view is. My point of view is simply that the assertion that curl only holds in 3D is true as a matter of fact. The problem with the article was that it attempted to give a reason for this assertion, that reason being,

because in 3 dimensions, bivectors (2-vectors) can be identified with vectors (1-vectors)

This so-called reason is a true statement in its own right. But it is not a reason for the original assertion because this statement would be true even in 7 dimensions. David Tombe (talk) 14:17, 20 June 2010 (UTC)
 * As I've already written in 7D (or in anything other than 3D) bivectors and vectors have different dimensions as vector spaces, so there is no straightforward identification. In 3D both are 3D vector spaces, and are related by the Hodge dual which is an isomorphism and a linear map. So the statement you quote is only true in 3D.-- JohnBlackburne wordsdeeds 16:13, 20 June 2010 (UTC)
 * The reason removed seems correct to me. I suggest that we restore it unless some clear objection is raised.   Sławomir Biały  (talk) 20:34, 1 July 2010 (UTC)

John, Even if we take the statement to be uniquely true for 3 dimensions, it is still not a reason for the true assertion that curl only exists in 3 dimensions. David Tombe (talk) 22:47, 20 June 2010 (UTC)
 * As noted above I don't see anything wrong with it. But as the purpose of the talk page is to improve the article how do you think it could be improved?-- JohnBlackburne wordsdeeds 23:03, 20 June 2010 (UTC)

John, The statement may well be true, but it is not a reason for the assertion, which is also true. You yourself have already admitted above that the assertion is a true fact in its own right and that we should accept it without a specific reason. Hence we remove that sentence as being a reason for the assertion, and I have already done that. David Tombe (talk) 09:47, 21 June 2010 (UTC)


 * David, the original statement was correct, and I think it should be restored. Here is a perhaps more intuitive, but longer, explanation of what it was saying (maybe this, or a refined version of this should go on the page). The curl of a vector field at a point measures the degree to which the vector field is rotating near that point in every plane through the point. As such it is a function on oriented planes through the point (bi-vectors), and this can be defined in other than three dimensions. However, only in three dimensions is it possible to identify the set of oriented planes through a point with the set of unit vectors, and thus define the curl of a vector field as another vector field.


 * Addendum: You should note that by "identify" I mean, "define a bijection between." Note that in 7D your product does give a map from oriented planes to unit vectors, but this map is not invertible (ie. there would still be no unique way to define the curl as a vector field). Holmansf (talk) 19:24, 24 June 2010 (UTC)

paddle wheels
Im slightly troubled of this wording in the first example of a curled field

If we stick a paddle wheel anywhere, we see immediately its tendency to rotate clockwise.

Now, sure, it is true, that field will rotate a paddle wheel and has a curl. However, it gives a slightly false impression. Since we're talking in terms of fluids we may call it the vorticity, and the vorticity of a fluid is really not a global property of the flow field but a local one. I mean many potential flow fields would move paddles here and there, but they are by definition curl free.

I know it is a common example to explain the curl, but it should perhaps be noted to be more of a mnemonic than anything directly out of the definition of a curl.

There is a plot of a curl free field in http://en.wikipedia.org/wiki/Conservative_vector_field#Irrotational_vector_fields which surely would make most people think a real life macroscopic paddle wheel would be rotated.

62.78.155.244 (talk) 18:36, 7 October 2010 (UTC)

Assessment comment
Substituted at 14:36, 1 May 2016 (UTC)