# Curl of a vector in a NON-orthogonal curvilinear coordinate system

1. Feb 11, 2009

### MasterD

Hi,

I have a certain NON-orthogonal curvilinear coordinate system in 3D (in the metric only $$g_{13}=g_{23}=g_{31}=g_{32}=0$$) and I want to take the curl ($$\nabla\times\mathbf{v}$$) of a vector.

Any idea on how to do this? The only information I can find is about taking the curl of a vector in an orthogonal curvilinear coordinate system.

Very much thanks in advance for any insights.

Dirk

2. Feb 11, 2009

### StatusX

The thing that's more natural to define than curl (ie, it's defined in any dimension, where as curl is only defined in 3D) is whats called the exterior derivative of a (covariant) vector. It's defined by (in any coordinate system):

$$(dv)_{ij} = \partial_i v_j - \partial_j v_i$$

The curl is then the "Hodge dual" of this, defined by:

$$(\nabla \times v)^i = \sqrt{g} \epsilon^{ijk} (dv)_{jk}$$

where g is the determinant of the metric, and $\epsilon^{ijk}$ is the Levi-civita symbol. I might have some factors missing, but you can check this by computing simple cases.

3. Feb 12, 2009

### MasterD

Ok; thanks a lot; I will look further into this.

One more question: Shouldn't there be any scaling WITHIN the exterior derivative?

Something like: $$(dv)_{ij}=\frac{1}{h_1}\partial_i h_1 v_j - \frac{1}{h_2}\partial_j h_2 v_i$$ ?

4. Feb 12, 2009

### StatusX

No, the definition of the exterior derivative takes the same form in all coordinate systems. That's what's nice about it. But you need to be careful about covariant vs. contravariant vectors, and there are some constant factors associated with the Hodge dual (these depend partly on convention, but in this case you'll need to pick them to match up with the usual definition of curl). Look up the wikipedia article on curl, it has a more explicit version of the formula I mentioned.

5. Feb 16, 2009

### victorbomba

$$(\nabla \times v)^i = \frac{1}{\sqrt{g}} \epsilon^{ijk} (dv)_{jk}$$

- in a coordinate basis.