# Generalized curl operator

Tags:
1. Feb 7, 2016

### Jianphys17

• Hi, i now studying vector calculus, and for sheer curiosity i would like know if there exist a direct fashion to generalize the rotor operator, to more than 3 dimensions!!
On wiki there exist a voice https://en.wikipedia.org/wiki/Curl_(mathematics)#Generalizations , but I do not know how you could do...

2. Feb 8, 2016

### Orodruin

Staff Emeritus
In order to understand the generalisation given in wikipedia you will need to learn a certain amount of tensor calculus. The generalisation is dealing not only with vectors and scalars, but also higher order tensors. The reason the three dimensional case is formulated with vectors only is that in three dimensions there is a direct connection between anisymmetric rank two tensors and vectors.

3. Feb 8, 2016

### Jianphys17

From what I understand, the generalized curl, makes it use of skewsymmetric (0,k)-rank (the k-differential form)!
But i since barely know the tensor calculus formalism..., i've read something on differential forms, therefore it can be expressed through the exterior algebra with differential k-form ( k>3), right?

4. Feb 8, 2016

### Hornbein

Funny you should mention that.

http://hi.gher.space/forum/viewtopic.php?f=27&t=2155

Here's N D curl done with geometric algebra, which is similar to differential forms.

Curl is traditionally a pseudovector. This works only in 3D. A pseudovector is the dual of a plane. So one recasts curl in terms of planes, ie bivectors or 2-forms. It's a chunk of work to get used to, but is essential to extending to N D. Curl is naturally expressed as a bivector or tensor.

Now only in 3D does a bivector define a surface. Surfaces are (N-1)D, planes are 2D. Only in 3D is N-1=2. So all those theorems that involve surface integrals of curl become more involved. To proceed I suspect it is necessary to go back to the roots of EM in special relativity.

5. Feb 8, 2016

### Jianphys17

Sorry, maybe I'm wrong... but it can be merely defined with an appropriate volume form ( a k-differential form), that is in a k-1 dimens. planes?

6. Feb 8, 2016

### Hornbein

You have to be careful about using the word "plane" in N D. I define a plane as always 2D. A surface is N-1 D. A surface may or may not be planar.

Using this definition, rotation is always in orthogonal planes.

7. Feb 8, 2016

### lavinia

Last edited: Feb 8, 2016