Talk:Helmholtz decomposition/Archive 1

Something wrong?
Something's wrong here. As $$\mathcal{G} $$ is defined in the article on the Newtonian potential operator, $$\mathcal{G} (\nabla \times \mathbf{F})$$ is a scalar field. How can you take the curl of it! --unsigned anon


 * I guess that in the formula


 * $$\mathbf{F} = - \nabla\,\mathcal{G} (\nabla \cdot \mathbf{F}) + \nabla \times \mathcal{G}(\nabla \times \mathbf{F})$$
 * the quantity $$\nabla \times \mathbf{F}$$ is a vector, therefore, the quantity $$\mathcal{G}(\nabla \times \mathbf{F})$$ is also a vector (applied componentwise to the components of $$\nabla \times \mathbf{F}$$). Oleg Alexandrov (talk) 04:32, 1 December 2007 (UTC)

recently added sentence in lead
I just removed the following sentence from the lead:


 * If $$\mathbf{F}$$ does not extend to infinity, but ends at a boundary, then its  normal component at the boundary must be specified in addition to $$\varphi$$ and $$\mathbf{A}$$ in order for $$\mathbf{F}$$ to be unique.

This doesn't quite make sense: $$\mathbf{F}$$ is a given. It isn't "unique", it's specified --- it's an assumption. On the other hand, there isn't a unique scalar potential $$\phi$$ or vector potential $$\mathbf{A}$$. Modulo constants and potential fields, though, there is uniqueness. Or, in the case of a compactly supported $$\mathbf{F}$$, one can specify BCs for $$\phi$$ and $$\mathbf{A}$$ as an alternative. Is this what you're thinking? Lunch (talk) 20:37, 18 June 2008 (UTC)