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

 P: 13 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
 HW Helper P: 2,566 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.
 P: 13 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: $$(dv)_{ij}=\frac{1}{h_1}\partial_i h_1 v_j - \frac{1}{h_2}\partial_j h_2 v_i$$ ?
 P: 3 $$(\nabla \times v)^i = \frac{1}{\sqrt{g}} \epsilon^{ijk} (dv)_{jk}$$ - in a coordinate basis.